Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

Number of solution(s) to the equation$$ $$ $${\cos ^{ - 1}}x + {\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) = \dfrac{\pi }{6}\,$$ is/are

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Hence,
option $$(B)$$ is correct answer.
Question 2
1 / -0

$$\displaystyle \sum^{\infty}_{r=1}\tan^{-1}\left(\dfrac {3}{r^{2}-r+9}\right)$$ is -

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

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

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

D
$$\dfrac {\pi}{12}$$
Question 3
1 / -0

$$If\,\cos \left( {2{{\sin }^{ - 1}}x} \right) = \frac{1}{9},\,then\,\,x\,\,is\,\,equal\,\,to$$

A
$$Only\,\frac{2}{3}\,$$

B
$$Only - \frac{2}{{3\,}}$$

C
$$\,\frac{2}{3}, - \frac{2}{3}$$

D
$$\frac{1}{3}$$

Solution
Question 4
1 / -0

If $${ cos }^{ -1 }\dfrac { 3 }{ 5 } -{ sin }^{ -1 }\dfrac { 4 }{ 5 } ={ cos }^{ -1 }x$$, then x is equal to -

A
0

B
1

C
-1

D
None of these

Solution

Given,

$$\cos ^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{4}{5}=\cos ^{-1}x$$

apply $$\cos$$ on both sides, we get,

$$\cos \left ( \cos ^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{4}{5} \right )=\cos (\cos ^{-1}x)$$

$$\mathrm{Use\:the\:following\:identity}:\quad \cos \left(s+t\right)=\cos \left(s\right)\cos \left(t\right)-\sin \left(s\right)\sin \left(t\right)$$

$$\cos \left(\cos ^{-1} \left(\dfrac{3}{5}\right)\right)\cos \left(\sin^{-1} \left(\dfrac{4}{5}\right)\right)-\sin \left(\cos ^{-1} \left(\dfrac{3}{5}\right)\right)\sin \left(\sin^{-1} \left(\dfrac{4}{5}\right)\right)=x$$

$$\dfrac{3}{5}\left ( \sqrt{1-\left (\dfrac{4}{5} \right )^2} \right )-\dfrac{4}{5}\left ( \sqrt{1-\left (\dfrac{3}{5} \right )^2} \right )=x$$

$$\dfrac{3}{5}\times \dfrac{3}{5}-\dfrac{4}{5}\times \dfrac{4}{5}=x$$

$$\dfrac{9}{25}-\dfrac{16}{25}=x$$

$$\therefore x=-\dfrac{7}{25}$$
Question 5
1 / -0

Number of solution of the equation $$\tan^{-1}\left(\dfrac {1}{x-1}+\dfrac {1}{x-2}+\dfrac {1}{x-3}+\dfrac {1}{x-4} \right)+\cos^{-1}(x)=\dfrac {3\pi}{4}-\sin^{-1}(x)$$ are

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution
Question 6
1 / -0

The range of the function $$f(x)=\ell n\ (\sin^{-1}(x^{2}+x))$$ is

A
$$\left[-\ell n \left(\sin^{-1}\dfrac {1}{4}\right),\ell n \dfrac {\pi}{4}\right]$$

B
$$\left[-\ell n \dfrac {\pi}{4},\ell n \dfrac {\pi}{2}\right]$$

C
$$\left(0,\ell n \dfrac {\pi}{2}\right]$$

D
$$\left(-\infty ,\ell n \dfrac {\pi}{2}\right]$$

Solution
Question 7
1 / -0

If $$f(x)=\cos^{-1}\left(\dfrac {\sqrt {2x^{2}+1}}{x^{2}+1}\right)$$, then range of $$f(x)$$ is

A
$$[0,\pi]$$

B
$$\left(0,\dfrac {\pi}{4}\right]$$

C
$$\left(0,\dfrac {\pi}{3}\right]$$

D
$$\left[0,\dfrac {\pi}{2}\right)$$

Solution

Given,

$$f\left(x\right)=\cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)$$

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

$$0< f(x)\leq1$$

$$\cos^{-1}(1)\leq \cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)<\cos^{-1}(0)$$

$$0\leq \cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)< \dfrac{\pi}{2}$$

Therefore the range is,

$$0\leq f(x)< \dfrac{\pi}{2}$$
Question 8
1 / -0

$$cos^{-1}\left(\dfrac{\pi}{3}+sec^{-1}(-2)\right)$$=

A
-1

B
1

C
0

D
None of these

Solution

As we know that
$$\text{sec}^{-1}(-2)=\dfrac{2\pi}{3}$$
$$\therefore \cos \bigg(\dfrac{\pi}{3}+\text{sec}^{-1}(-2)\bigg)=\cos \bigg(\dfrac{\pi}{3}+\dfrac{2\pi}{3}\bigg)=\cos \pi=-1$$
Question 9
1 / -0

The value of $$\cot\left(cosec^{-1}\dfrac{5}{3}+\tan^{-1}\dfrac{2}{3}\right)$$ is equal to-

A
$$\dfrac{6}{17}$$

B
$$\dfrac{3}{17}$$

C
$$\dfrac{4}{17}$$

D
$$\dfrac{5}{17}$$

Solution
Question 10
1 / -0

$$\int _{ 0 }^{ \pi }{ \left[ cotx \right] dx,where\left[ \cdot \right] } $$ denotes the greatest integer function, is equal to:

A
1

B
-1

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

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

Solution