If $$\displaystyle \sin ^{-1}x-\sin ^{-1}y= \frac{\pi }{2}$$ ,then

$$y= -\sqrt{1-x^{2}}, \:0\leq x\leq 1,-1\leq y\leq 0$$

$$y= \sqrt{1-x^{2}}, \: \left | x \right |< 1$$

MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 2

Range of

$\sin^{-1}x-\cos^{-1}x$ is

0%
$[\displaystyle \frac{-3\pi}{2}, \displaystyle \frac{\pi}{2}]$
0%
$[\displaystyle \frac{-5\pi}{3}, \displaystyle \frac{\pi}{3}]$
0%
$[\displaystyle \frac{-3\pi}{2}, \pi]$
0%
$[0, \pi]$

Explanation

range of

$\displaystyle sin ^{-1}x \epsilon [-\frac{\pi }{2} , \frac{\pi }{2}]$
range of

$cos ^{-1}x \epsilon [0, \pi ]$

$\therefore f(x)=sin ^{-1}x-cos ^{-1}x=sin ^{-1}x+cos ^{-1}x-2 cos ^{-1}x$

$\displaystyle \quad\quad\quad=\frac{\pi }{2}-2cos ^{-1}x\quad\quad [\because sin ^{-1}x+cos ^{-1}x=\frac{\pi }{2}] identity$
range of

$\displaystyle f(x) \epsilon [\frac{\pi }{2} -2(\pi ) , \frac{\pi }{2}-2(0)]$

$\therefore f(x)\displaystyle \epsilon [-\frac{3\pi }{2} ,\frac{\pi }{2}]$

$ABC$ is a triangular park with

$AB=AC=100$ cm.

$A$ clock tower is situated at the midpoint of

$BC$ . The angles of elevation of the top of the tower from

$A$ and

$B$ are

$cot ^{-1}3.2$ and

$cosec ^{-1} 2.6$ . The height of the tower is

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$25$ mt
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$50$ mt
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$100$ mt
0%
$50\sqrt{2}$ mt

Explanation

$y^{2}=AD^{2}=100^{2}-BD^{2}=10000-x^{2}$

$\cot^{-1}(3.2)=\alpha,cosec^{-1}(2.6)=\beta)$

$\cot \alpha =3.2$ &

$cosec \beta =2.6$

$\cot \alpha =\dfrac{y}{h}$

$\Rightarrow y=3.2 \times h$
Now,

$\cot \beta =\dfrac{x}{h}$

$x=\cot \beta \times h$

$x=h\sqrt{2.6^{2}-1}$

$x= 2.4h$

$x^{2}+y^{2}=100000$

$\Rightarrow (2.4h)^{2}+(3.2h)^{2}=100^{2}$

$\Rightarrow h^{2}(16)=100^{2}$

$\Rightarrow h=25$

A vertical pole more than

$100ft$ high consists of two portions, the lower being

$1/3$ of the whole. lf the upper portion subtends an angle

$\tan^{-1}(1/2)$ at a point distant 40 ft. from the foot of the pole, the height of the pole is

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$105$
0%
$120$
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$135$
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$150$

Explanation

$tan=1/2$

$tan\alpha =\frac{h}{3+40}$

$tan(\alpha +\theta )=\frac{h}{40}$

$\dfrac{tan\alpha }{3}=\dfrac{tan\alpha +tan\theta }{1-tan\alpha tan\theta }$

$\dfrac{1+2tan\alpha }{2-tan\alpha }=\dfrac{tan\alpha }{3}$

$3+6tan\alpha =2tan\alpha -tan\alpha$

$tan^{2}\alpha +4tan\alpha +3 =0$

$tan\alpha=-1,tan\alpha=-3$

$1=\frac{h}{120}$

$h=120$

$\theta\equiv sin^{-1}x+cos^{-1}x-tan^{-1}x,\ 0\leq x\leq 1$ , then the smallest interval in which

$\theta$ lies is given by ?

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$\displaystyle \frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}$
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$-\displaystyle \frac{\pi}{4}\leq\theta\leq 0$
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$0\displaystyle \leq\theta\leq\frac{\pi}{4}$
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$\displaystyle \frac{\pi}{2}\leq\theta\leq\frac{3\pi}{4}$

Explanation

Given

$0\le x\le 1\implies \tan^{-1} x\in [\tan^{-1}0,\tan^{-1}1]\implies \tan^{-1} x\in [0,\frac{\pi}{4}]\implies -\tan^{-1} x\in [-\frac{\pi}{4},0]$

$\theta=\sin^{-1} x+\cos^{-1} x-\tan^{-1} x=\dfrac{\pi}{2} -\tan^{-1} x$

$(\because \sin^{-1} x+\cos^{-1} x=\dfrac{\pi}{2})$

$-\tan^{-1} x\in [-\frac{\pi}{4},0]$

$\implies \dfrac{\pi}{2}-\tan^{-1} x\in [\frac{\pi}{4},\frac{\pi}{2}]$

$\implies \dfrac{\pi}{4}\le\theta\le \dfrac{\pi}{2}$

The value of sin

$\sin \left\{ {{{\tan }^{ - 1}}\left( {\tan \frac{{7\pi }}{6}} \right) + {{\cos }^{ - 1}}\left( {\cos \frac{{7\pi }}{3}} \right)} \right\}$ is

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0
0%
1
0%
-1
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None of these

Explanation

$Sin \left \{ \underbrace{tan^-1 \left (tan \frac{7\pi}{6} \right )} + \underbrace{cos^{-1}\left ( cos \frac{7\pi}{3} \right )} \right \}$

$A$

$B$

$A=tan^{-1} \left ( tan \left ( \pi + \frac{\pi}{6} \right ) \right )$

$=tan^{-1} \left ( tan \frac{\pi}{6} \right ) =\frac{\pi}{6}$

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

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

$\therefore sin \left ( \frac{\pi}{6}+ \frac{\pi}{3} \right ) = sin \left ( \frac{\pi}{2} \right ) =1$

$\alpha \epsilon \left ( 0,\dfrac{\pi}{2}\right )$ , then the value of

$\tan ^{-1}(\cot \alpha )-\cot ^{-1}(\tan \alpha )+\sin ^{-1}(\sin \alpha )-\cos ^{-1}(\cos \alpha )$ is equal to

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$2\alpha$
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$\pi +\alpha$
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$0$
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$\pi -2\alpha$

Explanation

$tan^{ -1 }\left( cot\alpha \right) -cot^{ -1 }\left( tan\alpha \right) +sin^{ -1 }\left( sin\alpha \right) -cos^{ -1 }\left( cos\alpha \right) \\ =tan^{ -1 }\left( tan\left( \cfrac { \pi }{ 2 } -\alpha \right) \right) -cot^{ -1 }\left( cot\left( \cfrac { \pi }{ 2 } -\alpha \right) \right) +sin^{ -1 }sin\alpha -cos^{ -1 }cos\alpha \\ =\left( \cfrac { \pi }{ 2 } -\alpha \right) -\left( \cfrac { \pi }{ 2 } -\alpha \right) +\alpha -\alpha =0$

$\tan ^{-1}\dfrac{\sqrt{1+x^{2}}-1}{x}=4^{\circ}$ ,then

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$x=\tan 2^{\circ}$
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$x=\tan 4^{\circ}$
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$x=\tan (1/4)^{\circ}$
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$x=\tan 8^{\circ}$

Explanation

Substituting

$x=tanA$

$\tan^{-1}(\dfrac{\sec A-1}{\tan A})$

$=\tan^{-1}(\dfrac{1-\cos A}{\sin A})$

$=\tan^{-1}(\tan(\dfrac{A}{2}))$ ............Using

$\cos A=1-2\sin^2\dfrac A2$

$4^{0}$

$=\dfrac{A}{2}$

$\Rightarrow A=8^{0}$

$\tan^{-1}(x)=8^{0}$

$x=\tan(8^{0})$

${\cot ^{ - 1}}x + {\cot ^{ - 1}}y + {\cot ^{ - 1}}z = \dfrac{\pi }{2}$ then

$x+y+z$ equals

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$xyz$
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$xy+yz+zx$
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$2xyz$
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None of these

Explanation

$\cot^{-1}x+\cot^{-1}y=\dfrac{\pi}{2} -\cot^{-1}z$

$\tan(\cot^{-1}x+\cot^{-1}y)=\tan \left ( \dfrac{\pi}{2} - \cot^{-1}z \right )$

$\displaystyle \dfrac{\tan(\cot^{-1}x)+\tan (\cot^{-1}y)}{1- \tan (\cot^{-1}x)\tan (\cot^{-1}y)}=z$

$\dfrac{\dfrac{1}{x} + \dfrac{1}{y}}{1-\dfrac{1}{xy}}=z$

$\dfrac{x+y}{xy-1}=z$

$x+y+z=xyz$

$f(x)=tan^{-1}(sin x+ cos x)$ is an increasing function in

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$\displaystyle (0, \frac{\pi}{4})$
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$\displaystyle (0, \frac{\pi}{2})$
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$\displaystyle (\frac{-\pi}{4}, \frac{\pi}{4})$
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None of these

Explanation

$f(x)=tan^{-1} (sin x + cos x)$

$f^1(x) \displaystyle =\frac{1}{1 +(sin x + cos x)^2}$

$(cos x -sin x)$
always positive g(x)

$g(x) =cos x -sin x$

$g(x) =0 \Rightarrow tan x=1$

$\therefore x=\frac{\pi}{4}$

$\therefore \left [ \frac{-3\pi}{4}, \frac{\pi}{4} \right ]$ increasing function
So, answers are

$\left ( 0, \frac{\pi}{4} \right ), \left ( \frac{-\pi}{4},\frac{\pi}{4} \right )$

$sin\left\{ \sin ^{ -1 }{ \cfrac { 1 }{ 5 } } +\cos ^{ -1 }{ x } \right\} =1$ , then

$x$ is equal to

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$1$
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$0$
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$\displaystyle\frac{ 4 }{ 5 }$
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$\displaystyle\frac{ 1 }{ 5 }$

Explanation

Given,

$sin \left(sin^{-1} \left(\displaystyle\frac{1}{5} \right)+cos^{-1}(x) \right)=1$

$sin^{-1} \left(\displaystyle\frac{1}{5} \right)+cos^{-1}(x)=sin^{-1}(1)$

$sin^{-1} \left(\displaystyle\frac{1}{5} \right)+cos^{-1}(x)=\dfrac{\pi}{2}$

$cos^{-1}(x)=\displaystyle\frac{\pi}{2}-sin^{-1} \left(\displaystyle\frac{1}{5} \right)$

$cos^{-1}(x)=cos^{-1} \left(\displaystyle\frac{1}{5} \right)$

$\therefore x=\displaystyle\frac{1}{5}$

$0\le x\le 1$ , then

$\tan { \left\{ \cfrac { 1 }{ 2 } \sin ^{ -1 }{ \cfrac { 2x }{ 1+{ x }^{ 2 } } +\cfrac { 1 }{ 2 } \cos ^{ -1 }{ \cfrac { 2x }{ 1+{ x }^{ 2 } } } } \right\} }$

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1
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$0$
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$\cfrac { 2x }{ 1+{ x }^{ 2 } }$
0%
$x$

Explanation

$tan[\frac{1}{2}(sin^{-1}(\frac{2x}{1+x^2})+cos^{-1}(\frac{2x}{1+x^2}))]$

$=tan[\frac{1}{2}(sin^{-1}(\theta)+cos^{-1}(\theta))]$

$=tan(\frac{\pi}{2(2)})$

$=tan(\frac{\pi}{4})$

$=1$

$\tan { \left( \cot ^{ -1 }{ x } \right) }$ is equal to

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$\cfrac { \pi }{ 2 } -x$
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$\cot { \left( \tan ^{ -1 }{ x } \right) }$
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$\tan { x }$
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none of these

Explanation

Given,

$\tan(cot^{-1}(x))$

$=\tan(\tan^{-1}(\dfrac{1}{x}))$

$=\dfrac{1}{x}$ .
Also

$\tan(\cot^{-1}(x))$

$=\tan(\dfrac{\pi}{2}-\tan^{-1}(x))$

$=\cot(\tan^{-1}(x))$

$\displaystyle x$ takes negative permissible value, then

$\displaystyle \sin ^{-1}x$ is equal to

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$\displaystyle \cos ^{-1}\sqrt{1-x^{2}}$
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$\displaystyle -\cos ^{-1}\sqrt{1-x^{2}}$
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$\displaystyle \cos ^{-1}\sqrt{x^{2}-1}$
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$\displaystyle \pi -\cos ^{-1}\sqrt{1-x^{2}}$

Explanation

$sin^{-1}(x)$

$=cos^{-1}(\sqrt{1-x^2})$ , for

$x>0$
Since x takes negative permissible value.

$sin^{-1}(x)=-cos^{-1}(\sqrt{1-x^2})$

$\sec ^{ 2 }{ \left( \tan ^{ -1 }{ 2 } \right) +cosec ^{ 2 }{ \left( \cot ^{ -1 }{ 3 } \right) } }$ =

0%
5
0%
10
0%
15
0%
20

Explanation

The given question can be re-written as

$1+\tan^{2}(\tan^{-1}(2))+1+\cot^{2}(\cot^{-3})$

$=2+2^{2}+3^{2}$

$=2+4+9$

$=15$

The range of the function,

$\displaystyle f\left ( x \right ) = \left ( 1 + \sec^{-1} x \right ) \left ( 1 + \cos^{-1} x \right )$ is

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$\displaystyle \left ( -\infty ,\: \infty \right )$
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$\displaystyle \left (-\infty, \: 0 \right ] \cup \left [4, \: \infty \right )$
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$\displaystyle \left \{ 0, \: \left ( 1 + \pi \right )^{2} \right \}$
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$\displaystyle \left [ 1, \: \left ( 1 + \pi \right )^{2} \right ]$

Explanation

$f(x)=(1+sec^{-1}(x))(1+cos^{-1}(x))$
Here the limiting component is

$cos^{-1}(x)$ , since the domain of

$cos^{-1}(x)$ is

$[-1,1]$ .
Therefore,

$f(1)=(1+0)(1+0)$

$=1$

$f(-1)=(1+\pi)(1+\pi)$

$=(1+\pi)^{2}$
Hence range of

$f(x)=[1,(1+\pi)^{2}]$

Number of real value of

$x$ satisfying the equation,

$\displaystyle \arctan \sqrt{x \left ( x+1 \right )} + \arcsin \sqrt{x \left ( x+1 \right ) + 1} = \frac{\pi}{2}$ is

0%
0
0%
1
0%
2
0%
more than 2

Explanation

Let

$f(x)=tan^{-1}(\sqrt{x(x+1)})+sin^{-1}(\sqrt{x(x+1)+1})-\dfrac{\pi}{2}$
Hence

$f(x)=0$ for x={-1,0}.
Where {} denotes singleton set.
Hence we have two solutions for the above given equation.

Range of

$\displaystyle f\left( x \right)=\sin ^{ -1 }{ x } +\tan ^{ -1 }{ x } +\cos ^{ -1 }{ x }$ is

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$\displaystyle \left[ 0,\pi \right]$
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$\displaystyle \left[ \frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } \right]$
0%
$\displaystyle \left[ -\pi ,2\pi \right]$
0%
None of these

Explanation

Given

$f(x)=\sin^{−1}(x)+\cos^{−1}(x)+\tan^{−1}(x)$

First we will calculate domain of function

$f(x)$

function

$\sin^{−1}(x)$ is defined in

$x∈[−1,1]$

Similarly function

$\cos^{−1}(x)$ is defined in

$x∈[−1,1]$

and function

$\tan^{−1}(x)$ is defined in

$x∈(−∞,+∞)$

$f(x)=\sin^{−1}(x)+\cos^{−1}(x)+\tan^{−1}(x)$ is defined in

$x∈[−1,1]$

$f(x)=\sin^{−1}(x)+\cos^{−1}(x)+\tan^{−1}(x)$ is defined in

$x∈[−1,1]$

$f(x)=\dfrac π2+\tan^{−1}(x)$

Now

$f′(x)=\dfrac {1}{1+x^2}>0 \, ∀x∈[−1,1]$

$f(x)$ is Strictly Increasing function.

$f(−1)=\dfrac π2+\tan^{−1}(−1)=\dfrac π2−\dfrac π4=\dfrac π4$

and

$f(+1)=\dfrac π2+\tan^{−1}(1)=\dfrac π2+\dfrac π4=\dfrac {3π}4$

$f(x)∈[\dfrac π4,\dfrac {3π}4]$

$\displaystyle f\left ( x \right ) = \sin^{-1}x + \sec^{-1} x$ is defined, then which of the following value/values is/are in its range?

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$\displaystyle \dfrac{- \pi}{ 2}$
0%
$\displaystyle \dfrac{\pi}{ 2}$
0%
$\displaystyle \pi$
0%
$\displaystyle \dfrac{3\pi}{ 2}$

Explanation

The domain for

$\sec^{-1}(x)$ is

$(-\infty,-1]\cup[1,\infty)$

The domain for

$\sin^{-1}(x)$ is

$[-1,1]$

Hence the domain for

$f(x)=\sin^{-1}(x)+\sec^{-1}(x)$ will be

$-1$ and

$1$ .

Therefore the range of values of

$f(x)$ will be

$f(-1)=\dfrac{-\pi}{2}+\pi$

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

$f(1)=\dfrac{\pi}{2}+0$

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

Hence the range of

$f(x)$ includes a single element that is

$\dfrac{\pi}{2}$

$\displaystyle \alpha = \sin^{-1} \left ( \cos \left ( \sin^{-1} x \right ) \right )$ and

$\displaystyle \beta = \cos^{-1} \left ( \sin \left ( \cos^{-1} x \right ) \right )$ then:

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$\displaystyle \tan \alpha = \cot \beta$
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$\displaystyle \tan \alpha = - \cot \beta$
0%
$\displaystyle \tan \alpha = \tan \beta$
0%
$\displaystyle \tan \alpha = - \tan \beta$

Explanation

$\alpha=sin^{-1}(cos(sin^{-1}(x))))$

$=\dfrac{\pi}{2}-cos^{-1}(cos(\dfrac{\pi}{2}-cos^{-1}(x)))$

$=\dfrac{\pi}{2}-cos^{-1}(sin(cos^{-1}(x)))$

$=\dfrac{\pi}{2}-\beta$
Hence

$\alpha=\dfrac{\pi}{2}-\beta$
Hence

$tan(\alpha)=cot(\beta)$

The value of

$\displaystyle sin^{-1}(cos(cos^{-1}(cos\:x)+sin^{-1}(sin\:x)))$ , where

$\displaystyle x\:\epsilon\:\left ( \frac{\pi}{2},\pi \right )$ , is equal to

0%
$\displaystyle \frac{\pi}{2}$
0%
$-\pi$
0%
$\pi$
0%
$\displaystyle -\frac{\pi}{2}$

Explanation

$sin^{-1}[coscos^{-1}(cos(x)+sin^{-1}(sin(x))]$

$=sin^{-1}[cos(x+\pi-x)]$

$=sin^{-1}[cos(\pi)]]$

$=sin^{-1}(-1)$

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

There exists a positive real number

$x$ satisfying

$\displaystyle \cos \left ( \tan^{-1} x \right ) = x$ . The value of

$\displaystyle \cos^{-1} \left ( \frac{x^{2}}{2} \right )$ is

0%
$\displaystyle \frac{\pi}{10}$
0%
$\displaystyle \frac{\pi}{5}$
0%
$\displaystyle \frac{2 \pi}{5}$
0%
$\displaystyle \frac{4 \pi}{5}$

Explanation

$\cos(\tan^{-1}(x))=x$

$\cos(\cos^{-1}(\frac{1}{\sqrt{1+x^{2}}}))=x$
Therefore

$\frac{1}{\sqrt{1+x^{2}}}=x$
Squaring on both the sides we get

$x^2(x^2+1)=1$

$x^4+x^2-1=0$
Hence

$x^2=\frac{-1+\sqrt{5}}{2}$

$\frac{x^2}{2}=\frac{\sqrt{5}-1}{4}$

$\cos^{-1}(\frac{x^2}{2})$

$=\cos^{-1}(\frac{\sqrt{5}-1}{4})$

$=72^{0}$

$=\frac{2\pi}{5}$

$\displaystyle \alpha , \: \beta \left ( \alpha \: < \: \beta \right )$ are the roots of the equation

$\displaystyle 6x^{2} + 11x + 3 = 0$ then which of the following are real?

0%
$\displaystyle \cos^{-1} \alpha$
0%
$\displaystyle \sin^{-1} \beta$
0%
$\displaystyle cosec^{-1} \alpha$
0%
Both $\displaystyle \cot^{-1} \alpha$ and $\displaystyle \cot^{-1} \beta$

Explanation

Given,

$6x^{2}+11x+3=0$

$6x^{2}+9x+2x+3=0$

$3x(2x+3)+1(2x+3)=0$

$(3x+1)(2x+3)=0$

$x=\dfrac{-1}{3}=\beta$ and

$x=\dfrac{-3}{2}=\alpha$

Hence

$\sin^{-1}(\beta)$ and

$\cos^{-1}(\beta)$ is real.

$cosec^{-1}(\alpha)$ is real.

$\tan^{-1}(\alpha)$

$\tan^{-1}(\beta)$

$\cot^{-1}(\alpha)$ and

$\cot^{-1}(\beta)$ are real.

The value of

$\sin ^{ -1 }{ \left( \sin { 10 } \right) }$ is

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$10$
0%
$10-3\pi$
0%
$3\pi -10$
0%
none of these

Explanation

Here

$\theta =10$ rad doesnt lie between

$-\dfrac \pi 2$ and

$\dfrac \pi 2$

But ,

$3\pi -\theta$ lies between

$-\dfrac \pi 2$ and

$\dfrac \pi 2$

Also ,

$\sin (3\pi -10 )=\sin 10$

$\implies \sin ^ {-1} (\sin 10) =\sin ^ {-1} (\sin (3\pi -10))=3\pi -10$

Which one of the following statement is meaningless?

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$\displaystyle \cos^{-1} \left ( ln \left ( \frac{2e + 4}{3} \right ) \right )$
0%
$\displaystyle cosec^{-1} \left ( \frac{\pi}{3} \right )$
0%
$\displaystyle \cot^{-1} \left ( \frac{\pi}{2} \right )$
0%
$\displaystyle \sec^{-1} \left ( \pi \right )$

Explanation

The domain for

$f(x)=cos^{-1}(x)$ is

$[-1,1]$

However, clearly

$ln \left(\dfrac{2e+4}{3}\right)>1$

Since

$f(x)$ for

$|x|>1$ is not possible.

Option A is meaningless.

The value of

$\displaystyle tan(sin^{-1}(cos(sin^{-1}x)))tan(cos^{-1}(sin(cos^{-1}x)))$ , where

$x\:\epsilon\:(0,1)$ , is equal to

0%
$0$
0%
$1$
0%
$-1$
0%
$none\:of\:these$

Explanation

$cos sin^{-1} x = \sqrt{1-x^2}$

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

$sin (cos^{-1}x ) = \sqrt{1-x^2} )$

$tan cos^{-1} (\sqrt{1-x^2}) = \dfrac{x}{\sqrt{1-x^2}}$

Hence, option B is correct.

$\displaystyle sec^{2}(tan^{-1}2)+cosec^{2}(cot^{-1}3)$ is equal to

0%
$5$
0%
$13$
0%
$15$
0%
$6$

Explanation

Given,

$\sec^{2}(\tan^{-1}(2))+cosec^{2}(\cot^{-1}(3))$

$=2+\tan^{2}(\tan^{-1}(2))+\cot^{2}(\cot^{-1}(3))$

$=2+2^{2}+3^{2}$

$=2+4+9$

$=15$

Which of the following is the solution set of the equation

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

0%
$(0,1)$
0%
$(-1,1)-\left\{0\right\}$
0%
$(-1,0)$
0%
$[-1,1]$

Explanation

$2{ cos }^{ -1 }x={ cos }^{ -1 }\left( { 2 }x^{ 2 }-1 \right)$ for

$x\epsilon \left[ 0,1 \right]$
And

${ cos }^{ -1 }\left( { 2 }x^{ 2 }-1 \right) ={ cot }^{ -1 }\left( \cfrac { { 2 }x^{ 2 }-1 }{ 2x\sqrt { 1-{ x }^{ 2 } } } \right)$ for

$x\epsilon \left( 0,1 \right)$

The value of

$\displaystyle sin^{-1}\left[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^{2}}\right]$ is equal to

0%
$sin^{-1}x+sin^{-1}\sqrt{x}$
0%
$sin^{-1}x-sin^{-1}\sqrt{x}$
0%
$sin^{-1}\sqrt {x}-sin^{-1}x$
0%
$0$

Explanation

$sin^{-1}(x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^{2}})$

$=sin^{-1}(x)-sin^{-1}(\sqrt{x})$ ...( by applying the formula of

$sin^{-1}(x)-sin^{-1}(y)$ )
Hence answer is Option B

The number of integer

$x$ satisfying

$\displaystyle sin^{-1}|x-2|+cos^{-1}(1-|3-x|)=\frac{\pi}{2}$ is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$x<2$

$sin^{-1}(2-x)+cos^{-1}(1-(3-x))$

$=sin^{-1}(2-x)+cos^{-1}(-2+x)$

$=sin^{-1}(2-x)+\pi-cos^{-1}(2-x)$

$=\frac{\pi}{2}$

$-\frac{pi}{2}+sin^{-1}(2-x)+\pi+sin^{-1}(2-x)=\frac{\pi}{2}$

$2sin^{-1}(2-x)=0$

$x=2$ ...(i)
For

$2<x<3$

$sin^{-1}(x-2)+cos^{-1}(-2+x)$

$=sin^{-1}(x-2)+cos^{-1}(x-2)$

$=\frac{\pi}{2}$
For

$x>3$

$sin^{-1}(x-2)+cos^{-1}(4-x)=\frac{\pi}{2}$

$x-2=4-x$

$2x=6$

$x=3$
Hence there are in total 2, solutions.

$x_{ 1 }=2\: tan^{ -1 }\left( \frac { 1+x }{ 1-x } \right) ,x_{ 2 }=sin^{ -1 }\left( \frac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right)$ , where

$x_1,x_2\:\epsilon\:(0,1)$ , then

$2\left( x_{ 1 }+x_{ 2 } \right)$ is equal to

0%
$0$
0%
$2\pi$
0%
$\pi$
0%
$none\:of\:these$

Explanation

${ x }_{ 1 }=2\tan^{ -1 }\left( \cfrac { 1+x }{ 1-x } \right) =\cos^{ -1 }\left( \cfrac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) \\ { x }_{ 2 }=\sin^{ -1 }\left( \cfrac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right)$
Therefore,

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

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

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$\displaystyle f\left ( x \right )$ has the least value of $\displaystyle \frac{\pi^{2}}{8}$
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$\displaystyle f\left ( x \right )$ has the greatest value of $\displaystyle \frac{5 \pi^{2}}{8}$
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$\displaystyle f\left ( x \right )$ has the least value of $\displaystyle \frac{\pi^{2}}{16}$
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$\displaystyle f\left ( x \right )$ has the greatest value of $\displaystyle \frac{5 \pi^{2}}{4}$

Explanation

The least value for the above function occurs at

$x=\dfrac{1}{\sqrt{2}}$

$\Rightarrow$

$=\dfrac{\pi^{2}}{16}+\dfrac{\pi^{2}}{16}$

$=\dfrac{\pi^{2}}{8}$ ...(i)
The greatest value for the above function occurs at

$x={-1}$

$\Rightarrow$

$=\dfrac{\pi^{2}}{4}+\pi^{2}$

$=\dfrac{5\pi^{2}}{4}$ ...(ii)

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Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1
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Both the statements are TRUE and STATEMENT 2 is NOT the correct explanation of STATEMENT 1
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STATEMENT 1 is TRUE and STATEMENT 2 is FALSE
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STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

Explanation

For

$f(x)=\tan^{-1}(x)+\sin^{-1}(x)+\cos^{-1}(x)$

The domain is

$[-1,1]$

In this domain we can re-write the expression as

$f(x)=\dfrac{\pi}{2}+\tan^{-1}(x)$

$f(-1)=\dfrac{\pi}{2}-\dfrac{\pi}{4}$

$=\dfrac{\pi}{4}$

$f(1)=\dfrac{\pi}{2}+\dfrac{\pi}{4}$

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

Therefore range of

$f(x)$ is

$[\dfrac{\pi}{4},\dfrac{3\pi}{4}]$

$\displaystyle \tan^{-1}\left [ \cfrac{\cos\:x}{1+\sin\:x} \right ]$ is equal to

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$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(-\frac{\pi}{2},\frac{3\pi}{2}\right)$
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$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
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$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(\frac{3\pi}{2},\frac{5\pi}{2}\right)$
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$\displaystyle \frac{\pi}{4}-\frac{x}{2},for\:x\:\epsilon\:\left(-\frac{3\pi}{2},-\frac{3\pi}{2}\right)$

Explanation

$\displaystyle \tan^{-1}\left[\frac{\sin(\frac{\pi}{2}-x)}{1+\cos(\frac{\pi}{2}-x)}\right]$

$\displaystyle = \tan^{-1}\left[\cfrac{2\sin(\frac{\pi}{4}-\frac{x}{2}).\cos(\frac{\pi}{4}-\frac{x}{2})}{2\cos^{2}(\frac{\pi}{4}-\frac{x}{2})}\right]$

$\displaystyle =\tan^{-1}\left[\tan(\frac{\pi}{4}-\frac{x}{2})\right]$

$\displaystyle =\cfrac{\pi}{4}-\frac{x}{2}$
where,

$\displaystyle -\frac { \pi }{ 2 } < \frac { \pi }{ 4 } -\frac { x }{ 2 } < \frac { \pi }{ 2 } \\ \displaystyle \Rightarrow -\frac { \pi }{ 2 } < x < \frac { 3\pi }{ 2 }$

The value of

$\displaystyle sin^{-1}(x^{2}-4x+6)+cos^{-1}(x^{2}-4x+6)$ for all

$x\:\epsilon\:R$ is

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$\displaystyle \frac{\pi}{2}$
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$\pi$
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$0$
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none of these

Explanation

$\sin^{-1}(x)\epsilon[-1,1]$

Therefore

$-1\leq x^{2}-4x+6\leq 1$

$-1\leq (x-2)^{2}-4+6\leq 1$

$-1\leq (x-2)^{2}+2\leq 1$

$-3\leq (x-2)^{2}\leq -1$

Now,

$(x-2)^{2}>0$ for all

$x\epsilon R$ .

Hence the inequality

$-3\leq (x-2)^{2}\leq -1$ yields no real real values of x.

Therefore,

$|x^{2}-4x+6|$ cannot be less than or equal to 1.

Hence

$\sin^{-1}(x^{2}-4x+6)$ is no possible.

The, same goes for

$\cos x$ .

Hence the answer is none of these.

Number of solutions of the equation

$\sin \left ( \displaystyle \frac{1}{3}\cos^{-1}x \right )=1$ are

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only one
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no solution
0%
only three
0%
at least two

Explanation

$\displaystyle \sin \left ( \frac{1}{3}\cos ^{-1}x \right )=1$

$\displaystyle \Rightarrow \frac { 1 }{ 3 } \cos ^{ -1 } x=\frac { \pi }{ 2 }$

$\displaystyle \Rightarrow \cos ^{-1}x=\frac{3\pi }{2}$

$\displaystyle \Rightarrow \exists$ no solution, as

$\displaystyle x \epsilon \left [ 0,\pi \right ]$

The value of

$\displaystyle \sin ^{-1}\left ( \sin 2 \right )$ is?

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$2+n\pi$
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$2-\pi$
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$-2+\pi$
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$2-2n\pi$

Explanation

To find value of

$\sin^{-1}(\sin(2))$

$\pi \space rads=180^{0}$

$1 \space rad=\dfrac{180^{0}}{\pi}$

$1 \space rad=57.29^{0}$
Hence

$2\space rads=114.59^{0}$
Hence it is in the second quadrant.
Therefore

$\sin^{-1}(\sin(2))$

$=\sin^{-1}(\sin(\pi-2))$

$=\pi-2$

The equation

$\displaystyle 3cos^{-1}x-\pi\:x-\frac{\pi}{2}=0$ has

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one negative solution
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one positive solution
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no solution
0%
more than one solution

Explanation

Substituting

$x=\frac{1}{2}$ , we get

$3cos^{-1}(\frac{1}{2})-\frac{\pi}{2}-\frac{\pi}{2}$

$=3(\frac{\pi}{3})-\pi$

$=0$
Hence there is one positive real solution to the above equation since range of

$cos^{-1}(x)\epsilon[\pi,0]$ .

$\sin^{-1}\left ( x^{2}-7x+12 \right )=m\pi ,\:\forall \:n\:\:\in \:I$ , then

$x =$

0%
-4
0%
4
0%
3
0%
-3

Explanation

${ \sin }^{ -1 }\left( { x }^{ 2 }-7x+12 \right) =m\pi$

$\Rightarrow { x }^{ 2 }-7x+12=\sin\left( m\pi \right)$

$\Rightarrow { x }^{ 2 }-7x+12=0$

$\Rightarrow \left( x-3 \right) \left( x-4 \right) =0$

$x=3,4$

The value of

$\tan^{-1}\left (\displaystyle \frac{1}{2}\tan 2A \right )+\tan^{-1}\left ( \cot A\right )+\tan^{-1}\left ( \cot ^{3}A\right )$ , for

$0< A< \pi /4$ , is :

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$\tan^{-1}2$
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$\tan^{-1}\left ( \cot A \right )$
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$4\:tan^{-1}\:\left ( 1 \right )$
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$2\tan^{-1}\:\left ( 2 \right )$

Explanation

Given, $\tan^{-1}\left (\displaystyle \frac{1}{2}\tan 2A \right )+\tan^{-1}\left ( \cot A\right )+\tan^{-1}\left ( \cot ^{3}A\right )$

$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)+\tan^{-1}\left(\dfrac{\cot A+\cot^{3}A}{1-\cot^{4}A}\right)$

$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)+\tan^{-1}\left(\dfrac{\cot A}{1-\cot^{2}A}\right)$

$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)+\tan^{-1}\left(\dfrac{\tan A}{\tan^{2}-1}\right)$

$=\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)-\tan^{-1}\left(\dfrac{\tan A}{1-\tan^{2}A}\right)$

$=0$

$=\pi$

$=4\left(\dfrac{\pi}{4}\right)$

$=4\tan^{-1}\left(1\right)$

There exists a positive real number

$x$ satisfying

$\displaystyle \cos(\tan^{-1}x)=x$ . Then the value of

$\displaystyle \cos^{-1}\left(\frac{x^{2}}{2}\right)$ is

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$\displaystyle \frac{\pi}{10}$
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$\displaystyle \frac{\pi}{5}$
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$\displaystyle \frac{2\pi}{5}$
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$\displaystyle \frac{4\pi}{5}$

Explanation

$\displaystyle \cos(\tan^{-1}x)=x\Rightarrow \tan^{-1}x = \cos^{-1}x$

$\displaystyle \Rightarrow \cos^{-1}\frac{1}{\sqrt{1+x^2}} = \cos^{-1}x\Rightarrow x^2(1+x^2) = 1$

$\Rightarrow x^2 = \cfrac{-1+\sqrt{5}}{2}\Rightarrow \cfrac{x^2}{2} = \cfrac{\sqrt{5}-1}{4} = \cos\cfrac{2\pi}{5}$

$\therefore \cos\left(\cfrac{x^2}{2}\right) = \cfrac{2\pi}{5}$

The number of real solution of the equation

$\displaystyle \sqrt{1+cos2x}=\sqrt{2}sin^{-1}(sin\:x),-\pi< x\leq \pi$ , is

0%
$0$
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$1$
0%
$2$
0%
$infinite$

Explanation

Given equation is,

$\\ \sqrt { 1+\cos 2x } =\sqrt { 2 } sin^{ -1 }\left( \sin\: x \right) \\ \Rightarrow \sqrt { 1+{ \cos }^{ 2 }x-{\sin }^{ 2 }x } =\sqrt { 2 } x\\ \Rightarrow \sqrt { 2{ \cos }^{ 2 }x } =\sqrt { 2 } x\\ \Rightarrow \pm \sqrt { 2 } \cos x=\sqrt { 2 } x\\ \Rightarrow \cos x=\pm 1\\ x=0,\pi$

The equation

$\sin ^{-1}x=2\sin ^{-1}a$ , has a solution for

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$\forall \:R$
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$\displaystyle \left | a \right |< \frac{1}{2}$
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$\displaystyle \left | a \right |\geq \frac{-1}{2}$
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$\displaystyle \frac{-1}{\sqrt{2}}\leq a\leq \frac{1}{\sqrt{2}}$

Explanation

Given equation

$\sin ^{-1}x= 2\sin ^{-1}a$

$\displaystyle \Rightarrow -\frac{\pi}{2}\leq 2\sin ^{-1}a\leq \frac{\pi}{2}\left \{ \because \sin ^{-1}x= 2\sin ^{-1}a \:and -\frac{\pi}{2}\leq 2\sin ^{-1}x\leq \frac{\pi}{2} \right\}$

$\displaystyle \Rightarrow -\frac{\pi}{4}\leq \sin ^{-1}a\leq \frac{\pi}{4}$

$\displaystyle \Rightarrow \sin \left ( -\frac{\pi}{4}\right )\leq a\leq \sin \left ( \frac{\pi}{4} \right )$

$\displaystyle \Rightarrow -\frac{1}{\sqrt{2}}\leq a\leq \frac{1}{\sqrt{2}}$

$\displaystyle \sin ^{-1}x-\sin ^{-1}y= \frac{\pi }{2}$ ,then

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$x^{2}+y^{2}= 1$
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$y= -\sqrt{1-x^{2}}, \:0\leq x\leq 1,-1\leq y\leq 0$
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$y= \sqrt{1-x^{2}}, \: \left | x \right |< 1$
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None of these

Explanation

$\sin ^{ -1 }{ x } -\sin ^{ -1 }{ y } =\frac { \pi }{ 2 } \\ \Rightarrow \sin ^{ -1 }{ y } =-\left( \frac { \pi }{ 2 } -\sin ^{ -1 }{ x } \right) \\ \Rightarrow \sin { \sin ^{ -1 }{ y } } =-\sin { \left( \frac { \pi }{ 2 } -\sin ^{ -1 }{ x } \right) } \\ \Rightarrow y=-\cos { \sin ^{ -1 }{ x } } =-\cos { \cos ^{ -1 }{ \sqrt { 1-{ x }^{ 2 } } } }$

$for\quad 0\le x\le 1$

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

$\Rightarrow -1\le y\le 0$

Ans: B

Let

$f(x) =e^{\displaystyle\cos^{-1}\sin \left ( x+\displaystyle\frac{\pi }{3} \right )}$ , then

$f\left (\displaystyle \frac{8\pi }{9} \right )$ equals

0%
$e^{\displaystyle\frac{7\pi }{12}}$
0%
$e^{\displaystyle\frac{13\pi }{18}}$
0%
$e^{\displaystyle\frac{5\pi }{18}}$
0%
$e^{\displaystyle\frac{\pi }{12}}$

Explanation

$f(x)=e^{\cos ^{-1}\sin\left ( x+\displaystyle \frac{\pi }{3} \right )}$

$\therefore f\left (\displaystyle \frac{8\pi }{9} \right )$

$f(x)=e^{\cos ^{-1}\sin\left ( \dfrac{\pi}{3}+\displaystyle \frac{8\pi }{9} \right )}$

$=e^{\cos ^{-1}\sin\left ( \displaystyle \frac{11\pi }{9} \right )}$

$=e^{\cos ^{-1}\sin\left ( \displaystyle \frac{22\pi }{18} \right )}$

$=e^{\cos ^{-1}\sin\left ( \displaystyle \frac{9\pi }{18}+\displaystyle \frac{13\pi }{18} \right )}$

$=e^{\cos ^{-1}\sin\left (\displaystyle \frac{\pi }{2}+\displaystyle \frac{13\pi }{18} \right )}$

$=e^{\cos ^{-1}\cos\left (\displaystyle \frac{13\pi }{18} \right )}$

$=e^{\displaystyle \frac{13\pi }{18}}$

$\displaystyle \cot ^{-1}\left [ \left ( \cos \alpha \right )^{1/2} \right ]+\left [ \tan ^{-1}\left ( \cos \alpha \right )^{1/2} \right ]=x$ , then

$\sin x$ equals

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$1$
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$\displaystyle \cot ^{2}\left ( \frac{\alpha }{2} \right )$
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$\tan \alpha$
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$\displaystyle \cot\left ( \frac{\alpha }{2} \right )$

Explanation

${ \tan }^{ -1 }\theta +{ \cot }^{ -1 }\theta =\cfrac { \pi }{ 2 }$

$\therefore x=\cfrac { \pi }{ 2 }$
Hence,

$\sin x=\sin\cfrac { \pi }{ 2 } =1$

$\displaystyle \sin ^{-1}\left ( a-\frac{a^{2}}{3}+\frac{a^{3}}{9}\cdots \infty \right )+\cos ^{-1}\left ( 1+b+b^{2}+b^{3}+\cdots \infty \right )= \frac{\pi }{2}$ , when

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$\displaystyle a= 1, b= -\frac{1}{3}$
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$\displaystyle a= -\frac{1}{6}, b= \frac{1}{2}$
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$\displaystyle a= \frac{1}{6}, b= \frac{1}{2}$
0%
None of these

Explanation

$a-\dfrac{a^{2}}{3}+\dfrac{a^{3}}{9}...\infty$ is an infinite G.P with a common ration of

$\dfrac{-a}{3}$ .

Hence

$S_{a}=\dfrac{a}{1+\dfrac{a}{3}}$

$=\dfrac{3a}{a+3}$ ...(i)

Similarly

$S_{b}=1+b+b^{2}...\infty$ .

$S_{b}=\dfrac{1}{1-b}$ ...(ii)

Now

$sin^{-1}(A)+cos^{-1}(B)=\dfrac{\pi}{2}$ implies

$A=B$ , then

$i=ii$

$\dfrac{3a}{a+3}=\dfrac{1}{1-b}$

$3a-3ab=a+3$

$2a-3ab=3$

$2a-3=3ab$

$b=\dfrac{2a-3}{3a}$

Substituting

$b=\dfrac{1}{2}$ , we get

$\dfrac{3a}{2}-2a=-3$

$\dfrac{a}{2}=3$

$a=6$ .

However

$|a|<1$ and

$|b|<1$ for the above infinite G.P.

Hence answer is none of the above.

The domain of

$\sin ^{-1}[x]$ , where

$[x]$ is greatest integer function, given by

0%
$\left [ -1,1 \right ]$
0%
$\left [ -1,2 \right )$
0%
$\left \{ -1,0,1 \right \}$
0%
None of these

Explanation

Domain of

$\sin^{-1}(x)$ is

$[-1,1]$

Now

$[x]$ is greatest integer function, lesser than or equal to

$x.$

Hence

$-1\leq [x]\leq 1$

Hence

$x\in[-1,2)$

$\cos^{-1}x+\cos^{-1}y+\cos^{-1}z= 3\pi$ , then value of

$\displaystyle \sum xy$ equals

0%
-3
0%
0
0%
3
0%
-1

Explanation

$0\le cos^{ -1 }\theta \le \pi$
For

$cos^{ -1 }x+cos^{ -1 }y+cos^{ -1 }z=3\pi$ this to true

$\Rightarrow cos^{ -1 }x=cos^{ -1 }y=cos^{ -1 }z=\pi \\ \Rightarrow x=y=z=-1\\ \therefore \sum { xy } =xy+yz+zx=(-1)^2+(-1)^2+(-1)^2=3$

The domain of

$f(x) =\displaystyle \frac{\sin ^{-1}x}{x}$ is

0%
$\left [ -1,1 \right ]$
0%
$\left \{ 0 \right \}$
0%
$\left [-1,0 \right )$
0%
None of these

Explanation

$f(x)=\displaystyle\frac{\sin ^{-1}x}{x}=\frac{g(x)}{t(x)},{t(x)}\neq 0$

${t(x)}\neq 0$

$\Rightarrow x\neq 0$

Domain of

$t(x)$ is

$R-\{0\}$

Again domain of

$g(x)$ is

$\left [ -1,1 \right ]$

Now using concept domain of

$f(x) =\displaystyle\frac{g(x)}{t(x)}$

domain of

$f(x) =$ Domain of

$g(x)$ - set of those points for which

$t(x)=0$ if

$g(x), \ t(x)$ is defined

$\forall x\epsilon R$

$\therefore$ Domain of

$f(x) =\left [ -1,1 \right ]-\left \{ 0 \right \}$

Number of triplets

$\left ( x, y, z \right )$ satisfying

$\sin ^{-1}x+\sin ^{-1}y+\cos ^{-1}z=2\pi$ is

0%
$1$
0%
$0$
0%
$2$
0%
$\infty$

Explanation

Let

$f(x, y, z)=\sin ^{-1}x+\sin ^{-1}y+\cos^{-1}z$
It will attain the value

$2\pi$ only if

$\sin ^{-1}x=\sin ^{-1}y=\displaystyle\frac{\pi }{2}$ and

$\cos^{-1}z=\pi$ .
This is possible only if

$x = y = 1$ and

$z=-1$ .
Hence there is only one solution

$f(1,1,-1)=2\pi$

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 2 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 2 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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