Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

If $$|x| \le 1$$, then $$2 \tan ^{-1}x +\sin ^{-1} \left(\dfrac{2x}{1+x^2} \right)$$ is equal to

A
$$4\ \tan^{-1}x$$

B
$$0$$

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

D
$$\pi$$

Solution

We have,$$|x| \le 1$$, then $$2 \tan ^{-1}x +\sin ^{-1} \left(\dfrac{2x}{1+x^2} \right)$$

Let $$x=\tan \theta$$
$$2 \tan^{-1} \tan \theta +\sin^{-1}\dfrac{2 \tan \theta}{1+ \tan^2 \theta}$$$$[\because \tan^{-1}(\tan x)=x]$$
$$2 \theta +\sin^{-1} \sin 2 \theta \ \left[\because \sin 2 \theta =\dfrac{2 \tan \theta}{1+ \tan^2 \theta}\right]$$

$$2 \theta +2 \theta \left[\because \sin^{-1} (\sin x)=x \right] $$
$$4 \theta \left[\because \theta \tan^{-1} \right]4 \tan^{-1} x$$
Question 2
1 / -0

The value of the expression $$2 \sec ^{-1}2+ \sin^{-1}\left(\dfrac{1}{2}\right)$$ is

A
$$\dfrac{\pi}{6}$$

B
$$\dfrac{5\pi}{6}$$

C
$$\dfrac{7\pi}{6}$$

D
$$1$$

Solution

We have $$2 \sec ^{-1}2+ \sin^{-1}\left(\dfrac{1}{2}\right)$$
$$2 \sec^{-1}(\sec \dfrac{\pi}{3})+\sin^{-1}(\sin \dfrac{\pi}{6})$$
$$=2.\dfrac{\pi}{3}+\dfrac{\pi}{6} \ \ [\because \sec^{-1} (\sec x)=x$$ and $$\sin^{-1} (\sin x)=x]$$
$$=\dfrac{4 \pi+\pi}{6}=\dfrac{5\pi}{6}$$
Question 3
1 / -0

The value of $$\cot (\sin^{-1}x)$$ is

A
$$\dfrac{\sqrt{1+x^2}}{x}$$

B
$$\dfrac{x}{\sqrt{1+x^2}}$$

C
$$\dfrac{1}{x}$$

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

Solution

(D) is the correct answer.

Let $$\sin^{-1}x=\theta $$ ,
Then $$\sin \theta =x$$
$$\Rightarrow \csc \theta =\dfrac 1x $$
$$\Rightarrow \csc^2 \theta =\dfrac {1}{x^2}$$
$$\Rightarrow 1+\cot^2\theta =\dfrac {1}{x^2}$$
$$\Rightarrow \cot \theta =\dfrac {\sqrt{1-x^2}}{x}$$
Question 4
1 / -0

The value of $$ \sin^{-1}\left\{\cos \left( \dfrac{43\pi}{5}\right)\right\}$$ is

A
$$\dfrac{3\pi}{5}$$

B
$$\dfrac{-7\pi}{5}$$

C
$$\dfrac{\pi}{10}$$

D
$$-\dfrac{\pi}{10}$$

Solution

(D) is the correct answer.

We have
$$\sin^{-1}\left( \cos \dfrac{40\pi +3\pi}{5}\right)$$
$$=\sin^{-1}\left\{\cos \left( 8\pi+\dfrac{3\pi}{5}\right\}\right\}$$
$$=\sin^{-1}\left( \cos \dfrac{3\pi}{5}\right)$$
$$=\sin^{-1}\left\{ \sin \left( \dfrac {\pi}{2}-\dfrac{3\pi}{5}\right)\right\}$$
$$=\sin^{-1}\left( \sin \left( -\dfrac{\pi}{10}\right)\right)$$
$$=-\dfrac{\pi}{10}$$
Question 5
1 / -0

The domain of the function defined by $$f(x)=\sin^{-1}x+\cos x$$ is

A
$$[-1, 1]$$

B
$$[-1, \pi +1]$$

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

D
$$\phi$$

Solution

(A) is the correct answer.

The domain of $$\cos x$$ is $$R$$ and the domain of $$\sin^{-1}$$ is $$[-1, 1]$$.
$$\therefore$$ The domain of $$f(x)=\cos x+\sin^{-1}x$$ is $$R \cap [-1, 1]$$ i.e., $$[-1, 1]$$
Question 6
1 / -0

If $$\sin^{-1}x+\sin^{-1}y=\dfrac{\pi}{2}$$, then the value of $$\cos^{-1}x+\cos^{-1}y$$ is

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

B
$$\pi$$

C
$$0$$

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

Solution

(A) is the correct answer.

Given $$\sin^{-1}x+\sin^{-1}y=\dfrac{\pi}{2}$$
We know that $$\sin^{-1}x+\cos^{-1}y=\dfrac{\pi}{2}$$
$$\therefore\ \cos^{-1}x+\cos^{-1}y\\=\left( \dfrac {\pi}{2}-\sin^{-1}x\right) + \left( \dfrac {\pi}{2}-\sin^{-1}y \right) \\=\pi-(\sin^{-1}x+\sin^{-1}y)\\=\pi-\dfrac {\pi}{2}=\dfrac {\pi}{2}$$

Hence, $$ \cos^{-1}x+\cos^{-1}y=\dfrac {\pi}{2}$$.
Question 7
1 / -0

If $$\tan^{-1}x=\dfrac{\pi}{10}$$ for some $$x\in R$$, then the value of $$\cot^{-1}x$$ is

A
$$\dfrac{\pi}{5}$$

B
$$\dfrac{2\pi}{5}$$

C
$$\dfrac{3\pi}{5}$$

D
$$\dfrac{4\pi}{5}$$

Solution

(B) is the correct answer.

We know that
$$\tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2}$$.
$$\Rightarrow \cot^{-1}x=\dfrac {\pi}{2}-\tan^{-1}x$$
$$\Rightarrow \cot^{-1}x=\dfrac{\pi}{2}-\dfrac{\pi}{10}=\dfrac{4\pi}{10}=\dfrac{2\pi}{5}$$
Question 8
1 / -0

The value of the expression $$\sin [ \cot^{-1}\{ \cos (\tan^{-1}1)\}]$$ is

A
$$0$$

B
$$1$$

C
$$\dfrac{1}{\sqrt 3}$$

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

Solution

(D) is the correct answer.

Let $$P=\sin [ \cot^{-1}\{ \cos (\tan^{-1}1)\}]$$
$$\\=\sin \left[ \cot^{-1}\left( \cos \dfrac {\pi}{4}\right) \right] \qquad\left[\because \tan^{-1}1=\dfrac{\pi}4\right]\\=\sin \left[ \cot^{-1}\dfrac{1}{\sqrt 2}\right] \qquad\left[\because \cos\dfrac{\pi}4=\dfrac1{\sqrt2}\right]\\$$

Let, $$\cot^{-1}\dfrac1{\sqrt2}=y\Rightarrow \cot y=\dfrac1{\sqrt2}$$
$$\\ \Rightarrow \sin y=\sqrt{\dfrac23}\Rightarrow y=\sin^{-1}\sqrt{\dfrac23}$$

$$\therefore P=\sin \left[ \sin^{-1}\sqrt{\dfrac 23}\right] =\sqrt{\dfrac 23}$$
Question 9
1 / -0

If $$\alpha \le 2\sin^{-1}x+\cos^{-1}x \le \beta$$, then

A
$$\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{\pi}{2}$$

B
$$\alpha =0, \beta =\pi$$

C
$$\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{3\pi}{2}$$

D
$$\alpha =0, \beta =2\pi$$

Solution

(B) is the correct answer.

We know that
$$\dfrac {-\pi}{2}\le \sin^{-1}x \le \dfrac {\pi}{2}\\$$
$$\Rightarrow \dfrac {-\pi}{2}+\dfrac {\pi}{2} \le \sin^{-1}x + \dfrac {\pi}{2} \le \dfrac {\pi}{2}+\dfrac {\pi}{2}\qquad [$$Addidng $$\dfrac{\pi}2 ]$$
$$\\\Rightarrow 0 \le \sin^{-1}x+ (\sin^{-1}x+\cos^{-1}x) \le \pi\qquad\left[\because(\sin^{-1}x+\cos^{-1}x)= \dfrac{\pi}2 \right]\\$$
$$\Rightarrow 0 \le 2\sin^{-1}x+\cos^{-1}x \le \pi$$
Question 10
1 / -0

The domain of $$y=\cos^{-1}(x^2-4)$$ is

A
$$[3, 5]$$

B
$$[0, \pi]$$

C
$$[-\sqrt 5, -\sqrt 3] \cap [ -\sqrt 5, \sqrt 3]$$

D
$$[ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, 5]$$

Solution

(D) is the correct answer.

Given $$y=\cos^{-1}(x^2-4) \Rightarrow \cos y =x^2-4$$
$$\therefore \ -1 \le x^2 -4 \le 1 $$ ( since $$-1 \le \cos y \le 1)$$
$$\Rightarrow 3 \le x^2 \le 5$$
$$\Rightarrow \sqrt 3 \le |x| \le \sqrt 5$$
$$\Rightarrow x \in [ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, \sqrt 5]$$

Hence the domain of $$y $$ is $$[ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, 5]$$

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 48

Result

Inverse Trigonometric Functions Test - 48

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SHARING IS CARING

Question 1

$$4\ \tan^{-1}x$$

$$0$$

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

$$\pi$$

Solution

Question 2

$$\dfrac{\pi}{6}$$

$$\dfrac{5\pi}{6}$$

$$\dfrac{7\pi}{6}$$

$$1$$

Solution

Question 3

$$\dfrac{\sqrt{1+x^2}}{x}$$

$$\dfrac{x}{\sqrt{1+x^2}}$$

$$\dfrac{1}{x}$$

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

Solution

Question 4

$$\dfrac{3\pi}{5}$$

$$\dfrac{-7\pi}{5}$$

$$\dfrac{\pi}{10}$$

$$-\dfrac{\pi}{10}$$

Solution

Question 5

$$[-1, 1]$$

$$[-1, \pi +1]$$

$$( -\infty, \infty)$$

$$\phi$$

Solution

Question 6

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

$$\pi$$

$$0$$

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

Solution

Question 7

$$\dfrac{\pi}{5}$$

$$\dfrac{2\pi}{5}$$

$$\dfrac{3\pi}{5}$$

$$\dfrac{4\pi}{5}$$

Solution

Question 8

$$0$$

$$1$$

$$\dfrac{1}{\sqrt 3}$$

$$\sqrt{\dfrac{2}{3}}$$

Solution

Question 9

$$\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{\pi}{2}$$

$$\alpha =0, \beta =\pi$$

$$\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{3\pi}{2}$$

$$\alpha =0, \beta =2\pi$$

Solution

Question 10

$$[3, 5]$$

$$[0, \pi]$$

$$[-\sqrt 5, -\sqrt 3] \cap [ -\sqrt 5, \sqrt 3]$$

$$[ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, 5]$$

Solution

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