Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

$${ sec\quad h }^{ -1 }\left( sin\quad \theta \right) =$$

A
$$log\left( tan\frac { \theta }{ 2 } \right) $$

B
$$log\left( sin\frac { \theta }{ 2 } \right) $$

C
$$log\left( cos\frac { \theta }{ 2 } \right)$$

D
$$log\left( cot\frac { \theta }{ 2 } \right) $$

Solution
Question 2
1 / -0

$$\cos^{ -1 }\left[\cos\left( -\frac { 17 }{ 15 } \pi \right) \right] $$ is equal to

A
$$-\frac { 17\pi }{ 15 } $$

B
$$\frac { 17\pi }{ 15 } $$

C
$$\frac { 13\pi }{ 15 } $$

D
$$\frac { -2\pi }{ 15 } $$

Solution
Question 3
1 / -0

Find $$\displaystyle \int x.\sin xdx$$

A
$$\sin x+x\cos x=C$$

B
$$\sin x-x\cos x+C$$

C
$$\sin x+x\sin x+C$$

D
$$\sin x-x\sin x+C$$
Question 4
1 / -0

If $$A=\tan^{1-}\left(\dfrac {x\sqrt {3}}{2k-x}\right)$$ and $$B=\tan^{-1}\left(\dfrac {2x-k}{k\sqrt {3}}\right)$$ then $$A.B=$$

A
$$0^{o}$$

B
$$\pi /6$$

C
$$\pi /4$$

D
$$\pi /3$$

Solution
Question 5
1 / -0

If $$\cos^{-1}x-\cos^{-1}(\dfrac {y}{2})=\alpha$$ $$ax^{2}-4xy\cos \alpha +y^{2}=$$

A
$$-4\sin^{2}\alpha$$

B
$$4\sin^{2}\alpha$$

C
$$4$$

D
$$2\sin 2\alpha$$

Solution
Question 6
1 / -0

if x$$>0\quad then\quad { tanh }^{ -1 }\left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right)$$

A
$${ log }_{ e }\left( 2x \right)$$

B
$${ log }_{ e }x$$

C
$${ log }_{ e }\left( 3x \right)$$

D
$${ log }_{ e }\left( 5x \right)$$

Solution
Question 7
1 / -0

$$\cos ^{ -1 }{ \left\{ \dfrac { 1 }{ 2 } { x }^{ 2 }+\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\dfrac { { x }^{ 2 } }{ 4 } } \right\} =\cos ^{ -1 }{ \dfrac { x }{ 2 } -\cos ^{ -1 }{ x } } }$$ holds for:

A
$$|x|\le 1$$

B
$$x\in R$$

C
$$0\le x\le 1$$

D
$$-1\le x\le 0$$

Solution

$$clearly, x/2\epsilon[-1,1]$$&$$x\epsilon[-1,1]$$
$$ x\epsilon [-2,2]$$&$$ x\epsilon[-1,1]$$
Intersection of above sells is [-1,1]
Thus $$,[\left | x \right |\leq 1]$$
Question 8
1 / -0

$$if\quad x>0\quad then\quad { tanh }^{ -1 }\left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right) $$

A
$${ log }_{ e }\left( 2x \right) $$

B
$${ log }_{ e }x$$

C
$${ log }_{ e }\left( 3x \right) $$

D
$${ log }_{ e }\left( 5x \right) $$

Solution
Question 9
1 / -0

If $$\cot^{-1}{x}+\tan^{-1}\left (\dfrac{1}{3}\right)=\dfrac{\pi}{2}$$, then $$x$$ will be

A
$$1$$

B
$$3$$

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

D
None of these

Solution

Given $$\text{cot}^{-1} x+\text{tan}^{-1}\bigg(\dfrac{1}{3}\bigg)=\dfrac{\pi}{2}$$
$$\implies \text{cot}^{-1} x=\dfrac{\pi}{2}-\text{tan}^{-1} \bigg(\dfrac{1}{3}\bigg)$$
$$\implies \text{cot}^{-1} x=\text{cot}^{-1}\bigg(\dfrac{1}{3}\bigg)$$
$$\implies x=\dfrac{1}{3}$$
Question 10
1 / -0

If $$x={ \sin }^{ -1 }(\sin10) $$ and $$y={ \cos }^{ -1 }(\cos10)$$, then find $$y - x$$.

A
$$\pi $$

B
$$7\pi $$

C
$$0$$

D
$$10$$

Solution

Given, $$x=sin^{-1}(sin 10)$$
$$x=3\pi -10$$

Also,
$$y=cos^{-1}(cos 10)$$
$$y=4\pi - 10$$

Therefore,
$$y-x=4\pi-10-3\pi+10 = \pi$$