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stringclasses 106
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float64 1
3
⌀ | solution
stringlengths 0
29.4k
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stringlengths 2
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|---|---|---|---|---|---|---|---|---|---|---|
f529d0580bd96ff0263ead7e4173b082
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The foot of the perpendicular drawn from the point (1,8,4) on the line joining the point (0,-11,4) and (2,-3,1) is
|
singleCorrect
| 2 |
Equation of line joining points (0,-11,4) and (2,-3,1) is<br/>$\frac{x-2}{2}=\frac{y+3}{8}=\frac{z-1}{-3}=\lambda$<br/>Let $P$ is any point of the above line then coordinate of $P$ is $(2 \lambda+2,8 \lambda-3,-3 \lambda+1)$<br/>$\therefore \mathrm{DR}^{\prime} \mathrm{s}$ of $P Q$ is $(2 \lambda+1,8 \lambda-11,-3 \lambda-3)$<br/>Now, $(2 \lambda+1)(4+(8 \lambda-11)(8)+(-3 \lambda-3) (-3)=0$<br/>$\Rightarrow \quad 4 \lambda+2+64 \lambda-88+9 \lambda+9=0$<br/>$77 \lambda-77=0$<br/>$\therefore$ Required foot of perpendicular.<br/>$$<br/>P(4,5,-2)<br/>$$<br/><img src="https://cdn.quizrr.in/question-assets/wbjee/2018_mat_a50.png">
|
["(4,5,2)", "(-4,5,2)", "(4,-5,2)", "(4,5,-2)"]
|
[3]
| null |
Practise Problem
|
09a4660f93655199764ebc3b810547c2
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If two lines $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $L_2: \frac{x-3}{1}=\frac{y-k}{2}=z$ intersect at a point, then $2 k$ is equal to
|
singleCorrect
| 2 |
Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$
Now, any point $P$ that lies on the lines $L_1$ has the form $(1+2 \lambda,-1+3 \lambda, 1+4 \lambda)$.
Now, on putting $x=1+2 \lambda, y=-1+3 \lambda$ and $z=1+4 \lambda$ into the equation of lines $L_2$, we get
$\begin{aligned}
& \frac{1+2 \lambda-3}{1}=\frac{-1+3 \lambda-k}{2}=1+4 \lambda \\
& \Rightarrow \quad \frac{1+2 \lambda-3}{1}=1+4 \lambda \\
& -2 \lambda=3 \Rightarrow \lambda=\frac{-3}{2} \\
& \frac{-1+3 \lambda-k}{2}=1+4 \lambda \\
& -1+3 \lambda-k=2+8 \lambda \\
& -5 \lambda=3+k \\
& -5\left(-\frac{3}{2}\right)=3+k \quad[\because \lambda=-3 / 2] \\
& k=\frac{15}{2}-3 \\
& k=\frac{9}{2} \\
& 2 k=9 \\
&
\end{aligned}$
|
["9", "$\\frac{1}{2}$", "$\\frac{9}{2}$", "1"]
|
[0]
| null |
Practise Problem
|
2ba69d8cf1d91fa226e32ac39004ae78
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The image of the point \((1,6,3)\) in the line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\) is
|
singleCorrect
| 2 |
Given line<br>\(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} \rightarrow(1)\)<br>Point \(P(1,6,3)\) and reflection \(Q\).<br><img src="https://cdn.quizrr.in/question-assets/kcet/py54ggf1c/sol_429.png"><br>Let \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda\)<br>and point \(R\) on line. So, coordinates of \(R\) is<br>\((\lambda, 1+2 \lambda, 2+3 \lambda)\)<br>Direction ratios of line is \((1,2,3)\).<br>Since PR is perpendicular to the given line. Then, direction ratios of<br>\(\overline{P R}=(\lambda-1,2 \lambda-5,3 \lambda-1)\)<br>So, \(\lambda-1+2(2 \lambda-5)+3(3 \lambda-1)=0\)<br>\(\Rightarrow 14 \lambda=14 \Rightarrow \lambda=1\)<br>Therefore, \(R(1,3,5)\) and \(Q(1,0,7)\)
|
["\\((1,0,7)\\)", "\\((7,0,1)\\)", "\\((2,7,0)\\)", "\\((-1,-6,-3)\\)"]
|
[0]
| null |
Practise Problem
|
136bd3d1bfe8b5ab3e55edf0493d0c1d
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is
|
singleCorrect
| 2 |
Equation of the line through $(1,-2,3)$ parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r$ (say)<br>Then any point on (1) is $(2 r+1,3 r-2,-6 r$ $+3$ )<br>If this point lies on the plane $x-y+z=5$ then<br>$$<br>(2 r+1)-(3 r-2)+(-6 r+3)=5 \Rightarrow r=\frac{1}{7}<br>$$<br>Hence the point is $\left(\frac{9}{7},-\frac{11}{7}, \frac{15}{7}\right)$<br>Distance between $(1,-2,3)$ and<br>$\begin{array}{l}<br>\left(\frac{9}{7},-\frac{11}{7}, \frac{15}{7}\right) \\<br>=\sqrt{\left(\frac{4}{49}+\frac{9}{49}+\frac{36}{49}\right)}=\sqrt{\left(\frac{49}{49}\right)}=1<br>\end{array}$
|
["1", "2", "4", "$2 \\sqrt{3}$"]
|
[0]
| null |
Practise Problem
|
b5b34e15c42b70c4ddd9c47b48dd5350
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The acute angle between the line $\bar{r}=(\hat{\imath}+2 \hat{\jmath}+\hat{k})+\lambda(\hat{\imath}+\hat{\jmath}+\hat{k})$ and the plane
$\bar{r} \cdot(2 \hat{\imath}-\hat{\jmath}+\hat{k})=5$ is
|
singleCorrect
| 2 |
Given $\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ and the plane $\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})=5$
The angle between the line $\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$ and the plane $\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=p$ is given by $\sin \theta=\frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}}{|\overline{\mathrm{b}}| \overline{\mathrm{n}} \mid}$
Here $\bar{b}=\hat{i}+\hat{j}+\hat{k} \Rightarrow|\bar{b}|=\sqrt{3}$ and $\bar{n}=2 \hat{i}-\hat{j}+\hat{k} \Rightarrow|\bar{n}|=\sqrt{6}$
Here $\bar{b} \cdot \bar{n}=(\hat{i}+\hat{j}+\hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})=2-1+1=2$
$\therefore \sin \theta=\frac{2}{\sqrt{3} \sqrt{6}}=\frac{2}{\sqrt{3} \times \sqrt{3} \times \sqrt{2}} \Rightarrow \sin \theta=\frac{\sqrt{2}}{3} \Rightarrow \theta=\sin ^{-1} \frac{\sqrt{2}}{3}$
|
["$\\sin ^{-1}\\left(\\frac{\\sqrt{2}}{3}\\right)$", "$\\sin ^{-1}\\left(\\frac{2}{3}\\right)$", "$\\sin ^{-1}\\left(\\sqrt{\\frac{2}{3}}\\right)$", "$\\sin ^{-1}\\left(\\frac{2}{\\sqrt{3}}\\right)$"]
|
[0]
| null |
Practise Problem
|
f9c80d19732f48d5bbf80f3afff19abc
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The lines $\frac{x-1}{2}=\frac{y-4}{4}=\frac{z-2}{3}$ and $\frac{1-x}{1}=\frac{y-2}{5}=\frac{3-z}{a}$ are perpendicular to each other, then $a$ equals to
|
singleCorrect
| 2 |
Let $L_1: \frac{x-1}{2}=\frac{y-4}{4}=\frac{z-2}{3}$
and $L_2=\frac{1-x}{1}=\frac{y-2}{5}=\frac{3-z}{a}$
the line $L_2$ can be written as $\frac{x-1}{-1}=\frac{y-2}{5}=\frac{z-3}{-a}$
Now, the DR's of lines $L_1$ and $L_2$ are $(2,4,3)$ and $(-1,5,-a)$ respectively.
Since, $L_1$ and $L_2$ are perpendicular to each other.
$\begin{array}{lrl}
\therefore & 2(-1)+4)(5)+3(-a)=0 \\
\Rightarrow & -2+20-3 a=0 \\
\Rightarrow & -3 a=-18 \Rightarrow a=6
\end{array}$
|
["-6", "6", "$\\frac{22}{3}$", "$-\\frac{22}{3}$"]
|
[1]
| null |
Practise Problem
|
7b9ff5f6108d30ee3e015edaaea5555a
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The acute angle between the two lines whose direction ratios are given by $l+m-n=0$ and $l^2+m^2-n^2=0$, is
|
singleCorrect
| 1 |
Given that
$$
l+m-n=0
$$
and
$$
l^2+m^2-n^2=0
$$
From equaion (i) $l=-(m-n)$
Putting in equation (ii), we get
$$
\begin{aligned}
\Rightarrow & (m-n)^2+m^2-n^2 & =0 \\
\Rightarrow & m^2+n^2-2 m n+m^2-n^2 & =0 \\
\Rightarrow & 2 m^2-2 m n & =0 \\
& 2 m(m-n) & =0 \\
\Rightarrow & m=0, m & =n
\end{aligned}
$$
From equation (i), if $m=0$, then
$$
l=n \Rightarrow l: m: n=1: 0: 1
$$
If $m=n$, then $l=0 \Rightarrow l: m: n=0: 1: 1$
If $\theta$ is the acute angle, then
$$
\begin{aligned}
\cos \theta & =\left|\frac{1.0+0.1+1.1}{\sqrt{1+1} \sqrt{1+1}}\right|=\frac{1}{2}=\cos \frac{\pi}{3} \\
\Rightarrow \quad \theta & =\frac{\pi}{3}
\end{aligned}
$$
|
["$0$", "$\\frac{\\pi}{6}$", "$\\frac{\\pi}{4}$", "$\\frac{\\pi}{3}$"]
|
[3]
| null |
Practise Problem
|
83a2c29d1ca9b4c5eabcbdebaa155058
|
BITSAT_MATH
|
Three Dimensional Geometry
|
Two intersecting lines lying in plane <math><msub><mrow><mi>P</mi></mrow><mrow><mn>1</mn></mrow></msub></math> have equations <math><mfrac><mrow><mi>x</mi><mo>-</mo><mn>1</mn></mrow><mrow><mn>1</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>3</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>4</mn></mrow><mrow><mn>3</mn></mrow></mfrac></math> and <math><mfrac><mrow><mi>x</mi><mo>-</mo><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>3</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>4</mn></mrow><mrow><mn>1</mn></mrow></mfrac><mo>.</mo></math> If the equation of plane <math><msub><mrow><mi>P</mi></mrow><mrow><mn>2</mn></mrow></msub></math> is <math><mn>7</mn><mi>x</mi><mo>-</mo><mn>5</mn><mi>y</mi><mo>+</mo><mi>z</mi><mo>-</mo><mn>6</mn><mo>=</mo><mn>0</mn></math> , then the distance between planes <math><msub><mrow><mi>P</mi></mrow><mrow><mn>1</mn></mrow></msub></math> and <math><msub><mrow><mi>P</mi></mrow><mrow><mn>2</mn></mrow></msub></math> is
|
singleCorrect
| null |
Equation of plane <math><msub><mrow><mi>P</mi></mrow><mrow><mn>1</mn></mrow></msub></math> is <math><mfenced close="|" open="|" separators="|"><mrow><mtable columnalign="left"><mtr><mtd><mi>x</mi><mo>-</mo><mn>1</mn></mtd><mtd><mi>y</mi><mo>-</mo><mn>3</mn></mtd><mtd><mi>z</mi><mo>-</mo><mn>4</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mn>2</mn></mtd><mtd><mn>3</mn></mtd></mtr><mtr><mtd><mn>2</mn></mtd><mtd><mn>3</mn></mtd><mtd><mn>1</mn></mtd></mtr></mtable></mrow></mfenced><mo>=</mo><mn>0</mn></math><br /><math><mfenced separators="|"><mrow><mi>x</mi><mo>-</mo><mn>1</mn></mrow></mfenced><mfenced separators="|"><mrow><mo>-</mo><mn>7</mn></mrow></mfenced><mo>-</mo><mfenced separators="|"><mrow><mi>y</mi><mo>-</mo><mn>3</mn></mrow></mfenced><mfenced separators="|"><mrow><mo>-</mo><mn>5</mn></mrow></mfenced><mo>+</mo><mfenced separators="|"><mrow><mi>z</mi><mo>-</mo><mn>4</mn></mrow></mfenced><mfenced separators="|"><mrow><mo>-</mo><mn>1</mn></mrow></mfenced><mo>=</mo><mn>0</mn></math><br /><math><mn>7</mn><mi>x</mi><mo>-</mo><mn>5</mn><mi>y</mi><mo>+</mo><mi>z</mi><mo>+</mo><mn>4</mn><mo>=</mo><mn>0</mn></math><br />So, distance between planes <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mrow><mn>1</mn><mo> </mo></mrow></msub><mo>&</mo><mo> </mo><msub><mi>P</mi><mn>2</mn></msub></math> is <math><mo>=</mo><mfenced close="|" open="|" separators="|"><mrow><mfrac><mrow><mn>10</mn></mrow><mrow><msqrt><mn>49</mn><mo>+</mo><mn>25</mn><mo>+</mo><mn>1</mn></msqrt></mrow></mfrac></mrow></mfenced><mo>=</mo><mfrac><mrow><mn>10</mn></mrow><mrow><mn>5</mn><msqrt><mn>3</mn></msqrt></mrow></mfrac></math><br /><math><mo>=</mo><mfrac><mrow><mn>2</mn></mrow><mrow><msqrt><mn>3</mn></msqrt></mrow></mfrac></math>
|
["<math><mfrac><mrow><mn>11</mn></mrow><mrow><mn>5</mn><msqrt><mn>3</mn></msqrt></mrow></mfrac></math>", "<math><mfrac><mrow><mn>2</mn></mrow><mrow><msqrt><mn>3</mn></msqrt></mrow></mfrac></math>", "<math><mfrac><mrow><mn>1</mn></mrow><mrow><msqrt><mn>3</mn></msqrt></mrow></mfrac></math>", "<math><mfrac><mrow><mn>7</mn></mrow><mrow><mn>5</mn><msqrt><mn>3</mn></msqrt></mrow></mfrac></math>"]
|
[1]
| null |
Practise Problem
|
d8a1a9f3dc8bc40458368ca9d9307aa8
|
BITSAT_MATH
|
Three Dimensional Geometry
|
A point $P$ lies on a line through $Q(1,-2,3)$ and is parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5},$ If $P$ lies on the plane $2 x+3 y-4 z+22=0,$ then segment PQ equals
|
singleCorrect
| 2 |
Equation of line through $Q(1,-2,3)$ and<br/>parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is $\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda \quad$ (say)<br/>since, point $P$ lies on above line.<br/>$\therefore$<br/>$$<br/>P(\lambda+1,4 \lambda-2,5 \lambda+3)<br/>$$<br/>since, $P$ lies on the given plane. $\therefore \quad 2(\lambda+1)+3(4 \lambda-2)-4(5 \lambda+3)+22=0$<br/>$\Rightarrow \quad 2 \lambda+2+12 \lambda-6-20 \lambda-12+22=0$<br/>$$<br/>\Rightarrow \quad-6 \lambda+6=0<br/>$$<br/>$\Rightarrow$<br/>$\lambda=1$<br/>$\therefore$<br/>$P(2,2,8)$<br/>$\therefore \quad P Q=\sqrt{(2-1)^{2}+(2+2)^{2}+(3-8)^{2}}$<br/>$\Rightarrow \quad P Q=\sqrt{1+16+25}=\sqrt{42}$
|
["$\\sqrt{42}$ units", "$\\sqrt{32}$ units", "4 units", "5 units"]
|
[0]
| null |
Practise Problem
|
ebc55a9a429450b9a5039cbd761afc79
|
BITSAT_MATH
|
Three Dimensional Geometry
|
A plane meets the axes in $A, B$ and $C$ such that centroid of the $\Delta A B C$ is $(1,2,3)$. The equation of the plane is
|
singleCorrect
| 3 |
Let the plane meets the coordinate axes at $A, B$ and $C .$
$$
\text { Let } \quad O A=a, O B=b \text { and } O C=c
$$
<img src="https://cdn-question-pool.getmarks.app/pyq/mht_cet/KsCvCV_8vyXPKxr4KpjY5jFPF7ilbTRx9XW4aw5d9aI.original.fullsize.png"><br>
Then, the centroid of $\Delta A B C$
is
$$
\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)=(1,2,3)
$$
(given)
$$
\Rightarrow \quad a=3, b=6, c=9
$$
Then, the equation of plane, meet the coordinate axes is
$$
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
$$
i.e.
$$
\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1
$$
|
["$x+y / 2+z / 3=1$", "$x / 3+y / 6+z / 9=1$", "$x+2 y+3 z=1$", "None of the above"]
|
[1]
| null |
Practise Problem
|
b8602197de5b52c7bb6f4f523a35b7be
|
BITSAT_MATH
|
Three Dimensional Geometry
|
From a point $P(a, b, c)$ perpendicular $P A, P B$ are drawn to $y z$ and $z x$ planes. Find the equation of the plane $O A B$, where $O$ is the origin.
|
singleCorrect
| 3 |
$P(a, b, c)$ and $P A$ and $P B$ are perpendicular to $Y Z$ and $Z X$ planes. Hence, coordinate of $A$ and $B$ are $(0, b, c)$ and $(a, 0, c)$ respectively.
Equation of plane passing through $(0,0,0),(0, b, c)$ and
$$
\begin{aligned}
&(a, 0, c) \text { is }\left|\begin{array}{lll}
x & y & z \\
0 & b & c \\
a & 0 & c
\end{array}\right|=0 \\
&\Rightarrow \quad x(b c-0)-y(0-a c)+z(0-a b)=0 \\
&\Rightarrow \quad b c x+a c y-a b z=0
\end{aligned}
$$
|
["$b c x+c a y+a b z=0$", "$b c x+c a y-a b z=0$", "$b c x-c a y+a b z=0$", "$-b c x+c a y+a b z=0$"]
|
[1]
| null |
Practise Problem
|
08cdb0e70dd74fcaf86ede0fa5a9a95e
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The point of intersection of the plane <math><mn>3</mn><mi>x</mi><mo>-</mo><mn>5</mn><mi>y</mi><mo>+</mo><mn>2</mn><mi>z</mi><mo>=</mo><mn>6</mn></math> with the straight line passing through the origin and perpendicular to the plane <math><mn>2</mn><mi>x</mi><mo>-</mo><mi>y</mi><mo>-</mo><mi>z</mi><mo>=</mo><mn>4</mn></math> is
|
singleCorrect
| null |
A vector parallel to line is <math><mn>2</mn><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>-</mo><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>-</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></math><br />So equation of line is<br /><math><mfrac><mrow><mi>x</mi><mo>-</mo><mn>0</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>0</mn></mrow><mrow><mo>-</mo><mn>1</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>0</mn></mrow><mrow><mo>-</mo><mn>1</mn></mrow></mfrac></math><br />Any point on the line is <math><mfenced separators="|"><mrow><mn>2</mn><mi>λ</mi><mo>,</mo><mo>-</mo><mi>λ</mi><mo>,</mo><mo>-</mo><mi>λ</mi></mrow></mfenced></math><br />It must satisfy the given plane<br /><math><mn>6</mn><mi>λ</mi><mo>+</mo><mn>5</mn><mi>λ</mi><mo>-</mo><mn>2</mn><mi>λ</mi><mo>=</mo><mn>6</mn></math><br /><math><mn>9</mn><mi>λ</mi><mo>=</mo><mn>6</mn><mo>⇒</mo><mi>λ</mi><mo>=</mo><mfrac><mrow><mn>2</mn></mrow><mrow><mn>3</mn></mrow></mfrac></math><br />Point is <math><mfenced separators="|"><mrow><mfrac><mrow><mn>4</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mo>,</mo><mfrac><mrow><mo>-</mo><mn>2</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mo>,</mo><mfrac><mrow><mo>-</mo><mn>2</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></mfenced></math>
|
["<math><mfenced separators=\"|\"><mrow><mn>1</mn><mo>,</mo><mo>-</mo><mn>1</mn><mo>,</mo><mo>-</mo><mn>1</mn></mrow></mfenced></math>", "<math><mfenced separators=\"|\"><mrow><mo>-</mo><mn>1</mn><mo>,</mo><mo>-</mo><mn>1,2</mn></mrow></mfenced></math>", "<math><mfenced separators=\"|\"><mrow><mn>4,2</mn><mo>,</mo><mn>2</mn></mrow></mfenced></math>", "<math><mfenced separators=\"|\"><mrow><mfrac><mrow><mn>4</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mo>,</mo><mfrac><mrow><mo>-</mo><mn>2</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mo>,</mo><mfrac><mrow><mo>-</mo><mn>2</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></mfenced></math>"]
|
[3]
| null |
Practise Problem
|
9b57500e012b5490ea93dd2a6ba6e004
|
BITSAT_MATH
|
Three Dimensional Geometry
|
A variable plane is at a constant distance $h$ from the origin and meets the coordinate axes in $A, B, C$. Locus of centroid of $A B C$ is
|
singleCorrect
| 3 |
The centroid of the triangle is $C\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$
<img src="https://cdn-question-pool.getmarks.app/pyq/ap_eamcet/YUboTmZu-Svj4usG9XluNQ1R6tEjoCQAgcTiRN0j-zE.original.fullsize.png"><br>
The direction coines of the line $O C$ are $\left(\frac{h}{a}, \frac{h}{b}, \frac{h}{c}\right)$.
As we know that,
$\begin{array}{rlrl} & l^2+m^2+n^2 & =1 \\ \Rightarrow & & \frac{h^2}{a^2}+\frac{h^2}{b^2}+\frac{h^2}{c^2} & =1 \\ \Rightarrow & & \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} & =\frac{1}{h^2} \\ \Rightarrow & & \left(\frac{3}{a}\right)^2+\left(\frac{3}{b}\right)^2+\left(\frac{3}{c}\right)^2 & =\frac{9}{h^2}\end{array}$
The locus of the centroid is
$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{9}{h^2}$
|
["$x^2+y^2+z^2=h^{-2}$", "$x^2+y^2+z^2=4 h^{-2}$", "$x^2+y^2+z^2=16 h^2$", "$\\frac{1}{x^2}+\\frac{1}{y^2}+\\frac{1}{z^2}=\\frac{9}{h^2}$"]
|
[3]
| null |
Practise Problem
|
a48c0dde55542759bccff3bf01ba7045
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If a plane meets the coordinate axes at $A, B$ and $C$ such that the centroid of the triangle is $(1,2,4)$, then the equation of the plane is
|
singleCorrect
| 2 |
Let the equation of the plane is<br>$\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=1$<br>Then, $\mathrm{A}(\alpha, 0,0), \mathrm{B}(0, \beta, 0)$ and $\mathrm{C}(0,0, \gamma)$ are the points on the coordinate axes, The centroid of the triangle is $(1,2,4)$.<br>$\begin{array}{l}<br>\therefore \frac{\alpha}{3}=1 \Rightarrow \alpha=3 \\<br>\frac{\beta}{3}=2 \Rightarrow \beta=6<br>\end{array}$<br>and $\frac{\gamma}{3}=4 \Rightarrow \gamma=12$<br>$\therefore \quad$ The equation of the plane is<br>$\begin{aligned}<br>& \frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1 \\<br>\Rightarrow & 4 x+2 y+z=12<br>\end{aligned}$
|
["$x+2 y+4 z=12$", "$4 x+2 y+z=12$", "$x+2 y+4 z=3$", "$4 x+2 y+z=3$"]
|
[1]
| null |
Practise Problem
|
66bcdc99a2028dbaa289891ba9185710
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The equation of the plane passing through the points $(2,3,1),(4,-5,3)$ and parallel
to $\mathrm{y}$ -axis is
|
singleCorrect
| 2 |
Equation of plane passing through the point $(2,3,1)$ is
$a(x-2)+b(y-3)+c(z-1)=0$ $\ldots(1)$
Given point $(4,-5,3)$ lies on plane
$2 a-8 b+2 c=0$ $\ldots(2)$
Since plane is parallel to $\mathrm{Y}$-axis, having d.r. $(0,1,0)$
(a) $(0)+(\mathrm{b})(1)+(\mathrm{c})(0)=0 \Rightarrow \mathrm{b}=0$
Putting in equation (2) we get
$2 a+2 c=0 \Rightarrow a=-c$
Putting values of $\mathrm{a}, \mathrm{b}$ in equation (1)
$\begin{aligned}
&-c(x-2)+c(z-1)=0 \\
\therefore &(x-2)-(z-1)=0 \\
\therefore & x-z=1
\end{aligned}$
|
["$x+z=3$", "$x+z=1$", "$x-z=1$", "$z-x+2=0$"]
|
[2]
| null |
Practise Problem
|
770d907797bc2fbff9a1aac593b089db
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$ lies in the plane $a x+b y+z=7$, then $a+b=$
|
singleCorrect
| 1 |
Given that, line
$\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$
Line is passing through point $P(4,+2,7)$ and direction ratios $(1,1,2)$
and lies on the given plane $a x+b y+z=7$
So, point will satisfy the equation of plane $a(4)+b(+2)+(7)=7$
$4 a+2 b=0$ $\ldots(i)$
Now direction ratios of the line and direction ratios of the plane will be perpendicular so their product must be zero
$a_1 a_2+b_1 b_2+c_1 c_2=(1)(a)+(1)(b)+(2)(1)$
$0=a+b+2$
or $\quad a+b=-2$
|
["-2", "3", "5", "7"]
|
[0]
| null |
Practise Problem
|
757193967a33aee68ca7f8b83d54831c
|
BITSAT_MATH
|
Three Dimensional Geometry
|
Equation of line passing through the point \((2,3,1)\) and parallel to the line of intersection of the<br>plane \(x-2 y-z+5=0\) and \(x+y+3 z=6\) is
|
singleCorrect
| 2 |
Given equation of planes,<br>\(\begin{array}{l}
P_{1}: x-2 y-z+5=0 \rightarrow(1) \\
P_{2}: x+y+3 z=6 \rightarrow(2)
\end{array}\)<br>and point \(P(2,3,1)\).<br>Normal vector of Eq. (1) is given by,<br>\(\vec{N}_{1}=\hat{i}-2 \hat{j}+\hat{k} \rightarrow(3)\)<br>Similarly, normal vector of Eq. (2) is given by<br>\(\vec{N}_{2}=\hat{i}+\hat{j}+3 \hat{k} \rightarrow(4)\)<br>Vector perpendicular to the normal to the plans are:<br>\(\begin{aligned}
\vec{b} &=\vec{N}_{1} \times \vec{N}_{2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 1 \\
1 & 1 & 3
\end{array}\right| \\
&=\hat{i}(-6+1)-\hat{j}(3+1)+\hat{k}(1+2)=-5 \hat{i}-4 \hat{j}+3 \hat{k}
\end{aligned}\)<br>So, required equation of line passing through the point \((2,3,1)\) is,<br>\(\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
|
["\\(\\frac{x-2}{5}=\\frac{y-3}{-4}=\\frac{z-1}{3}\\)", "\\(\\frac{x-2}{-5}=\\frac{y-3}{-4}=\\frac{z-1}{3}\\)", "\\(\\frac{x-2}{5}=\\frac{y-3}{4}=\\frac{z-1}{3}\\)", "\\(\\frac{x-2}{4}=\\frac{y-3}{3}=\\frac{z-1}{2}\\)"]
|
[1]
| null |
Practise Problem
|
1d9656a2563ced23d85e17141864dfcf
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The distance of the point $(-5,-5,-10)$ from the point of intersection of the line $\mathrm{r} .=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$ and the plane $r \cdot(\hat{i}-\hat{j}+\hat{k})=5$ is
|
singleCorrect
| 2 |
Given equation of line is<br>$\begin{gathered}<br>r=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k}) \\<br>(x \hat{i}+y \hat{j}+z \hat{k})=(2+3 \lambda) \hat{i}+(-1+4 \lambda) \hat{j}+(2+2 \lambda) \hat{k}<br>\end{gathered}$<br>Any point on the line is<br>$$<br>(2+3 \lambda,-1+4 \lambda, 2+2 \lambda)<br>$$<br>Since it also lie on the plane $r \cdot(\hat{i}-\hat{j}+\hat{k})$<br>$\begin{aligned}<br>& \text { So, }[(2+3 \lambda) \hat{i}+(-1+4 \lambda) \hat{j}+(2+2 \lambda) \hat{k}] \cdot(\hat{i}-\hat{j}+\hat{k})=5 \\<br>& \Rightarrow 2+3 \lambda+1-4 \lambda+2+2 \lambda=5 \Rightarrow \lambda=0<br>\end{aligned}$<br>Therefore, coordinate of the point of intersection of line and plane is $(2,-1,2)$.<br>$\therefore$ Distance<br>$$<br>\mathrm{d}=\sqrt{(2+1)^2+(-1+5)^2+(2+10)^2}=13<br>$$
|
["13", "12", "$4 \\sqrt{15}$", "$10 \\sqrt{2}$"]
|
[0]
| null |
Practise Problem
|
58abdb656ab9ad41dac445a848d142aa
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If the points having the position vectors $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}},-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ are coplanar, then $\lambda=$
|
singleCorrect
| 2 |
Let $A, B, C$ and $D$ are the position vectors of four points given as
$$
\begin{aligned}
& A(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}}), B(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}), C(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\
& \text { and } D(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}})
\end{aligned}
$$
Here,
$$
\begin{aligned}
& \mathrm{AB}=-\hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\
& \mathrm{BC}=-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}
\end{aligned}
$$
and
$$
C D=5 \hat{i}+4 \hat{j}+(\lambda-2) \hat{k}
$$
As given that, these points are coplanar, therefore $[\mathrm{AB} \quad \mathrm{BC} \mathrm{CD}]=0$
$$
\begin{aligned}
& \left|\begin{array}{ccc}
-1 & 5 & -3 \\
-3 & -2 & 6 \\
5 & 4 & \lambda-2
\end{array}\right|=0 \\
\Rightarrow \quad & -1(-2 \lambda+4-24)-5(-3(\lambda-2)-30) \\
\Rightarrow \quad & -3(-12+10)=0 \\
\Rightarrow \quad & 17 \lambda+20+20+15(\lambda-2)+150+36-30=0 \\
\Rightarrow \quad & 17 \lambda+146=0 \Rightarrow \lambda=-\frac{146}{17}
\end{aligned}
$$
|
["$\\frac{46}{17}$", "8", "-8", "$\\frac{146}{17}$"]
|
[0]
| null |
Practise Problem
|
18eb5fd95fd8f8245d51ebfc696b55be
|
BITSAT_MATH
|
Three Dimensional Geometry
|
Lying in the plane <math><mi>x</mi><mo>+</mo><mi>y</mi><mo>+</mo><mi>z</mi><mo>=</mo><mn>6</mn></math> is a line <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>L</mi></math> passing through <math><mfenced separators="|"><mrow><mn>1,2</mn><mo>,</mo><mn>3</mn></mrow></mfenced></math> and perpendicular to the line of intersection of planes <math><mi>x</mi><mo>+</mo><mi>y</mi><mo>+</mo><mi>z</mi><mo>=</mo><mn>6</mn></math> and <math><mn>2</mn><mi>x</mi><mo>-</mo><mi>y</mi><mo>+</mo><mi>z</mi><mo>=</mo><mn>4</mn><mo>,</mo></math> then the equation of <math><mi>L</mi></math> is
|
singleCorrect
| null |
Required line lies in the plane <math><mi>x</mi><mo>+</mo><mi>y</mi><mo>+</mo><mi>z</mi><mo>=</mo><mn>6</mn></math> so it is perpendicular to <math><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></math> and it is also perpendicular to line of intersection of the two given planes.<br />Let, <math><msub><mrow><mover accent="true"><mrow><mi>n</mi></mrow><mo>→</mo></mover></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover><mo>,</mo><mi mathvariant="normal"></mi><msub><mrow><mover accent="true"><mrow><mi>n</mi></mrow><mo>→</mo></mover></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mn>2</mn><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>-</mo><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></math> & <math><mover accent="true"><mrow><mi>ν</mi></mrow><mo>→</mo></mover></math> be a vector along the required line<br />So, <math><mover accent="true"><mrow><mi>ν</mi></mrow><mo>→</mo></mover><mo>=</mo><msub><mrow><mover accent="true"><mrow><mi>n</mi></mrow><mo>→</mo></mover></mrow><mrow><mn>1</mn></mrow></msub><mo>×</mo><mfenced separators="|"><mrow><msub><mrow><mover accent="true"><mrow><mi>n</mi></mrow><mo>→</mo></mover></mrow><mrow><mn>1</mn></mrow></msub><mo>×</mo><msub><mrow><mover accent="true"><mrow><mi>n</mi></mrow><mo>→</mo></mover></mrow><mrow><mn>2</mn></mrow></msub></mrow></mfenced></math><br /><math><mo>=</mo><mn>2</mn><mfenced separators="|"><mrow><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></mrow></mfenced><mo>-</mo><mn>3</mn><mfenced separators="|"><mrow><mn>2</mn><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>-</mo><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></mrow></mfenced></math><br /><math><mo>=</mo><mo>-</mo><mn>4</mn><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>+</mo><mn>5</mn><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>-</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></math><br />Hence, equation of the line is <math><mfrac><mrow><mi>x</mi><mo>-</mo><mn>1</mn></mrow><mrow><mn>4</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>2</mn></mrow><mrow><mo>-</mo><mn>5</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>3</mn></mrow><mrow><mn>1</mn></mrow></mfrac></math>
|
["<math><mfrac><mrow><mi>x</mi><mo>-</mo><mn>1</mn></mrow><mrow><mn>4</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>2</mn></mrow><mrow><mo>-</mo><mn>7</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>3</mn></mrow><mrow><mn>3</mn></mrow></mfrac></math>", "<math><mfrac><mrow><mi>x</mi><mo>-</mo><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>2</mn></mrow><mrow><mn>1</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>3</mn></mrow><mrow><mo>-</mo><mn>3</mn></mrow></mfrac></math>", "<math><mfrac><mrow><mi>x</mi><mo>-</mo><mn>1</mn></mrow><mrow><mn>4</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>2</mn></mrow><mrow><mo>-</mo><mn>5</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>3</mn></mrow><mrow><mn>1</mn></mrow></mfrac></math>", "<math><mfrac><mrow><mi>x</mi><mo>-</mo><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>2</mn></mrow><mrow><mn>1</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>3</mn></mrow><mrow><mo>-</mo><mn>4</mn></mrow></mfrac></math>"]
|
[2]
| null |
Practise Problem
|
07f976eef95599b842898ee40d275675
|
BITSAT_MATH
|
Three Dimensional Geometry
|
A plane $\mathrm{E}_{1}$ makes intercepts $1,-3,4$ on the co-ordinate axes. The equation of a plane
parallel to plane $\mathrm{E}_{1}$ and passing through $(2,6,-8)$ is
|
singleCorrect
| 2 |
A plane $\mathrm{E}_{1}$ makes intercepts $1,-3,4$ on the coordinate axes
Equation of plane is $\frac{x}{1}+\frac{y}{-3}+\frac{z}{4}=1 \Rightarrow 12 x-4 y+3 z=12$
d.r.s. are $12,-4,3$
Since required plane is parallel to given plane, normal vector $\bar{n}$ to required plane is
$\overline{\mathrm{n}}=12 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
The vector equation of the plane passing through $(2,6,-8)$ is
$\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}$, where $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$
$\bar{a} \cdot \bar{n}=(12)(2)-(4)(6)+(3)(-8)=24-24-24=-24$
$\therefore$ Required equation is $\overline{\mathrm{r}} \cdot(12 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=-24$
$\therefore$ Cartesion form of equation is $12 x-4 y+3 z+24=0$
$\therefore \frac{x}{1}-\frac{y}{3}+\frac{z}{4}+2=0$
|
["$\\frac{x}{2}-\\frac{y}{3}+\\frac{z}{4}+3=0$", "$\\frac{x}{1}-\\frac{y}{3}+\\frac{z}{4}+12=0$", "$\\frac{x}{1}-\\frac{y}{3}+\\frac{z}{4}+2=0$", "$\\frac{x}{3}-\\frac{y}{6}+\\frac{z}{2}+\\frac{13}{3}=0$"]
|
[2]
| null |
Practise Problem
|
c9e25eaa89e9f86d9e9a6ec5c5835de7
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The combined equation for a pair of planes is <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mo>≡</mo><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>-</mo><mn>6</mn><msup><mi>y</mi><mn>2</mn></msup><mo>-</mo><mn>12</mn><msup><mi>z</mi><mn>2</mn></msup><mo>+</mo><mn>18</mn><mi>y</mi><mi>z</mi><mo>+</mo><mn>2</mn><mi>z</mi><mi>x</mi><mo>+</mo><mi>x</mi><mi>y</mi><mo>=</mo><mn>0</mn><mo>.</mo></math> If one of the planes is parallel to<br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>+</mo><mn>2</mn><mi>y</mi><mo>-</mo><mn>2</mn><mi>z</mi><mo>=</mo><mn>5</mn><mo>,</mo></math> then the acute angle between the planes <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mo>=</mo><mn>0</mn></math> is<gwmw style="display:none;"></gwmw>
|
singleCorrect
| 2 |
<p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mo>=</mo><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>-</mo><mn>6</mn><msup><mi>y</mi><mn>2</mn></msup><mo>-</mo><mn>12</mn><msup><mi>z</mi><mn>2</mn></msup><mo>+</mo><mn>18</mn><mi>y</mi><mi>z</mi><mo>+</mo><mn>2</mn><mi>z</mi><mi>x</mi><mo>+</mo><mi>x</mi><mi>y</mi><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mo>=</mo><mo>(</mo><mi>x</mi><mo>+</mo><mn>2</mn><mi>y</mi><mo>-</mo><mn>2</mn><mi>z</mi><mo>)</mo><mo>(</mo><mn>2</mn><mi>x</mi><mo>-</mo><mn>3</mn><mi>y</mi><mo>+</mo><mn>6</mn><mi>z</mi><mo>)</mo><mo>=</mo><mn>0</mn></math></p><p>Hence, <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>(</mo><mi>x</mi><mo>+</mo><mn>2</mn><mi>y</mi><mo>-</mo><mn>2</mn><mi>z</mi><mo>)</mo></math> is parallel to one plane.</p><p>So, <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mo>-</mo><mn>2</mn></mrow></mfenced><mo>,</mo><mo> </mo><mo>(</mo><mn>2</mn><mo>,</mo><mo>-</mo><mn>3</mn><mo>,</mo><mn>6</mn><mo>)</mo></math> will be the required direction ratios.</p><p>Here, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>cos</mi><mi>θ</mi><mo>=</mo><mfrac><mrow><mn>2</mn><mo>-</mo><mn>6</mn><mo>-</mo><mn>12</mn></mrow><mrow><msqrt><mn>9</mn></msqrt><mo>×</mo><msqrt><mn>49</mn></msqrt></mrow></mfrac><mo>=</mo><mfrac><mn>16</mn><mrow><mn>3</mn><mo>×</mo><mn>7</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mo>-</mo><mn>16</mn></mrow><mn>21</mn></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>θ</mi><mo>=</mo><msup><mi>cos</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup><mfenced><mfrac><mrow><mo>-</mo><mn>16</mn></mrow><mn>21</mn></mfrac></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>{</mo><mi>cos</mi><mo>(</mo><mo>-</mo><mi>θ</mi><mo>)</mo><mo>=</mo><mi>cos</mi><mi>θ</mi><mo>}</mo></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>θ</mi><mo>=</mo><msup><mi>cos</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup><mfenced><mfrac><mn>16</mn><mn>21</mn></mfrac></mfenced></math></p>
|
["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><msup><mi>cos</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup><mfenced><mfrac><mn>16</mn><mn>21</mn></mfrac></mfenced></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mfrac><mi>π</mi><mn>2</mn></mfrac></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mfrac><mrow><mn>2</mn><mi>π</mi></mrow><mn>3</mn></mfrac></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><msup><mi>sin</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup><mfenced><mfrac><mn>7</mn><mn>15</mn></mfrac></mfenced></math>"]
|
[0]
| null |
Practise Problem
|
ec850c37fad610aeeca6625d1e491870
|
BITSAT_MATH
|
Three Dimensional Geometry
|
$P, Q, R$ and $S$ are four points with the position vectors $3 \mathbf{i}-4 \mathbf{j}+5 \mathbf{k},-4 \mathbf{i}+5 \mathbf{j}+\mathbf{k}$ and $-3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}$, respectively. Then, the line $P Q$ meets the line $R S$ at the point
|
singleCorrect
| 2 |
Let the coordinates of four points $P, Q, R$ and $S$ be $(3,-4,5), \quad(0,0,4), \quad(-4,5,1) \quad$ and $\quad(-3,4,3)$ respectively.
Now, equation of line $P Q$ is
$$
\begin{aligned}
& \frac{x-3}{0-3}=\frac{y+4}{0+4}=\frac{z-5}{4-5} \\
\Rightarrow \quad & \frac{x-3}{-3}=\frac{y+4}{4}=\frac{z-5}{-1}=r_1
\end{aligned}
$$
Equation of line $R S$ is
$$
\frac{x+4}{-3+4}=\frac{y-5}{4-5}=\frac{z-1}{3-1}
$$
$$
\Rightarrow \quad \frac{x+4}{1}=\frac{y-5}{-1}=\frac{z-1}{2}=r_2
$$
Let $\left(-3 r_1+3,4 r_1-4,-r_1+5\right)$ and $\left(r_2-4,-r_2+5\right.$, $\left.2 r_2+1\right)$ be the points on line (i) and (ii), respectively. Since, both lines intersect at a common point, then
$$
\begin{aligned}
& & -3 r_1+3 & =r_2-4 \\
\Rightarrow & & 3 r_1+r_2 & =7 \\
\text { and } & & -r_2+5 & =4 r_1-4 \\
\Rightarrow & & 4 r_1+r_2 & =9
\end{aligned}
$$
On subtracting Eq. (iv) from Eq. (iii), we get
$$
r_1=2
$$
On putting the value of $r_1$ in Eq. (iii), we get
$$
3(2)+r_2=7 \Rightarrow r_2=1
$$
So, required point of intersection is $(-3,4,3)$ i.e.,
$$
-3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}
$$
|
["$3 \\mathbf{i}+4 \\mathbf{j}+3 \\mathbf{k}$", "$-3 \\mathbf{i}+4 \\mathbf{j}+3 \\mathbf{k}$", "$-\\mathbf{i}+4 \\mathbf{j}+\\mathbf{k}$", "$\\mathbf{i}+\\mathbf{j}+\\mathbf{k}$"]
|
[1]
| null |
Practise Problem
|
79abb8b126fb3c574d4342796412cfca
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The equation of the plane, which bisects the line joining the points (1,2,3) and (3,4,5) at right angles is
|
singleCorrect
| 2 |
The given points are $A(1,2,3)$ and $B(3,4,5)$ The direction ratios of line segment $A B$ is given by<br/>$\begin{array}{l}<br/>\left(x_{2}-x_{1}\right),\left(y_{2}-y_{1}\right),\left(z_{2}-z_{1}\right) \\<br/>(3-1),(4-2),(5-3)=(2,2,2)=(1,1,1)<br/>\end{array}$<br/>Since, the plane bisects $A B$ at right angles, $\mathbf{A B}$ is the normal to the plane. Which is n Therefore, $\mathbf{n}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$<br/>Let $c$ be the mid-point of $A B$<br/>$\begin{aligned}<br/>\left(\frac{x_{1}+x_{2}}{2},\right.&\left.\frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right) \\<br/>=& c\left(\frac{1+3}{2}, \frac{2+4}{2}, \frac{3+5}{2}\right) \\<br/>=& c(2,3,4)<br/>\end{aligned}$<br/>Let this vector be $\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$<br/>Hence, the required equation of vector<br/>$$<br/>\{\mathbf{r}-(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \mathbf{k})\}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\mathbf{k})=0<br/>$$<br/>We know that, $\mathbf{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$<br/>$\begin{array}{l}<br/>\Rightarrow\{(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-(2\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})\}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \\<br/>\Rightarrow \quad(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\<br/>=(2\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\<br/>\Rightarrow x+y+z=2+3+4 \\<br/>\Rightarrow x+y+z=9<br/>\end{array}$
|
["$x+y+z=0$", "$x+y-z=9$", "$x+y+z=9$", "$x+y-z+9=0$"]
|
[2]
| null |
Practise Problem
|
bacaf1ca8c59065c8545c43a022d94b1
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The distance of the point $(3,4,5)$ from the point of intersection of the line
$\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ and plane $x+y+z=2$ is
|
singleCorrect
| 2 |
Given $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda \quad \ldots$ (say) ...(1)
Hence coordinates of any point on this line are $\therefore x=\lambda+3, y=2 \lambda+4, z=2 \lambda+5$
Since this point lies on the plane, we write
$\begin{array}{l}
(\lambda+3)+(2 \lambda+4)+(2 \lambda+5)=2 \\
5 \lambda+12=2 \Rightarrow \lambda=-2
\end{array}$
Hence coordinates of point of inter section are $\equiv(-2+3,-4+4,-4+5)$ i.e. $(1,0,1)$ $\therefore$ Distance between $(3,4,5)$ and $(1,0,1)$ is
$=\sqrt{(3-1)^{2}+(4-0)^{2}+(5-1)^{2}}=6$
|
["6 units", "13 units", "10 units", "7 units"]
|
[0]
| null |
Practise Problem
|
4519397a9195b9ee2cbbbb63fac05a1b
|
BITSAT_MATH
|
Three Dimensional Geometry
|
Determine the plane through the intersection of the planes $x+2 y+3 z-4=0$ and
$2 x+y-z+5=0$ and perpendicular to the plane
$5 x+3 y+6 z+8=0$
|
singleCorrect
| 3 |
Equation of plane through the intersection of planes $x+2 y+3 z-4=0$ and $2 x+y-z+5=0$ is
$(x+2 y+3 z-4)+k(2 x+y-z+5)=0$
or $(1+2 k) x+(2+k) y+(3-k) z+(5 k-4)=0 \quad \text{...(i)}$
D.R.' $s$ of normal of plane (i) are
$$
= < (1+2 k),(2+k),(3-k)>
$$
Given, $5 x+3 y+6 z+8=0 \quad \text{...(ii)}$
D.R.'s of plane (ii) are $ < 5,3,6>$.
Since Eq. (i) is Perpendicular to the plane (ii),
$$
\begin{aligned}
&\therefore \quad 5(1+2 k)+(2+k) 3+6(3-k)=0 \\
&\Rightarrow \quad 5+10 k+6+3 k+18-6 k=0 \\
&\Rightarrow 7 k+29=0 \Rightarrow k=\frac{-29}{7}
\end{aligned}
$$
$\therefore$ Required equation of plane is
$$
\begin{aligned}
&(x-2 y+3 z-4)+\left(-\frac{29}{7}\right)(2 x+y-z+5)=0 \\
&\Rightarrow 7 x+14 y+21 z-28-58 x-29 y+29 z-145=0 \\
&\Rightarrow-51 x-15 y+50 z-173=0 \\
&\Rightarrow 51 x+15 y-50 z+173=0
\end{aligned}
$$
|
["$-51 x-15 y-50 z-173=0$", "$51 x+15 y-50 z+173=0$", "$51 x-15 y+50 z-173=0$", "$51 x+50 y+15 z+173=0$"]
|
[1]
| null |
Practise Problem
|
c509f9ef8940d222bce9f3639b9bd857
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If the planes <math><mover accent="true"><mrow><mi>r</mi></mrow><mo>→</mo></mover><mo>⋅</mo><mfenced separators="|"><mrow><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></mrow></mfenced><mo>=</mo><mn>1</mn><mo>,</mo><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mover accent="true"><mrow><mi>r</mi></mrow><mo>→</mo></mover><mo>⋅</mo><mfenced separators="|"><mrow><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>+</mo><mn>2</mn><mi>a</mi><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></mrow></mfenced><mo>=</mo><mn>2</mn></math> and <math><mover accent="true"><mrow><mi>r</mi></mrow><mo>→</mo></mover><mo>⋅</mo><mfenced separators="|"><mrow><mi>a</mi><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>+</mo><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></mrow></mfenced><mo>=</mo><mn>3</mn></math> intersect in a line, then the possible number of real values of <math><mi>a</mi></math> is
|
singleCorrect
| null |
<math><mo>∵</mo></math> all <math><mn>3</mn></math> planes intersect in a line<br /><math><mo>⇒</mo><mi mathvariant="normal"></mi><mfenced close="|" open="|" separators="|"><mrow><mtable columnalign="left"><mtr><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mn>2</mn><mi>a</mi></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mi>a</mi></mtd><mtd><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup></mtd><mtd><mn>1</mn></mtd></mtr></mtable></mrow></mfenced><mo>=</mo><mn>0</mn></math><br /><math><mo>⇒</mo><mi mathvariant="normal"></mi><mfenced separators="|"><mrow><mn>2</mn><mi>a</mi><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced><mo>-</mo><mn>1</mn><mfenced separators="|"><mrow><mn>1</mn><mo>-</mo><mi>a</mi></mrow></mfenced><mo>+</mo><mn>1</mn><mfenced separators="|"><mrow><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><mn>2</mn><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfenced><mo>=</mo><mn>0</mn></math><br /><math><mo>⇒</mo><mi mathvariant="normal"></mi><mn>2</mn><mi>a</mi><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><mn>1</mn><mo>+</mo><mi>a</mi><mo>-</mo><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mn>0</mn></math><br /><math><mo>⇒</mo><mi mathvariant="normal"></mi><mo>-</mo><mn>2</mn><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mn>3</mn><mi>a</mi><mo>-</mo><mn>1</mn><mo>=</mo><mn>0</mn><mi mathvariant="normal"></mi><mo>⇒</mo><mi>a</mi><mo>=</mo><mn>1</mn><mo>,</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></math><br />for both values of <math><mi>a</mi><mo>=</mo><mn>1</mn><mo>,</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>,</mo></math> two of the given <math><mn>3</mn></math> planes are parallel (rejected)
|
["<math><mn>1</mn></math>", "<math><mn>2</mn></math>", "<math><mn>4</mn></math>", "<math><mn>0</mn></math>"]
|
[3]
| null |
Practise Problem
|
d278aec7fcdb03df2badf3fe7bdaf6b4
|
BITSAT_MATH
|
Three Dimensional Geometry
|
A plane passing through $(-1,2,3)$ and whose normal makes equal angles with the coordinate axes is
|
singleCorrect
| 2 |
A plane passing through the point $(-1,2,3)$, then its equation is
$$
a(x+1)+b(y-2)+c(z-3)=0
$$
where $\langle a, b, c\rangle$ are direction ratios of normal to the plane $A B C$.
So, the normal makes equal angles with coordinate axes i.e.,
$$
(a, b, c)=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)
$$
Now, from Eq. (i),
$$
\begin{array}{cc}
\frac{1}{\sqrt{3}}(x+1)+\frac{1}{\sqrt{3}}(y-2)+\frac{1}{\sqrt{3}}(z-3)=0 \\
\Rightarrow \quad x+y+z-4 & =0
\end{array}
$$
|
["$x+y+z+4=0$", "$x-y+z+4=0$", "$x+y+z-4=0$", "$x+y+z=0$"]
|
[2]
| null |
Practise Problem
|
e24d2264601ea59bb1d5a3266d069ace
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}$ $=\frac{z-2}{2}$ and the plane $2 x-y+\sqrt{\lambda} z+4=0$ is such that $\sin \theta=\frac{1}{3}$ then the value of $\lambda$ is
|
singleCorrect
| 2 |
If $\theta$ is the angle between line and plane then $\left(\frac{\pi}{2}-\theta\right)$ is the angle between line and normal to plane given by<br>$\begin{aligned}<br>& \cos \left(\frac{\pi}{2}-\theta\right)=\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(2 \hat{i}-\hat{j}+\sqrt{\lambda} \hat{k})}{3 \sqrt{4+1+\lambda}} \\<br>& \cos \left(\frac{\pi}{2}-\theta\right)=\frac{2-2+2 \sqrt{\lambda}}{3 \times \sqrt{5}+\lambda} \\<br>& \Rightarrow \sin \theta=\frac{2 \sqrt{\lambda}}{3 \sqrt{5}+\lambda}=\frac{1}{3} \Rightarrow 4 \lambda=5+\lambda \Rightarrow \lambda=\frac{5}{3} .<br>\end{aligned}$
|
["$\\frac{5}{3}$", "$\\frac{-3}{5}$", "$\\frac{3}{4}$", "$\\frac{-4}{3}$"]
|
[0]
| null |
Practise Problem
|
b60a96eeab40296f637fe00deb343b75
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The co-ordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line
$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ are
|
singleCorrect
| 2 |
Let $\alpha$ be the foot of the Ler drawn from the point $\mathrm{P}(0,2,3)$ to the given line.
Co-ordinates of any point on given line are $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda$...say
Let $Q \equiv(5 \lambda-3,2 \lambda+1,3 \lambda-4)$ and $P \equiv(0,2,3)$
$\therefore$ d.r. of PQ are $5 \lambda-3,2 \lambda-1,3 \lambda-7$
d.r. of given line are $5,2,3$
$\therefore 5(5 \lambda-3)+2(2 \lambda-1)+3(3 \lambda-7)=0$
$\Rightarrow 25 \lambda-15+4 \lambda-2+9 \lambda-21=0 \Rightarrow 38 \lambda-38=0 \Rightarrow \lambda=1$
$\therefore \mathrm{Q} \equiv(2,1,-4)$
<img src="https://cdn.quizrr.in/question-assets/mhtcet/py35bdsq/Vr91t5xwTXxj_2fEmuNJ_tX7mEp7b8xBMizE-vAXG0A.original.fullsize.png">
|
["$(-2,-3,1)$", "$(2,1,-4)$", "$(2,3,1)$", "(-2,-3,-1)"]
|
[1]
| null |
Practise Problem
|
686174517236b1cef234425d2d5fb735
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If the point of intersection of the lines $\mathbf{r}=\hat{\mathbf{i}}-6 \hat{\mathbf{j}}+(p \sec \alpha) \hat{\mathbf{k}}+t(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ and $\mathbf{r}=4 \hat{\mathbf{j}}+\hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+(p \tan \alpha) \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ is $8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}$, (where $0 < \alpha < \frac{\pi}{2}$ ), then $p=$
|
singleCorrect
| 1 |
It is given that the lines
$\mathbf{r}=\hat{\mathbf{i}}-6 \hat{\mathbf{j}}(p \sec \alpha) \hat{\mathbf{k}}+t(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
and $\mathbf{r}=4 \hat{\mathbf{j}}+\hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+(p \tan \alpha) \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$
Intersected at point $8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}$, so $1+t=8,-6+2 t=8$ and $p \sec \alpha+t=9$
$\therefore \quad t=7$ and $p \sec \alpha=2$ $\ldots(\mathrm{i})$
and $2 \lambda=8,4+\lambda p \tan \alpha=8$ and $1+2 \lambda=9$
$\therefore \quad \lambda=4$ and $p \tan \alpha=1$ $\ldots(\mathrm{ii})$
From Eqs. (i) and (ii), we get
$p^2 \sec ^2 \alpha-p^2 \tan ^2 \alpha=4-1 \Rightarrow p^2=3 \Rightarrow p= \pm \sqrt{3}$
|
["$\\sqrt{5}$", "$\\sqrt{3}$", "$\\sqrt{2}$", "0"]
|
[1]
| null |
Practise Problem
|
1b25ff32a4d52eb7fbfff0cdd083f533
|
BITSAT_MATH
|
Three Dimensional Geometry
|
Foot of perpendicular drawn from the origin to the plane \(2 x-3 y+4 z=29\) is
|
singleCorrect
| 2 |
Given equation of plane<br>\(2 x-3 y+4 z=29 \rightarrow(1)\)<br>Normal vector of this plane is given by<br>\(\vec{N}=2 \hat{i}-3 \hat{j}+4 \hat{k}\)<br>Now, equation of line which passes through the origin and has the same direction as normal vectors \(N \rightarrow\) is<br>\(\begin{array}{l}
\frac{x}{2}=\frac{y}{-3}=\frac{z}{4}=\lambda \rightarrow(2) \\
\Rightarrow x=2-x, y=-3 \lambda, z=4 \lambda \rightarrow(3)
\end{array}\)<br>The coordinates of the foot of the perpendicular is<br>\(\begin{array}{l}
4 \lambda+9 \lambda+16 \lambda=29 \\
\Rightarrow 29 \lambda=29 \Rightarrow \lambda=1
\end{array}\)<br>Putting the value of \(\lambda\) in Eq. (3), we have<br>\(x=2, y=-3, z=4\) or \((2,-3,4)\)
|
["\\((5,-1,4)\\)", "\\((2,-3,4)\\)", "\\((7,-1,3)\\)", "\\((5,-2,3)\\)"]
|
[1]
| null |
Practise Problem
|
9a2df78b37674e9784f65e2d9256072a
|
BITSAT_MATH
|
Three Dimensional Geometry
|
Let $P(1,-2,5)$ be the foot of the perpendicular drawn from the origin to the plane $\pi_1$ and the same $P$ be the foot of the perpendicular from $(1,2,-1)$ to the plane $\pi_2$. Then the acute angle between the planes $\pi_1$ and $\pi_2$ is
|
singleCorrect
| 2 |
According to given informations, the direction
ratios of normal to the plane $\pi_1$ are $1-0,-2-0$, $5-0$ i.e. $1,-2,5$.
And similarly direction ratios of normal to the plane $\pi_2$ are
$1-1,2-(-2),-1-5$ i.e. $0,4,-6$
So, the acute angle between planes $\pi_1$ and $\pi_2$ is
$\cos ^{-1}\left|\left(\frac{(1)(0)+(-2)(4)+(5)(-6)}{\sqrt{1+4+25} \sqrt{0+16+36}}\right)\right|$
$\begin{aligned}=\cos ^{-1}\left|\frac{-8-30}{\sqrt{30} \sqrt{52}}\right| & =\cos ^{-1}\left(\frac{38}{2 \sqrt{30 \times 13}}\right) \\ & =\cos ^{-1} \frac{19}{\sqrt{390}}\end{aligned}$
Hence, option (a) is correct.
|
["$\\cos ^{-1}\\left(\\frac{19}{\\sqrt{390}}\\right)$", "$\\cos ^{-1}\\left(\\frac{19}{\\sqrt{340}}\\right)$", "$\\cos ^{-1}\\left(\\frac{19}{\\sqrt{370}}\\right)$", "$\\cos ^{-1}\\left(\\frac{19}{\\sqrt{350}}\\right)$"]
|
[0]
| null |
Practise Problem
|
d1d5df0f605429345f5653d1dc2ee720
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The shortest distance between the lines $\mathbf{r}=3 \mathbf{i}+5 \mathbf{j}+7 \mathbf{k}+\lambda(\mathbf{i}+2 \mathbf{j}+\mathbf{k}) \quad$ and $\mathbf{r}=-\mathbf{i}-\mathbf{j}-\mathbf{k}+\mu(7 \mathbf{i}-6 \mathbf{j}+\mathbf{k})$ is
|
singleCorrect
| 2 |
The given lines are $\mathbf{r}=\mathbf{a}_1+\lambda \mathbf{b}_1, \mathbf{r}=\mathbf{a}_2+\mu \mathbf{b}_2$ where,
$$
\begin{aligned}
& \mathbf{a}_1=3 \mathbf{i}+5 \mathbf{j}+7 \mathbf{k}, \mathbf{b}_1=\mathbf{i}+2 \mathbf{j}+\mathbf{k} \\
& \mathbf{a}_2=-\mathbf{i}-\mathbf{j}-\mathbf{k}, \mathbf{b}_2=7 \mathbf{i}-6 \mathbf{j}+\mathbf{k} \\
& \quad\left|\mathbf{b}_1 \times \mathbf{b}_2\right|=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 1 \\
7 & -6 & 1
\end{array}\right| \\
\Rightarrow \quad & |\mathbf{i}(2+6)-\mathbf{j}(1-7)+\mathbf{k}(-6-14)| \\
\Rightarrow & |8 \mathbf{i}+6 \mathbf{j}-20 \mathbf{k}| \\
\Rightarrow & \sqrt{64+36+400}=\sqrt{500}=10 \sqrt{5}
\end{aligned}
$$
Now,
$$
\begin{aligned}
& {\left[\left(\mathbf{a}_2-\mathbf{a}_1\right) \mathbf{b}_1 \mathbf{b}_2\right]=\left(\mathbf{a}_2-\mathbf{a}_1\right) \cdot\left(\mathbf{b}_1 \times \mathbf{b}_2\right)} \\
& =(-4 \mathbf{i}-6 \mathbf{j}-8 \mathbf{k}) \cdot(8 \mathbf{i}+6 \mathbf{j}-20 \mathbf{k}) \\
& =-32-36+160 \\
& =160-68=92
\end{aligned}
$$
Shortest distance
$$
\begin{aligned}
& =\frac{\left[\left(\mathbf{a}_2-\mathbf{a}_1\right) \cdot\left(\mathbf{b}_1 \times \mathbf{b}_2\right)\right]}{\left|\mathbf{b}_1 \times \mathbf{b}_2\right|} \\
& =\frac{92}{10 \sqrt{5}}=\frac{46}{5 \sqrt{5}}
\end{aligned}
$$
|
["$\\frac{16}{5 \\sqrt{5}}$", "$\\frac{26}{5 \\sqrt{5}}$", "$\\frac{36}{5 \\sqrt{5}}$", "$\\frac{46}{5 \\sqrt{5}}$"]
|
[3]
| null |
Practise Problem
|
931a1c14aa47a6fb7b84cc0c9704b451
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The perpendicular bisector of a line segment with end points <math><mfenced separators="|"><mrow><mn>1,2</mn><mo>,</mo><mn>6</mn></mrow></mfenced></math> and <math><mfenced separators="|"><mrow><mo>-</mo><mn>3,6</mn><mo>,</mo><mn>2</mn></mrow></mfenced></math> passes through <math><mfenced separators="|"><mrow><mo>-</mo><mn>6,2</mn><mo>,</mo><mn>4</mn></mrow></mfenced></math> and has the equation of the form <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>x</mi><mo>+</mo><mn>6</mn></mrow><mi>l</mi></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>2</mn></mrow><mi>m</mi></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>4</mn></mrow><mi>n</mi></mfrac></math> (Where <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi><mo>,</mo><mi>m</mi><mo>,</mo><mi>n</mi></math> are integers, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi></math> is a prime number and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi><mo>></mo><mn>0</mn></math>), then the value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi><mi>m</mi><mi>n</mi><mo>-</mo><mfenced separators="|"><mrow><mi>l</mi><mo>+</mo><mi>m</mi><mo>+</mo><mi>n</mi></mrow></mfenced></math> equals to
|
singleCorrect
| null |
Midpoint of the line segment is <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mrow><mfrac><mrow><mn>1</mn><mo>-</mo><mn>3</mn></mrow><mn>1</mn></mfrac><mo>,</mo><mfrac><mrow><mn>2</mn><mo>+</mo><mn>6</mn></mrow><mn>2</mn></mfrac><mo>,</mo><mfrac><mrow><mn>6</mn><mo>+</mo><mn>2</mn></mrow><mn>2</mn></mfrac></mrow></mfenced><mo>≡</mo><mfenced separators="|"><mrow><mo>-</mo><mn>1</mn><mo>,4</mo><mo>,</mo><mn>4</mn></mrow></mfenced></math><math xmlns="http://www.w3.org/1998/Math/MathML"/><br />Parallel vector to the required line <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfenced separators="|"><mrow><mo>-</mo><mn>1</mn><mo>+</mo><mn>6</mn></mrow></mfenced><mover><mi>i</mi><mo>^</mo></mover><mo>+</mo><mfenced separators="|"><mrow><mn>4</mn><mo>-</mo><mn>2</mn></mrow></mfenced><mover><mi>j</mi><mo>^</mo></mover><mo>+</mo><mfenced separators="|"><mrow><mn>4</mn><mo>-</mo><mn>4</mn></mrow></mfenced><mover><mi>k</mi><mo>^</mo></mover></math><br /><math><mo>=</mo><mn>5</mn><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>+</mo><mn>2</mn><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>+</mo><mn>0</mn><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></math><br />Hence, equation of the line is<br /><math><mfrac><mrow><mi>x</mi><mo>+</mo><mn>6</mn></mrow><mrow><mn>5</mn></mrow></mfrac><mo>-</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>2</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>4</mn></mrow><mrow><mn>0</mn></mrow></mfrac></math><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>l</mi><mo>=</mo><mn>5</mn><mo>,</mo><mo> </mo><mi>m</mi><mo>=</mo><mn>2</mn><mo>,</mo><mo> </mo><mi>n</mi><mo>=</mo><mn>0</mn></math>
|
["<math><mo>-</mo><mn>3</mn></math>", "<math><mo>-</mo><mn>5</mn></math>", "<math><mo>-</mo><mn>7</mn></math>", "<math><mo>-</mo><mn>9</mn></math>"]
|
[2]
| null |
Practise Problem
|
05788e3ad78de582cfad386345a8217c
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The shortest distance between the lines $\quad \bar{r}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-2 t) \hat{k}$ and
$\bar{r}=(p+1) \hat{\imath}+(2 p-1) \hat{\jmath}+(2 p+1) \hat{k} \quad$ is
|
singleCorrect
| 2 |
$\begin{aligned} \ell_{1}: \overline{\mathrm{r}} &=(1-\mathrm{t}) \hat{\mathrm{i}}+(\mathrm{t}-2) \hat{\mathrm{j}}+(3-2 \mathrm{t}) \hat{\mathrm{k}} \\ &=(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mathrm{t}(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \\ \ell_{2}: \overline{\mathrm{r}} &=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mathrm{p}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\ \text { Here } & \overline{\mathrm{a}}_{2}-\overline{\mathrm{a}}_{1}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})-(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\ & \overline{\mathrm{b}}_{1} \times \overline{\mathrm{b}}_{2}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1 & 1 & -2 \\ 1 & 2 & 2\end{array}\right|=\hat{\mathrm{i}}(2+4)-\hat{\mathrm{j}}(0)-3 \hat{\mathrm{k}}=6 \hat{\mathrm{i}}-3 \hat{\mathrm{k}} \end{aligned}$
$\therefore\left|\bar{b}_{1} \times \bar{b}_{2}\right| \quad=\sqrt{36+9}=3 \sqrt{5}$
shortest distance $=\left|\frac{\left(\bar{b}_{1} \times \bar{b}_{2}\right) \cdot\left(\bar{a}_{2}-\bar{a}_{1}\right)}{\left(\bar{b}_{1} \times \bar{b}_{2}\right)}\right|=\left|\frac{(6 \hat{i}-3 \hat{k}) \cdot(\hat{j}-2 \hat{k})}{3 \sqrt{5}}\right|=\frac{6}{3 \sqrt{5}}=\frac{2}{\sqrt{5}}$
|
["$\\frac{8}{\\sqrt{29}}$ units", "$\\frac{4}{\\sqrt{29}}$ units", "$\\frac{2}{\\sqrt{5}}$ units", "$\\frac{4}{\\sqrt{19}}$ units"]
|
[2]
| null |
Practise Problem
|
3b35f4d63f414a9f439c24979a26b964
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If the vector $19 \hat{\mathbf{i}}+22 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ bisects an angle between the vectors a and $6 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}$, then the unit vector in the direction of $\mathbf{a}$ is
|
singleCorrect
| 2 |
Let vector $\mathbf{a}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
Now, vector bisector of angle between $\mathbf{a}$ and $6 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}$ is
$\begin{aligned} & \lambda\left(\frac{\mathbf{a}}{|\mathbf{a}|}+\frac{6 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}}{\sqrt{6^2+8^2}}\right)=\lambda\left(\frac{\mathbf{a}}{|\mathbf{a}|}+\frac{6 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}}{10}\right) \\ & =19 \hat{\mathbf{i}}+22 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\end{aligned}$
(given)
$\Rightarrow \frac{\mathbf{a}}{|\mathbf{a}|}=\frac{19 \hat{\mathbf{i}}+22 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}}{\lambda}-\left(\frac{3}{5} \hat{\mathbf{i}}+\frac{4}{5} \hat{\mathbf{j}}\right)$
$=\left(\frac{19}{\lambda}-\frac{3}{5}\right) \hat{\mathbf{i}}+\left(\frac{22}{\lambda}-\frac{4}{5}\right) \hat{\mathbf{j}}+\frac{5}{\lambda} \hat{\mathbf{k}}$
$\because \frac{\mathbf{a}}{|\mathbf{a}|}$ is unit vector along $\mathbf{a}$, so
$\begin{aligned} & \left(\frac{19}{\lambda}-\frac{3}{5}\right)^2+\left(\frac{22}{\lambda}-\frac{4}{5}\right)^2+\left(\frac{5}{\lambda}\right)^2=1 \\ & \frac{361}{\lambda^2}+\frac{484}{\lambda^2}+\frac{25}{\lambda^2}+\frac{9}{25}+\frac{16}{25}-\frac{114}{5 \lambda}-\frac{176}{5 \lambda}=1 \\ & \Rightarrow \frac{870}{\lambda^2}-\frac{290}{5 \lambda}=0 \Rightarrow \lambda=15 \\ & \text { So, } \frac{\mathbf{a}}{|\mathbf{a}|}=\left(\frac{19}{15}-\frac{3}{5}\right) \hat{\mathbf{i}}+\left(\frac{22}{15}-\frac{4}{5}\right) \hat{\mathbf{j}}+\frac{\hat{\mathbf{k}}}{3} \\ & =\frac{10}{15} \hat{\mathbf{i}}+\frac{10}{15} \hat{\mathbf{j}}+\frac{\hat{\mathbf{k}}}{3}=\frac{1}{3}(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\end{aligned}$
Hence, option (b) is correct.
|
["$\\frac{1}{5}(4 \\hat{\\mathbf{i}}+3 \\hat{\\mathbf{k}})$", "$\\frac{1}{3}(2 \\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+\\hat{\\mathbf{k}})$", "$\\frac{1}{3}(\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+2 \\hat{\\mathbf{k}})$", "$\\frac{1}{3}(2 \\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}-\\hat{\\mathbf{k}})$"]
|
[1]
| null |
Practise Problem
|
c3b563ff99440d88332b8e7de1fd96ac
|
BITSAT_MATH
|
Three Dimensional Geometry
|
$\triangle A B C$ is formed by $A(1,8,4), B(0,-11,4)$ and $C(2,-3,1)$. If $D$ is the foot of the perpendicular from $A$ to $B C$. Then the coordinates of $D$ are
|
singleCorrect
| 2 |
The vertices of $\triangle A B C$ are given as $A(1,8,4)$, $B(0,-11,4)$ and $C(2,-3,1)$.
Equation of line $B C$,
$$
\frac{x}{2}=\frac{y+11}{8}=\frac{z-4}{-3}=\lambda \text { (say) }
$$
<img src="https://cdn-question-pool.getmarks.app/pyq/ap_eamcet/6fOC6-FbwF8qnqaxdEqwPLZMDAtGDuFEeUiw2Bs2MaM.original.fullsize.png"><br>
As point $D$ is on line $B C$, so coordinates of $D$ are $(2 \lambda, 8 \lambda-11,-3 \lambda+4)$
Since, $A D \perp B C$
$$
\begin{aligned}
\therefore \quad & A D \cdot B C=0 \\
& (2 \lambda-1,8 \lambda-19,-3 \lambda) \cdot(2,8,-3)=0 \\
\therefore \quad & 2(2 \lambda-1)+8(8 \lambda-19)+(-3)(-3 \lambda)=0 \\
& 4 \lambda-2+64 \lambda-152+9 \lambda=0 \\
& 77 \lambda=154 \\
& \lambda=2
\end{aligned}
$$
Hence, the coordinates of $D$ are $(4,5,-2)$
|
["(\u2013 4, 5, 2)", "(4, 5, \u2013 2)", "(4, \u2013 5, 2)", "(4, \u2013 5, \u2013 2)"]
|
[1]
| null |
Practise Problem
|
935d8a8def4388db615f099cfd0f4626
|
BITSAT_MATH
|
Three Dimensional Geometry
|
A line with positive direction cosines passes through the point $\mathrm{P}(2,-1,2)$ and makes equal angle with co-ordinate axes. The line meets the plane $2 x+y+z=9$ at point $Q$. The length of the line segment $P Q$ equals.
|
singleCorrect
| 2 |
Let d.c. of line is $(\mathrm{k}, \mathrm{k}, \mathrm{k})$<br/>$\begin{array}{l}<br/>\Rightarrow 3 \mathrm{k}^{2}=1 \Rightarrow \mathrm{k}=\frac{1}{\sqrt{3}} \\<br/>\Rightarrow \text { d.c. of line is }\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)<br/>\end{array}$<br/>Equation of a line passing through point $p$ is<br/>$\begin{array}{l}<br/>\frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y+1}{\frac{1}{\sqrt{3}}}=\frac{z-2}{\frac{1}{\sqrt{3}}}=k \\<br/>\Rightarrow x=k+2, y=k-1, z=k+2 \\<br/>\therefore 2(k+2)+(k-1)+(k+2)=9 \\<br/>4 k=4 \Rightarrow k=1<br/>\end{array}$<br/>$\therefore$ Point $\mathrm{Q}$ is $(3,0,3)$<br/>$P Q=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$ units.
|
["1 unit", "$\\sqrt{2}$ unit", "$\\sqrt{3}$ unit", "2 unit"]
|
[2]
| null |
Practise Problem
|
3863a79d9ba7df4d6dc96f5726de4843
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then $k=$
|
singleCorrect
| 2 |
The coordinates of any point on the first line are given by<br>\(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda\)<br>i.e \(x=2 \lambda+1 y=3 \lambda-1 z=4 \lambda+1\)<br>Thus, the coordinates of any point on this line are \((2 \lambda+1,3 \lambda-1,4 \lambda+1)\).<br>The coordinates of any point on the second line are given by<br>\(\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu\)<br>i.e.<br>\(\begin{aligned}<br>& x=\mu+3 \\<br>& y=2 \mu+k \\<br>& z=\mu<br>\end{aligned}\)<br>Thus, the coordinates of any point on this line are \((\mu+3,2 \mu+k, \mu)\).<br>If these two lines intersect each other, then<br>\(\begin{aligned}<br>& 2 \lambda+1=\mu+3,3 \lambda-1=2 \mu+k, 4 \lambda+1=\mu \\<br>& \Rightarrow 2 \lambda-\mu=2,3 \lambda-2 \mu=k+1,4 \lambda-\mu=-1<br>\end{aligned}\)<br>Solving \(2 \lambda-\mu=2\) and \(4 \lambda-\mu=-1\), we get<br>\(\lambda=-3 / 2 \text { and } \mu=-5\)<br>By substituting the values \(\lambda=-3 / 2\) and \(\mu=-5\) in \(3 \lambda-2 \mu=k+1\), we get<br>\(k=9/ 2 `\)
|
["$\\frac{9}{2}$", "$\\frac{2}{9}$", "$\\frac{-9}{2}$", "$\\frac{-2}{9}$"]
|
[0]
| null |
Practise Problem
|
685ba3528b8dbb7821306adaee6152c5
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The equation of a plane passing through the line of intersection of the planes $x+2 y+3 z=2$, $x-y+z=3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3,1,-1)$ is
|
singleCorrect
| 2 |
Equation of plane passing through intersection of two planes<br>$x+2 y+3 z=2$ and $x-y+z=3$ $\Rightarrow \quad(x+2 y+3 z-2)+\lambda(x-y+z-3)=0$ $\Rightarrow(1+\lambda) x+(2-\lambda) y+(3+\lambda) z-(2+3 \lambda)=0$ whose distance from $(3,1,-1)$ is $\frac{2}{\sqrt{3}} .$ $\therefore \frac{|3(1+\lambda)+1 \cdot(2-\lambda)-1(3+\lambda)-(2+3 \lambda)|}{\sqrt{(1+\lambda)^{2}+(2-\lambda)^{2}+(3+\lambda)^{2}}}=\frac{2}{\sqrt{3}}$ $\Rightarrow \lambda=-\frac{7}{2}$<br>Hence, equation of plane is $5 x-11 y+z-17=0$.
|
["$5 x-11 y+z=17$", "$\\sqrt{2} x+y=3 \\sqrt{2}-1$", "$x+y+z=\\sqrt{3}$", "$x-\\sqrt{2} y=1-\\sqrt{2}$"]
|
[0]
| null |
Practise Problem
|
bd4db2e68e2a9017c48a7eac8f56c0c7
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The length of the perpendicular (in units) from the point <math><mfenced separators="|"><mrow><mn>1,2</mn><mo>,</mo><mn>4</mn></mrow></mfenced></math> on the straight line <math><mfrac><mrow><mi>x</mi><mo>-</mo><mn>2</mn></mrow><mrow><mn>1</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>-</mo><mn>7</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>z</mi><mo>-</mo><mn>3</mn></mrow><mrow><mo>-</mo><mn>1</mn></mrow></mfrac></math> lies in the interval
|
singleCorrect
| null |
<img src="https://cdn-question-pool.getmarks.app/nta_abhyas/jee_main/images/1fe8f10b-b7a2-4b14-bc16-6fabb1dccf6a.png" style="width:2.48056in;height:1.16667in" /><br />Let, point <math><mi>A</mi></math> is <math><mfenced separators="|"><mrow><mn>1,2</mn><mo>,</mo><mn>4</mn></mrow></mfenced></math> and any general point on the line is <math><mi>B</mi><mfenced separators="|"><mrow><mi>λ</mi><mo>+</mo><mn>2,2</mn><mi>λ</mi><mo>+</mo><mn>7</mn><mo>,</mo><mo>-</mo><mi>λ</mi><mo>+</mo><mn>3</mn></mrow></mfenced></math>, <br /><math><mover accent="true"><mrow><mi>A</mi><mi>B</mi></mrow><mo>→</mo></mover><mo>=</mo><mfenced separators="|"><mrow><mi>λ</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>+</mo><mfenced separators="|"><mrow><mn>2</mn><mi>λ</mi><mo>+</mo><mn>5</mn></mrow></mfenced><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>+</mo><mfenced separators="|"><mrow><mo>-</mo><mi>λ</mi><mo>-</mo><mn>1</mn></mrow></mfenced><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></math><br />and a vector parallel to the line is <math><mover accent="true"><mrow><mi>b</mi></mrow><mo>→</mo></mover><mo>=</mo><mover accent="true"><mrow><mi>i</mi></mrow><mo>^</mo></mover><mo>+</mo><mn>2</mn><mover accent="true"><mrow><mi>j</mi></mrow><mo>^</mo></mover><mo>-</mo><mover accent="true"><mrow><mi>k</mi></mrow><mo>^</mo></mover></math><br /><math><mover accent="true"><mrow><mi>A</mi><mi>B</mi></mrow><mo>→</mo></mover><mo>⋅</mo><mover accent="true"><mrow><mi>b</mi></mrow><mo>→</mo></mover><mo>=</mo><mn>0</mn><mo>⇒</mo><mi>λ</mi><mo>+</mo><mn>1</mn><mo>+</mo><mn>4</mn><mi>λ</mi><mo>+</mo><mn>10</mn><mo>+</mo><mi>λ</mi><mo>+</mo><mn>1</mn><mo>=</mo><mn>0</mn></math><br /><math><mo>⇒</mo><mi>λ</mi><mo>=</mo><mo>-</mo><mn>2</mn></math><br />The length of the perpendicular <math><mo>=</mo><mfenced close="|" open="|" separators="|"><mrow><mover accent="true"><mrow><mi>A</mi><mi>B</mi></mrow><mo>→</mo></mover></mrow></mfenced></math><br /><math><mo>=</mo><msqrt><msup><mrow><mfenced separators="|"><mrow><mo>-</mo><mn>2</mn><mo>+</mo><mn>1</mn></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow><mfenced separators="|"><mrow><mo>-</mo><mn>4</mn><mo>+</mo><mn>5</mn></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow><mfenced separators="|"><mrow><mn>2</mn><mo>-</mo><mn>1</mn></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></msqrt><mo>=</mo><msqrt><mn>3</mn></msqrt></math> units
|
["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mfenced separators=\"|\"><mrow><mn>1</mn><mo>,</mo><mfrac><mn>3</mn><mn>2</mn></mfrac></mrow></mfenced></math>", "<math><mfenced separators=\"|\"><mrow><mn>2,3</mn></mrow></mfenced></math>", "<math><mfenced close=\"]\" separators=\"|\"><mrow><mn>0,2</mn></mrow></mfenced></math>", "<math><mfenced open=\"[\" separators=\"|\"><mrow><mn>4,5</mn></mrow></mfenced></math>"]
|
[2]
| null |
Practise Problem
|
8d51d2ee81df53e5e494e452130f410d
|
BITSAT_MATH
|
Three Dimensional Geometry
|
In $\triangle A B C, \mathrm{~L}, \mathrm{M}, \mathrm{N}$ are points on $B C, C A, A B$ respectively, dividing them in the ratio $1: 2,2: 3,3: 5$. If the point $K$ divides $A B$ in the ratio $5: 3$, then
|
singleCorrect
| 2 |
Given that, in $\triangle A B C, L, M, N$ are points on BC, $\mathrm{CA}$ and $\mathrm{AB}$ respectively, dividing in the ratio $1: 2,2: 3$ and $3: 5$. Also, point $K$ divides $A B$ in the ratio $5: 3$.
<img src="https://cdn-question-pool.getmarks.app/pyq/ap_eamcet/AS_kAO-9eD1d0_xN8680x5GPdwZGgC0MPr-IbH3BNA0.original.fullsize.png"><br>
Now, according to the section formula,
$$
\begin{aligned}
& L=\frac{2 \mathbf{b}+\mathbf{c}}{3}, M=\frac{2 \mathbf{a}+3 \mathbf{c}}{5} \\
& N=\frac{3 \mathbf{b}+5 \mathbf{a}}{8}, K=\frac{5 \mathbf{b}+3 \mathbf{a}}{8} \\
& \overrightarrow{\mathrm{AL}}=\frac{2 \mathbf{b}+\mathbf{c}}{3}-\mathbf{a}=\frac{2 \mathbf{b}+\mathbf{c}-3 \mathbf{a}}{3} \\
& \overrightarrow{\mathrm{BM}}=\frac{2 \mathbf{a}+3 \mathbf{c}}{5}-\mathbf{b}=\frac{2 \mathbf{b}+3 \mathbf{c}-5 \mathbf{b}}{5} \\
& \overrightarrow{\mathrm{CN}}=\frac{3 \mathbf{b}+5 \mathbf{a}}{8}-\mathbf{c}=\frac{3 \mathbf{b}+5 \mathbf{a}-8 \mathbf{c}}{8} \\
& \overrightarrow{\mathrm{CK}}=\frac{5 \mathbf{b}+3 \mathbf{a}}{8}-\mathbf{c}=\frac{5 \mathbf{b}+3 \mathbf{a}-8 \mathbf{c}}{8} \\
& \because \quad \frac{\overrightarrow{\mathrm{AL}}+\overrightarrow{\mathrm{BM}}+\overrightarrow{\mathrm{CN}}}{\overrightarrow{\mathrm{CK}}} \mid=
\end{aligned}
$$
$\begin{aligned} & \left|\frac{\frac{2 \mathbf{b}+\mathbf{c}-3 \mathbf{a}}{3}+\frac{2 \mathbf{a}+3 \mathbf{c}-5 \mathbf{b}}{5}+\frac{3 \mathbf{b}+5 \mathbf{a}-8 \mathbf{c}}{8}}{\frac{5 \mathbf{b}+3 \mathbf{a}-8 \mathbf{c}}{8}}\right| \\ = & \left|\frac{(5 \mathbf{b}+3 \mathbf{a}-8 \mathbf{c})}{15(5 \mathbf{b}+3 \mathbf{a}-8 \mathbf{c})}\right|=\frac{1}{15}\end{aligned}$
|
["$\\frac{5}{8}$", "$\\frac{2}{5}$", "$\\frac{3}{5}$", "$\\frac{1}{15}$"]
|
[3]
| null |
Practise Problem
|
e8a7637686c0f08635ec7f2db553e043
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If a plane passing through the point $(2,2,1)$ and is perpendicular to the planes $3 x+2 y+4 z+1=0$ and $2 x+y+3 z+2=0$. Then, the equation of the plane is
|
singleCorrect
| 2 |
Equation of plane passing through $(2,2,1)$ is<br>$\begin{array}{ll} & \mathrm{a}(\mathrm{x}-2)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-1)=0 \\ & \text { Since, above plane is perpendicular to } \\ & 3 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}+1=0 \\ & \text { and } 2 \mathrm{x}+\mathrm{y}+3 \mathrm{z}+2=0 \\ \therefore & 3 \mathrm{a}+2 \mathrm{~b}+4 \mathrm{c}=0 \\ & \text { and } 2 \mathrm{a}+\mathrm{b}+3 \mathrm{c}=0 \\ & {\left[\because \text { for perpendicular, } \mathrm{a}_{1} \mathrm{a}_{2}\right.} \\ & \left.\quad+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}=0\right]\end{array}$<br>On multiplying eq. (iii) by 2 , we get $4 a+2 b+6 c=0$<br>On subtracting eq. (iv) from eq. (ii), we get<br>$\Rightarrow \quad c=\frac{-a}{2}$<br>On putting $\mathrm{c}=\frac{-\mathrm{a}}{2}$ in eq. (iii), we get $\mathrm{b}$<br>$=\frac{-\mathrm{a}}{2}$<br>On putting $b=\frac{-a}{2}$ and $c=\frac{-a}{2}$ in eq. (i),<br>$\begin{aligned}<br>& \text { we get } a(x-2)-\frac{a}{2}(y-2)-\frac{a}{2}(z-1)=0 \\<br>\Rightarrow & \frac{a}{2}[2(x-2)-(y-2)-(z-1)]=0 \\<br>\Rightarrow & 2 x-4-y+2-z+1=0 \\<br>\Rightarrow & 2 x-y-z-1=0<br>\end{aligned}$
|
["$2 x-y-z-1=0$", "$2 x+3 y+z-1=0$", "$2 x+y+z+3=0$", "$x-y+z-1=0$"]
|
[0]
| null |
Practise Problem
|
47e8bac85f8367d870ab5570c0dc959d
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The cartesian equation of the plane passing through the point $(3,-2,-1)$ and parallel to the vectors $\mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $\mathbf{c}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ is
|
singleCorrect
| 2 |
Vector b and c are given as,
$\mathrm{b}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
and $c=3 \hat{i}+2 \hat{j}-5 \hat{k}$
Cartesian equation of plane passing through the point $(3,-2,-1)$
$\begin{aligned}
& \left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
x-3 & y+2 & z+1 \\
1 & -2 & 4 \\
3 & 2 & -5
\end{array}\right|=0 \\
& (x-3)(10-8)-(y+2)(-5-12)+(z+1)(2+6)=0 \\
& (x-3)(2)-(y+2)(-17)+(z+1) 8=0 \\
& 2 x-6+17 y+34+8 z+8=0 \\
& 2 x+17 y+8 z+36=0
\end{aligned}$
|
["(a) $2 x-17 y-8 z+63=0$", "$3 x+17 y+8 z-36=0$", "$2 x+17 y+8 z+36=0$", "$3 x-16 y+8 z-63=0$"]
|
[2]
| null |
Practise Problem
|
835213393502771d7e4739120daae0d6
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If the direction ratios of two lines are given by $a+2 b+c=0$ and $11 b c+6 c a-14 a b=0$. then the angle between these lines is
|
singleCorrect
| 1 |
We have,
$$
\begin{aligned}
& a+2 b+c=0 \\
& 11 b c+6 c a-14 a b=0
\end{aligned}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
& 11\left(\frac{-a-c}{2}\right) c+6 a c-14 a\left(\frac{-a-c}{2}\right)=0 \\
\Rightarrow & -11 a c-11 c^2+12 a c+14 a^2+14 a c=0 \\
\Rightarrow & 14 a^2+15 a c-11 c^2=0 \\
\Rightarrow & 14 a^2+22 a c-7 a c-11 c^2=0 \\
\Rightarrow & (7 a+11 c)(2 a-c)=0 \\
\Rightarrow & a=\frac{-11}{7} c \text { and } a=\frac{c}{2}
\end{aligned}
$$
$\therefore$ Direction ratio of line are $(-11,2,7)$ and $(2,-3,4)$ Angle between lines are
$$
\begin{aligned}
& \therefore \cos \theta=-22-6+28=0 \\
& \because \quad \theta=\frac{\pi}{2}
\end{aligned}
$$
|
["$\\frac{\\pi}{3}$", "$\\cos ^{-1}\\left(\\frac{1}{3}\\right)$", "$\\cos ^{-1}\\left(\\frac{2}{3}\\right)$", "$\\frac{\\pi}{2}$"]
|
[3]
| null |
Practise Problem
|
c140e344c82e7d42d345a162dd62129f
|
BITSAT_MATH
|
Three Dimensional Geometry
|
Let \(a=i+j+k, b=i-j+k\) and \(c=i-j-k\) be three vectors. A vector \(v\) in the plane of a and \(b\), whose projection on \(c\) is \(1 / \sqrt{3}\), is given by
|
singleCorrect
| 2 |
Given that,<br>\(\begin{array}{l}
\vec{a}=\hat{i}+2 \hat{j}+\hat{k} \rightarrow(1) \\
\vec{b}=\hat{i}-\hat{j}+\hat{k} \rightarrow(2) \\
\vec{C}=\hat{i}+\hat{j}-\hat{k} \rightarrow(3)
\end{array}\)<br>A vector in the plane of \(\vec{a}\) and \(\vec{b}\) can be written as,<br>\(\begin{array}{l}
\vec{r}=m \vec{a}+n \vec{b} \\
=m(\hat{i}+2 \hat{j}+\hat{k})+n(\hat{i}-\hat{j}+\hat{k}) \\
\vec{r}=(m+n) \hat{i}+(2 m-n) \hat{j}+(m+n) \hat{k}
\end{array}\)<br>So, projection on \(\vec{C}\) by \(\vec{r}\) is given by<br>\(\begin{array}{l}
\frac{\vec{r} \cdot \vec{C}}{|\vec{C}|}=\frac{(m+n)+(2 m-n)-(m+n)}{\sqrt{1+1+1}} \\
\Rightarrow \frac{2 m-n}{\sqrt{3}}=\frac{1}{\sqrt{3}}
\end{array}\)<br>As use can see from Eq. (4), the \(^{\wedge} i\) component and the \(\hat{k}\) component must be equal and \(\hat{j}\) can be either \(-1\) or \(1 .\)<br>Therefore, option (4) is correct, that is \(4 \hat{i}-\hat{j}+4 \hat{k}\)
|
["\\(i-3 j-3 k\\)", "\\(-3 i-3 j+k\\)", "\\(3 i-j+3 k\\)", "\\(i+3 j-3 k\\)"]
|
[3]
| null |
Practise Problem
|
c4f11a7060414c5ee542313c06ab72b1
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The plane $\ell \mathrm{x}+\mathrm{my}=0$ is rotated about its line of intersection with the plane $\mathrm{z}=0$ through an angle $\alpha$. The equation changes to
|
singleCorrect
| 2 |
$P_{1}: \ell x+m y=0, \quad P_{2}=z=0$<br/>Plane through common line of $P_{1}$ and $P_{2}$ <br/>$\mathrm{P}_{3}: \ell \mathrm{x}+\mathrm{my}+\mathrm{nz}=0$<br/>angle between $P_{1}$ and $P_{3}=\alpha$ <br/>$\therefore \cos \alpha=\hat{n}_{1}, \hat{n}_{2}$<br/>$=\frac{\ell^{2}+\mathrm{m}^{2}}{\sqrt{\ell^{2}+\mathrm{m}^{2}} \sqrt{\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}}}=\sqrt{\frac{\ell^{2}+\mathrm{m}^{2}}{\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}}}$<br/>$\Rightarrow \cos ^{2} \alpha=\frac{\ell^{2}+\mathrm{m}^{2}}{\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}}$<br/>$\Rightarrow \mathrm{n}^{2}=\left(\ell^{2}+\mathrm{m}^{2}\right) \tan ^{2} \alpha$<br/>$\Rightarrow \mathrm{n}=\pm \sqrt{\ell^{2}+\mathrm{m}^{2}} \tan \alpha$<br/>$\mathrm{P}_{3}: \ell \mathrm{x}+\mathrm{my} \pm \sqrt{\ell^{2}+\mathrm{m}^{2}}$ tan $\alpha=0$
|
["$\\ell \\mathrm{x}+\\mathrm{my} \\pm \\tan \\alpha \\sqrt{\\ell^{2}+\\mathrm{m}^{2}}=0$", "$\\ell \\mathrm{x}+\\mathrm{my} \\pm \\mathrm{z} \\tan \\alpha \\sqrt{\\ell^{2}+\\mathrm{m}^{2}+1}=0$", "$\\ell \\mathrm{x}+\\mathrm{my} \\pm \\mathrm{z} \\tan \\alpha \\sqrt{\\ell^{2}+1}=0$", "$\\ell \\mathrm{x}+\\mathrm{my} \\pm \\mathrm{z} \\tan \\alpha \\sqrt{\\ell^{2}+\\mathrm{m}^{2}}=0$"]
|
[3]
| null |
Practise Problem
|
905ef14b715327e170a6c60738e5206a
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If the line joining the points $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ meets the plane that passes through the point $2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$ and parallel to the vectors $3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}-\hat{\mathbf{k}}$ at, $P$, then the position vector of the point $P$ is
|
singleCorrect
| 2 |
Equation of line joining points $(\hat{\mathbf{i}}+\hat{\mathbf{j}})$ and $(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})$ in cartesian form is
$$
\frac{x-1}{2}=\frac{y-1}{0}=\frac{z}{-1}=r(\text { let })
$$
Then, point on line Eq. (i) is $p(2 r+1,1,-r)$.
Now, equation of plane passes through the point $(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})$ and parallel to the vectors $(3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})$ and $(3 \hat{\mathbf{i}}-\hat{\mathbf{k}})$ in cartesian form is
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-4 & z \\
0 & 3 & 5 \\
3 & 0 & -1
\end{array}\right| & =0 \\
\Rightarrow & -3(x-2)+15(y-4)-9 z & =0 \\
\Rightarrow & 3 x-15 y+9 z+54 & =0
\end{aligned}
$$
Let the point $P(2 r+1,1,-r)$ on the plane Eq. (ii) itself, so $r=14$
So, position vector of the point $P$ is
$$
\mathbf{O P}=29 \hat{\mathbf{i}}+\hat{\mathbf{j}}-14 \hat{\mathbf{k}} .
$$
|
["$-27 \\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+14 \\hat{\\mathbf{k}}$", "$29 \\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}-14 \\hat{\\mathbf{k}}$", "$-14 \\hat{\\mathbf{i}}+89 \\hat{\\mathbf{j}}+3 \\hat{\\mathbf{k}}$", "$2 \\hat{\\mathbf{i}}+5 \\hat{\\mathbf{j}}-7 \\hat{\\mathbf{k}}$"]
|
[1]
| null |
Practise Problem
|
8ddf76c988ad360d624ad91907d7949c
|
BITSAT_MATH
|
Three Dimensional Geometry
|
Let <math><mi mathvariant="normal">A</mi></math> be the foot of the perpendicular from the origin to the plane <math><mi mathvariant="normal">x</mi><mo>-</mo><mn>2</mn><mi mathvariant="normal">y</mi><mo>+</mo><mn>2</mn><mi mathvariant="normal">z</mi><mo>+</mo><mn>6</mn><mo>=</mo><mn>0</mn></math> and <math><mi mathvariant="normal">B</mi></math><math><mfenced separators="|"><mrow><mn>0</mn><mo>,</mo><mi mathvariant="normal"></mi><mo>-</mo><mn>1</mn><mo>,</mo><mi mathvariant="normal"></mi><mo>-</mo><mn>4</mn></mrow></mfenced></math> be a point on the plane. Then, the length of <math><mi mathvariant="normal">A</mi><mi mathvariant="normal">B</mi></math> is
|
singleCorrect
| null |
<img src="https://cdn-question-pool.getmarks.app/nta_abhyas/jee_main/images/7a32efef-ae41-4d2b-9d0a-51c06979edcc.png" style="width:2.23333in;height:1.92014in" /><br />From the diagram<br /><math><mi mathvariant="normal">O</mi><mi mathvariant="normal">B</mi><mo>=</mo><msqrt><mn>0</mn><mo>+</mo><mn>1</mn><mo>+</mo><mn>16</mn></msqrt><mo>=</mo><msqrt><mn>17</mn></msqrt></math><br /><math><mi mathvariant="normal">O</mi><mi mathvariant="normal">A</mi><mo>=</mo><mfenced close="|" open="|" separators="|"><mrow><mfrac><mrow><mn>0</mn><mo>-</mo><mn>0</mn><mo>+</mo><mn>0</mn><mo>+</mo><mn>6</mn></mrow><mrow><msqrt><mn>1</mn><mo>+</mo><mn>4</mn><mo>+</mo><mn>4</mn></msqrt></mrow></mfrac></mrow></mfenced><mo>=</mo><mn>2</mn></math><br /><math><mo>∵</mo><mo>∆</mo><mi mathvariant="normal">O</mi><mi mathvariant="normal">A</mi><mi mathvariant="normal">B</mi></math> is right angled triangle<br /><math><mi mathvariant="normal">A</mi><mi mathvariant="normal">B</mi><mo>=</mo><msqrt><msup><mrow><mfenced separators="|"><mrow><mi mathvariant="normal">O</mi><mi mathvariant="normal">B</mi></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><msup><mrow><mfenced separators="|"><mrow><mi mathvariant="normal">O</mi><mi mathvariant="normal">A</mi></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup></msqrt><mo>=</mo><msqrt><mn>13</mn></msqrt></math> units
|
["<math><msqrt><mn>13</mn></msqrt></math> units", "<math><msqrt><mn>15</mn></msqrt></math> units", "<math><mn>4</mn></math> units", "<math><msqrt><mn>17</mn></msqrt></math> units"]
|
[0]
| null |
Practise Problem
|
4d8a8e97ba0f63b36737b06bae9ee484
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The length of the perpendicular to the plane $\bar{r} \cdot(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})=14$ from the origin is
|
singleCorrect
| 2 |
Length of $\perp$ er from the point $A(\bar{a})$ to the plane $\bar{r} \cdot \bar{n}=p$ is $\frac{|\bar{a} \cdot \bar{n}-p|}{|\bar{n}|}$
Here $\overline{\mathrm{a}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}$ and $\overline{\mathrm{n}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$\therefore \overline{\mathrm{a}} \cdot \overline{\mathrm{n}}=0 \quad$ and $|\overline{\mathrm{n}}|=\sqrt{1+4+9}=\sqrt{14}$
Hence required distance is $\frac{|0-14|}{\sqrt{14}}=\sqrt{14}$
|
["$\\sqrt{7}$ units", "7 units", "14 units", "$\\sqrt{14}$ units"]
|
[3]
| null |
Practise Problem
|
1c142b91392d6ca45aae4fc0ed72f9bb
|
BITSAT_MATH
|
Three Dimensional Geometry
|
If $L_1$ is a line through the point $5 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}$ and parallel to the vector $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $L_2$ Is a line through the point $4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}$ and parallel to the vector $3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$, then the point of intersection of $L_1$ and $L_2$ is
|
singleCorrect
| 2 |
Line $L_1$ is passing through $5 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}$ and parallel to the vector $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$.
$$
\therefore \quad L_1 \equiv \mathbf{r}=5 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})
$$
line $L_2$ is passing through $4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}$ and parallel to the vector $3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$
$$
\therefore \quad L_2 \equiv \mathbf{r}=4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})
$$
$L_1$ and $L_2$ are intersecting.
$$
\begin{aligned}
& \therefore \quad 5 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}} \\
& +8 \hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \\
& \Rightarrow \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}=(3 \mu-2 \lambda) \hat{\mathbf{i}}+(4 \mu-3 \lambda) \hat{\mathbf{j}}+(5 \mu-4 \lambda) \hat{\mathbf{k}} \\
& \Rightarrow \quad 3 \mu-2 \lambda=1 \\
& 4 \mu-3 \lambda=2 \\
&
\end{aligned}
$$
On solving Eqs. (i) and (ii), we get
$$
\mu=-1, \lambda=-2
$$
Put $\quad \lambda=-2 \operatorname{in} L_1$
We get $\quad r=\hat{i}+2 \hat{\mathbf{j}}+3 \hat{k}$
$\therefore$ Point of intersection of $L_1$ and $L_2$ is $\hat{\mathbf{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathbf{k}}$
|
["$\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}}$", "$\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+3 \\hat{\\mathbf{k}}$", "$2 \\hat{\\mathbf{i}}+3 \\hat{\\mathbf{j}}+\\hat{\\mathbf{k}}$", "$\\hat{\\mathbf{i}}-2 \\hat{\\mathbf{j}}+2 \\hat{\\mathbf{k}}$"]
|
[1]
| null |
Practise Problem
|
e0cdc530bbc9b77abb7d9f7271215f31
|
BITSAT_MATH
|
Three Dimensional Geometry
|
The acute angle between the lines whose direction cosines are given by the equations $l+m+n=0$ and $2 l m+2 l n-m n=0$ is
|
singleCorrect
| 2 |
Given equations are,
<img src="https://cdn-question-pool.getmarks.app/pyq/ap_eamcet/K-8jPqn7zYIL1mjPm7yQ3KwzrMitn-uFn0bQka0HSAw.original.fullsize.png"><br>
On squaring both side of the Eq. (i).
$$
\begin{array}{cc}
& l^2+m^2+n^2+2 l m+2 m n+2 n l=0 \\
\because & l^2+m^2+n^2=1,
\end{array}
$$<img src="https://cdn-question-pool.getmarks.app/pyq/ap_eamcet/GLgpvobrW5D3lUKSMX6OrvQGi013870-gP1gb4zmb9Q.original.fullsize.png"><br>
From Eqs. (ii) and (iv),
$$
3 m n=-1 \Rightarrow m n=-\frac{1}{3}
$$
From Eq. (iii)
$6 l^2-1=0$, let root of this equation is $l_1$ and $l_2$, so
$l_1 l_2=-\frac{1}{6}$.
Now, from Eqs. (i) and (ii),
$$
2 l m+(2 l-m)(-l-m)=0
$$
$$
\begin{array}{rlrl}
\Rightarrow & 2 l^2-l m-m^2 =0 \\
\Rightarrow & 2\left(\frac{l}{m}\right)^2-\left(\frac{l}{m}\right)-1 =0 \\
\Rightarrow & \frac{l_1 l_2}{m_1 m_2} =\frac{-1}{2} \\
\Rightarrow & \frac{l_1 l_2}{-1 / 6} =\frac{m_1 m_2}{2 / 6} \\
& \text { Similarly, } \frac{l_1 l_2}{-1 / 6} =\frac{n_1 n_2}{2 / 6}
\end{array}
$$
So, $\quad \cos \theta=\left|l_1 l_2+m_1 m_2+n_1 n_2\right|=\frac{3}{6}=\frac{1}{2}$
$$
\Rightarrow \quad \theta=\frac{\pi}{3}
$$
|
["$\\frac{\\pi}{6}$", "$\\frac{\\pi}{4}$", "$\\frac{\\pi}{3}$", "$\\frac{2 \\pi}{5}$"]
|
[2]
| null |
Practise Problem
|
e5e8e56a31beb62bb583331d4a963c17
|
BITSAT_MATH
|
Trigonometric Equations
|
The general solution of $\tan x-\sin x=1-\tan x \cdot \sin x$
|
singleCorrect
| 2 |
$\tan x-\sin x=1-\tan x \sin x$
$\Rightarrow \tan x+\tan x \sin x=1+\sin x$
$\tan x(1+\sin x)-(1+\sin x)=0$
$(1+\sin x)(\tan x-1)=0$
$\Rightarrow \quad \tan x-1=0$ or $1+\sin x=0$
$\tan x=1=\tan \frac{\pi}{4} \quad$ or $\sin x=-1=\sin \left(-\frac{\pi}{2}\right)$
$x=n \pi+\frac{\pi}{4} \quad$ or $\quad x=n \pi+(-1)^{n}\left(-\frac{\pi}{2}\right)$
|
["$x=n \\pi+\\frac{\\pi}{4}$ and or $x=n \\pi+(-1)^{n}\\left(\\frac{-\\pi}{2}\\right)$", "$x=\\frac{n \\pi}{4}-\\frac{\\pi}{4}$ and or $x=n \\pi+(-1)^{n}\\left(-\\frac{\\pi}{2}\\right)$", "$x=n \\pi+\\frac{\\pi}{4}$", "$x=n \\pi+\\frac{\\pi}{6}$ and or $\\pi=n \\pi+(-1)^{n}\\left(-\\frac{\\pi}{2}\\right)$"]
|
[0]
| null |
Practise Problem
|
1c92ac0fbe3fe0741f601f6ff2c0cd2d
|
BITSAT_MATH
|
Trigonometric Equations
|
The solutions of $\sin x+\sin 5 x=\sin 3 x$ in $\left(0, \frac{\pi}{2}\right)$ are
|
singleCorrect
| 2 |
$\begin{aligned} & \sin x+\sin 5 x=\sin 3 x \\ & \Rightarrow 2 \sin 3 x \cos 2 x=\sin 3 x \\ & \Rightarrow \sin 3 x(2 \cos 2 x-1)=0 \\ & \Rightarrow \sin 3 x=0 \text { or } \cos 2 x=\frac{1}{2}=\cos \frac{\pi}{3} \\ & \Rightarrow 3 x=\mathrm{n} \pi \text { or } 2 x=2 \mathrm{n} \pi \pm \frac{\pi}{3} \\ & \Rightarrow x=\frac{\mathrm{n} \pi}{3} \text { or } x=\mathrm{n} \pi \pm \frac{\pi}{6}\end{aligned}$
$\Rightarrow x=\frac{\pi}{3}, \frac{\pi}{6} \quad \quad \ldots\left[\because x \in\left(0, \frac{\pi}{2}\right)\right]$
|
["$\\frac{\\pi}{4}, \\frac{\\pi}{10}$", "$\\frac{\\pi}{6}, \\frac{\\pi}{3}$", "$\\frac{\\pi}{4}, \\frac{\\pi}{12}$", "$\\frac{\\pi}{8}, \\frac{\\pi}{16}$"]
|
[1]
| null |
Practise Problem
|
4d6f2b566d7329dc9aae5ca14879e682
|
BITSAT_MATH
|
Trigonometric Equations
|
If $\cos x+\cos y-\cos (x+y)=\frac{3}{2}$, then
|
singleCorrect
| 2 |
$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$
$\begin{array}{r}
\therefore \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-\left(2 \cos ^2\left(\frac{x+y}{2}\right)-1\right)=\frac{3}{2} \\
\quad \cdots\left[\begin{array}{l}
\because \cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right) \text { and } \\
\cos \theta=2 \cos ^2\left(\frac{\theta}{2}\right)-1
\end{array}\right]
\end{array}$
$\begin{aligned} & \therefore \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-2 \cos ^2\left(\frac{x+y}{2}\right)=\frac{3}{2}-1 \\ & \therefore \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-2 \cos ^2\left(\frac{x+y}{2}\right)=\frac{1}{2} \\ & \therefore \quad 4 \cos ^2\left(\frac{x+y}{2}\right)-4 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)+1=0\end{aligned}$
Substituting $\cos \left(\frac{x+y}{2}\right)=\mathrm{t}$, we get
$4 \mathrm{t}^2-4 \mathrm{t} \cos \left(\frac{x-y}{2}\right)+1=0$
As $t$ is real, we get $b^2-4 a c \geq 0$
$\begin{aligned}
& \Rightarrow\left[-4 \cos \left(\frac{x-y}{2}\right)\right]^2-4 \times 4 \times 1 \geq 0 \\
& \Rightarrow 16 \cos ^2\left(\frac{x-y}{2}\right)-16 \geq 0 \\
& \Rightarrow \cos ^2\left(\frac{x-y}{2}\right) \geq 1 \\
& \Rightarrow \cos ^2\left(\frac{x-y}{2}\right)=1
\end{aligned}$
$\ldots[\because-1 \leq \cos \theta \leq 1$, for all values of $\theta]$
$\begin{aligned}
& \Rightarrow \frac{x-y}{2}=0 \\
& \Rightarrow x=y
\end{aligned}$
|
["$x+y=0$", "$x=2 y$", "$x=y$", "$2 x=y$"]
|
[2]
| null |
Practise Problem
|
c2e69484dd8f9ad390fb5098be27f379
|
BITSAT_MATH
|
Trigonometric Equations
|
The equation $\sqrt{3} \sin x+\cos x=4$ has
|
singleCorrect
| 2 |
Hints : $\sqrt{3} \sin x+\cos x=2 \sin \left(x+\frac{\pi}{6}\right) \leq 2$. Therefore
$\sqrt{3} \sin x+\cos x=4 \quad$ cannot have a solution
|
["only one solution", "two solutions", "infinitely many solutions", "no solution"]
|
[3]
| null |
Practise Problem
|
599e9d678585ac3187a4fdcce7655d74
|
BITSAT_MATH
|
Trigonometric Equations
|
If $\sin 3 \theta=\sin \theta$, how many solutions exist such that $-2 \pi < \theta < 2 \pi$ ?
|
singleCorrect
| 2 |
We have, $\sin 3 \theta=\sin \theta$
$$
\begin{aligned}
&\Rightarrow \quad \sin 3 \theta-\sin \theta=0 \\
&\Rightarrow 2 \cos \left(\frac{3 \theta+\theta}{2}\right) \sin \left(\frac{3 \theta-\theta}{2}\right)=0 \\
&\Rightarrow \cos 2 \theta \cdot \sin \theta=0 \quad \cos 2 \theta=0 \quad \text { or } \sin \theta=0, \pi, 2 \pi \\
&\Rightarrow \quad \cos 2 \theta=\cos \left(\frac{\pi}{2}\right) \text { or } \quad \theta=0, \pi=2 \pi \\
&\Rightarrow \quad \quad \quad \quad \quad \quad \theta=0, \pi, 2 \pi \\
&\Rightarrow \quad \theta=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \quad \text { or } \quad \theta=0, \pi, 2 \pi \\
&\because \quad \theta=0, \pi, 2 \pi
\end{aligned}
$$
Thus, total number of solutions $=7$
|
["8", "9", "5", "7"]
|
[2]
| null |
Practise Problem
|
27873a6914f48561222554d21bc55a8e
|
BITSAT_MATH
|
Trigonometric Equations
|
If $0 < x < \pi$ and $\cos x+\sin x=\frac{1}{2}$, then $\tan x$ is
|
singleCorrect
| 2 |
$\cos x+\sin x=\frac{1}{2} \Rightarrow 1+\sin 2 x=\frac{1}{4} \Rightarrow \sin 2 x=-\frac{3}{4}$, so $x$ is obtuse and $\frac{2 \tan x}{1+\tan ^2 x}=-\frac{3}{4} \Rightarrow 3 \tan ^2 x+8 \tan x+3=0$
$\therefore \tan x=\frac{-8 \pm \sqrt{64-36}}{6}=\frac{-4 \pm \sqrt{7}}{3}$
$\because \tan x < 0 \quad \therefore \tan x=\frac{-4-\sqrt{7}}{3}$
|
["\n$\\frac{(1-\\sqrt{7})}{4}$\n", "\n$\\frac{(4-\\sqrt{7})}{3}$\n", "\n$-\\frac{(4+\\sqrt{7})}{3}$\n", "\n$\\frac{(1+\\sqrt{7})}{4}$"]
|
[2]
| null |
Practise Problem
|
bea922e12c222ac685a189f5608dece4
|
BITSAT_MATH
|
Trigonometric Equations
|
The sum of all values of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi></math> in <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>[</mo><mn>0</mn><mo>,</mo><mn>2</mn><mi>π</mi><mo>]</mo></math>, for which <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>sin</mi><mi>x</mi><mo>+</mo><mi>sin</mi><mn>2</mn><mi>x</mi><mo>+</mo><mi>sin</mi><mn>3</mn><mi>x</mi><mo>+</mo><mi>sin</mi><mn>4</mn><mi>x</mi><mo>=</mo><mn>0</mn></math>, is equal to :
|
singleCorrect
| 2 |
<p>Given,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>sin</mi><mi>x</mi><mo>+</mo><mi>sin</mi><mn>2</mn><mi>x</mi><mo>+</mo><mi>sin</mi><mn>3</mn><mi>x</mi><mo>+</mo><mi>sin</mi><mn>4</mn><mi>x</mi><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo>(</mo><mi>sin</mi><mi>x</mi><mo>+</mo><mi>sin</mi><mn>4</mn><mi>x</mi><mo>)</mo><mo>+</mo><mo>(</mo><mi>sin</mi><mn>2</mn><mi>x</mi><mo>+</mo><mi>sin</mi><mn>3</mn><mi>x</mi><mo>)</mo><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mn>2</mn><mi>sin</mi><mfrac><mrow><mn>5</mn><mi>x</mi></mrow><mn>2</mn></mfrac><mi>cos</mi><mfrac><mrow><mn>3</mn><mi>x</mi></mrow><mn>2</mn></mfrac><mo>+</mo><mn>2</mn><mi>sin</mi><mfrac><mrow><mn>5</mn><mi>x</mi></mrow><mn>2</mn></mfrac><mi>cos</mi><mfrac><mi>x</mi><mn>2</mn></mfrac><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mn>2</mn><mi>sin</mi><mfrac><mrow><mn>5</mn><mi>x</mi></mrow><mn>2</mn></mfrac><mfenced open="{" close="}"><mrow><mi>cos</mi><mfrac><mrow><mn>3</mn><mi>x</mi></mrow><mn>2</mn></mfrac><mo>+</mo><mi>cos</mi><mfrac><mi>x</mi><mn>2</mn></mfrac></mrow></mfenced><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mn>2</mn><mi>sin</mi><mfrac><mrow><mn>5</mn><mi>x</mi></mrow><mn>2</mn></mfrac><mfenced close="}" open="{"><mrow><mn>2</mn><mi>cos</mi><mi>x</mi><mi>cos</mi><mfrac><mi>x</mi><mn>2</mn></mfrac></mrow></mfenced><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mi>sin</mi><mfrac><mrow><mn>5</mn><mi>x</mi></mrow><mn>2</mn></mfrac><mo>=</mo><mn>0</mn><mo>⇒</mo><mfrac><mrow><mn>5</mn><mi>x</mi></mrow><mn>2</mn></mfrac><mo>=</mo><mn>0</mn><mo>,</mo><mi>π</mi><mo>,</mo><mn>2</mn><mi>π</mi><mo>,</mo><mn>3</mn><mi>π</mi><mo>,</mo><mn>4</mn><mi>π</mi><mo>,</mo><mn>5</mn><mi>π</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>x</mi><mo>=</mo><mn>0</mn><mo>,</mo><mfrac><mrow><mn>2</mn><mi>π</mi></mrow><mn>5</mn></mfrac><mo>,</mo><mfrac><mrow><mn>4</mn><mi>π</mi></mrow><mn>5</mn></mfrac><mo>,</mo><mfrac><mrow><mn>6</mn><mi>π</mi></mrow><mn>5</mn></mfrac><mo>,</mo><mfrac><mrow><mn>8</mn><mi>π</mi></mrow><mn>5</mn></mfrac><mo>,</mo><mn>2</mn><mi>π</mi></math></p><p>and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>cos</mi><mfrac><mi>x</mi><mn>2</mn></mfrac><mo>=</mo><mn>0</mn><mo>⇒</mo><mfrac><mi>x</mi><mn>2</mn></mfrac><mo>=</mo><mfrac><mi>π</mi><mn>2</mn></mfrac><mo>⇒</mo><mi>x</mi><mo>=</mo><mi>π</mi></math></p><p>and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>cos</mi><mi>x</mi><mo>=</mo><mn>0</mn><mo>⇒</mo><mi>x</mi><mo>=</mo><mfrac><mi>π</mi><mn>2</mn></mfrac><mo>,</mo><mfrac><mrow><mn>3</mn><mi>π</mi></mrow><mn>2</mn></mfrac></math></p><p>So sum <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>6</mn><mi>π</mi><mo>+</mo><mi>π</mi><mo>+</mo><mn>2</mn><mi>π</mi><mo>=</mo><mn>9</mn><mi>π</mi></math></p>
|
["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>8</mn><mi>π</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>11</mn><mi>π</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>12</mn><mi>π</mi></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>9</mn><mi>π</mi></math>"]
|
[3]
| null |
Practise Problem
|
799025a015adf61066af8bfa1a97f5bd
|
BITSAT_MATH
|
Trigonometric Equations
|
If the equation <math><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mn>4</mn><mo>+</mo><mn>3</mn><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mfenced separators="|"><mrow><mi>a</mi><mi>x</mi><mo>+</mo><mi>b</mi></mrow></mfenced></mrow></mrow><mo>-</mo><mn>2</mn><mi>x</mi><mo>=</mo><mn>0</mn></math> has atleast one real solution, where <math><mi>a</mi></math>, <math><mi>b</mi><mo>∈</mo><mo>[</mo><mn>0, 2</mn><mi>π</mi><mo>]</mo></math>, then one possible value of <math><mo>(</mo><mi>a</mi><mo>+</mo><mi>b</mi><mo>)</mo></math> can be equal to
|
singleCorrect
| null |
<math><msup><mrow><mfenced separators="|"><mrow><mi>x</mi><mo>-</mo><mn>1</mn></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mn>3</mn><mo>+</mo><mn>3</mn><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mfenced separators="|"><mrow><mi>a</mi><mi>x</mi><mo>+</mo><mi>b</mi></mrow></mfenced><mo>=</mo><mn>0</mn></mrow></mrow></math><br /><br /><math><msup><mrow><mfenced separators="|"><mrow><mi>x</mi><mo>-</mo><mn>1</mn></mrow></mfenced></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mn>3</mn><mo>=</mo><mo>-</mo><mn>3</mn><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mo>(</mo><mi>a</mi><mi>x</mi><mo>+</mo><mi>b</mi><mo>)</mo></mrow></mrow></math><br /><br /><math><mi>L</mi><mo>.</mo><mi>H</mi><mo>.</mo><mi>S</mi><mi> </mi><mo>≥</mo><mn>3</mn><mo>,</mo><mi> </mi> <mrow> <mi>R</mi><mi>H</mi><mi>S</mi><mtext> </mtext><mo>∈</mo><mrow><mo>[</mo> <mrow> <mo>−</mo><mn>3</mn><mo>,</mo><mn>3</mn> </mrow> <mo>]</mo></mrow> </mrow><mi> </mi><mrow><mrow></mrow></mrow></math> now <math><mi mathvariant="normal">s</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">n</mi><mo></mo><mo>(</mo><mi>a</mi><mi>x</mi><mo>+</mo><mi>b</mi><mo>)</mo><mo>=</mo><mo>-</mo><mn>1</mn></math><br /><br /><math><mo>∴</mo><mi>x</mi><mo>=</mo><mn>1</mn><mi> </mi><mi> </mi><mi> </mi><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mfenced separators="|"><mrow><mi>b</mi><mo>+</mo><mi>a</mi></mrow></mfenced></mrow></mrow><mo>=</mo><mo>-</mo><mn>1</mn></math>
|
["<math><mfrac><mrow><mn>7</mn><mi>\u03c0</mi></mrow><mrow><mn>2</mn></mrow></mfrac></math>", "<math><mfrac><mrow><mn>5</mn><mi>\u03c0</mi></mrow><mrow><mn>2</mn></mrow></mfrac></math>", "<math><mfrac><mrow><mn>9</mn><mi>\u03c0</mi></mrow><mrow><mn>2</mn></mrow></mfrac></math>", "None of these"]
|
[0]
| null |
Practise Problem
|
99e2919ce006168c179cdbc0f7d12a2a
|
BITSAT_MATH
|
Trigonometric Equations
|
If $\sec x+\tan x=3, x \in\left(0, \frac{\pi}{2}\right)$ then, $\sin x=$
|
singleCorrect
| 2 |
We have $\sec x+\tan x=3$ ...(1)
We know that $\sec ^{2} x-\tan ^{2} x=1$
$\therefore(\sec x-\tan x)(\sec x+\tan x)=1 \Rightarrow \sec x-\tan x=\frac{1}{3}$ ...(2)
Adding (1) and (2)
$2 \sec x=\frac{10}{3} \Rightarrow \sec x=\frac{5}{3} \Rightarrow \cos x=\frac{3}{5} \Rightarrow \sin x=\sqrt{1-\cos ^{2} x}=\frac{4}{5}$
|
["$\\frac{3}{5}$", "$\\frac{4}{5}$", "$-1$", "$\\frac{1}{5}$"]
|
[1]
| null |
Practise Problem
|
9ad8eb67fed86acb93071c71f9d15d5d
|
BITSAT_MATH
|
Trigonometric Equations
|
If $2 \sin ^{2} x+7 \cos x=5$, then permissible value of $\cos x$ is
|
singleCorrect
| 1 |
We have $2 \cos ^{2} \theta+3 \cos \theta=2$
$2 \cos ^{2} \theta+4 \cos \theta-\cos \theta-2=0 \Rightarrow 2 \cos \theta(\cos \theta+2)-1(\cos \theta+2)=0$
$(2 \cos \theta-1)(\cos \theta+2)=0$
$\therefore \cos \theta=\frac{1}{2},-2$ (Impossible) $\Rightarrow \cos \theta=\frac{1}{2}$
|
["$\\frac{1}{2}$", "0", "1", "$-\\frac{1}{2}$"]
|
[0]
| null |
Practise Problem
|
f320ac4fc8acd2e68aab961e90bc187c
|
BITSAT_MATH
|
Trigonometric Equations
|
If $\cos \alpha+2 \cos \beta+3 \cos \gamma=0$,
$\sin \alpha+2 \sin \beta+3 \sin \gamma=0$ and $\alpha+\beta+\gamma=\pi$, then $\sin 3 \alpha+8 \sin 3 \beta+27 \sin 3 \gamma$ is equal to
|
singleCorrect
| 2 |
Let $a=\cos \alpha+i \sin \alpha$
$\quad b=\cos \beta+i \sin \beta$
and $\mathrm{c}=\cos \gamma+i \sin \gamma$
Then, $a+2 b+3 c=(\cos \alpha+2 \cos \beta+3 \cos \gamma)$
$\quad+i(\sin \alpha+2 \sin \beta+3 \sin \gamma)=0$
$\Rightarrow \quad a^{3}+8 b^{3}+27 c^{3}=18 a b c$
$\left(\because\right.$ if $\left.x+y+z=0 \Rightarrow x^{3}+y^{3}+z^{3}=3 x y z\right)$
$\Rightarrow \quad \cos 3 \alpha+8 \cos 3 \beta+27 \cos 3 \gamma$
and $\quad \sin 3 \alpha+8 \sin 3 \beta+27 \sin 3 \gamma$
$\quad=18 \sin (\alpha+\beta+\gamma)$
$=18 \sin \pi=0$
|
["$-18$", "0", "3", "9"]
|
[1]
| null |
Practise Problem
|
e5ff746e8b1c72dd4e73950b2726f0c0
|
BITSAT_MATH
|
Trigonometric Equations
|
$\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$
|
singleCorrect
| 2 |
Let $\frac{2 \pi r}{7}=\theta$
$\Rightarrow \quad 2 \pi r=3 \theta+4 \theta$
$\Rightarrow \quad 4 \theta=2 \pi r-3 \theta$
$\Rightarrow \quad \sin 4 \theta=\sin (2 \pi r-3 \theta)$
$\Rightarrow \quad \sin 4 \theta=-\sin 3 \theta$
$\Rightarrow 2 \sin 2 \theta \cos 2 \theta=-\left[3 \sin \theta-4 \sin ^{3} \theta\right.$
$\Rightarrow 2 \times 2 \sin \theta \cos \theta\left(2 \cos ^{2} \theta-1\right)$
$=-3 \sin \theta+4 \sin ^{3} \theta$
$\Rightarrow \sin \theta\left[8 \cos ^{3} \theta-4 \cos \theta+3-4\left(1-\cos ^{2} \theta\right)\right]=0$
$\Rightarrow 8 \cos ^{3} \theta+4 \cos ^{2} \theta-4 \cos \theta-1=0$ (i)
Thus, $\cos \frac{2 \pi}{7}, \cos \frac{4 \pi}{7}$ and $\cos \frac{6 \pi}{7}$ are the roots of
the above equation. $\therefore \quad \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}=-\frac{1}{2}$
|
["is equal to zero", "lies between 0 and 3", "is a negative number", "lies between 3 and 6"]
|
[2]
| null |
Practise Problem
|
d7f341997a61f0d019734fca02491084
|
BITSAT_MATH
|
Trigonometric Equations
|
For $x \in I R, 3 \cos (4 x-5)+4$ lies in the interval
|
singleCorrect
| 1 |
Since, the value of $\cos \theta$ lies between -1 and +1
$$
\begin{aligned}
& \text { i.e. }-1 \leq \cos \theta \leq 1 \\
& \therefore \quad-1 \leq \cos (4 x-5) \leq 1 \\
& \Rightarrow \quad-3 \leq 3 \cos (4 x-5) \leq 3 \\
& \Rightarrow \quad 4-3 \leq 3 \cos (4 x-5)+4 \leq 3+4 \\
& \Rightarrow \quad 1 \leq 3 \cos (4 x-5)+4 \leq 7 \\
&
\end{aligned}
$$
|
["$[1,7]$", "$[4,7]$", "$[0,7]$", "$[2,7]$"]
|
[0]
| null |
Practise Problem
|
cf405c89def06c53817e1721a37aa94f
|
BITSAT_MATH
|
Trigonometric Equations
|
If $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}, 0 \leq \alpha \leq \frac{\pi}{2}$, then the value of $\cos 2 \theta$ is
|
singleCorrect
| 2 |
$\begin{array}{ll} & \tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha} \\ & \frac{\sin \theta}{\cos \theta}=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha} \\ \therefore \quad & \sin \alpha \sin \theta+\cos \alpha \sin \theta=\sin \alpha \cos \theta-\cos \alpha \cos \theta \\ \therefore \quad & \cos \alpha \cos \theta+\sin \alpha \sin \theta=\sin \alpha \cos \theta-\cos \alpha \sin \theta \\ \therefore \quad & \cos (\alpha-\theta)=\sin (\alpha-\theta) \\ \therefore \quad & \alpha-\theta=\frac{\pi}{4} \\ \therefore \quad & \theta=\alpha-\frac{\pi}{4} \\ \therefore \quad & 2 \theta=2 \alpha-\frac{\pi}{2} \\ \therefore \quad & \cos 2 \theta=\cos \left(2 \alpha-\frac{\pi}{2}\right) \\ & =\cos \left[\because 0 \leq \alpha \leq \frac{\pi}{2}\right] \\ & \left.=\cos \left(\frac{\pi}{2}-2 \alpha\right)\right] \quad \ldots[\because \cos (-\theta)=\cos \theta] \\ \therefore \quad & \cos 2 \theta=\sin 2 \alpha\end{array}$
|
["$\\cos 2 \\alpha$", "$\\sin \\alpha$", "$\\cos \\alpha$", "$\\sin 2 \\alpha$"]
|
[3]
| null |
Practise Problem
|
cceab3a59f0e0fedbe341a76b200700e
|
BITSAT_MATH
|
Trigonometric Equations
|
The general solution of the system of equations <math> <mrow> <msup> <mtext>sin</mtext> <mn>3</mn> </msup> <mtext>x</mtext> <mo>+</mo> <msup> <mtext>sin</mtext> <mn>3</mn> </msup> <mfenced close=")" open="(" separators=","> <mrow> <mfrac> <mrow> <mn>2</mn> <mi>π</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mtext>x</mtext> </mrow> </mfenced> <mo>+</mo> <msup> <mtext>sin</mtext> <mn>3</mn> </msup> <mfenced close=")" open="(" separators=","> <mrow> <mfrac> <mrow> <mn>4</mn> <mi>π</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mtext>x</mtext> </mrow> </mfenced> <mo>+</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mtext>cos </mtext> <mn>2</mn> <mtext>x</mtext> <mo>=</mo> <mn>0</mn> </mrow> </math> and <math><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo></mo><mrow><mi mathvariant="normal">x</mi></mrow></mrow><mo>≠</mo><mn>0</mn></math> is
|
singleCorrect
| null |
<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∵</mo><mi>sin</mi><mn>3</mn><mi mathvariant="normal">x</mi><mo>=</mo><mn>3</mn><mi>sinx</mi><mo>-</mo><mn>4</mn><mi mathvariant="normal"> </mi><msup><mi>sin</mi><mn>3</mn></msup><mi mathvariant="normal">x</mi><mo>  </mo><mo>⇒</mo><msup><mi>sin</mi><mn>3</mn></msup><mi mathvariant="normal">x</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac><mfenced close="]" open="[" separators="|"><mrow><mn>3</mn><mi>sinx</mi><mo>-</mo><mi>sin</mi><mn>3</mn><mi mathvariant="normal">x</mi><mi mathvariant="normal"> </mi></mrow></mfenced><mi mathvariant="normal"> </mi></math><br /><br />The given equation reduces to<br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>4</mn></mfrac><mfenced separators="|"><mrow><mn>3</mn><mi>sinx</mi><mo>-</mo><mi>sin</mi><mn>3</mn><mi mathvariant="normal">x</mi></mrow></mfenced><mo>+</mo><mfrac><mn>1</mn><mn>4</mn></mfrac><mfenced separators="|"><mrow><mn>3</mn><mi>sin</mi><mfenced separators="|"><mrow><mfrac><mrow><mn>2</mn><mi mathvariant="normal">π</mi></mrow><mn>3</mn></mfrac><mo>+</mo><mi mathvariant="normal">x</mi></mrow></mfenced><mo>-</mo><mi mathvariant="normal"> </mi><mi>sin</mi><mfenced separators="|"><mrow><mn>2</mn><mi mathvariant="normal">π</mi><mo>+</mo><mn>3</mn><mi mathvariant="normal">x</mi></mrow></mfenced></mrow></mfenced><mo>+</mo><mfrac><mn>1</mn><mn>4</mn></mfrac><mo>(</mo><mn>3</mn><mi>sin</mi><mfenced separators="|"><mrow><mfrac><mrow><mn>4</mn><mi mathvariant="normal">π</mi></mrow><mn>3</mn></mfrac><mo>+</mo><mi mathvariant="normal">x</mi></mrow></mfenced><mo>-</mo><mi>sin</mi><mfenced separators="|"><mrow><mn>4</mn><mi mathvariant="normal">π</mi><mo>+</mo><mn>3</mn><mi mathvariant="normal">x</mi></mrow></mfenced><mo>+</mo><mfrac><mn>3</mn><mn>4</mn></mfrac><mi>cos</mi><mn>2</mn><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn><mi mathvariant="normal"> </mi></math><br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mfenced close="}" open="{" separators="|"><mrow><mi>sin</mi><mo>⁡</mo><mi mathvariant="normal">x</mi><mo>+</mo><mi>sin</mi><mfenced separators="|"><mrow><mfrac><mrow><mn>2</mn><mi mathvariant="normal">π</mi></mrow><mn>3</mn></mfrac><mo>+</mo><mi mathvariant="normal">x</mi></mrow></mfenced><mo>+</mo><mi>sin</mi><mo>⁡</mo><mfenced separators="|"><mrow><mfrac><mrow><mn>4</mn><mi mathvariant="normal">π</mi></mrow><mn>3</mn></mfrac><mo>+</mo><mi mathvariant="normal">x</mi></mrow></mfenced></mrow></mfenced><mo>-</mo><mfenced close="}" open="{" separators="|"><mrow><mi>sin</mi><mo>⁡</mo><mn>3</mn><mi mathvariant="normal">x</mi><mo>+</mo><mi>sin</mi><mo>⁡</mo><mfenced separators="|"><mrow><mn>2</mn><mi mathvariant="normal">π</mi><mo>+</mo><mn>3</mn><mi mathvariant="normal">x</mi></mrow></mfenced><mo>+</mo><mi>sin</mi><mo>⁡</mo><mfenced separators="|"><mrow><mn>4</mn><mi mathvariant="normal">π</mi><mo>+</mo><mn>3</mn><mi mathvariant="normal">x</mi></mrow></mfenced></mrow></mfenced><mo>+</mo><mn>3</mn><mi>cos</mi><mo>⁡</mo><mn>2</mn><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn></math><br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mfenced close="}" open="{" separators="|"><mrow><mi mathvariant="normal"> </mi><mi>sinx</mi><mo>+</mo><mn>2</mn><mi>sinx</mi><mo> </mo><mi>cos</mi><mo>⁡</mo><mi mathvariant="normal"> </mi><mfrac><mrow><mn>2</mn><mi mathvariant="normal">π</mi></mrow><mn>3</mn></mfrac></mrow></mfenced><mo>-</mo><mn>3</mn><mi>sin</mi><mn>3</mn><mi mathvariant="normal">x</mi><mo>+</mo><mn>3</mn><mi>cos</mi><mn>2</mn><mi mathvariant="normal">x</mi><mo>=</mo><mn>0</mn></math><br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒    </mo><mtext>sin </mtext><mn>3</mn><mtext>x</mtext><mo>=</mo><mtext>cos </mtext><mn>2</mn><mtext>x </mtext><mo>⇒ </mo><mtext>cos</mtext><mfenced><mrow><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac><mo>-</mo><mn>3</mn><mtext>x</mtext></mrow></mfenced><mo>=</mo><mtext>cos </mtext><mn>2</mn><mtext>x</mtext></math><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mn>2</mn><mi mathvariant="normal">x</mi><mo>=</mo><mn>2</mn><mi>kπ</mi><mo>±</mo><mfenced separators="|"><mrow><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac><mo>-</mo><mn>3</mn><mi mathvariant="normal">x</mi></mrow></mfenced></math><br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒    </mo><mn>5</mn><mtext>x</mtext><mo>=</mo><mn>2</mn><mtext>k</mtext><mi mathvariant="normal">π</mi><mo>+</mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac><mo>=</mo><mfenced><mrow><mn>4</mn><mtext>k</mtext><mo>+</mo><mn>1</mn></mrow></mfenced><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac></math> <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∀</mo><mi mathvariant="normal">k</mi><mo>∈</mo><mi>Z</mi></math> or <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal"> </mi><mi mathvariant="normal">x</mi><mo>=-2</mo><mi>kπ</mi><mo>+</mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac><mo>,</mo><mi mathvariant="normal"> </mi><mo>∀</mo><mi mathvariant="normal">k</mi><mo>∈</mo><mi>Z</mi></math><br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒    </mo><mtext>x</mtext><mo>=</mo><mfenced><mrow><mn>4</mn><mtext>k</mtext><mo>+</mo><mn>1</mn></mrow></mfenced><mfrac><mi mathvariant="normal">π</mi><mn>10</mn></mfrac><mtext>, </mtext><mo>∀</mo><mi mathvariant="normal">k</mi><mo>∈</mo><mi>Z</mi></math><math><mi mathvariant="normal"> </mi><mi mathvariant="normal"> </mi><mi mathvariant="normal"> </mi><mi mathvariant="normal"> </mi><mfenced close="}" open="{" separators="|"><mrow><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo>⁡</mo><mrow><mi mathvariant="normal">x</mi></mrow></mrow><mo>≠</mo><mn>0</mn></mrow></mfenced></math>
|
["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mi mathvariant=\"normal\">x</mi><mo>=</mo><mfrac><mrow><mfenced><mrow><mn>2</mn><mtext>k</mtext><mo>+</mo><mn>1</mn></mrow></mfenced><mi mathvariant=\"normal\">π</mi></mrow><mn>10</mn></mfrac><mtext>, k</mtext><mo>∈</mo><mtext>Z</mtext></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mi mathvariant=\"normal\">x</mi><mo>=</mo><mfrac><mrow><mfenced><mrow><mn>2</mn><mtext>k</mtext><mo>+</mo><mn>1</mn></mrow></mfenced><mi mathvariant=\"normal\">π</mi></mrow><mn>5</mn></mfrac><mtext>, k</mtext><mo>∈</mo><mtext>Z</mtext></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mi mathvariant=\"normal\">x</mi><mo>=</mo><mfrac><mrow><mfenced><mrow><mn>4</mn><mtext>k</mtext><mo>+</mo><mn>1</mn></mrow></mfenced><mi mathvariant=\"normal\">\u03c0</mi></mrow><mn>10</mn></mfrac><mtext>, k</mtext><mo>\u2208</mo><mtext>Z</mtext></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mi mathvariant=\"normal\">x</mi><mo>=</mo><mfenced><mfrac><mrow><mn>4</mn><mtext>k</mtext><mo>+</mo><mn>1</mn></mrow><mn>5</mn></mfrac></mfenced><mi mathvariant=\"normal\">π</mi><mtext>, k</mtext><mo>∈</mo><mtext>Z</mtext></math>"]
|
[2]
| null |
Practise Problem
|
ae9557ee75be9fe018f2cd6c81b28216
|
BITSAT_MATH
|
Trigonometric Equations
|
If \(2 y \cos \theta=x \sin \theta\) and \(2 x \sec \theta-y \operatorname{cosec} \theta=3\), then \(x^2+4 y^2=\)
|
singleCorrect
| 2 |
We have \(2 \mathrm{y} \cos \theta=\mathrm{x} \sin \theta\)
or \(\frac{\cos \theta}{\mathrm{x}}=\frac{\sin \theta}{2 \mathrm{y}}=\mathrm{k}\) (say)
Then \(\cos \theta=\mathrm{kx}\) and \(\sin \theta=2 \mathrm{ky}\) ...(1)
Again \(2 \mathrm{x} \sec \theta-\mathrm{y} \operatorname{cosec} \theta=3\)
or \(\frac{2 \mathrm{x}}{\cos \theta}-\frac{\mathrm{y}}{\sin \theta}=3\)
or \(\frac{2 \mathrm{x}}{\mathrm{kx}}-\frac{\mathrm{y}}{2 \mathrm{ky}}=3 \quad\) [from(1)]
or \(\frac{2}{\mathrm{k}}-\frac{1}{2 \mathrm{k}}=3\), giving \(\mathrm{k}=\frac{1}{2}\)
We now get \(\cos \theta=\frac{\mathrm{x}}{2}\) and \(\sin \theta=\mathrm{y}\)
Squaring and adding we get
\(\begin{aligned}
& \cos ^2 \theta+\sin ^2 \theta=\frac{x^2}{4+y^2} \\
& \Rightarrow \frac{x^2}{4}+y^2=1, \text { or } x^2+4 y^2=4
\end{aligned}\)
|
["2", "4", "1", "none"]
|
[1]
| null |
Practise Problem
|
40c085fe24da024424d6f80620ce038d
|
BITSAT_MATH
|
Trigonometric Equations
|
If $3 \cos x \neq 2 \sin x$, then the general solution of $\sin ^{2} x-\cos 2 x=2-\sin 2 x$ is
|
singleCorrect
| 1 |
$$
\begin{array}{l}
\sin ^{2} x-\cos 2 x=2-\sin 2 x \\
\Rightarrow 1-\cos ^{2} x-\left(2 \cos ^{2}-1\right)=2-2 \sin x \cos x \\
\Rightarrow-3 \cos ^{2} x+2 \sin x \cos x=0 \\
\Rightarrow \cos x(2 \sin x-3 \cos x)=0 \\
\Rightarrow \cos x=0, \quad(\because 2 \sin x-3 \cos x \neq 0) \\
\Rightarrow x=2 n \pi \pm \frac{\pi}{2} \\
\Rightarrow x=(4 n \pm 1) \frac{\pi}{2}, n \in Z
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow x=2 n \pi \pm \frac{\pi}{2} \\
\Rightarrow x=(4 n \pm 1) \frac{\pi}{2}, n \in Z \\
\end{array}
$$
|
["$n \\pi+(-1)^{n} \\frac{\\pi}{2}, n \\in Z$", "$\\frac{n \\pi}{2}, n \\in Z$", "$(4 n \\pm 1) \\frac{\\pi}{2}, n \\in Z$", "$(2 n-1) \\pi, n \\in Z$"]
|
[2]
| null |
Practise Problem
|
18acd60ae6bb7221aa2e9e0dfdc029e8
|
BITSAT_MATH
|
Trigonometric Equations
|
The values of <math><mi>x</mi></math> in <math><mfenced separators="|"><mrow><mn>0</mn><mo>,</mo><mi></mi><mi></mi><mfrac><mrow><mi>π</mi></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></mfenced></math> satisfying the equation <math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>x</mi><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo></mo><mrow><mi>x</mi><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>4</mn></mrow></mfrac></mrow></mrow></mrow></mrow></math> are ________
|
singleCorrect
| 1 |
Given: Equation <math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>x</mi></mrow></mrow><mo>.</mo><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo></mo><mrow><mi>x</mi></mrow></mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>4</mn></mrow></mfrac></math> <br/>then <math><mi mathvariant="normal">s</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">n</mi><mn>2</mn><mi>x</mi><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></math> <br/><math><mn>2</mn><mi>x</mi><mo>=</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac><mo>,</mo><mn>5</mn><mfrac><mrow><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac><mo>⇒</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>12</mn></mrow></mfrac><mo>,</mo><mi mathvariant="normal"></mi><mfrac><mrow><mn>5</mn><mi>π</mi></mrow><mrow><mn>12</mn></mrow></mfrac></math>
|
["<math><mfrac><mrow><mi>\u03c0</mi></mrow><mrow><mn>6</mn></mrow></mfrac><mo>,</mo><mfrac><mrow><mi>\u03c0</mi></mrow><mrow><mn>12</mn></mrow></mfrac></math>", "<math><mfrac><mrow><mi>\u03c0</mi></mrow><mrow><mn>12</mn></mrow></mfrac><mo>,</mo><mfrac><mrow><mn>5</mn><mi>\u03c0</mi></mrow><mrow><mn>12</mn></mrow></mfrac></math>", "<math><mfrac><mrow><mi>\u03c0</mi></mrow><mrow><mn>8</mn></mrow></mfrac><mo>,</mo><mfrac><mrow><mn>3</mn><mi>\u03c0</mi></mrow><mrow><mn>8</mn></mrow></mfrac></math>", "<math><mfrac><mrow><mi>\u03c0</mi></mrow><mrow><mn>8</mn></mrow></mfrac><mo>,</mo><mfrac><mrow><mi>\u03c0</mi></mrow><mrow><mn>4</mn></mrow></mfrac></math>"]
|
[1]
| null |
Practise Problem
|
a5ed23af04b75936c508fe053701461c
|
BITSAT_MATH
|
Trigonometric Equations
|
The number of solutions of \(2 \sin x+\cos x=3\) is
|
singleCorrect
| 1 |
Hints: \(\sqrt{5} < 3\) No solution
|
["1", "2", "infinite", "No solution"]
|
[3]
| null |
Practise Problem
|
43ded358f2a1b84c0169b3cafe706b29
|
BITSAT_MATH
|
Trigonometric Equations
|
The number of solutions for the equation $\sin 2 x+\cos 4 x=2$ is
|
singleCorrect
| 1 |
Given, $\sin 2 x+\cos 4 x=2$
$$
\begin{array}{ll}
\Rightarrow & \sin 2 x+1-2 \sin ^{2} 2 x=2 \\
\Rightarrow & 2 \sin ^{2} 2 x-\sin 2 x+1=0
\end{array}
$$
Now, Discriminant, $\mathrm{D}=(-1)^{2}-4 \cdot 2 \cdot 1=-7 < 0$
Hence, it is an imaginary equation, so the real roots does not exist.
|
["0", "1", "2", "$\\infty$"]
|
[0]
| null |
Practise Problem
|
faf3ab273ef5d3dd5739a2857d060c2a
|
BITSAT_MATH
|
Trigonometric Equations
|
The number of solutions for the equation $\sin 2 x+\cos 4 x=2$ is
|
singleCorrect
| 2 |
We have,
$$
\begin{aligned}
\sin 2 x+\cos 4 x=2 \\
\Rightarrow \quad \sin 2 x+1 &-2 \sin ^{2} 2 x=2 \\
\Rightarrow \quad 2 \sin ^{2} 2 x-\sin 2 x+1=0 \\
\text { Now, Discriminant } &=b^{2}-4 a c \\
&=(-1)^{2}-4(2)(1) \\
&=1-8=-7 \\
& < 0
\end{aligned}
$$
So, there is no real root of the given equation.
|
["0", "1", "2", "Infinite"]
|
[0]
| null |
Practise Problem
|
09f951a6a2ae51fb3492f4fe74601f30
|
BITSAT_MATH
|
Trigonometric Equations
|
Find the general solution of $\sin 2 x+\cos x=0$.
|
singleCorrect
| 2 |
$$
\begin{array}{ll}
\text { } \sin 2 x+\cos x=0 \\
\Rightarrow & 2 \sin x \cos x+\cos x=0 \\
\Rightarrow & \cos x(2 \sin x+1)=0 \\
\Rightarrow & \cos x=0 \text { or } 2 \sin x+1=0 \\
\Rightarrow & \cos x=\cos \frac{\pi}{2} \Rightarrow x=n \pi \pm \frac{\pi}{2}, n \in z \\
\text { or } & \sin x=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}\right) \\
& =\sin \left(\pi+\frac{\pi}{6}\right) \\
\Rightarrow & \sin x=\sin \left(\frac{7 \pi}{6}\right) \\
\therefore & x=n \pi+(-1)^n \frac{7 \pi}{6}, n \in Z
\end{array}
$$
|
["$(n \\pm 1) \\frac{\\pi}{2}$", "$n \\pi \\pm \\frac{\\pi}{2}$", "$n \\pi+(-1)^n \\frac{7 \\pi}{6}$ or $(2 n \\pm 1) \\frac{\\pi}{2}$", "None of the above"]
|
[2]
| null |
Practise Problem
|
de16dd4a2e25dc755ea275a1c5146a0d
|
BITSAT_MATH
|
Trigonometric Equations
|
The principal solutions of $\cos 2 x=\frac{-1}{2}$ are
|
singleCorrect
| 1 |
(B)
$\begin{aligned} & \cos 2 x=\frac{-1}{2} \\ \therefore & \frac{-1}{2}=\cos \left(\pi-\frac{\pi}{3}\right)-\cos \left(\pi+\frac{\pi}{3}\right) \Rightarrow \frac{-1}{2}-\cos \frac{2 \pi}{3}-\cos \frac{4 \pi}{3} \\ \therefore & 2 x=\frac{2 \pi}{3} \text { or } 2 x=\frac{4 \pi}{3} \Rightarrow x=\frac{\pi}{3} \text { or } \frac{2 \pi}{3} \end{aligned}$
|
["$x=\\frac{-2 \\pi}{3}, x=\\frac{4 \\pi}{3}$", "$x=\\frac{\\pi}{3}, \\quad x=\\frac{2 \\pi}{3}$", "$x=\\frac{-\\pi}{3}, \\quad x=\\frac{5 \\pi}{6}$", "$x=\\frac{\\pi}{3}, \\quad x=\\frac{7 \\pi}{6}$"]
|
[1]
| null |
Practise Problem
|
0967423eeb2e3b564eecb807d1e69601
|
BITSAT_MATH
|
Trigonometric Equations
|
The general solution of $|\sin x|=\cos x$ is (when $\mathrm{n} \in \mathrm{I}$ ) given by
|
singleCorrect
| 1 |
Given, $|\sin x|=\cos x$
$$
\begin{array}{lr}
\therefore & \sin ^{2} \mathrm{x}=\cos ^{2} \mathrm{x} \\
\Rightarrow & 1-\cos ^{2} \mathrm{x}=\cos ^{2} \mathrm{x} \\
\Rightarrow & 2 \cos ^{2} \mathrm{x}=1
\end{array}
$$
$\Rightarrow \quad \cos x=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \quad \cos x=+\frac{1}{\sqrt{2}} \quad(\because \cos x$ cannot be negative $)$
$\Rightarrow \quad \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{4}$
|
["$n \\pi+\\frac{\\pi}{4}$", "$2 \\mathrm{n} \\pi \\pm \\frac{\\pi}{4}$", "$n \\pi \\pm \\frac{\\pi}{4}$", "$n \\pi-\\frac{\\pi}{4}$"]
|
[1]
| null |
Practise Problem
|
27a3b6d5215bba3cb866439973468d3c
|
BITSAT_MATH
|
Trigonometric Equations
|
The number of solution of $\tan x+\sec x=2 \cos x$ in $[0,2 \pi)$ is
|
singleCorrect
| 1 |
The given equation is $\tan x+\sec x=2 \cos x \Rightarrow \sin x+1=2 \cos ^2 x$
$$
\begin{aligned}
& \Rightarrow \sin x+1=2\left(1-\sin ^2 x\right) \Rightarrow 2 \sin ^2 x+\sin x-1=0 \\
& \Rightarrow(2 \sin x-1)(\sin x+1)=0 \Rightarrow \sin x=\frac{1}{2},-1 \Rightarrow x=30^{\circ}, 150^{\circ}, 270^{\circ}
\end{aligned}
$$
|
["\n2\n", "\n3\n", "\n0\n", "\n1"]
|
[1]
| null |
Practise Problem
|
b21c5394e7c30b1e1f576a9a55b53c3d
|
BITSAT_MATH
|
Trigonometric Equations
|
If <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>tan</mi><mo></mo><mfenced separators="|"><mrow><mi mathvariant="normal">k</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mi>θ</mi><mo>=</mo><mi>tan</mi><mo></mo><mi>θ</mi></math>, then the set of all the values of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>θ</mi></math> is
|
singleCorrect
| null |
As we know the general solution of the equation<br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>tan</mi><mi>x</mi><mo>=</mo><mi>tan</mi><mi>α</mi><mtext> </mtext><mtext> </mtext><mtext>is</mtext><mtext> </mtext><mtext> </mtext><mtext>given</mtext><mtext> </mtext><mtext> </mtext><mtext>by</mtext><mtext> </mtext><mtext> </mtext><mi>x</mi><mo>=</mo><mi mathvariant="normal">n</mi><mi>π</mi><mo>+</mo><mi>α</mi><mo>,</mo><mo> </mo><mi mathvariant="normal">n</mi><mo>∈</mo><mi>I</mi></math><br /><br />Hence the solution of the equation, <math> <mrow> <mtext>tan</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>k</mtext> <mo>+</mo> <mn>1</mn> </mrow> </mfenced> <mi>θ</mi> <mo>=</mo> <mtext>tan</mtext> <mi>θ</mi> </mrow> </math><br /><br />is given by <math> <mrow> <mo></mo> <mfenced close=")" open="(" separators=","> <mrow> <mtext>k</mtext> <mo>+</mo> <mn>1</mn> </mrow> </mfenced> <mi>θ</mi> <mo>=</mo> <mtext>n</mtext> <mi>π</mi> <mo>+</mo> <mi>θ</mi> <mo> ⇒ </mo> <mtext>k</mtext> <mi>θ</mi> <mo>=</mo> <mtext>n</mtext> <mi>π, </mi> </mrow></math><math><mrow><mtext> n</mtext> <mo>∈</mo> <mi>I</mi> </mrow></math><br /><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mi>θ</mi><mo> </mo><mo>∈</mo><mo> </mo><mfrac><mrow><mi mathvariant="normal">n</mi><mi>π</mi></mrow><mi mathvariant="normal">k</mi></mfrac><mo> </mo><mo> </mo><mo>:</mo><mo> </mo><mo> </mo><mi mathvariant="normal">n</mi><mo> </mo><mo>∈</mo><mo> </mo><mi>I</mi></math><span style="display: none;"> </span>
|
["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mo>{</mo><mi mathvariant=\"normal\">n</mi><mi>π</mi><mi> </mi><mo>:</mo><mi mathvariant=\"normal\">n</mi><mo>∈</mo><mi>I</mi><mo>}</mo></math>", "<math> <mrow> <mfenced close=\"}\" open=\"{\" separators=\",\"> <mrow> <mfrac> <mrow> <mtext>n</mtext> <mi>π</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mtext> : </mtext> <mtext>n</mtext> <mo>∈</mo> <mtext mathvariant=\"italic\">I</mtext> </mrow> </mfenced> </mrow> </math>", "<math> <mrow> <mfenced close=\"}\" open=\"{\" separators=\",\"> <mrow> <mfrac> <mrow> <mtext>n</mtext> <mi>\u03c0</mi> </mrow> <mrow> <mtext>k</mtext> </mrow> </mfrac> <mtext> : </mtext> <mtext>n</mtext> <mo>\u2208</mo> <mtext mathvariant=\"italic\">I</mtext> </mrow> </mfenced> </mrow> </math>", "<math> <mrow> <mfenced close=\"}\" open=\"{\" separators=\",\"> <mrow> <mfrac> <mrow> <mtext>n</mtext> <mi>π</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mtext> : </mtext> <mtext>n</mtext> <mo>∈</mo> <mtext mathvariant=\"italic\">I</mtext> </mrow> </mfenced> </mrow> </math>"]
|
[2]
| null |
Practise Problem
|
109c98902c15572f2aa20377b7a54388
|
BITSAT_MATH
|
Trigonometric Equations
|
The equation $\sqrt{3} \sin x+\cos x=4$, has
|
singleCorrect
| 1 |
We have,
$\begin{array}{cc} & \sqrt{3} \sin x+\cos x=4 \\ \Rightarrow & \frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=2 \\ \Rightarrow & \sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6}=2 \\ \Rightarrow & \sin \left(x+\frac{\pi}{6}\right)=2, \text { which is not possible. }\end{array}$
Therefore, this equation has no solution.
|
["only one solution", "two solution", "infinitely many solution", "no solution"]
|
[3]
| null |
Practise Problem
|
80cad1d19a235b90cc2a6426e5a93ca3
|
BITSAT_MATH
|
Trigonometric Equations
|
The solution of $\sin x=-\frac{\sqrt{3}}{2}$ is
|
singleCorrect
| 1 |
We have, $\sin x=-\frac{\sqrt{3}}{2}=-\sin \frac{\pi}{3}$
Hence, $\sin x=\sin \frac{4 \pi}{3}$, which gives $\mathrm{x}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{4 \pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$
|
["$x=\\mathrm{n} \\pi+(-1)^{\\mathrm{n}} \\frac{4 \\pi}{3}$, where $\\mathrm{n} \\in \\mathrm{Z}$", "$x=\\mathrm{n} \\pi+(-1)^{\\mathrm{n}} \\frac{2 \\pi}{3}$, where $\\mathrm{n} \\in \\mathrm{Z}$", "$x=\\mathrm{n} \\pi+(-1)^{\\mathrm{n}} \\frac{3 \\pi}{3}$, where $\\mathrm{n} \\in Z$", "None of the these"]
|
[0]
| null |
Practise Problem
|
beb15b8c91819d285681f2c8bb2346f8
|
BITSAT_MATH
|
Trigonometric Equations
|
The period of \(\tan 3 \theta\) is
|
singleCorrect
| 1 |
\(\tan \theta\) is of period \(\pi\) so that \(\tan 3 \theta\) is of period \(\pi / 3\).
|
["\\(\\pi\\)", "\\(3 \\pi / 4\\)", "\\(\\pi / 2\\)", "None of these"]
|
[3]
| null |
Practise Problem
|
6bbf29c1380d13ec9009db054b3610f8
|
BITSAT_MATH
|
Trigonometric Equations
|
If $5 \cos 2 \theta+2 \cos ^2 \frac{\theta}{2}+1=0$, when $(0 < \theta < \pi)$, then the values of $\theta$ are :
|
singleCorrect
| 1 |
Hints : $5 \cos 2 \theta+1+\cos \theta+1=0$
$$
\begin{aligned}
& 5\left(2 \cos ^2 \theta-1\right)+\cos \theta+2=0 \\
& 10 \cos ^2 \theta+\cos \theta-3=0 \\
& (5 \cos \theta+3)(2 \cos \theta-1)=0 \\
& \cos \theta=\frac{1}{2} \\
& \theta=\frac{\pi}{3}
\end{aligned}
$$
$$
\mid \begin{aligned}
& \cos \theta=-\frac{3}{5} \\
& \theta=\cos ^{-1}\left(-\frac{3}{5}\right) \\
& =\pi-\cos ^{-1}\left(\frac{3}{5}\right)
\end{aligned}
$$
|
["$\\frac{\\pi}{3} \\pm \\pi$", "$\\frac{\\pi}{3}, \\cos ^{-1}\\left(\\frac{3}{5}\\right)$", "$\\cos ^{-1}\\left(\\frac{3}{5}\\right) \\pm \\pi$", "$\\frac{\\pi}{3}, \\pi-\\cos ^{-1}\\left(\\frac{3}{5}\\right)$"]
|
[3]
| null |
Practise Problem
|
0a771ee964a1319d5ee59bf9b28bf496
|
BITSAT_MATH
|
Trigonometric Equations
|
$\{x \in R: |\cos x | \geq \sin x\} \cap\left[0, \frac{3 \pi}{2}\right]$ is equal to
|
singleCorrect
| 2 |
Given, $\{x \in R:|\cos x| \geq \sin x\} \cap\left[0, \frac{3 \pi}{2}\right]$<br/>If we draw the graphs of $|\cos x|$ and $\sin x,$ clearly $|\cos x| \geq \sin x$ when<br/><img src="https://cdn.quizrr.in/question-assets/wbjee/2015_mat_a109.png"><br/>$x \in\left[0, \frac{\pi}{4}\right] \cup\left[\frac{3 \pi}{4} \cdot \frac{3 \pi}{2}\right]$<br/>$\therefore \quad x \in\left[0, \frac{\pi}{4}\right] \cup\left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right] \cap\left[0, \frac{3 \pi}{2}\right]$<br/>$\Rightarrow \quad x \in\left[0, \frac{\pi}{4}\right] \cup\left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right]$
|
["$\\left[0, \\frac{\\pi}{4}\\right] \\cup\\left[\\frac{3 \\pi}{4}, \\frac{3 \\pi}{2}\\right]$", "$\\left[0, \\frac{\\pi}{4}\\right] \\cup\\left[\\frac{\\pi}{2} \\cdot \\frac{3 \\pi}{2}\\right]$", "$\\left[0, \\frac{\\pi}{4}\\right] \\cup\\left[\\frac{5 \\pi}{4} \\cdot \\frac{3 \\pi}{2}\\right]$", "$\\left[0, \\frac{3 \\pi}{2}\\right]$"]
|
[0]
| null |
Practise Problem
|
49da95710975d24d220b04b704f66936
|
BITSAT_MATH
|
Trigonometric Equations
|
The general solution of the equation <math><mrow><mrow><msup><mrow><mi mathvariant="normal">tan</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo></mo><mrow><mi>x</mi></mrow></mrow><mo>=</mo><mn>1</mn></math> is
|
singleCorrect
| 1 |
<math><mrow><mrow><msup><mrow><mi mathvariant="normal">tan</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo></mo><mrow><mi>x</mi></mrow></mrow><mo>=</mo><mn>1</mn></math> <br/> <math><mo>∴</mo><mi> </mi><mrow><mrow><mi mathvariant="normal">tan</mi></mrow><mo></mo><mrow><mi>x</mi></mrow></mrow><mo>=</mo><mo>±</mo><mi> </mi><mn>1</mn></math> <br/> <math><mo>=</mo><mi>x</mi><mo>=</mo><mi>n</mi><mi>π</mi><mo>±</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>4</mn></mrow></mfrac></math>
|
["<math><mi>n</mi><mi>\u03c0</mi><mo>+</mo><mfrac><mrow><mi>\u03c0</mi></mrow><mrow><mn>4</mn></mrow></mfrac></math>", "<math><mi>n</mi><mi>\u03c0</mi><mo>-</mo><mfrac><mrow><mi>\u03c0</mi></mrow><mrow><mn>4</mn></mrow></mfrac></math>", "<math><mi>n</mi><mi>\u03c0</mi><mo>\u00b1</mo><mfrac><mrow><mi>\u03c0</mi></mrow><mrow><mn>4</mn></mrow></mfrac></math>", "<math><mn>2</mn><mi>n</mi><mi>\u03c0</mi><mo>\u00b1</mo><mfrac><mrow><mi>\u03c0</mi></mrow><mrow><mn>4</mn></mrow></mfrac></math>"]
|
[2]
| null |
Practise Problem
|
685daab65c901622c13dfb2a7b7f8423
|
BITSAT_MATH
|
Trigonometric Equations
|
If $\sin x+\sin ^{2} x=1$, then $\cos ^{8} x+2 \cos ^{6} x+\cos ^{4} x$ is
|
singleCorrect
| 3 |
Given $\sin x+\sin ^{2} x=1 \Rightarrow \sin x=1-\sin ^{2} x$
$\begin{aligned} \therefore \sin x=\cos ^{2} x \Rightarrow \sin ^{2} x &=\cos ^{4} x \\ \cos ^{8} x+2 \cos ^{6} x+\cos ^{4} x &=\left(\cos ^{4} x+\cos ^{2} x\right)^{2} \\ &=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}=1 \end{aligned}$
|
["3", "2", "1", "4"]
|
[2]
| null |
Practise Problem
|
f40beb85f480baa95579d490bc287186
|
BITSAT_MATH
|
Trigonometric Equations
|
If \( \cos x=|\sin x| \) then, the general solution is
|
singleCorrect
| 2 |
(A)
$\cos x=|\sin x|$
$\Rightarrow \pm \cos x=\sin x$
$\Rightarrow \tan x=\pm 1$
$x=n \pi \pm \frac{\Pi}{4}, n \in z$, but $\cos x$ is positive so $x=2 n \pi \pm \frac{\Pi}{4}, n \in z$
|
["\\( x=2 n \\pi \\pm \\frac{\\Pi}{4}, n \\in z \\)", "\\( x=(2 n+1) \\pi \\pm \\frac{I}{4}, n \\in Z \\)", "\\( x=n \\pi \\pm \\frac{\\Pi}{4}, n \\in Z \\)", "\\( x=n \\pi \\pm(-1)^{n} \\frac{\\Pi}{4}, n \\in Z \\)"]
|
[0]
| null |
Practise Problem
|
bc8c0177e1ca650e3fa4676e0219f5b0
|
BITSAT_MATH
|
Trigonometric Equations
|
Number of roots of the equation <math><mrow><mrow><msup><mrow><mi mathvariant="normal">cos</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo></mo><mrow><mi>x</mi><mo>+</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mo>+</mo><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>x</mi></mrow></mrow><mo>-</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt></mrow><mrow><mn>4</mn></mrow></mfrac><mo>-</mo><mn>1</mn><mo>=</mo><mn>0</mn></mrow></mrow></math> which lie in the interval <math><mfenced close="]" open="[" separators="|"><mrow><mo>-</mo><mi>π</mi><mo>,</mo><mi>π</mi></mrow></mfenced></math> is
|
singleCorrect
| null |
Given equation is<br /><br /><math><mn>1</mn><mo>-</mo><mrow><mrow><msup><mrow><mi mathvariant="normal">sin</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo></mo><mrow><mi>x</mi><mo>+</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mo>+</mo><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>x</mi><mo>-</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt></mrow><mrow><mn>4</mn></mrow></mfrac><mo>-</mo><mn>1</mn><mo>=</mo><mn>0</mn></mrow></mrow></mrow></mrow></math><br /><br /><math><mo>⇒ </mo><mrow><mrow><msup><mrow><mi mathvariant="normal">sin</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo></mo><mrow><mi>x</mi><mo>-</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mo>+</mo><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>x</mi><mo>+</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt></mrow><mrow><mn>4</mn></mrow></mfrac><mo>=</mo><mn>0</mn><mo>;</mo><mn>4</mn><mrow><mrow><msup><mrow><mi mathvariant="normal">sin</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo></mo><mrow><mi>x</mi><mo>-</mo><mn>2</mn><msqrt><mn>3</mn></msqrt><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>x</mi><mo>-</mo><mn>2</mn><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>x</mi><mo>+</mo><msqrt><mn>3</mn></msqrt><mo>=</mo><mn>0</mn></mrow></mrow></mrow></mrow></mrow></mrow></mrow></mrow></mrow></mrow></math><br /><br /><math> <mn>2</mn><mi>sin</mi><mi>x</mi><mrow><mo>(</mo> <mrow> <mn>2</mn><mi>sin</mi><mi>x</mi><mo>−</mo><msqrt> <mn>3</mn> </msqrt> </mrow> <mo>)</mo></mrow><mo>−</mo><mrow><mo>(</mo> <mrow> <mn>2</mn><mi>sin</mi><mi>x</mi><mo>−</mo><msqrt> <mn>3</mn> </msqrt> </mrow> <mo>)</mo></mrow><mo>=</mo><mn>0</mn> </math><br /><br /><math> <mo>⇒</mo><mrow><mo>(</mo> <mrow> <mn>2</mn><mi>sin</mi><mi>x</mi><mo>−</mo><mn>1</mn> </mrow> <mo>)</mo></mrow><mrow><mo>(</mo> <mrow> <mn>2</mn><mi>sin</mi><mi>x</mi><mo>−</mo><msqrt> <mn>3</mn> </msqrt> </mrow> <mo>)</mo></mrow><mo>=</mo><mn>0</mn> </math><br /><br />On solving we get <math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>x</mi></mrow></mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></math> <math><mo>;</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt></mrow><mrow><mn>2</mn></mrow></mfrac><mo></mo></math><br /><br /><math><mi>x</mi><mo>=</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac><mo>,</mo><mfrac><mrow><mn>5</mn><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac><mi> </mi><mo>;</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>3</mn></mrow></mfrac><mo>,</mo><mfrac><mrow><mn>2</mn><mi>π</mi></mrow><mrow><mn>3</mn></mrow></mfrac></math>
|
["2", "4", "6", "8"]
|
[1]
| null |
Practise Problem
|
bde96ff5af424e9d46f252457dd1f288
|
BITSAT_MATH
|
Trigonometric Equations
|
\(\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}\) is equal to
|
singleCorrect
| 1 |
\(\begin{aligned}
& \frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta} \\
& =\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}+\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}} \\
& =\frac{\cos ^2 \theta}{\cos \theta-\sin \theta}-\frac{\sin ^2 \theta}{\cos \theta-\sin \theta} \\
& =\frac{\cos ^2 \theta-\sin ^2 \theta}{\cos \theta-\sin \theta}=\cos \theta+\sin \theta
\end{aligned}\)
|
["\\(\\sin \\theta-\\cos \\theta\\)", "\\(\\sin \\theta+\\cos \\theta\\)", "\\(\\tan \\theta+\\cot \\theta\\)", "\\(\\tan \\theta-\\cot \\theta\\)"]
|
[1]
| null |
Practise Problem
|
619b14682af1b76c27844f02370eb33f
|
BITSAT_MATH
|
Trigonometric Equations
|
If \(\frac{1}{6} \sin \theta, \cos \theta, \tan \theta\) are in G.P., then the solution set of \(\theta\) is
|
singleCorrect
| 2 |
\(\begin{aligned}
& \text {Hint : } 6 \cos ^3 \theta+\cos ^2 \theta-1=0 \\
& \Rightarrow \quad(2 \cos \theta-1)\left(3 \cos ^2 \theta+2 \cos \theta+1\right)=0 \\
& \Rightarrow \quad \cos \theta=\frac{1}{2}\left(3 \cos ^2 \theta+2 \cos \theta+1 \neq 0\right)
\end{aligned}\)
|
["\\(2 \\mathrm{n} \\pi \\pm \\frac{\\pi}{6}\\)", "\\(2 \\mathrm{n} \\pi \\pm \\frac{\\pi}{3}\\)", "\\(\\mathrm{n} \\pi+(-1)^{\\mathrm{n}} \\frac{\\pi}{3}\\)", "\\(\\mathrm{n} \\pi+\\frac{\\pi}{3}\\)"]
|
[1]
| null |
Practise Problem
|
96231fc9e1bd5f3a3a56ffc6100b7bf8
|
BITSAT_MATH
|
Trigonometric Equations
|
If $2 \cos ^{2} \theta+3 \cos \theta=2$, then permissible value of $\cos \theta$ is
|
singleCorrect
| 1 |
(C)
We have $2 \cos ^{2} \theta+3 \cos \theta=2$
$2 \cos ^{2} \theta+4 \cos \theta-\cos \theta-2=0 \Rightarrow 2 \cos \theta(\cos \theta+2)-1(\cos \theta+2)=0$
$(2 \cos \theta-1)(\cos \theta+2)=0$
$\therefore \cos \theta=\frac{1}{2},-2($ Impossible $) \Rightarrow \cos \theta=\frac{1}{2}$
|
["0", "1", "$\\frac{1}{2}$", "$\\frac{-1}{2}$"]
|
[2]
| null |
Practise Problem
|
c113782b13ebe3231ce380218785608b
|
BITSAT_MATH
|
Trigonometric Equations
|
$\cos \alpha \sin (\beta-\gamma)+\cos \beta \sin (\gamma-\alpha)$ $+\cos \gamma \sin (\alpha-\beta)$ is equal to
|
singleCorrect
| 2 |
$\cos \alpha \sin (\beta-\gamma)+\cos \beta \sin (\gamma-\alpha)$$+\cos \gamma \sin (\alpha-\beta)$
$=\cos \alpha[\sin \beta \cos \gamma-\cos \beta \sin \gamma]$$+\cos \beta[\sin \gamma \cos \alpha-\cos \gamma \sin \alpha]$$+\cos \gamma[\sin \alpha \cos \beta-\cos \alpha \sin \beta]=0$
|
["0", "$\\frac{1}{2}$", "1", "$4 \\cos \\alpha \\cos \\beta \\cos \\gamma$"]
|
[0]
| null |
Practise Problem
|
43e4ba6ec8e8dc3461bf5cd8bf03ab5c
|
BITSAT_MATH
|
Trigonometric Equations
|
The general solution of $1+\sin ^{2} x=3 \sin x \cdot \cos x, \tan x \neq \frac{1}{2}$, is
|
singleCorrect
| 2 |
$$
1+\sin ^{2} x=3 \sin x \cdot \cos x, \tan x \neq \frac{1}{2}
$$
Divided by $\cos ^{2} \mathrm{x}$ on both sides,
$$
\begin{gathered}
\frac{1}{\cos ^{2} \mathrm{x}}+\frac{\sin ^{2} \mathrm{x}}{\cos ^{2} \mathrm{x}}=3 \frac{\sin \mathrm{x} \cdot \cos \mathrm{x}}{\cos \mathrm{x} \cdot \cos \mathrm{x}} \\
\sec ^{2} \mathrm{x}+\tan ^{2} \mathrm{x}=3 \tan \mathrm{x} \\
1+\tan ^{2} \mathrm{x}+\tan ^{2} \mathrm{x}=3 \tan \mathrm{x} \\
2 \tan ^{2} \mathrm{x}-3 \tan \mathrm{x}+1=0 \\
2 \tan ^{2} \mathrm{x}-2 \tan \mathrm{x}-\tan \mathrm{x}+1=0 \\
2 \tan \mathrm{x}(\tan \mathrm{x}-1)-1(\tan \mathrm{x}-1)=0 \\
(\tan \mathrm{x}-1)(2 \tan \mathrm{x}-1)=0 \\
\tan \mathrm{x}=1, \frac{1}{2} \\
\mathrm{We take,} \quad \tan \mathrm{x}=1 \\
\left(\because \tan \mathrm{x} \neq \frac{1}{2}\right) \\
\mathrm{x}=\mathrm{tan}(\pi / 4)
\end{gathered}
$$
$\mathrm{x}=\mathrm{n} \pi+\pi / 4, \mathrm{n} \in \mathrm{Z}$
|
["$2 \\mathrm{n} \\pi+\\frac{\\pi}{4}, \\mathrm{n} \\in \\mathrm{Z}$", "$2 \\mathrm{n} \\pi-\\frac{\\pi}{4}, \\mathrm{n} \\in \\mathrm{Z}$", "$\\mathrm{n} \\pi-\\frac{\\pi}{4}, \\mathrm{n} \\in \\mathrm{Z}$", "$\\mathrm{n} \\pi+\\frac{\\pi}{4}, \\mathrm{n} \\in \\mathrm{Z}$"]
|
[3]
| null |
Practise Problem
|
c0ee087c257ec660f569912e27a3e685
|
BITSAT_MATH
|
Trigonometric Ratios & Identities
|
Let <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>10</mn></math> vertical poles standing at equal distances on a straight line, subtend the same angle of elevation <math><mi>α</mi></math> at a point <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>O</mi></math> on this line and all the poles are on the same side of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>O</mi></math>. If the height of the longest pole is <math><mi>h</mi></math> and the distance of the foot of the smallest pole from <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>O</mi></math> is <math><mi>a</mi></math>; then the distance between two consecutive poles, is
|
singleCorrect
| null |
<br /><img alt="" src="https://cdn-question-pool.getmarks.app/nta_abhyas/jee_main/images/454564545.JPG" style="width: 200px; height: 132px;" /><br /><br />from diagram<br /><br /><math><mrow><mrow><mi mathvariant="normal">tan</mi></mrow><mo></mo><mrow><mi>α</mi></mrow></mrow><mo>=</mo><mfrac><mrow><mi>h</mi></mrow><mrow><mi>a</mi><mo>+</mo><mn>9</mn><mi>b</mi></mrow></mfrac></math><br /><br /><math><mi>a</mi><mo>+</mo><mn>9</mn><mi>b</mi><mo>=</mo><mfrac><mrow><mi>h</mi><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo></mo><mrow><mi>α</mi></mrow></mrow></mrow><mrow><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>α</mi></mrow></mrow></mrow></mfrac></math><br /><br /><math><mn>9</mn><mi>b</mi><mo>=</mo><mi> </mi><mfrac><mrow><mi>h</mi><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo></mo><mrow><mi>α</mi></mrow></mrow><mo>-</mo><mi>a</mi><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>α</mi></mrow></mrow></mrow><mrow><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>α</mi></mrow></mrow></mrow></mfrac></math><br /><br /><math><mi>b</mi><mo>=</mo><mfrac><mrow><mi>h</mi><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo></mo><mrow><mi>α</mi></mrow></mrow><mo>-</mo><mi>a</mi><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>α</mi></mrow></mrow></mrow><mrow><mn>9</mn><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo></mo><mrow><mi>α</mi></mrow></mrow></mrow></mfrac></math>
|
["<math><mfrac><mrow><mi>h</mi><mi>sin</mi><mi>\u03b1</mi><mo>+</mo><mi>a</mi><mtext> cos</mtext><mi>\u03b1</mi></mrow><mrow><mn>9</mn><mi> </mi><mi>cos</mi><mi>\u03b1</mi></mrow></mfrac></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mfrac><mrow><mi>h</mi><mi>cos</mi><mi>\u03b1</mi><mo>-</mo><mi>a</mi><mi mathvariant=\"normal\"> s</mi><mi mathvariant=\"normal\">i</mi><mi mathvariant=\"normal\">n</mi><mi mathvariant=\"normal\">\u03b1</mi></mrow><mrow><mn>9</mn><mi>sin</mi><mi>\u03b1</mi></mrow></mfrac></math>", "<math><mfrac><mrow><mi>h</mi><mi>sin</mi><mi>\u03b1</mi><mo>+</mo><mi>a</mi><mtext> cos</mtext><mi>\u03b1</mi></mrow><mrow><mn>9</mn><mi>sin</mi><mi>\u03b1</mi></mrow></mfrac></math>", "<math><mfrac><mrow><mi>h</mi><mi>cos</mi><mi>\u03b1</mi><mo>-</mo><mi>a</mi><mtext> sin</mtext><mi>\u03b1</mi></mrow><mrow><mn>9</mn><mi>cos</mi><mi>\u03b1</mi></mrow></mfrac></math>"]
|
[1]
| null |
Practise Problem
|
2ca7c521da781757c319e28d8c071b13
|
BITSAT_MATH
|
Trigonometric Ratios & Identities
|
$\tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]$ is equal to
|
singleCorrect
| 3 |
Hints : Let $\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)=\theta$, then $\cos 2 \theta=\frac{a}{b}$
$$
\begin{aligned}
& \tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right] \\
& =\tan \left(\frac{\pi}{4}+\theta\right)+\tan \left(\frac{\pi}{4}-\theta\right)=2\left(\frac{1+\tan ^2 \theta}{1-\tan ^2 \theta}\right)=\frac{2}{\cos 2 \theta}=\frac{2}{a / b}=\frac{2 b}{a}
\end{aligned}
$$
|
["$\\frac{2 a}{b}$", "$\\frac{2 b}{a}$", "$\\frac{a}{b}$", "$\\frac{b}{a}$"]
|
[1]
| null |
Practise Problem
|
4ff217d1fdd7e8b34bbc8d91a9aaa33e
|
BITSAT_MATH
|
Trigonometric Ratios & Identities
|
If $x=\tan 15^{\circ}, y=\operatorname{cosec} 75^{\circ}$ and $z=4 \sin 18^{\circ}$, then :
|
singleCorrect
| 2 |
$\because \quad x=\tan 15^{\circ}=\tan\left(45^{\circ}-30^{\circ}\right)$
$=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{3-1}$
$=\frac{3+1-2 \sqrt{3}}{2}=2-\sqrt{3}$
and $y=\operatorname{cosec} 75^{\circ}$
$=\frac{1}{\sin \left(45^{\circ}+30^{\circ}\right)}$
$=\frac{1}{\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}}$
$=\frac{1}{\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}}=\frac{2 \sqrt{2}}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{2 \sqrt{2}(\sqrt{3}-1)}{3-1}=\sqrt{6}-\sqrt{2}$
and $\quad z=4 \sin 18^{\circ}=4\left(\frac{\sqrt{5}-1}{4}\right)=\sqrt{5}-1$
It is clear from above that
$(2-\sqrt{3}) < (\sqrt{6}-\sqrt{2}) < (\sqrt{5}-1)$
$\Rightarrow \quad x < y < z$
|
["$x < y < z$", "$y < z < x$", "$z < x < y$", "$x < z < y$"]
|
[0]
| null |
Practise Problem
|
c15f8e7aca897abfc19f11850bafb18f
|
BITSAT_MATH
|
Trigonometric Ratios & Identities
|
The value of $\frac{\tan 330^{\circ} \sec 420^{\circ} \sin 300^{\circ}}{\tan 135^{\circ} \sin 210^{\circ} \sec 315^{\circ}}$ is equal to
|
singleCorrect
| 2 |
We have, $\frac{\tan 330^{\circ} \sec 420^{\circ} \sin 300^{\circ}}{\tan 135^{\circ} \sin 210^{\circ} \sec 315^{\circ}}$
$=\frac{\tan (360-30)^{\circ} \sec (360+60)^{\circ} \sin (360-60)^{\circ}}{\tan (180-45)^{\circ} \sin (180+30)^{\circ} \sec (360-45)^{\circ}}$
$=\frac{-\tan 30^{\circ} \times \sec 60^{\circ} \times\left(-\sin 60^{\circ}\right)}{\left(-\tan 45^{\circ}\right)\left(-\sin 30^{\circ}\right) \sec 45^{\circ}}$
${[\tan (2 \pi-\theta)=-\tan \theta, \sec (2 \pi \pm \theta)=\sec \theta}$
$\sin (2 \pi-\theta)=-\sin \theta, \sin (\pi+\theta)=-\sin \theta$
$\tan (\pi-\theta)=-\tan \theta]$
$=\frac{1 \times 2 \times \sqrt{3} \times 2}{\sqrt{3} \times 2 \times 1 \times \sqrt{2}}=\sqrt{2}$
|
["$\\frac{1}{\\sqrt{2}}$", "$\\sqrt{2}$", "$\\frac{1}{\\sqrt{3}}$", "$\\sqrt{3}$"]
|
[1]
| null |
Practise Problem
|
8b8aa6683d56e09cd31c7e92287ac333
|
BITSAT_MATH
|
Trigonometric Ratios & Identities
|
The value of $\tan 67 \frac{1^{\circ}}{2}+\cot 67 \frac{1^{\circ}}{2}$ is
|
singleCorrect
| 2 |
Now, $\tan 67 \frac{1 \circ}{2}+\cot 67 \frac{1}{2} \circ$
$$
\begin{aligned}
&=\tan \left(90^{\circ}-22 \frac{1}{2}^{\circ}\right)+\cot \left(90^{\circ}-22 \frac{1}{2}^{\circ}\right) \\
&=\tan 22 \frac{1}{2}^{\circ}+\cot 22 \frac{1}{2} \circ \\
&=\sqrt{2}-1+\sqrt{2}+1 \\
&=2 \sqrt{2}
\end{aligned}
$$
|
["$\\sqrt{2}$", "$3 \\sqrt{2}$", "$2 \\sqrt{2}$", "$2-\\sqrt{2}$"]
|
[2]
| null |
Practise Problem
|
ecc75c1ed8dbe4aae12e492dc5e2661e
|
BITSAT_MATH
|
Trigonometric Ratios & Identities
|
For $0 \leq P, Q \leq \frac{\pi}{2},$ if $\sin P+\cos Q=2$, then the
value of $\tan \left(\frac{\vec{P}+Q}{2}\right)$ is equal to
|
singleCorrect
| 2 |
Given, $0 \leq P, Q \leq \frac{\pi}{2}$
and $\sin P+\cos Q=2$
This equation hold only when, $\quad P=\frac{\pi}{2}$
and $\quad Q=0$
$\mathrm{LHS}=\sin P+\cos Q=\sin \frac{\pi}{2}+\cos 0$
$$
=1+1=2=\mathrm{RHS}
$$
$\therefore \quad \tan \left(\frac{P+Q}{2}\right)=\tan \left(\frac{\frac{\pi}{2}+0}{2}\right)=\tan \frac{\pi}{4}=1$
|
["1", "$\\frac{1}{\\sqrt{2}}$", "$\\frac{1}{2}$", "$\\frac{\\sqrt{3}}{2}$"]
|
[0]
| null |
Practise Problem
|
3f87e9e9e3a78b77dfaf20dc578b0ed0
|
BITSAT_MATH
|
Trigonometric Ratios & Identities
|
If <math><mi>p</mi><mo>=</mo><mrow><mrow><msup><mrow><mi mathvariant="normal">sin</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo></mo><mrow><mi>x</mi><mo>+</mo><mrow><mrow><msup><mrow><mi mathvariant="normal">cos</mi></mrow><mrow><mn>4</mn></mrow></msup></mrow><mo></mo><mrow><mi>x</mi></mrow></mrow></mrow></mrow></math>, then
|
singleCorrect
| null |
<p><math><mi>p</mi><mo>=</mo><mrow><mrow><msup><mrow><mi mathvariant="normal">sin</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>⁡</mo><mrow><mi>x</mi><mo>+</mo><mrow><mrow><msup><mrow><mi mathvariant="normal">cos</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>⁡</mo><mrow><mi>x</mi></mrow></mrow><mo>(</mo><mn>1</mn><mo>-</mo><mrow><mrow><msup><mrow><mi mathvariant="normal">sin</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>⁡</mo><mrow><mi>x</mi></mrow></mrow><mo>)</mo></mrow></mrow></math><br /><br /><math><mo>⇒</mo><mi>p</mi><mo>=</mo><mo>(</mo><mrow><mrow><msup><mrow><mi mathvariant="normal">sin</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>⁡</mo><mrow><mi>x</mi><mo>+</mo><mrow><mrow><msup><mrow><mi mathvariant="normal">cos</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>⁡</mo><mrow><mi>x</mi><mo>)</mo><mo>-</mo><mrow><mrow><msup><mrow><mi mathvariant="normal">sin</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>⁡</mo><mrow><mi>x</mi></mrow></mrow><mrow><mrow><msup><mrow><mi mathvariant="normal">cos</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>⁡</mo><mrow><mi>x</mi></mrow></mrow></mrow></mrow></mrow></mrow></math><br /><br /> <math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnalign="left"><mtr><mtd><mo>⇒</mo><mi>p</mi><mo>=</mo><mn>1</mn><mo>−</mo><mfrac><mn>1</mn><mn>4</mn></mfrac><msup><mrow><mo>(</mo><mn>2</mn><mi>sin</mi><mi>x</mi><mi>cos</mi><mi>x</mi><mo>)</mo></mrow><mn>2</mn></msup></mtd></mtr><mtr><mtd><mo>⇒</mo><mi>p</mi><mo>=</mo><mn>1</mn><mo>−</mo><mfrac><mn>1</mn><mn>4</mn></mfrac><msup><mi>sin</mi><mn>2</mn></msup><mrow><mo>(</mo><mn>2</mn><mi>x</mi><mo>)</mo></mrow></mtd></mtr></mtable></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnalign="left"><mtr><mtd><mi>Max</mi><mo>.</mo><mtext> </mtext><mi>value</mi><mtext> </mtext><mo>=</mo><mtext> </mtext><mn>1</mn></mtd></mtr><mtr><mtd><mi>Min</mi><mo>.</mo><mtext> </mtext><mi>value</mi><mtext> </mtext><mo>=</mo><mtext> </mtext><mfrac><mn>3</mn><mn>4</mn></mfrac></mtd></mtr></mtable></math></p>
|
["<math><mfrac><mrow><mn>3</mn></mrow><mrow><mn>4</mn></mrow></mfrac><mo>\u2264</mo><mi>p</mi><mo>\u2264</mo><mn>1</mn></math>", "<math><mfrac><mrow><mn>3</mn></mrow><mrow><mn>16</mn></mrow></mfrac><mo>\u2264</mo><mi>p</mi><mo>\u2264</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>4</mn></mrow></mfrac></math>", "<math> <mrow> <mfrac> <mn>1</mn> <mn>4</mn> </mfrac> <mo>≤</mo><mi>p</mi><mo>≤</mo><mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow></math>", "None of these"]
|
[0]
| null |
Practise Problem
|
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