Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 20

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Question 1
1 / -0

The largest interval lying in $$\left ( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right )$$ for which the function $$\left [ f(x)=4^{-x^{2}}+\cos^{-1}\left ( \dfrac{x}{2}-1 \right )+\log (\cos x) \right ]$$ is defined, is-

A
$$[0,\pi ]$$

B
$$\left ( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right )$$

C
$$\left [- \dfrac{\pi }{4},\dfrac{\pi }{2} \right )$$

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

Solution

For given function to be defined $$-1 \leq \cfrac{x}{2}-1 \leq 1$$ and $$\cos x > 0$$
$$\Rightarrow 0 \leq \cfrac{x}{2} \leq 2$$ and $$ \cfrac{-\pi}{2} \leq x < \cfrac{\pi}{2}$$
$$\Rightarrow 0 \leq x \leq 4$$ and $$ \cfrac{-\pi}{2} \leq x < \cfrac{\pi}{2}$$
Taking common of both we get $$x \in \left[0,\cfrac{\pi}{2}\right)$$
Question 2
1 / -0

Statement I : The equation $$(sin^{-1}x)^3+(cos^{-1}x)^3-a\pi^3=0$$ has a solution for all $$a\geqslant \dfrac {1}{32}.$$
Statement II : For any $$x\epsilon R, sin^{-1}x+cos^{-1}x=\dfrac {\pi}{2}$$ and $$0\leq (sin^{-1}x-\dfrac {\pi}{4})^2\leq \dfrac {9\pi^2}{16}$$.

A
Both statements I and II are true.

B
Both statements I and II are true but I is not an explanation of II

C
Statement I is true and statement II is false

D
Statement I is false and statement II is true.

Solution

Say $$f\left( x \right) ={ \left( \sin ^{ -1 }{ x } \right) }^{ 3 }+{ \left( \cos ^{ -1 }{ x } \right) }^{ 3 }$$
$$f'\left( x \right) =0$$ at $$x=\dfrac { \pi }{ 4 } $$
$$f''\left(x\right) \ge0 $$ at $$x=\dfrac{\pi}{4}$$
so $$f\left( x \right) =\dfrac { { \pi }^{ 3 } }{ 32 } $$ This is least value.
$$\therefore f\left( x \right) \ge a{ \pi }^{ 3 }$$ has a solution.
$$\therefore \dfrac{1}{32}\ge$$ $$a$$
Statement $$ I $$ is incorrect.
Now $$\dfrac { -\pi }{ 2 } \le \sin ^{ -1 }{ x } \le \dfrac { \pi }{ 2 } $$
$$0\le { \left( \sin ^{ -1 }{ x } -\dfrac { \pi }{ 4 } \right) }^{ 2 }\le \dfrac { 9{ \pi }^{ 2 } }{ 16 } $$
Min at $$x=\dfrac { \pi }{ 4 } $$ and max at $$x=\dfrac { \pi }{ 2 } $$. So, Statement $$I$$ is incorrect and $$II$$ is correct.
Question 3
1 / -0

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

A
$$|x|\leq 1$$

B
$$x\epsilon R$$

C
$$0\leq x\leq 1$$

D
$$-1\leq x\leq 0$$

Solution

$$\frac { x }{ 2 } \epsilon (-1,1)$$, $$x\epsilon (-1,1)$$
Intersection for $${ cos }^{ -1 }\left( \frac { x }{ 2 } \right) $$ and $${ cos }^{ -1 }(x)$$ to be valid
Thus x lies from (-1, 1)
for $$cos\left( { cos }^{ -1 }\left( \frac { x }{ 2 } \right) \right) =\frac { x }{ 2 } $$
$$\frac { x }{ 2 } $$ should be from $$(2n\pi ,(2n+1))$$
Similarly for $$cos({ cos }^{ -1 }x)=x$$
x should be from $$(2n\pi ,(2n+1)\pi )$$
Thus, $$x\epsilon (0,1)$$
Question 4
1 / -0

Solve: $$\displaystyle { \tan }^{ -1 }\left( \frac { x }{ y } \right) -{ \tan }^{ -1 }\frac { x-y }{ x+y } $$ is equal to

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

B
$$\displaystyle \frac { \pi }{ 3 } $$

C
$$\displaystyle \frac { \pi }{ 4 } $$

D
$$\displaystyle \frac { -3\pi }{ 4 } $$

Solution

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

={ \tan }^{ -1 }\left[ \frac { \frac { x }{ y } -\frac { x-y }{ x+y }

}{ 1+\left( \frac { x }{ y } \right) \left( \frac { x-y }{ x+y }

\right) } \right] $$
$$\displaystyle ={ \tan }^{

-1 }\left[ \frac { \frac { x\left( x+y \right) -y\left( x-y \right) }{

y\left( x+y \right) } }{ \frac { y\left( x+y \right) +x\left( x-y

\right) }{ y\left( x+y \right) } } \right] $$
$$\displaystyle ={ \tan }^{ -1 }\left( \frac { { x }^{ 2 }+xy-xy+{ y }^{ 2 } }{ xy+{ y }^{ 2 }+{ x }^{ 2 }-xy } \right) $$
$$\displaystyle

={ \tan }^{ -1 }\left( \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { x }^{ 2 }+{ y

}^{ 2 } } \right) ={ \tan }^{ -1 }1=\frac { \pi }{ 4 } $$
Question 5
1 / -0

Value of $$\tan^{-1}\begin{Bmatrix}\dfrac{\sin\,2-1}{\cos\,2}\end{Bmatrix}$$ is

A
$$\dfrac{\pi}{2}-1$$

B
$$1-\dfrac{\pi}{4}$$

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

D
$$\dfrac{\pi}{4}-1$$

Solution

$$tan^{-1}\begin{Bmatrix}\dfrac{\sin\,2-1}{\cos\,2}\end{Bmatrix}$$
$$tan^{-1}\begin{Bmatrix}\dfrac{2\sin1\,cos\,1-[sin^21+\cos^21]}{\cos^21-sin^21}\end{Bmatrix}$$
$$tan^{-1}\begin{Bmatrix}\dfrac{-[\cos1-\sin1]^2}{(\cos 1-\sin 1)(cos1+sin1)}\end{Bmatrix}$$
$$tan^{-1}\begin{Bmatrix}\dfrac{\cos 1-\sin 1}{\cos 1+\sin 1}\end{Bmatrix}$$ divide by $$cos1=tan^{-1}(tan(1-\dfrac{\pi}{4}))=1-\dfrac{\pi}{4}$$
Question 6
1 / -0

The number of triplets $$\left ( x,y,z \right )$$ satisfies the equation $$\displaystyle f\left ( x,y,z \right )= \sin ^{-1}x+\sin ^{-1}y+\sin ^{-1}z= \dfrac{3\pi }{2}$$ is

A
$$1$$

B
$$2$$

C
$$0$$

D
Infinite

Solution

As $$-\cfrac { \pi }{ 2 } \le { sin }^{ -1 }\theta \le \cfrac { \pi }{ 2 } $$

And for $${ sin }^{ -1 }x+{ sin }^{ -1 }y+{ sin }^{ -1 }z=\cfrac { 3\pi }{ 2 } $$

$${ sin }^{ -1 }x={ sin }^{ -1 }y={ sin }^{ -1 }z=\cfrac { \pi }{ 2 } $$

$$\Rightarrow x=y=z=1$$

Hence option 'A' is correct.
Question 7
1 / -0

Calculate the value of $$\displaystyle \sin^{-1} \cos \left ( \sin^{-1} x\right ) + \cos^{-1} \sin \left ( \cos^{-1} x \right ) $$. where $$\displaystyle\left | x \right | \leq 1$$

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

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

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

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

Solution

Given, $$\displaystyle \sin^{-1} \cos \left ( \sin^{-1} x\right ) + \cos^{-1} \sin \left ( \cos^{-1} x \right ) $$

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

$$=\displaystyle \sin^{-1} \left(\sqrt{1-x^2}\right ) + \cos^{-1}\left ( \sqrt{1-x^2} \right ) =\cfrac{\pi}{2}$$

Since $$\sin^{-1}x+\cos^{-1}x = \cfrac{\pi}{2} \forall |x|\leq 1$$
Question 8
1 / -0

Sin-1 $$(\displaystyle \frac{3}{5})+\mathrm{S}\mathrm{i}\mathrm{n}^{-1}(\frac{5}{13})=\sin_{X}^{-1}$$ then$$\mathrm{x}=$$

A
$$(^{\frac{63}{65})}$$

B
$$\displaystyle \frac{56}{65}$$

C
$$\displaystyle \frac{56}{63}$$

D
$$\displaystyle \frac{16}{63}$$
Question 9
1 / -0

If $$\sec^{-1} x+ \sec^{-1}y + \sec^{-1}z = 3\pi$$, then $$xy + yz + zx =$$ _______.

A
$$0$$

B
$$-3$$

C
$$3$$

D
$$1$$

Solution

Range of $$\sec^{-1}x\space :\space [0,\pi/2)\cup(\pi/2,\pi]$$
So $$\sec^{-1}x+\sec^{-1}y+\sec^{-1}z=3\pi$$ only when all of them are equal to $$\pi$$
That is, $$\sec^{-1}x=\pi,\sec^{-1}y=\pi,\sec^{-1}z=\pi$$
$$\sec^{-1}x=\pi \rightarrow x=-1 $$
Similarly, $$y=-1,z=-1$$
$$\therefore \space xy+yz+zx=1+1+1=3$$
Hence, $$(C)$$
Question 10
1 / -0

Find value of $$\cos^{-1}\left(-\dfrac {1}{2}\right)$$.

A
$$\dfrac {7\pi}{3}$$

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

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

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

Solution

$$\begin{array}{l} { \cos ^{ -1 } }\left( { \dfrac { { -1 } }{ 2 } } \right) \\ =\pi -{ \cos ^{ -1 } }\left( { \dfrac { 1 }{ 2 } } \right) \\ =\pi -{ \cos ^{ -1 } }\left( { \cos \dfrac { \pi }{ 3 } } \right) \\ =\pi -\dfrac { \pi }{ 3 } \\ =\dfrac { { 2\pi } }{ 3 } \end{array}$$

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 20

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Inverse Trigonometric Functions Test - 20

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Question 1

$$[0,\pi ]$$

$$\left ( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right )$$

$$\left [- \dfrac{\pi }{4},\dfrac{\pi }{2} \right )$$

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

Solution

Question 2

Both statements I and II are true.

Both statements I and II are true but I is not an explanation of II

Statement I is true and statement II is false

Statement I is false and statement II is true.

Solution

Question 3

$$|x|\leq 1$$

$$x\epsilon R$$

$$0\leq x\leq 1$$

$$-1\leq x\leq 0$$

Solution

Question 4

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

$$\displaystyle \frac { \pi }{ 3 } $$

$$\displaystyle \frac { \pi }{ 4 } $$

$$\displaystyle \frac { -3\pi }{ 4 } $$

Solution

Question 5

$$\dfrac{\pi}{2}-1$$

$$1-\dfrac{\pi}{4}$$

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

$$\dfrac{\pi}{4}-1$$

Solution

Question 6

$$1$$

$$2$$

$$0$$

Infinite

Solution

Question 7

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

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

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

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

Solution

Question 8

$$(^{\frac{63}{65})}$$

$$\displaystyle \frac{56}{65}$$

$$\displaystyle \frac{56}{63}$$

$$\displaystyle \frac{16}{63}$$

Question 9

$$0$$

$$-3$$

$$3$$

$$1$$

Solution

Question 10

$$\dfrac {7\pi}{3}$$

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

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

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

Solution

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