Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$ \lim_{\left | x \right |\rightarrow \infty} $$ $$ \cos \left ( \tan^{-1}\left ( \sin \left ( \tan^{-1} x \right ) \right ) \right ) $$ is equal to :

A
$$ -1 $$

B
$$ \sqrt{2} $$

C
$$ \displaystyle -\frac{1}{\sqrt{2}} $$

D
$$ \displaystyle \frac{1}{\sqrt{2}} $$

Solution

Given, $$lim_{|x|\rightarrow \infty}\cos(\tan^{-1}(\sin(\tan^{-1}(x))))$$

$$=\cos \left(\tan^{-1} \left (\sin \left (\cfrac{\pi}{2} \right) \right) \right)$$ $$[\tan \dfrac{\pi}{2}=\infty]$$

$$=\cos(\tan^{-1}(1))$$

$$=\cos \left(\dfrac{\pi}{4} \right)$$

$$=\dfrac{1}{\sqrt{2}}$$
Question 2
1 / -0

$$ \tan^{-1}a +\tan^{-1}b $$, where $$ a> 0, b> 0, ab> 1 $$ is equal to :

A
$$ \tan^{-1}\left (\displaystyle \frac{a\:+\:b}{1\:-\:ab} \right ) $$

B
$$ \tan^{-1}\left (\displaystyle \frac{a\:+\:b}{1\:-\:ab} \right )-\pi $$

C
$$ \pi + \tan^{-1}\left (\displaystyle \frac{a\:+\:b}{1\:-\:ab} \right ) $$

D
none of these

Solution

The value of $$\tan^{-1}(a)+\tan^{-1}(b)$$

$$\tan^{-1}(a)+\tan^{-1}(b)$$$$=\tan^{-1}(\dfrac{a+b}{1-ab})$$

Now $$ab>1$$ and $$a>0,$$ $$b>0$$

Therefore, it is in first quadrant.

Hence, $$\tan^{-1}(a)+\tan^{-1}(b)=$$$$\tan^{-1}\left(\dfrac{a+b}{1-ab}\right)$$
Question 3
1 / -0

Let $$ f(x)=cosec^{-1}\:\left [ 1+\sin ^{2} x\right ] $$, where $$ \left [ . \right ] $$ denotes the greatest integer function. Then $$ f(x) $$ equals;

A
$$ \displaystyle \left \{ \frac{\pi }{2} \right \} $$

B
$$ \displaystyle \left \{ \frac{\pi }{2},cosec^{-1}\:2 \right \} $$

C
$$ \left \{ cosec^{-1}\:2 \right\} $$

D
none of these

Solution

$$-1\le \sin x\le 1\Rightarrow 0\le \sin^{ 2 } x\le 1$$

For $$\sin^{ 2 }x=0\quad f\left( x \right) =cosec^{ -1 }\left\{ 1+0 \right\} =cosec^{ -1 }1=\cfrac { \pi }{ 2 } $$

And for $$\sin^{ 2 }x=1\quad f\left( x \right) =cosec^{ -1 }\left\{ 1+1 \right\} =cosec^{ -1 }2$$
Question 4
1 / -0

If $$\displaystyle \tan \left ( \cos^{-1} x \right ) = \sin \left ( \cot^{-1} \frac{1}{2} \right )$$, then find the value of $$\displaystyle x$$

A
$$x=-\frac{\sqrt{5}}{3}$$

B
$$x=+\frac{\sqrt{5}}{3}$$

C
$$x=-\frac{2}{3}$$

D
$$x=+\frac{2}{3}$$

Solution

Given, $$\displaystyle \tan { \left( \cos ^{ -1 }{ x } \right) } =\sin { \left( \cot ^{ -1 }{ \frac { 1 }{ 2 } } \right) } $$
$$\displaystyle \Rightarrow \tan { \left( \tan ^{ -1 }{ \left( \frac { \sqrt { 1-{ x }^{ 2 } } }{ x } \right) } \right) } =\sin { \left( \sin ^{ -1 }{ \left( \frac { 2 }{ \sqrt { 5 } } \right) } \right) } $$
$$\displaystyle \Rightarrow \frac { \sqrt { 1-{ x }^{ 2 } } }{ x } =\frac { 2 }{ \sqrt { 5 } } \Rightarrow x=\frac { \sqrt { 5 } }{ 3 } $$
Question 5
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If $$\tan^{-1}\left ( \dfrac{x}{\sqrt{a^p-x^q}} \right )= \sin^{-1}\left ( \dfrac{x}{a} \right ),a> 0$$. Find the value of p and q.

A
$$p=1, q=1$$

B
$$p=1, q=2$$

C
$$p=2, q=1$$

D
$$p=2, q=2$$

Solution

Let $$\displaystyle \sin^{-1}\frac{x}{a} = \theta\Rightarrow \frac{x}{a} = \sin \theta$$

$$\displaystyle \Rightarrow \tan\theta = \frac{\sin \theta}{\cos\theta} = \frac{x/a}{\sqrt{1-x^2/a^2}}=\frac{x}{\sqrt{a^2-x^2}}$$

$$\therefore\displaystyle \theta = \tan^{-1} \frac{x}{\sqrt{a^2-x^2}}= \sin^{-1}\frac{x}{a}$$

$$\Rightarrow p=2, \ q=2$$
Question 6
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If $$\displaystyle \tan^{-1} \frac{1}{\sqrt{x^{a} - b}} = \frac{\pi}{c} - \sec^{-1} x$$, $$\displaystyle \left | x \right | > 1$$.
Find the value of a,b and c.

A
$$a=1, b=1, c=1$$

B
$$a=1, b=2, c=2$$

C
$$a=2, b=2, c=1$$

D
$$a=2, b=1, c=2$$

Solution

We know that, $$\displaystyle \tan^{-1} \frac{1}{\sqrt{x^{2} - 1}} = \frac{\pi}{2} - \sec^{-1} x$$
Comparing this with the given $$\displaystyle \tan^{-1} \frac{1}{\sqrt{x^{a} - b}} = \frac{\pi}{c} - \sec^{-1} x$$, $$\displaystyle \left | x \right | > 1$$.

We get the unknown parameter $$a=2,b=1,c=2$$
Question 7
1 / -0

The value of $$\displaystyle \sin ^{-1}\left ( \sin 2 \right )$$ is?

A
$$2+n\pi$$

B
$$2-\pi$$

C
$$-2+\pi$$

D
$$2-2n\pi$$

Solution

To find value of $$\sin^{-1}(\sin(2))$$
$$\pi \space rads=180^{0}$$
$$1 \space rad=\dfrac{180^{0}}{\pi}$$
$$1 \space rad=57.29^{0}$$
Hence
$$2\space rads=114.59^{0}$$
Hence it is in the second quadrant.
Therefore
$$\sin^{-1}(\sin(2))$$
$$=\sin^{-1}(\sin(\pi-2))$$
$$=\pi-2$$
Question 8
1 / -0

Find the minimum value the function $$\displaystyle f\left ( x \right )=\frac{\pi ^{2}}{16\cot^{-1}\left ( -x \right )}-\cot^{-1}x$$

A
$$-\cfrac{\pi }{4}$$

B
$$-\cfrac{\pi }{2}$$

C
$$0$$

D
$$\cfrac{\pi }{2}$$

Solution

$$\displaystyle f\left ( x \right )=\frac{\pi ^{2}}{16\cot^{-1}\left (

-x \right )}-\left ( \pi -\cot^{-1}\left ( -x \right ) \right )$$

$$\displaystyle =\cot^{-1}\left ( -x \right )+\frac{\pi ^{2}}{16\cot^{-1}\left ( -x \right )}-\pi $$

$$\displaystyle =\left ( \sqrt{\cot^{-1}\left ( -x \right )}+\frac{\pi

}{4\sqrt{\cot^{-1}\left ( -x \right )}} \right )^2+\frac{\pi }{2}-\pi $$,

Since $$\left [ \cot^{-1} \theta > 0\right ]$$$$\displaystyle \geq -\frac{\pi }{2}$$

$$\therefore $$   $$\displaystyle f_{min}=-\frac{\pi }{2}$$
Question 9
1 / -0

For $$\displaystyle \tan^{-1}\left ( \frac{1-x}{1+x} \right )$$, $$0\leq x\leq 1$$.
What is the sum of the smallest and the largest values of function.

A
$$ \dfrac{\pi }{4} $$

B
$$ \dfrac{\pi }{2} $$

C
$$ \dfrac{3\pi }{4} $$

D
$$ \dfrac{3\pi }{2} $$

Solution

Let $$\displaystyle f\left ( x \right )=\tan^{-1}\left ( \frac{1-x}{1+x} \right )$$, $$0\leq x\leq 1$$

Now $$\displaystyle \frac{1-x}{1+x}=\frac{2}{1+x}-1$$

Given $$0\leq x\leq 1$$

$$\Rightarrow $$   $$\displaystyle \frac{2}{1+x}-1\in \left [ 0, 1 \right ]$$

$$\Rightarrow

$$   $$\displaystyle \tan^{-1}\left ( \frac{1-x}{1+x} \right )\in

\left [ \tan^{-1}0, \tan^{-1} 1\right ]$$   or   $$\displaystyle \left [

0, \frac{\pi }{4} \right ]$$
Question 10
1 / -0

Find the number of real solutions :
$$ \tan^{-1}\sqrt{x\left ( x+1 \right )}+\sin^{-1}\sqrt{x^2+x+1}= \dfrac{\pi }{2} $$

A
$$1$$

B
$$3$$

C
$$2$$

D
$$4$$

Solution

Our requirement is $$x^{ 2 }+x\ge 0$$
Second requirement $$\sqrt { { x }^{ 2 }+x+1 } \le 1$$
$$\Rightarrow { x }^{ 2 }+x\le 0$$
Hence $$x=0,x=-1$$

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 30

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Inverse Trigonometric Functions Test - 30

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Question 1

$$ -1 $$

$$ \sqrt{2} $$

$$ \displaystyle -\frac{1}{\sqrt{2}} $$

$$ \displaystyle \frac{1}{\sqrt{2}} $$

Solution

Question 2

$$ \tan^{-1}\left (\displaystyle \frac{a\:+\:b}{1\:-\:ab} \right ) $$

$$ \tan^{-1}\left (\displaystyle \frac{a\:+\:b}{1\:-\:ab} \right )-\pi $$

$$ \pi + \tan^{-1}\left (\displaystyle \frac{a\:+\:b}{1\:-\:ab} \right ) $$

none of these

Solution

Question 3

$$ \displaystyle \left \{ \frac{\pi }{2} \right \} $$

$$ \displaystyle \left \{ \frac{\pi }{2},cosec^{-1}\:2 \right \} $$

$$ \left \{ cosec^{-1}\:2 \right\} $$

none of these

Solution

Question 4

$$x=-\frac{\sqrt{5}}{3}$$

$$x=+\frac{\sqrt{5}}{3}$$

$$x=-\frac{2}{3}$$

$$x=+\frac{2}{3}$$

Solution

Question 5

$$p=1, q=1$$

$$p=1, q=2$$

$$p=2, q=1$$

$$p=2, q=2$$

Solution

Question 6

$$a=1, b=1, c=1$$

$$a=1, b=2, c=2$$

$$a=2, b=2, c=1$$

$$a=2, b=1, c=2$$

Solution

Question 7

$$2+n\pi$$

$$2-\pi$$

$$-2+\pi$$

$$2-2n\pi$$

Solution

Question 8

$$-\cfrac{\pi }{4}$$

$$-\cfrac{\pi }{2}$$

$$0$$

$$\cfrac{\pi }{2}$$

Solution

Question 9

$$ \dfrac{\pi }{4} $$

$$ \dfrac{\pi }{2} $$

$$ \dfrac{3\pi }{4} $$

$$ \dfrac{3\pi }{2} $$

Solution

Question 10

$$1$$

$$3$$

$$2$$

$$4$$

Solution

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