Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

Sum of maximum and minimum values of (sin$$^{-1}$$x)$$^4$$ + (cos$$^{-1}$$x)$$^4$$ is:

A
$$\dfrac{137\pi^2}{128}$$

B
$$\dfrac{\pi^2}{17}$$

C
$$\dfrac{17\pi^4}{16}$$

D
$$\dfrac{137\pi^4}{128}$$
Question 2
1 / -0

$$\displaystyle \sum^{\infty}_{r=1}\tan^{-1}\left(\dfrac {1}{r^{2}+5r+7}\right)$$ equal to

A
$$\tan^{-1}3$$

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

C
$$\sin^{-1}\dfrac {1}{\sqrt {10}}$$

D
$$\cot^{-1}2$$

Solution
Question 3
1 / -0

if $${ tan }^{ -1 }\frac { \sqrt { 1+{ x }^{ 2 }-1 } }{ x } =4\quad then\quad x$$

A
tan8

B
tan4

C
$$tan\frac { 1 }{ 4 } $$

D
tan8

Solution
Question 4
1 / -0

Number of values of x for which $$(tan^{-1} x)^2 + (cot^{-1} x)^2 = \dfrac{\pi^2}{4}$$ is

A
2

B
4

C
1

D
3

Solution
Question 5
1 / -0

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

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

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

C
$$\pi$$

D
None

Solution
Question 6
1 / -0

The value of sin $$\left ( 3 sin^{-1}\left ( 0.8 \right ) \right )$$ is

A
sin(2)

B
sin(1.88)

C
-sin(0.88)

D
None of these

Solution
Question 7
1 / -0

If $$f(x)=2tan^{-1}x+sin^{-1}\left (\dfrac {2x}{1+x^2}\right )$$ then for x>1, f(x)=

A
$$sec^{-1}x$$

B
$$sin^{-1}x$$

C
$$\pi $$

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

Solution
Question 8
1 / -0

If $$(\cos^{-1}x)^{2}+(\cos^{-1}y)^{2}+2(\cos^{-1}x)(\cos^{-1}y)=4\pi^{2}$$ then $$x^{2}+y^{2}$$ is equal to

A
$$1$$

B
$$\dfrac{3}{2}$$

C
$$2$$

D
`Will depends on x and y

Solution
Question 9
1 / -0

If $${ \tan \text{h} }^{ -1 }\left( \dfrac { 1 }{ 3 } \right) =\dfrac { 1 }{ 2 } \log t ,$$ then $$t$$ is equal to

A
$$2$$

B
$$3$$

C
$$1$$

D
$$4$$

Solution
Question 10
1 / -0

If $$f\left( x \right) =\cos ^{ -1 }{ x } +\cos ^{ -1 }{ \left\{ \frac { x }{ 2 } +\dfrac { 1 }{ 2 } \sqrt { 3-3{ x }^{ 2 } } \right\} } $$ then

A
$$f\left( \dfrac { 2 }{ 3 } \right) =\dfrac { \pi }{ 3 } $$

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

C
$$f\left( \dfrac { 1 }{ 3 } \right) =\dfrac { \pi }{ 3 }$$

D
$$f\left( \dfrac { 1 }{ 3 } \right) =2\cos ^{ -1 }{ \dfrac { 1 }{ 3 } -\dfrac { \pi }{ 3 } } m$$

Solution