Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 52

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Question 1
1 / -0

The value of $$sin^{-1}(sin2010^{0})+cos^{-1}(cos2010^{0})+tan^{-1}(tan2010^{0})$$ is

A
$$\frac{\pi }{6}$$

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

C
$$\frac{2\pi }{3}$$

D
$$\frac{5\pi }{6}$$

Solution

$$2010^{0}$$
$$=11\pi+\frac{\pi}{6}$$ ...(i)
Hence
$$sin^{-1}(sin(2010^{0}))=sin^{-1}(sin(11\pi+\frac{\pi}{6}))$$ ...(from(i))
$$cos^{-1}(cos(2010^{0}))=cos^{-1}(cos(11\pi+\frac{\pi}{6}))$$
$$tan^{-1}(tan(2010^{0}))=tan^{-1}(tan(11\pi+\frac{\pi}{6}))$$
Therefore
$$sin^{-1}(sin(11\pi+\frac{\pi}{6}))+cos^{-1}(cos(11\pi+\frac{\pi}{6}))+tan^{-1}(tan(11\pi+\frac{\pi}{6}))$$
$$=-\frac{\pi}{6}+\pi-\frac{\pi}{6}+\frac{\pi}{6}$$
$$=\frac{5\pi}{6}$$
Question 2
1 / -0

The number of solutions of the equation $$1+x^{2}+2x\, sin\: (cos^{-1}y)=0$$ is

A
1

B
2

C
3

D
4

Solution

$$1+x^{2}+2x sin cos ^{1-}y=0$$

$$\Rightarrow 1+x^{2}+2x\sqrt{1-y^{2}}=0$$

$$x^{2}+2x\sqrt{1-y^{2}}+1=0$$

$$x^{2}+2x\sqrt{1-y^{2}}+1-y^{2}=-y^{2}$$

$$(x+\sqrt{1-y^{2}})=-y^{2} \Rightarrow -y^{2} \geq 0 $$

$$x+\sqrt{1-y^{2}}=0 \space and \space y=0 (x+\sqrt{1-y^{2}})^{2} \geq 0$$

$$\rightarrow y=0 , x=-1$$ only one solution
Question 3
1 / -0

The value of $$\displaystyle \sec^{-1}\left (\displaystyle \frac{1}{1-2x^{2}}\right)+4{\cos^{-1}}\sqrt{\displaystyle \frac{1+x}{2}}$$ is equal to

A
$$\pi $$

B
$$2\pi$$

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

D
None of these

Solution

Given

$$\sec ^{-1} \left(\dfrac 1{1-2x^2}\right)+4\cos ^{-1}\sqrt {\dfrac {1+x}2}$$

Let $$ x=\cos \theta $$

$$\implies \sec ^{-1} \dfrac 1{1-2\cos ^2 x} +4\cos ^{-1}\left( \sqrt {\dfrac {1+\cos \theta }2}\right)$$

$$= \sec ^{-1}\left( \dfrac {-1}{\cos 2\theta }\right) +4\cos ^{-1} (\sqrt {\cos ^2\theta /2})$$

$$ =\pi -\sec ^{-1} (\sec 2\theta )+4 \cos ^{-1} (\cos \theta/2)$$

$$ =\pi -2\theta +4 \left(\dfrac \theta 2\right)$$

$$=\pi -2\theta +2\theta$$

$$=\pi $$
Question 4
1 / -0

The value of $$\sin^{-1}$$$$\left \{ \tan\left ( \cos^{-1}\sqrt{\dfrac{2+\sqrt{3}}{4}}+\cos^{-1}\dfrac{\sqrt{12}}{4} -\text{cosec}^{-1}\sqrt{2}\right ) \right \}$$, is

A
$$0$$

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

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

D
$$\pi $$

Solution

Given:
$$E = \sin^{-1} \left(\tan \left(\cos^{-1} \dfrac {\sqrt {2 + \sqrt {3}}}{2} + \cos^{-1} \dfrac {\sqrt 3}{2} - \text{cosec}^{-1} \sqrt 2\right) \right)$$

We know that,
$$\cos 15^0 = \sqrt{\dfrac{2+\sqrt3}{4}}$$

Therefore,
$$\Rightarrow E= \sin^{-1} (\tan (15^0 + 30^0 - 45^0))$$
$$\Rightarrow E= \sin^{-1} (\tan0^0)$$
$$\Rightarrow E = \sin^{-1} 0 = 0$$
Question 5
1 / -0

If $$x> 0\, $$ and $$\, cos^{-1}\left ( \dfrac{12}{x} \right )+cos^{-1}\left ( \dfrac{35}{x} \right )=\dfrac{\pi }{2},$$ then x is

A
$$7$$

B
$$39$$

C
$$37$$

D
$$-37$$

Solution

$$cos^{-1}(a)+cos^{-1}(b)=\dfrac{\pi}{2}$$
$$\Rightarrow cos^{-1}(a)=\dfrac{\pi}{2}-cos^{-1}(b)$$
$$\Rightarrow cos^{-1}(a)=sin^{-1}(b)$$
Therefore, $$a^2+b^2=1$$
Substituting the value of a and b we get
$$\dfrac{12}{x}^{2}+\dfrac{35}{x}^{2}=1$$
$$x^2=1369$$
$$x=37,-37$$
However , since $$x>0$$
$$x=37$$
Question 6
1 / -0

If $$cosec ^{ -1 }\left(cosec (x) \right)$$ and $$cosec\left(cosec ^{ -1 }(x) \right) $$ are equal functions, then the maximum range of value of $$x$$ is

A
$$\displaystyle\:\left [ -\frac{\pi }{2},-1\right ]\cup \left [ 1,\frac{\pi }{2} \right ]$$

B
$$\displaystyle\:\left (-\frac{\pi }{2},-1\right )\cup \left (1,\frac{\pi }{2} \right )$$

C
$$\displaystyle\:\left ( -\infty ,-1]\cup [1,\infty \right )$$

D
$$\displaystyle\:\left ( -\infty ,-1)\cup (1,\infty \right )$$

Solution

The range of $$cosec^{-1}(cosec(x))$$
$$=\left[\dfrac{-\pi}{2},0\right)\cup\left(0,\dfrac{\pi}{2}\right]$$
The range of $$cosec(cosec^{-1}(x))$$
$$=(-\infty,-1]\cup[1,\infty)$$
Since it is given that both the functions are equal, then the range of value x common to both will be
$$=\left[\dfrac{-\pi}{2},-1\right]\cup\left[1,\dfrac{\pi}{2}\right]$$
Question 7
1 / -0

The range of values of p for which the equation $$\sin \cos ^{-1}(\cos (\tan ^{-1}x))= p$$ has a solution is

A
$$\left ( -\frac{1}{\sqrt{2}},\frac{2}{\sqrt{2}} \right )$$

B
$$[0,1)$$

C
$$\left ( \frac{1}{\sqrt{2}1} \right )$$

D
$$(-1,1)$$

Solution

Here The range of $$cos(tan^{-1}(x))$$ will be [-1,1] since the range of $$cos\theta$$ is always $$[-1,1]$$
Hence for $$cos(tan^{-1}(x))=1$$
$$sin(cos^{-1}(1))$$
$$=sin(0)$$
$$=0$$
For $$cos(tan^{-1}(x))=0$$
$$sin(cos^{-1}(0))$$
$$=sin(\frac{\pi}{2})$$
$$=1$$
For $$cos(tan^{-1}(x))=-1$$
$$sin(cos^{-1}(-1))$$
$$=sin(\pi)$$
$$=0$$
Howere p=1 for $$x\rightarrow \infty$$
Hence range of p=[0,1).
Question 8
1 / -0

The solution set of the equation $$\displaystyle \sin^{-1} \sqrt{1-x} + \cos^{-1}x = \cot^{-1} \left ( \frac{\sqrt{1-x^{2}}}{x} \right )-\sin^{-1}x$$

A
$$\displaystyle \left [ -1,\: 1 \right ] - \left \{ 0 \right \}$$

B
$$\displaystyle \left ( 0, \: 1 \right ] \cup \left \{ -1 \right \}$$

C
$$\displaystyle \left [ -1,\: 0 \right ) \cup \left \{ 1 \right \}$$

D
$$\displaystyle \{ 1 \}$$

Solution

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

$$\Rightarrow \sin ^{ -1 } \sqrt { 1-x } +\dfrac { \pi }{ 2 } =\cot ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right) $$

$$\Rightarrow \dfrac { \pi }{ 2 } -\cot ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right) =-\sin ^{ -1 } \sqrt { 1-x } $$

$$\Rightarrow \tan ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right) =-\sin ^{ -1 } \sqrt { 1-x } $$

Thus is only true when, $$x = \{1\}$$
Question 9
1 / -0

If $$\sin ^{ -1 }{ x } +\sin ^{ -1 }{ y } +\sin ^{ -1 }{ z } =\cfrac { 3\pi }{ 2 } $$ and $$f(1)=2,f(x+y)=f(x)f(y)$$ for all $$x,y\in R$$. Then $${ x }^{ { f(1) } }+{ y }^{ f(2) }+{ z }^{ f(3) }-\cfrac { x+y+z }{ { x }^{ f(1) }+{ y }^{ f(2) }+{ z }^{ f(3) } } $$ is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

We know that $$-\cfrac { \pi }{ 2 } \le \sin ^{ -1 }{ x } ,\sin ^{ -1 }{ y } ,\sin ^{ -1 }{ z } \le \cfrac { \pi }{ 2 } $$
$$\therefore \sin ^{ -1 }{ x } +\sin ^{ -1 }{ y } +\sin ^{ -1 }{ z } =\cfrac { 3\pi }{ 2 } $$
$$\Rightarrow \sin ^{ -1 }{ x } =\sin ^{ -1 }{ y } =\sin ^{ -1 }{ z } =\cfrac { \pi }{ 2 } $$
$$\Rightarrow x+y+z=1$$
It is given that $$f(x+y)=f(x)f(y)$$ for all $$x,y\in R$$
$$\therefore f(x)={ \left[ f(1) \right] }^{ x }$$ for all $$x\in R$$
$$\Rightarrow f(x)={ 2 }^{ x }$$ for all $$x\in R$$
$$\Rightarrow f(1)=2,\quad f(2)=4,\quad f(3)=8$$
$$\therefore \quad { x }^{ f(1) }+{ y }^{ f(2) }+{ z }^{ f(3) }=\cfrac { x+y+z }{ { x }^{ f(1) }+{ y }^{ f(2) }{ + }{ z }^{ f(3) } } $$
$$1+1+1-\cfrac { 1+1+1 }{ 1+1+1 } =3-1=2$$
Question 10
1 / -0

If $$\left[ \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } } \right] =1$$, where $$[.]$$ denotes the greatest integer function, the $$\theta$$ lies in the interval

A
$$[\tan { \sin { \cos { 1 } } } ,\sin { \tan { \cos { \sin { 1 } } } } ]$$

B
$$[\sin { \tan { \cos { 1 } } } ,\tan { \sin { \cos { \sin { 1 } } } } ]$$

C
$$[\tan { \sin { \cos { 1 } } } ,\tan { \sin { \cos { \sin { 1 } } } } ]$$

D
None of these

Solution

We have $$\left[ \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } } \right] =1$$    $$\Rightarrow 1\le \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } }\le \displaystyle\frac { \pi }{ 2 }$$
$$\Rightarrow \sin { 1 } \le \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } \le 1$$    $$\Rightarrow \cos { \sin { 1 } } \ge \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } \ge \cos { 1 }$$
$$\Rightarrow \sin { \cos { \sin { 1 } } } \ge \tan ^{ -1 }{ \theta } \ge \sin { \cos { 1 } }$$ $$\Rightarrow \tan { \sin { \cos { \sin { 1 } } } } \ge \theta \ge \tan { \sin { \cos { 1 } } } $$

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 52

Result

Inverse Trigonometric Functions Test - 52

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SHARING IS CARING

Question 1

$$\frac{\pi }{6}$$

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

$$\frac{2\pi }{3}$$

$$\frac{5\pi }{6}$$

Solution

Question 2

1

2

3

4

Solution

Question 3

$$\pi $$

$$2\pi$$

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

None of these

Solution

Question 4

$$0$$

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

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

$$\pi $$

Solution

Question 5

$$7$$

$$39$$

$$37$$

$$-37$$

Solution

Question 6

$$\displaystyle\:\left [ -\frac{\pi }{2},-1\right ]\cup \left [ 1,\frac{\pi }{2} \right ]$$

$$\displaystyle\:\left (-\frac{\pi }{2},-1\right )\cup \left (1,\frac{\pi }{2} \right )$$

$$\displaystyle\:\left ( -\infty ,-1]\cup [1,\infty \right )$$

$$\displaystyle\:\left ( -\infty ,-1)\cup (1,\infty \right )$$

Solution

Question 7

$$\left ( -\frac{1}{\sqrt{2}},\frac{2}{\sqrt{2}} \right )$$

$$[0,1)$$

$$\left ( \frac{1}{\sqrt{2}1} \right )$$

$$(-1,1)$$

Solution

Question 8

$$\displaystyle \left [ -1,\: 1 \right ] - \left \{ 0 \right \}$$

$$\displaystyle \left ( 0, \: 1 \right ] \cup \left \{ -1 \right \}$$

$$\displaystyle \left [ -1,\: 0 \right ) \cup \left \{ 1 \right \}$$

$$\displaystyle \{ 1 \}$$

Solution

Question 9

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 10

$$[\tan { \sin { \cos { 1 } } } ,\sin { \tan { \cos { \sin { 1 } } } } ]$$

$$[\sin { \tan { \cos { 1 } } } ,\tan { \sin { \cos { \sin { 1 } } } } ]$$

$$[\tan { \sin { \cos { 1 } } } ,\tan { \sin { \cos { \sin { 1 } } } } ]$$

None of these

Solution

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