Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The domain of $$f(x) =\displaystyle \frac{\sin ^{-1}x}{x}$$ is

A
$$\left [ -1,1 \right ]$$

B
$$\left \{ 0 \right \}$$

C
$$\left [-1,0 \right )$$

D
None of these

Solution

$$f(x)=\displaystyle\frac{\sin ^{-1}x}{x}=\frac{g(x)}{t(x)},{t(x)}\neq 0$$

$${t(x)}\neq 0$$

$$\Rightarrow x\neq 0$$

Domain of $$t(x)$$ is $$R-\{0\}$$

Again domain of $$g(x)$$ is $$\left [ -1,1 \right ]$$

Now using concept domain of $$f(x) =\displaystyle\frac{g(x)}{t(x)}$$

domain of $$f(x) =$$ Domain of $$g(x)$$ - set of those points for which $$t(x)=0$$ if $$g(x), \ t(x)$$ is defined $$\forall x\epsilon R$$

$$\therefore$$ Domain of $$f(x) =\left [ -1,1 \right ]-\left \{ 0 \right \}$$
Question 2
1 / -0

Number of triplets $$\left ( x, y, z \right )$$ satisfying $$\sin ^{-1}x+\sin ^{-1}y+\cos ^{-1}z=2\pi$$ is

A
$$1$$

B
$$0$$

C
$$2$$

D
$$\infty$$

Solution

Let $$f(x, y, z)=\sin ^{-1}x+\sin ^{-1}y+\cos^{-1}z$$
It will attain the value $$2\pi$$ only if $$\sin ^{-1}x=\sin ^{-1}y=\displaystyle\frac{\pi }{2}$$ and $$\cos^{-1}z=\pi$$.
This is possible only if $$x = y = 1$$ and $$z=-1$$.
Hence there is only one solution $$f(1,1,-1)=2\pi$$
Question 3
1 / -0

The equation $$2 cos^{-1} x = sin^{-1} (2 x \sqrt{1 - x^2}) $$ is valid for all values of x satisfying

A
$$-1 \leq x \leq 1$$

B
$$0 \leq x \leq 1$$

C
$$0 \leq x \leq 1/\sqrt{2}$$

D
$$1/\sqrt{2} \leq x \leq 1$$

Solution

If we denote $$cos^{-1}x$$ by $$y$$, then
since $$0 \leq cos^{-1} x \leq \pi \Rightarrow 0 \leq 2 y \leq 2 \pi$$ (1)
Also since $$-\pi/2 \leq sin^{-1} (2 x \sqrt{1 - x^2}) \leq \pi/2 \Rightarrow - \pi/2 \leq sin^{-1} sin (2y) \leq \pi/2$$
$$\Rightarrow - \pi/2 \leq 2y \leq \pi/2$$ (2)
From (1) and (2) we find $$0 \leq 2y \leq \pi /2$$
$$\Rightarrow 0 \leq y \leq \pi/4 \Rightarrow 0 \leq cos^{-1} x \leq \pi/4$$
which holds if $$1/\sqrt{2} \leq x \leq 1$$
Question 4
1 / -0

If $$\displaystyle \tan^{-1}{\left(\frac{a}{x}\right)}+\tan^{-1}{\left(\frac{b}{x}\right)}=\frac{\pi}{2}$$, then $$x$$ is equal to :

A
$$ab$$

B
$$\dfrac{a}{b}$$

C
$$\dfrac{b}{a}$$

D
$$\sqrt{ab}$$

Solution

$$\tan^{-1}\left(\dfrac{a}{x}\right)+\tan^{-1}\left( \dfrac{b}{x}\right)=\dfrac{\pi}{2}$$
$$\Rightarrow \tan^{-1}\left(\dfrac{a}{x}\right)=\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{b}{x} \right)=\cot^{-1}\left(\dfrac{b}{x} \right)$$
$${\left(\because \tan^{-1}\theta+\cot^{-1}\theta =\dfrac{\pi}{2}\right)} $$
$$\Rightarrow \tan^{-1}\left(\dfrac{a}{x}\right)=\cot^{-1}\left(\dfrac{b}{x} \right)=\tan^{-1}\left(\dfrac{x}{b}\right)$$
$$\Rightarrow \dfrac{a}{x}=\dfrac{x}{b}\Rightarrow x^2=ab$$
$$\Rightarrow x=\pm \sqrt {ab}$$
Question 5
1 / -0

If $$ 2 sin h^{-1}\left ( \dfrac{a}{\sqrt{1-a^{2}}} \right ) = log\left ( \dfrac{1+X}{1-X} \right )$$ then X=

A
a

B
$$\dfrac{1}{a}$$

C
$$\sqrt{1-a^{2}}$$

D
$$\dfrac{1}{\sqrt{1-a^{2}}} $$

Solution

As we know that
$$\sinh^{-1} x=\log(x+\sqrt{x^2+1})$$
$$2\sinh^{-1} \bigg(\dfrac{a}{\sqrt{1-a^2}}\bigg)=2\log\bigg(\dfrac{a}{\sqrt{1-a^2}}+\sqrt{\dfrac{a^2}{1-a^2}+1}\bigg)$$
$$=2\log \bigg(\dfrac{a+\sqrt{a^2+1-a^2}}{\sqrt{1-a^2}}\bigg)$$
$$=2\log \bigg(\dfrac{a+1}{\sqrt{1-a^2}}\bigg)$$
$$=\log\bigg(\dfrac{(1+a)^2}{(1-a)(1+a)}\bigg)$$
$$=\log \bigg(\dfrac{1+a}{1-a}\bigg)$$
But given $$2\sinh^{-1} \bigg(\dfrac{a}{\sqrt{1-a^2}}\bigg)=\log \bigg(\dfrac{1+X}{1-X}\bigg)$$
So $$X=a$$
Question 6
1 / -0

The sum of $$\cot ^{ -1 }{ 2 } +\cot ^{ -1 }{ 8 } +\cot ^{ -1 }{ 18 } ....\infty =\cfrac { \pi }{ \lambda } $$, then $$\lambda $$ is

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

$$\cot^{-1}(2)+\cot^{-1}(8)+\cot^{-1}(18).......\infty =\dfrac{\pi}{\lambda}$$
from L.H.S.
$$n^{th}$$ term is $$\Rightarrow \cot^{-1}(2n^2)$$
$$=\tan^{-1}\left(\dfrac{1}{2n^2}\right)$$
$$=\tan^{-1}\left(\dfrac{2}{4n^2}\right)$$
$$=\tan^{-1}\left(\dfrac{2}{4n^2-1+1}\right)$$
$$=\tan^{-1}\left(\dfrac{2}{(2n-1)(2n+1)+1}\right)$$
$$=\tan^{-1}\left(\dfrac{(2n+1)-(2n-1)}{(2n-1)(2n+1)+1}\right)$$
Now,
$$\tan^{-1}A-\tan^{-1}B=\tan^{-1}\left(\dfrac{A-B}{1+AB}\right)$$
$$A=2n+1$$; $$B=2n-1$$
$$=\tan^{-1}(2n+1)-\tan^{-1}(2n-1)$$
$$n=1, 2, 3$$...
$$\Rightarrow \tan^{-1}(3)-\tan^{-1}(1)+\tan^{-1}(5)-\tan^{-1}(3)+.....\tan^{-1}\infty$$
$$\Rightarrow \tan^{-1}(\infty)-\tan^{-1}(1)$$
$$\Rightarrow \dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$$
Comparing RHS to
$$\lambda =4$$.
Question 7
1 / -0

The domain of the function $$f(x)=\sqrt{cos^{-1}\begin{pmatrix}\dfrac{1-|x|}{2}\end{pmatrix}}$$ is

A
$$(-3,\,3)$$

B
$$[-3,\,3]$$

C
$$(-\infty,\,-3)\cup(-3,\,\infty)$$

D
$$(-\infty,\,-3)\cup(3,\,\infty)$$

Solution

$$f(x)=\sqrt{cos^{-1}\begin{pmatrix}\dfrac{1-|x|}{2}\end{pmatrix}}$$

for $$f(x)$$ to be defined $$\cos^{-1}{\left(\dfrac{1-|x|}{2}\right)}\ge0$$

$$\Rightarrow \dfrac{-\pi}{2}\le \cos^{-1}{\left(\dfrac{1-|x|}{2}\right)}\le \dfrac{\pi}{2}$$

$$\Rightarrow-1 \le \dfrac{1-|x|}{2} \le 1$$

$$\Rightarrow\;-2\le 1-|x| \le 2$$
$$\Rightarrow\;-2-1\le -|x| \le 2-1$$
$$\Rightarrow\;-3 \le -|x| \le 1$$.
$$\Rightarrow\;-1\le|x|\le3$$

$$\Rightarrow |x|\ge-1$$ but $$|x|$$ is always positive

$$\Rightarrow |x|\le3$$
$$\Rightarrow -3\le x\le3$$
$$\Rightarrow\;x\,\in[-3,\,3]$$
Question 8
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The trigonometric equation $$\sin^{-1}x=2\, \sin^{-1}2a$$ has a real solution, if

A
$$|a| > \dfrac{1}{\sqrt{2}}$$

B
$$\dfrac{1}{2\sqrt{2}} < |a| < \dfrac{1}{\sqrt{2}}$$

C
$$|a| > \dfrac{1}{2\sqrt{2}}$$

D
$$|a| \le \dfrac{1}{2\sqrt{2}}$$

Solution

We have that
$$\displaystyle \frac{-\pi}{2} \le \sin^{-1}x\le \frac{\pi}{2}$$

$$\Rightarrow \dfrac{-\pi}{2}\le 2\, \sin^{-1}2a\le \dfrac{\pi}{2}$$

$$\Rightarrow \dfrac{-\pi}{4}\le \sin^{-1}2a\le \dfrac{\pi}{4}$$

$$\Rightarrow \sin\left(\dfrac{-\pi}{4}\right)\le 2a \le \sin\left(\dfrac{\pi}{4}\right)$$

$$\Rightarrow \dfrac{-1}{\sqrt{2}}\le 2a \le \dfrac{1}{\sqrt{2}}$$

$$\Rightarrow \dfrac{-1}{2\sqrt{2}}\le a \le \dfrac{1}{2\sqrt{2}}\Rightarrow |a|\le \dfrac{1}{2\sqrt{2}}$$
Question 9
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$$\tan^{-1}(x + \sqrt{1 + x^{2}})$$ =

A
$$\dfrac{\pi}{4} - \dfrac{1}{2} \tan^{-1} x$$

B
$$\dfrac{1}{2} \tan^{-1} x$$

C
$$\dfrac{\pi}{2} - \dfrac{1}{2} \tan^{-1} x$$

D
$$\dfrac{\pi}{4} + \dfrac{1}{2} \tan^{-1} x$$

Solution

Let, $$x= \tan \theta$$
$$\therefore 1+x^2=\space 1+\tan^2 \theta = \sec^2 \theta$$ .....(i)
Given, $$ \tan^{-1} (x+ \sqrt{1+x^2})$$
$$ = \tan^{-1} (\tan \theta +\sqrt {\sec ^2 \theta})$$ ....{From (i)}
$$ = \tan ^{-1} (\tan \theta + \sec \theta)$$
$$ = \tan ^{-1} \left (\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos\theta}\right)$$
$$ =\tan ^{-1} \left (\dfrac{1+\sin \theta }{\cos \theta}\right)$$
$$= \tan ^{-1} \left (\dfrac{2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} +\sin^2\dfrac {\theta}{2}+\cos^2\dfrac{\theta}{2}}{\cos^2\dfrac{\theta}{2}-\sin ^2 \dfrac{\theta}{2}}\right)$$
$$ = \tan^-1 \left ( \dfrac{\left (\cos\dfrac{\theta}{2}+ \sin \dfrac{\theta}{2}\right)^2}{\left (\cos \dfrac{\theta}{2}-\sin\dfrac{\theta}{2}\right)(\cos \dfrac{\theta}{2}+\sin\dfrac{\theta}{2})}\right)$$
$$=\tan^{-1}\left (\dfrac {\cos \dfrac {\theta}{2}+\sin \dfrac {\theta}{2} }{\cos \dfrac {\theta}{2}-\sin \dfrac {\theta}{2}}\right)$$
Dividing by $$ \cos \dfrac{\theta}{2} $$ on both numerator and denominator, we get
$$=\tan ^{-1} \dfrac{1+\tan \dfrac{\theta}{2}}{1- \tan \dfrac{\theta}{2}} $$
$$= \tan ^{-1} \tan \left (\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)$$
$$=\dfrac {\pi}{4}+\dfrac {\theta}{2}$$
$$= \dfrac{\pi}{4}+\dfrac{1}{2}\tan^{-1}x$$ ....Resubstituting value of $$\theta$$
Question 10
1 / -0

The value of $$\cot^{-1} \left[ \dfrac{\sqrt{1 - \sin x} +\sqrt{1 + \sin x}}{\sqrt{(1 - \sin x)} - \sqrt{(1 + \sin x)}} \right]$$ is

A
$$\pi - x$$

B
$$2 \pi - x$$

C
$$x/2$$

D
$$\pi - \dfrac{1}{2} x$$

Solution

We have $$\cot^{-1} \, \left[ \dfrac{\sqrt{1 - \sin x} +\sqrt{1 + \sin x}}{\sqrt{(1 - \sin x)} - \sqrt{(1 + \sin x)}} \right]$$

$$= \cot^{-1} \left[ \dfrac{ \sqrt {\left( \cos \dfrac{x}{2} - \sin \dfrac{x}{2} \right)^2} + \sqrt {\left( \cos \dfrac{x}{2} + \sin \dfrac{x}{2} \right)^2} }{ \sqrt {\left( \cos \dfrac{x}{2} - \sin \dfrac{x}{2} \right)^2} - \sqrt {\left( \cos \dfrac{x}{2} + \sin \dfrac{x}{2} \right)^2 }} \right]$$

$$= \cot^{-1} \left[ \dfrac{ \left( \cos \dfrac{1}{2} x - \sin \dfrac{1}{2} x \right) + \left( \cos \dfrac{1}{2} x + \sin \dfrac{1}{2} x \right) }{ \left( \cos \dfrac{1}{2} x - \sin \dfrac{1}{2} x \right) - \left( \cos \dfrac{1}{2} x + \sin \dfrac{1}{2} x \right) } \right]$$

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

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 33

Result

Inverse Trigonometric Functions Test - 33

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SHARING IS CARING

Question 1

$$\left [ -1,1 \right ]$$

$$\left \{ 0 \right \}$$

$$\left [-1,0 \right )$$

None of these

Solution

Question 2

$$1$$

$$0$$

$$2$$

$$\infty$$

Solution

Question 3

$$-1 \leq x \leq 1$$

$$0 \leq x \leq 1$$

$$0 \leq x \leq 1/\sqrt{2}$$

$$1/\sqrt{2} \leq x \leq 1$$

Solution

Question 4

$$ab$$

$$\dfrac{a}{b}$$

$$\dfrac{b}{a}$$

$$\sqrt{ab}$$

Solution

Question 5

a

$$\dfrac{1}{a}$$

$$\sqrt{1-a^{2}}$$

$$\dfrac{1}{\sqrt{1-a^{2}}} $$

Solution

Question 6

$$2$$

$$4$$

$$6$$

$$8$$

Solution

Question 7

$$(-3,\,3)$$

$$[-3,\,3]$$

$$(-\infty,\,-3)\cup(-3,\,\infty)$$

$$(-\infty,\,-3)\cup(3,\,\infty)$$

Solution

Question 8

$$|a| > \dfrac{1}{\sqrt{2}}$$

$$\dfrac{1}{2\sqrt{2}} < |a| < \dfrac{1}{\sqrt{2}}$$

$$|a| > \dfrac{1}{2\sqrt{2}}$$

$$|a| \le \dfrac{1}{2\sqrt{2}}$$

Solution

Question 9

$$\dfrac{\pi}{4} - \dfrac{1}{2} \tan^{-1} x$$

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

$$\dfrac{\pi}{2} - \dfrac{1}{2} \tan^{-1} x$$

$$\dfrac{\pi}{4} + \dfrac{1}{2} \tan^{-1} x$$

Solution

Question 10

$$\pi - x$$

$$2 \pi - x$$

$$x/2$$

$$\pi - \dfrac{1}{2} x$$

Solution

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