MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 14

The set of values of

${x}$ for which

$\sin x\cos^{3}x>\cos x\sin^{3} x, 0\leq x\leq\pi$ is

0%
$(0,\pi)$
0%
$(0,\displaystyle \frac{\pi}{4})$
0%
$(\displaystyle \frac{\pi}{4},\pi)$
0%
$(0,\displaystyle \frac{\pi}{2})$

Explanation

since $\sin x > 0$
we can cancel it from both sides
$\cos ^{3}x > \cos x \sin^{2}x$
$\cos x (\cos ^{2}x - \sin^{2}x) > 0$
$\cos x \cos 2x > 0$
which holds only if $0 \leq x < \displaystyle \frac{\pi}{4}$

$\displaystyle {\frac{\sin^{3}\theta-\cos^{3}\theta}{\sin\theta-\cos\theta}-\frac{\cos\theta}{\sqrt{1+\cot^{2}\theta}}}-2\tan\theta\cot\theta=-1, \theta\in [0,2\pi]$ , then

0%
$\displaystyle \theta\in(0,\frac{\pi}{2})-\{\frac{\pi}{4}\}$
0%
$\displaystyle \theta\in(\frac{\pi}{2}, \pi)-\{\frac{3\pi}{4}\}$
0%
$\displaystyle \theta\in(\pi,\frac{3\pi}{2})-\{\frac{5\pi}{4}\}$
0%
$\displaystyle \theta\in(0, \pi)-\{\frac{\pi}{4},\frac{\pi}{2}\}$

Explanation

We know that

$\sin ^{ 3 }{ \theta } -\cos ^{ 3 }{ \theta } =(\sin { \theta } -\cos { \theta } )(\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +\sin { \theta } \cos { \theta } )=(\sin { \theta } -\cos { \theta } )(1+\sin { \theta } \cos { \theta } )$

$\Rightarrow \dfrac { \sin ^{ 3 }{ \theta } -\cos ^{ 3 }{ \theta } }{ \sin { \theta } -\cos { \theta } } =1+\sin { \theta } \cos { \theta }$

But for it to exist

$\sin \theta-\cos \theta \ne 0$ , which implies

$\theta \ne \frac{\pi}{4}$

The value of

$\dfrac { \cos { \theta } }{ \sqrt { 1+\cot ^{ 2 }{ \theta } } } =|\sin { \theta } |\cos { \theta }$

So the given equation becomes

$1+\sin \theta \cos \theta-|\sin \theta | \cos \theta-2\tan \theta \cot \theta =-1$

We know that

$\tan \theta \cot \theta=1$ , remember that for

$\theta = \dfrac{\pi}{2}$ ,

$\tan \theta$ does not exist.

So the given equation reduces to

$\sin \theta \cos \theta=|\sin \theta| \cos \theta$

$\Rightarrow |\sin \theta|= \sin \theta$

Which implies

$0<\theta <\pi$ and

$\theta \ne \frac{\pi}{4} , \frac{\pi}{2}$

So the correct option is

$D$

The number of the solutions of thc equation

$\cos(\pi\sqrt{\mathrm{x}-4})\cos(\pi\sqrt{\mathrm{x}})=1$ is

0%
$> 2$
0%
$2$
0%
$1$
0%
$0$

Explanation

the above is possible only and only when

$\cos(\pi \sqrt{x - 4}) = \cos(\pi \sqrt{x}) = 1$ or

$-1$
clearly

$x = 4$ is the only solution

$0<\mathrm{x}<2\pi;0<\mathrm{y}<2\pi$ and

$3^{\sin \mathrm{x}+\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{y}}=1$ and

$25^{\mathrm{s}\mathrm{i}{\mathrm{n}\mathrm{x}^{2}}+\cos^{2}\mathrm{y}}= 5$ then (x,y) is

0%
$(\displaystyle \frac{7\pi}{6},\frac{\pi}{3})$
0%
$(\displaystyle \frac{7\pi}{6},\frac{5\pi}{3})$
0%
$(\displaystyle \frac{11\pi}{6},\frac{\pi}{3})$
0%
$(\displaystyle \frac{11\pi}{6},\frac{5\pi}{3})$

Explanation

from first equation we get

$\sin x + \cos y = 0$
and from second equation we get

$\sin^{2}x + \cos^{2}y = \dfrac{1}{2}$
but from first equation ,

$\sin^{2}x = \cos^{2}y$
put this in second we get

$\sin^{2}x = \cos^{2}y = \dfrac{1}{4}$
so all of

$A,B,C,D$ are true

The number of values of

$\mathrm{x}$ in

$[0,2\pi]$ satisfying the equation

$|\cos x-\sin \mathrm{x}|\geq\sqrt{2}$ , is

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

Given inequality is,

$|\sqrt{2} \cos(x + \dfrac{\pi}{4})| \geq \sqrt{2}$

This is possible only when

$\cos (x + \dfrac{\pi}{4}) = \pm 1$

so there are

$2$ solutions

$\mathrm{n}$ be the number of solutions of the equation

$|\cot \mathrm{x}|= \cot x +\displaystyle \frac{1}{\sin \mathrm{x}}(0<\mathrm{x}<2\pi)$ , then n

$=$

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$\mid \cot x \mid=\cot x+\cfrac{1}{\sin x} \,\,\,\,\,\,\, 0<x<2\pi$

$\mid \cot x \mid$ is positive if

$\cot x>0$

Then,

$\cot x=\cot x+\cfrac{1}{\sin x}$

or,

$\cfrac{1}{\sin x}=0$

or,

$\sin x=\infty$ which is impossible.

$\mid \cot x \mid$ is -ve if

$\cot x<0$

$-\cot x=\cot x+\cfrac{1}{\sin x}$

or,

$-2\cot x=\cfrac{1}{\sin x}$

or,

$\cfrac{-2\cos x}{\sin x}=\cfrac{1}{\sin x}$

$\sin x \ne 0$ because

$x \ne 0$

or,

$-2\cos x=1$

or,

$\cos x=\cfrac{-1}{2}=\cos \cfrac{2\pi}{3}$

or,

$x=2n\pi \pm \cfrac{2\pi}{3}$

$n=1$

$x=2\pi-\cfrac{2\pi}{3}=\cfrac{4\pi}{3}$

which is in the range

$(0,2\pi)$

$\therefore x=\cfrac{2\pi}{3},\cfrac{4\pi}{3}$

$\therefore n=2$

$\mathrm{s}in(\pi \mathrm{c}\mathrm{o}s\theta)= \mathrm{c}os$

$(\pi\sin\theta)$ , then which of the following is correct

0%
$\cos \displaystyle \theta=\frac{3}{2\sqrt{2}}$
0%
$\cos (\theta-\dfrac{\pi}{2})=\dfrac{1}{2\sqrt{2}}$
0%
$\cos (\theta-\dfrac{\pi}{4})=\dfrac{1}{2\sqrt{2}}$
0%
$\cos \left(\dfrac{\pi\theta}{4}\right )=\dfrac{1}{2\sqrt{2}}$

Explanation

$\sin(\pi \cos\theta)= \cos (\pi\sin\theta)$

$\Rightarrow \sin(\pi \cos\theta)= \sin (\dfrac{\pi}{2}-\pi\sin\theta)$

$\Rightarrow (\pi \cos\theta)= (\dfrac{\pi}{2}-\pi\sin\theta)$ or

$\pi \cos\theta +\pi \sin \theta =\dfrac{\pi}{2}$

$\displaystyle \Rightarrow \cos\theta + \sin\theta = \frac{1}{2}$ or

$\cos\theta -\sin\theta = \dfrac{1}{2}$

Let us consider the first case,

$\displaystyle \Rightarrow \frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta = \frac{1}{2\sqrt{2}}$

$\displaystyle \Rightarrow cos\left(\frac{\pi}{4}-\theta \right)=\frac{1}{2\sqrt{2}}$

Similarly, we can show that,

$\cos \left(\theta +\dfrac{\pi}{4} \right) = \dfrac{1}{2\sqrt2}$

The total number of solutions of

$\cos x=\sqrt{1-\sin 2x}$ in

$\left [ 0,2\pi \right ]$ is equal to

0%
$2$
0%
$3$
0%
$5$
0%
None of these

Explanation

$\cos x=\sqrt{1-\sin 2x}=\left | \sin x -\cos x\right |$

(a)

$\sin x< \cos x\Rightarrow x\in \left [ 0,\frac{\pi }{4})\cup (\frac{5\pi }{4},2\pi \right ]$
Then the given equation is

$\cos x=\cos x-\sin x$

$\sin x=0\Rightarrow x=0,\pi ,2\pi \left [ from equ.(i) \right ]$

$x=\pi$ is not possible

(b)

$\sin x\geq \cos x\Rightarrow x\in \left [ \frac{\pi }{4},\frac{5\pi }{4} \right ]$

$\cos x=\sin x -\cos x$

$\tan x=2$

$x=\tan ^{-1}2$

Hence, there are three solutions.

$|\tan x + \sec x| = |\tan x| - |\sec x|, x \epsilon [0,2\pi]$ if and only if x belongs to the interval

0%
$[0,\pi]$
0%
$[0,\dfrac{\pi}{2})\cup (\dfrac{\pi}{2},\pi]$
0%
$[\pi,\dfrac{3\pi}{2})\cup (\dfrac{3\pi}{2},2\pi]$
0%
none of these

Explanation

$|\tan x + \sec x| = |\tan x| - |\sec x|$

Then $\tan x . \sec x \leq 0$ and $|\tan x| \geq |\sec x|$

$Now, |\tan x| \geq |\sec x|$

$\Rightarrow \tan ^2x \geq \sec ^2x$

$\Rightarrow \tan ^2x \geq 1+ \tan ^2x \Rightarrow 0 \geq 1$ This is not possible.

The difference between greatest and least solution of

$x$ is-

0%
$\dfrac {3\pi}{2}$
0%
$\dfrac {\pi}{2}$
0%
$\pi$
0%
$10\pi$

Explanation

Given equation is

$\sin\,x+3\,\sin\,2x+\sin\,3x=\cos\,x+3\,\cos\,2x+\cos\,3x$

$(\sin x+\sin\,3x)+3\sin\,2x=(\cos x+\cos\,3x)+3\cos\,2x$

$(2\sin\,2x\,\cos x+3\sin\,2x)-(2\cos\,2x\,\cos x+3\cos\,2x)=0$

$\sin\,2x(2\cos\,x+3)-\cos\,2x(2\cos\,x+3)=0$

$\Rightarrow (\sin 2x-\cos 2x)(2\cos x+3)=0$

$cos\,x=-\displaystyle\frac{3}{2}$
which is not possible

$\sin\,2x=\cos\,2x$

$\tan\,2x=1$

$\tan 2x=tan\displaystyle\frac{\pi}{4}, \tan\frac{5\pi}{4}$

$x=\displaystyle\frac{\pi}{8},\dfrac{5\pi}{8}$
Required difference

$\dfrac{5\pi}{8}-\dfrac{\pi}{8}=\dfrac{\pi}{2}$

Find the value of

$\cot^{-1}\left(\dfrac{\sqrt {1-\sin x}+\sqrt {1+\sin x}}{\sqrt {1-\sin x}-\sqrt {1+\sin x}}\right)$

0%
$\pi -\dfrac{x}{2}$
0%
$\pi$
0%
$\dfrac{x}{2}$
0%
$\dfrac{\pi}{2}$

The value of

$\dfrac{\tan \alpha}{1-\cot \alpha}+\dfrac{\cot \alpha}{1-\tan \alpha}$ is identically equal to

0%
$\sec \alpha.\csc \alpha$
0%
$\sin \alpha. \cos \alpha$
0%
$\sec \alpha. \csc \alpha +1$
0%
$\sin \alpha . \cos \alpha +1$

Explanation

$\dfrac{\tan\alpha}{1-\cot\alpha}+\dfrac{\cot\alpha}{1-\tan\alpha}$

$=\dfrac{\tan\alpha}{1-\dfrac{1}{\tan\alpha}}+\dfrac{\cot\alpha}{1-\tan\alpha}$

$=\dfrac{\tan^2\alpha}{\tan\alpha-1}-\dfrac{\cot\alpha}{\tan\alpha-1}$

$=\dfrac{\tan^2-\dfrac{1}{\tan\alpha}}{\tan\alpha-1}$

$=\dfrac{\tan^3\alpha-1}{(\tan\alpha)(\tan\alpha-1)}$

$=\dfrac{(\tan\alpha-1)(\tan^2\alpha+\tan\alpha+1)}{(\tan\alpha)(\tan\alpha-1)}$

$=\dfrac{\tan^2\alpha+\tan\alpha+1}{\tan\alpha}$

$={\tan\alpha+1+\cot\alpha}$

$=\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}+1$

$=\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos\alpha \sin\alpha}+1$

$=\dfrac{1}{\cos\alpha \sin\alpha}+1$

$=\sec\alpha\:\csc\alpha+1$

Hence answer is C

$\cos \theta - \sin \theta =\sqrt{2} \sin \theta$ , then

$\cos \theta + \sin \theta$ is

0%
$\sqrt{2}\cos \theta$
0%
$\sqrt{2} \sin \theta$
0%
$0$
0%
$1$

Explanation

Given,

$\cos \theta - \sin \theta =\sqrt{2} \sin \theta$

Squaring, we get

$\cos^2\theta+\sin^2\theta-2 \sin \theta \cos \theta=2 \sin^2 \theta$
or

$1-2 \sin \theta \cos \theta=2(1-\cos^2 \theta)$
or

$1+2 \sin \theta \cos \theta =2 \cos^2 \theta$
or

$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 2 \cos^2 \theta$
or

$(\sin \theta + \cos \theta)^2=2 \cos^2 \theta$

$\therefore \sin \theta + \cos \theta =\sqrt{2} \cos \theta$

The trigonometric equation is-

$\sin x+3 \sin 2x+\sin 3x=\cos x+3 \cos 2x+\cos 3x$
when

$x$ lies in first four quadrants. It means

$x\epsilon [0, 2\pi]$ , then-

How many solutions are there-

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$\sin x+3 \sin 2x+\sin 3x=\cos x+3 \cos 2x+\cos 3x$

$2\sin { 2x } \cos { x } +3\sin { 2x } =2\cos { 2x } \cos { x } +3\cos { 2x }$

$\Rightarrow (\sin { 2x-\cos { 2x } ) } (2\cos { x } +3)=0$

$\Rightarrow \tan 2x=1, \cos x=-\displaystyle \frac{3}{2}$ (not possible)

Since,

$0 \le x\le 2\pi$

$0 \le 2x \le 4\pi$

Hence, there are a number of solutions

$= 4$

$\sec$

$\beta=\alpha+\dfrac{1}{4a}$ ,then the value of

$\sec\beta+\tan\beta$ is

0%
$a$ or $\dfrac{1}{a}$
0%
$2a$ or $\dfrac{1}{2a}$
0%
$4a$ or $\dfrac{1}{4a}$
0%
$1$

Explanation

We know that

$\sec^2\beta-\tan^2\beta=1$

Therefore

$\tan\beta=\pm\sqrt{\sec^2\beta-1}$

$\tan\beta=\pm\sqrt{\alpha^2+(\dfrac{1}{4\alpha})\:^2+\dfrac{1}{2}}-1$

$\tan\beta=\pm\sqrt{\alpha^2+(\dfrac{1}{4\alpha})\:^2-\dfrac{1}{2}}$

$\tan\beta=\pm\sqrt{(\alpha-\dfrac{1}{4\alpha})\:^2}$

$\tan\beta=\pm(\alpha-\dfrac{1}{4\alpha})$ ...(i)

Hence $sec\alpha+\tan\alpha=2\alpha\:or\:\dfrac{1}{2\alpha}$

The answer is B

$|\tan\theta+\sec\theta|=|\tan\theta|+|\sec\theta|, 0\leq \theta \leq 2\pi$ is possible only if-

0%
$\theta \epsilon [0, \pi]-\left \{\dfrac {\pi}{2}\right \}$
0%
$\theta \epsilon [0, \pi]$
0%
$\theta \epsilon [0, \dfrac {\pi}{2})$
0%
$(0, \dfrac {\pi}{2}]$

Explanation

$\mid \tan\theta+\sec\theta \mid=\mid \tan\theta \mid +\sec\theta$

Case 1:-

$\theta \in [0, \cfrac{\pi}{2})$ then

$\tan\theta=+ve , \,\, \sec\theta=+ve$

$\tan\theta+\sec\theta=+ve$

$\therefore \mid \sec\theta+\tan\theta \mid =\mid \tan\theta \mid +\mid \sec\theta \mid$

Case 2:-

$\theta \in (\cfrac{\pi}{2}, \pi]$

$\tan\theta=-ve , \,\, \sec\theta=-ve$

$\mid \sec\theta+\tan\theta \mid=+ve$

$\mid \sec\theta \mid=+ve \,\,\, \mid \tan\theta \mid=+ve$

$\mid \sec\theta+\tan\theta \mid=\mid \sec\theta \mid +\mid \tan\theta \mid$

Case 3:-

$\theta \in [ \pi,\cfrac{3\pi}{2}]$

$\tan\theta=+ve , \,\, \sec\theta=-ve$

$\mid \sec\theta+\tan\theta \mid=\tan\theta-\sec\theta$

$\mid \tan\theta \mid +\mid \sec\theta \mid=\sec\theta+\tan\theta$

$\mid \sec\theta+\tan\theta \mid \ne \mid \sec\theta \mid +\mid \tan\theta \mid$

Case 4:-

$\theta \in (\cfrac{3\pi}{2}, 2\pi]$

$\tan\theta=-ve , \,\, \sec\theta=+ve$

$\mid \sec\theta+\tan\theta \mid=\sec\theta -\tan\theta$

$\mid \sec\theta \mid=\sec\theta \,\,\, \mid \tan\theta \mid=\tan\theta$

$\mid \sec\theta+\tan\theta \mid \ne \mid \sec\theta \mid +\mid \tan\theta \mid$

$\theta \ne \cfrac{\pi}{2}$ because in that case

$\tan\theta=\infty$

The sum of the solution of

$x$ is-

0%
$\frac {14\pi}{23}$
0%
$\frac {3\pi}{4}$
0%
$\frac {9\pi}{8}$
0%
$6\pi$

Explanation

Given equation is

$\sin\,x+3\,\sin\,2x+\sin\,3x=\cos\,x+3\,\cos\,2x+\cos\,3x$

$(\sin x+\sin\,3x)+3\sin\,2x=(\cos x+\cos\,3x)+3\cos\,2x$

$(2\sin\,2x\,\cos x+3\sin\,2x)-(2\cos\,2x\,\cos x+3\cos\,2x)=0$

$\sin\,2x(2\cos\,x+3)-\cos\,2x(2\cos\,x+3)=0$

$\Rightarrow (\sin 2x-\cos 2x)(2\cos x+3)=0$

$cos\,x=-\displaystyle\frac{3}{2}$
which is not possible

$\sin\,2x=\cos\,2x$

$\tan\,2x=1$

$\tan 2x=tan\displaystyle\frac{\pi}{4}, \tan\frac{5\pi}{4}$

$x=\displaystyle\frac{\pi}{8},\dfrac{5\pi}{8}$
Required sum

$=\dfrac{5\pi}{8}+\dfrac{\pi}{8}=\frac{3\pi}{4}$

Consider the system of equations

$\displaystyle \sin x \cos 2y= (a^{2}-1)^{2}+1,\ \cos x\sin 2y= a+1$ , then the number of values of

$\displaystyle y\in [0,2\pi]$ when the system has solution for permissible values of

$a$ are,

0%
$2$
0%
$3$
0%
$4$
0%
$5$

The value of

$\displaystyle \sin ^{2}1^{\circ}+\sin ^{2}2^{\circ}+\sin ^{2}2^{\circ}+...+\sin ^{2}89^{\circ}+\sin ^{2}90^{\circ}$

0%
1
0%
0
0%
45.5
0%
44

Explanation

$\displaystyle \sin ^{2}1^{\circ}+\sin ^{2}2^{\circ}+\sin ^{2}2^{\circ}+...+\sin ^{2}89^{\circ}+\sin ^{2}90^{\circ}$
=

$(\sin^2 1 + \sin^2 89) + (\sin^2 + \sin^288) + ....+ (\sin^2 44 + \sin^2 46) + (\sin^245 + \sin^ 90)$
=

$(\sin^2 1 + \cos^2 (90 - 1)) + (\sin^2 + \cos^2(90 - 2)) + ....+ (\sin^2 44 + \sin^2 (90 - 44)) + (\sin^245 + \sin^ 90)$
=

$(\sin^2 1 + \sin^2 1) + (\sin^2 + \sin^2 2) + ....+ (\sin^2 44 + \cos^2 44) + (\sin^245 + \sin^ 90)$
=

$1 + 1 ...44 times + (\dfrac{1}{\sqrt{2}})^2 +1$
=

$45 + \dfrac{1}{2}$
=

$45.5$

$\displaystyle \sin x+\mathrm{cosec}\: x=2,$ then

$\displaystyle \sin ^{n} x + \mathrm{cosec} ^{n} \: x$ is equal to

0%
$2$
0%
$\displaystyle 2^{n}$
0%
$\displaystyle 2^{n-1}$
0%
$\displaystyle 2^{n-2}$

Explanation

We have

$sin x + cosec x = 2$

$=> sin x + \dfrac {1}{sin x } = 2$

This means

$x = 90^0$ as maximum value of

$sin x = 1$ such that

$1 + 1 = 2$

Now, for any

$n$ , we have

$sin^n 90^0 + cosec ^n 90^0 = 1^2 + 1^2 = 2$

Consider the system of equations

$\displaystyle \sin x. \cos 2y= (a^{2}-1)^{2}+1,\ \cos x.\sin 2y= a+1$ , then the number of values of

$\displaystyle x\epsilon [0,2\pi]$ when the system has a solution for permissible values of a is/are,

0%
$1$
0%
$2$
0%
$3$
0%
$4$

The total number of solutions of

$\displaystyle \sin \left \{ x \right \}=\cos \left \{ x \right \}$ (where

$\displaystyle \left \{ . \right \}$ denotes the fractional part) in

$\displaystyle \left [ 0,2\pi \right ]$ is equal to

0%
$5$
0%
$6$
0%
$8$
0%
None of these

Explanation

$\sin\{x\} = \cos\{x\}$

$\Rightarrow \tan\{x\} = 1$

thus general solution is,

$\{x\} = x - [x] = n\pi + \dfrac{\pi}{4}$ , where n is any integer

hence solution in the given interval is,

$x = \dfrac{\pi}{4}, 1 + \dfrac{\pi}{4}, 2 + \dfrac{\pi}{4}, 3 + \dfrac{\pi}{4}, 4 + \dfrac{\pi}{4}, 5 + \dfrac{\pi}{4}$

Hence, option 'B' is correct.

In the given figure,

$\displaystyle \angle B =90^{\circ}$ and

$\displaystyle \angle ADB=x^{\circ}$ , then find

$\displaystyle \cos^{2} C^{\circ}+\sin^{2} C^{\circ}$ .

0%
1
0%
2
0%
0
0%
-1

Explanation

$\triangle CAB$ ,

$CA = 20$ and

$AB = 12$

Using Pythagoras theorem,

$CA^2 = AB^2 + BC^2$

$20^2 = 12^2 + BC^2$

$BC = 16$

$\cos^2 C + \sin^2 C = \left (\dfrac{BC}{AC}\right )^2 + \left (\dfrac{AB}{AC}\right )^2$

$\left (\dfrac{16}{20}\right )^2 + \left (\dfrac{12}{20}\right )^2$

$\dfrac{256 + 144}{400}$

$\dfrac{400}{400}$

$1$

The number of solutions of

$\displaystyle \sum_{r=1}^{5} \cos r x=5$ in the interval of

$\displaystyle \left [ 0,2\pi \right ]$ is

0%
$0$
0%
$2$
0%
$5$
0%
$10$

Explanation

$\cos x + \cos 2x + \cos 3x + \cos 4x + \cos 5x = 5$
we know, $\cos x, \cos 2x, \cos 3x, \cos 4x, \cos 5x \leq 1$
considering both,
$\cos x, \cos 2x, \cos 3x, \cos 4x, \cos 5x = 1$
hence general solutions in the given interval are,
$\cos x = 1 \Rightarrow x = 0, 2\pi$
$\cos 2x = 1 \Rightarrow 2x = 0, 2\pi, 4\pi \Rightarrow x = 0, \pi, 2\pi$
$\cos 3x = 1 \Rightarrow 3x = 0, 2\pi, 4\pi, 6\pi \Rightarrow x = 0, \dfrac{2\pi}{3}, 2\pi$
$\cos 4x = 1 \Rightarrow 4x = 0, 2\pi, 4\pi, 6\pi, 8\pi \Rightarrow x = 0, \dfrac{\pi}{2},\pi, \dfrac{3\pi}{2}, 2\pi$
$\cos 5x = 1 \Rightarrow 5x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi \Rightarrow x = 0, \dfrac{2\pi}{5}, \dfrac{4\pi}{5}, \dfrac{6\pi}{5}, \dfrac{8\pi}{5}, 2\pi$
thus common solutions are,
$x = 0, 2\pi$
Hence, option 'B' is correct.

$3x = \text{cosec } \theta$ and

$\dfrac {3}{x} = \cot \theta$ , then

$\left (x^{2} - \dfrac {1}{x^{2}}\right ) =$

0%
$\dfrac {1}{27}$
0%
$\dfrac {1}{81}$
0%
$\dfrac {1}{3}$
0%
$\dfrac {1}{9}$

Explanation

Since

$cosec^{2} \theta - \cot^{2}\theta = 1$

$\therefore (3x)^{2} - \left (\dfrac {3}{x}\right )^{2} = 1$ , substitute the given values

$\Rightarrow 9x^{2} - \dfrac {9}{x^{2}} = 1\Rightarrow 9 \left (x^{2} - \dfrac {1}{x^{2}}\right ) = 1$

$\Rightarrow x^{2} - \dfrac {1}{x^{2}} = \dfrac {1}{9}$ .

The value of

$\displaystyle \frac { \sin { \theta } \cos { \theta } .\sin { \left( { 90 }^{ o }-\theta \right) } }{ \cos { \left( { 90 }^{ o }-\theta \right) } } +\frac { \cos { \theta } .\sin { \theta } .\cos { \left( { 90 }^{ o }-\theta \right) } }{ \sin { \left( { 90 }^{ o }-\theta \right) } } +\frac { { \sin }^{ 2 }{ 27 }^{ o }+{ \sin }^{ 2 }{ 63 }^{ o } }{ { \cos }^{ 2 }{ 40 }^{ o }+{ \cos }^{ 2 }{ 50 }^{ o } }$ is :

0%
$1$
0%
$2$
0%
$3$
0%
$0$

Explanation

$\displaystyle \sin { \left( { 90 }^{ o }-\theta \right) } =\cos { \theta } ;\cos { \left( { 90 }^{ o }-\theta \right) } =\sin { \theta }$

$\displaystyle \sin { { 27 }^{ o } } =\sin { \left( { 90 }^{ o }-{ 63 }^{ o } \right) } ={ \cos63 }^{ o }$

$\displaystyle { \cos40 }^{ o }=\cos { \left( { 90 }^{ o }-{ 50 }^{ o } \right) } =\sin { { 50 }^{ o } }$

$\displaystyle \therefore \quad Given\quad exp.$

$\displaystyle =\frac { \sin { \theta } .\cos { \theta } .\cos { \theta } }{ \sin { \theta } } +\frac { \cos { \theta } .\sin { \theta } \sin { \theta } }{ \cos { \theta } } +\frac { \left( { \cos }^{ 2 }{ 63 }^{ o }+{ \sin }^{ 2 }{ 63 }^{ o } \right) }{ \left( { \sin }^{ 2 }{ 50 }^{ o }+{ \cos }^{ 2 }{ 50 }^{ o } \right) }$

$\displaystyle ={ \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta +\frac { 1 }{ 1 } =1+1=2$

$\displaystyle p=\sqrt{\frac{1-\sin x}{1+\sin x}},q=\frac{1-\sin x}{\cos x},r=\frac{\cos x}{1+\sin x}$
Which one of the following statement is correct ?

0%
$\displaystyle p=q\neq r$
0%
$\displaystyle q=r\neq p$
0%
$\displaystyle r=p\neq q$
0%
$\displaystyle p=q=r$

Explanation

Consider,

$\displaystyle p=\sqrt{\frac{1-\sin x}{1+\sin x}}$

$=\displaystyle \sqrt{\frac{1-\sin x}{1+\sin x}\times\frac{1-\sin x}{1+\sin x}}$

$=\displaystyle \sqrt{\frac{(1-\sin x)^2}{1-\sin^2 x}}$

$=\displaystyle \sqrt{\frac{(1-\sin x)^2}{\cos^2 x}}$

$q=\displaystyle \frac{1-\sin x}{\cos x}$

Again consider,

$r=\displaystyle \frac{\cos x}{1+\sin x}=\displaystyle \frac{\cos x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}$

$=\displaystyle \frac{\cos x(1-\sin x)}{1-\sin^2 x}$

$=\displaystyle \frac{\cos x(1-sinx)}{]cos^2 x}$

$=\displaystyle \frac{1-\sin x}{\cos x}$

$\text{Clearly, p=q=r}$

Option D is correct.

The value of

$\alpha \varepsilon (- \pi, 0)$ satisfying

$sin \alpha + \int_{\alpha}^{2 \alpha} . cos 2x dx = 0$ is

0%
$0$
0%
$- \dfrac{\pi}{3}$
0%
$-\pi$
0%
All of these

Explanation

Consider,

$sin\alpha + \int^{2\alpha}_\alpha cos2x\ dx = sin\alpha +\left [\dfrac{sin2\alpha}{2} \right]_{\alpha}^{2\alpha}$

$\Rightarrow sin \alpha+\dfrac{1}{2}[sin 4\alpha-sin 2\alpha]\\= sin 2\alpha(2cos 2\alpha-1)+sin\alpha\\= (4cos^3\alpha-3cos\alpha +1)sin\alpha$

And it is given that

$(\alpha \ne 0)$

$cos \alpha=1/2$ is satisfying the above equation

$\alpha =\dfrac{-\pi}{3}$

$\cos (2001) \pi + \cot (2001)\dfrac {\pi}{2} + \sec (2001) \dfrac {\pi}{3} + \tan (2001) \dfrac {\pi}{4} + cosec (2001) \dfrac {\pi}{6}$ equal to

0%
$0$
0%
$1$
0%
$-2$
0%
Not defined

$\dfrac12\sin{(2x)}(1+\cot ^{ 2 }{ (x) } )$ is equal to

0%
$\tan(x)$
0%
$\sin (x)$
0%
$\cos(x)$
0%
$\cot(x)$
0%
$\sec(x)$

Explanation

The given identity can be solved as:

$\left( \dfrac { 1 }{ 2 } \right) (\sin { 2x) } (1+\cot ^{ 2 }{ x } )$

$=\left( \dfrac { 1 }{ 2 } \right) (2\sin { x } \cos { x } )(cosec ^{ 2 }{ x } )$ .... (because

$\sin { 2x } =2\sin { x } \cos { x }$ and

$1+\cot ^{ 2 }{ x } =cosec ^{ 2 }{ x }$ )

$=(\sin { x } \cos { x } )\left( \dfrac { 1 }{ \sin ^{ 2 }{ x } } \right)$ ...... (because

$cosec ^{ 2 }{ x } =\dfrac { 1 }{ \sin ^{ 2 }{ x } }$ )

$=\dfrac { \cos { x } }{ \sin { x } }$

$=\cot { x }$

Hence,

$\left( \dfrac { 1 }{ 2 } \right) (\sin { 2x) } (1+\cot ^{ 2 }{ x } )=\cot { x }$ .

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 14 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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