If $$0 < x_{1} < x_{2}$$ which of following is true for $$y = \sec^{-1}x$$.

$$\sec^{-1}x_{1} + \sec^{-1}x_{2} < 2\sec^{-1} \left (\dfrac {x_{1} + x_{2}}{2}\right )$$

$$\sec^{-1}x_{1} + \sec^{-1}x_{2} > \sec^{-1} \left (\dfrac {x_{1} + x_{2}}{2}\right )$$

$$\sec^{-1}x_{1} > \sec^{-1}x_{2}$$

$$\sec^{-1}x_{1} = \sec^{-1}x_{2}$$

If $$3\cos ^{ -1 }{ x } +\sin ^{ -1 }{ x } =\pi $$, then $$x=.....$$

$$\cfrac { \sqrt { 3 } }{ 2 } $$

If $$0< x < 1$$ , then $$\tan^{-1}(\cfrac{\sqrt{1-x^{2}}}{1+x})$$ is equal to

$${\cos ^{ - 1}}{{\sqrt {1 + x} } \over 2}$$

$${\sin ^{ - 1}}\sqrt {{{1 - x} \over 2}} $$

$${1 \over 2}\sqrt {{{1 + x} \over {1 - x}}} $$

MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 8

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

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

Explanation

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

$x\in R$

$\Rightarrow f(1)=2,\quad f(2)=4,\quad f(3)=8$

$\therefore \quad { x }^{ f(1) }+{ y }^{ f(2) }+{ z }^{ f(3) }=\cfrac { x+y+z }{ { x }^{ f(1) }+{ y }^{ f(2) }{ + }{ z }^{ f(3) } }$

$1+1+1-\cfrac { 1+1+1 }{ 1+1+1 } =3-1=2$

$\displaystyle \cot ^{-1}x+\cot ^{-1}y+\cot ^{-1}z=\frac{\pi }{2}, x,y,z> 0$ and

$\displaystyle xy< 1$ , then

$\displaystyle x+y+z$ is also equal to

0%
$\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
0%
$\displaystyle xyz$
0%
$\displaystyle xy+yz+zx$
0%
none of these

Explanation

$cot^{-1}(x)+cot^{-1}(y)+cot^{-1}(z)=\frac{\pi}{2}$

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

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

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

$x+y=xyz-z$

$x+y+z=xyz$

Exhaustive set of values of parameter

$a$ so that

$\sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } =a\quad$ has a solution is

0%
$\left[ -\cfrac { \pi }{ 6 } ,\cfrac { \pi }{ 6 } \right]$
0%
$\left[ -\cfrac { \pi }{ 4 } ,\cfrac { \pi }{ 4 } \right]$
0%
$\left[ -\cfrac { \pi }{ 2 } ,\cfrac { \pi }{ 2 } \right]$
0%
none of these

Explanation

Let

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

$sin^{-1}x \space and \space \tan^{-1}x$ are increasong functions between x=[-1,1].

Minimum value of

$f(x) is -\pi/4$ and maximum value is

$\pi/4$ .

Hence the correct option is answer B.

The value of

$a$ for which

$\displaystyle ax^{2}+sin^{-1}(x^{2}-2x+2)+cos^{-1}(x^{2}-2x+2)=0$ has a real solution is

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

Explanation

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

$ax^{2}+\frac{\pi}{2}=0$

$x^{2}=\frac{-\frac{\pi}{2}}{a}$

$x^{2}=\frac{-\pi}{2a}$ ...(i)
And

$-1\leq(x-1)^{2}+1\leq1$

$-2\leq(x-1)^{2}\leq0$

$(x-1)^{2}\leq0$
There is only one possibility

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

$x=1$
Substituting in 1, we get

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

The number of solution of the equation

$1+x^{2}+2x\:\sin \left ( \cos^{-1}y \right )= 0$ is :

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

Explanation

Given,

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

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

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

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

$4x^2(1-y^2)=1+x^4+2x^2$

$4x^2-4x^2y^2=1+x^4+2x^2$

$-4x^{2}(y^2)=(1-x^2)^{2}$
Hence solution will be

$x=1$ and

$y=\dfrac{1}{2}$

The set of values of parameter

$a$ so that the equation

$\displaystyle (\sin^{-1}x)^{3}+(\cos^{-1}x)^{3}=a\pi^{3}$ has a solution.

0%
$\displaystyle \left [ \frac{-1}{32},\frac{7}{8} \right ]$
0%
$\displaystyle \left [ \frac{1}{32},,\frac{9}{8} \right ]$
0%
$\displaystyle \left [ 0,\frac{7}{8} \right ]$
0%
$\displaystyle \left [ \frac{1}{32},\frac{7}{8} \right ]$

Explanation

Let

$y = (\sin^{-1}x)^{3}+(\cos^{-1}x)^{3} = (\sin^{-1}x+\cos^{-1}x)[(\sin^{-1}x)^2+(\cos^{-1}x)^2-\sin^{-1}x.\cos^{-1}x]$

$\Rightarrow y = \cfrac{\pi}{2}[(\sin^{1-}x+\cos^{-1}x)^2-3\sin^{-1}x.\cos^{-1}x]$

$\Rightarrow y = \cfrac{\pi}{2}\left[\cfrac{\pi^2}{4}-3\sin^{-1}x\left(\cfrac{\pi}{2}-\sin^{-1}x\right)\right ]$

$\Rightarrow y = \cfrac{\pi}{2}\left[\cfrac{\pi^2}{16}+3\left(\cfrac{\pi}{4}-\sin^{-1}x\right)^2\right ]$
Now we know

$-\cfrac{\pi}{2} \leq \sin^{-1}x \leq \cfrac{\pi}{2}$
Thus Range of

$y$ is

$\left[\cfrac{\pi^3}{32}, \cfrac{7\pi^3}{8}\right]$
Thus given equation to have solution

$a \in \left[\cfrac{1}{32}, \cfrac{7}{8}\right]$

$\displaystyle sin^{-1}x+sin^{-1}y+sin^{-1}z=\pi$ , then

$x^{4}+y^{4}+z^{4}+4x^{2}y^{2}z^{2}=k(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})$ , where

$k$ is equal to

0%
$1$
0%
$2$
0%
$4$
0%
$none\:of\:these$

Explanation

$sin^{-1}(x)+sin^{-1}(y)=\pi-sin^{-1}(z)$

$sin^{-1}(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})=\pi-sin^{-1}(z)$

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

$x^{2}(1-y^{2})+y^{2}(1-x^{2})+2xy\sqrt{(1-x^{2})(1-y^{2})}=z^2$

$x^{2}+y^{2}-2x^{2}y^{2}+2xy\sqrt{(1-x^{2})(1-y^{2})}=z^{2}$

$x^{2}+y^{2}-z^{2}=2x^{2}y^{2}-2xy\sqrt{(1-x^{2})(1-y^{2})}$
Squaring and simplifying, we get

$x^4+y^4+z^4+4x^2y^2z^2=2(x^2y^2+z^2y^2+x^2z^2)$

The value of p for which system has a solution is

0%
$1$
0%
$2$
0%
$0$
0%
$-1$

Explanation

$cos^{ -1 }x+\left( sin^{ -1 }y \right) ^{ 2 }=\cfrac { p\pi ^{ 2 } }{ 4 }$ ...(1)

$\left( cos^{ -1 }x \right) \left( sin^{ -1 }y \right) ^{ 2 }=\cfrac { \pi ^{ 4 } }{ 16 }$
Therefore equation with roots

$cos^{ -1 }x$ and

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

$x^{ 2 }-\cfrac { p\pi ^{ 2 } }{ 4 } x+\cfrac { \pi ^{ 4 } }{ 16 } =0$
And for real values of x

$\cfrac { p^{ 2 }\pi ^{ 4 } }{ 16 } -\cfrac { 4\pi ^{ 4 } }{ 16 } \ge 0\Rightarrow p^{ 2 }-4\ge 0$

$\Rightarrow p^{ 2 }\ge 4\Rightarrow p\ge \pm 2$

If the equation

$\sin^{-1}\left ( x^{2}+x+1 \right )+\cos^{-1}\left ( ax+1 \right )=\displaystyle\frac{\pi }{2}$ has exactly two distinct solutions then value of

$a$ could not be

0%
-1
0%
0
0%
1
0%
2

Explanation

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

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

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

$x^2+x+1-ax-1=0$

$x^2+x(1-a)=0$

$x=\frac{a-1\pm(1-a)}{2}$
Therefore

$x=0$ and

$x=(a-1)$
Also

$-1\leq ax+1\leq 1$

$\frac{-2}{a} \leq x \leq 0$
Clearly

$a$ cannot be 1 as we would get coincident values of x.
If a=-1, x=-2 then

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

$cos^{-1}(3)$ does not exists.
If a=2, x=1

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

$=sin^{-1}(3)$
Which does not exists.
Hence only

$a=0$ and

$x=-1$ satisfies the above equation.

$\sin^{-1}a +\sin^{-1}b+\sin^{-1}c= \displaystyle \frac{3\pi }{2}$ and

$f\left ( 2 \right )=2,{f\left ( x+y \right )}= f\left ( x \right )\:f\left ( y \right )\:\:\forall \:\:x,\:y\:\epsilon \:R$ then

$a^{f\left ( 2 \right )}+\:b^{f\left ( 4 \right )}+\:c^{f\left ( 6 \right )}-\:\displaystyle \frac{3\left ( a^{f\left ( 2 \right )}. \:b^{f\left ( 4 \right )}.\:c^{f\left ( 6 \right )}\right )}{a^{f\left ( 2 \right )} +\:b^{f\left ( 4 \right )}+\:c^{f\left ( 6 \right )}}$ equals

0%
2
0%
4
0%
6
0%
8

Explanation

Since

$sin^{-1}(a)+sin^{-1}(b)+sin^{-1}(c)=\dfrac{3\pi}{2}$
Hence

$a=b=c=1$ .
Now

$f(x+y)=f(x).f(y)$ implies

$f(x)=a^{x}$
Now

$f(2)=2$
Or

$a^{2}=2$
Or

$a=\sqrt{2}$ .
Hence

$f(x)=2^{\frac{x}{2}}$ .
Substituting the values of f(x) and a,b,c in the above expression, we get

$1^{2}+1^{2^{2}}+1^{2^{3}}-\dfrac{3(1^{2}.1^{2^{2}}.1^{2^{3}})}{1^{2}+1^{2^{2}}+1^{2^{3}}}$

$=3-\dfrac{3.(1)}{3}$

$=2$ .

Indicate the relation which can hold in their respective domain for infinite values of

$\displaystyle x$ .

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$\displaystyle \tan \left | \tan^{-1}x \right | = \left | x \right |$
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$\displaystyle \cot \left | \cot^{-1}x \right | = \left | x \right |$
0%
$\displaystyle \tan^{-1} \left | \tan x \right | = \left | x \right |$
0%
$\displaystyle \sin \left | \sin^{-1}x \right | = \left | x \right |$

Explanation

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

for

$x<0$

$tan(tan^{-1}(x))$ implies

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

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

$=-x$ ...(i)
for

$x>0$

$tan(tan^{-1}(x))$ implies

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

$=x$ ...(ii)

Hence

$tan(|tan^{-1}(x)|)=|x|$

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

for

$x<0$

$cot(cot^{-1}(x))$ implies

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

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

$=-x$

for

$x>0$

$cot(cot^{-1}(x))$ implies

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

$=x$ ...(ii)

Hence

$cot(|cot^{-1}(x)|)=|x|$

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

for

$x<0$

$sin(sin^{-1}(x))$ implies

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

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

$=-x$

for

$x>0$

$sin(sin^{-1}(x))$ implies

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

$=x$

Hence

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

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

For

$x\epsilon(\dfrac{\pi}{2},\pi]\cup[\dfrac{-3\pi}{2},2\pi)$

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

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

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

$=-x$

For

$x\epsilon [0,\dfrac{\pi}{2})\cup[\pi,\dfrac{3\pi}{2})$

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

$=x$

Hence

$tan^{-1}|tan(x)|=|x|$

$\displaystyle \sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3\pi }{2}$ and

$f\left ( 2 \right )= 2, f\left ( a+b \right )=f\left ( a \right )f\left ( b \right ), \:\forall \:a,\:b \:\epsilon \:R$ , then

$x^{f\left ( 2 \right )},y^{f\left ( 4 \right )},z^{f\left ( 6 \right )}$ are in

0%
A.P.
0%
G.P
0%
H.P
0%
None

Explanation

we know if,

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

$\Rightarrow x = y = z = 1$ , because maximum value of

$sin^{-1}x$ is

$\pi/2$

$\Rightarrow$

$x =1$
Now,

$f(4) = f(2+2) = f(2). f(2) = 2.2 = 4$
and

$f(6) = f(4+2) = f(4). f(2) = 4.2 = 8$
Therefore,

$x^{f(2)} = 1, y^{f(4)} = 1, z^{f(6)} = 1$
Hence, numbers are both in A.P and in G.P and H.P.

Let

$\displaystyle f:A\rightarrow B$ be a function defined by

$\displaystyle y=f(x)$ where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping

$\displaystyle g:B\rightarrow A$ such that

$\displaystyle f(x)=y$ if and only if

$\displaystyle g(y)=x\forall x \epsilon A,y \epsilon B$ Then function g is said to be inverse of f and vice versa so we write

$\displaystyle g=f^{-1}:B\rightarrow A[\left \{ f(x),x \right \}:\left \{ x,f(x) \right \}\epsilon f^{-1}]$ when branch of an inverse function is not given (define) then we consider its principal value branch.

$\displaystyle -1<x<0$ ,then

$\displaystyle \tan^{-1}x$ equals?

0%
$\displaystyle \pi-\cos^{-1}(\sqrt{1-x^{2}})$
0%
$\displaystyle \sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right )$
0%
$\displaystyle -\cos^{-1}\left(\frac {\sqrt{1-x^{2}}}{x}\right)$
0%
$\displaystyle \ cosec ^{-1}x$

Explanation

Let

$\displaystyle \tan^{-1} x=\alpha$

$\Rightarrow \dfrac{-\pi}{4}<\alpha<0$

$\displaystyle \therefore \tan\alpha=x$ such that

$\displaystyle \frac{-\pi}{4}<\alpha <0$

$\displaystyle \therefore \sin\alpha=\frac{x}{\sqrt{1+x^{2}}}\Rightarrow \alpha=\sin^{-1}\left(\frac{x}{\sqrt{1+x^{2}}} \right)$

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

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

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

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

The number of solutions of

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

0%
$1$
0%
$2$
0%
$3$
0%
$\infty$

Explanation

Given :

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

The given equation will be valid if

$-1\le 1+b+b^{2}+\cdots \infty \leq 1$
and

$a-\displaystyle \frac{a^{2}}{3}+\displaystyle \frac{a^{3}}{9}-\cdots\infty \epsilon \left [ -1,1 \right ]$

$\sin ^{ -1 } \left( 1+b+b^{ 2 }+\cdots \infty \right) =\sin ^{ -1 } \left( a-\dfrac { a^{ 2 } }{ 3 } +\dfrac { a^{ 2 } }{ 9 } \cdots \infty \right)$

Also

$1+b+b^{2}+\cdots \infty=a-\displaystyle \frac{a^{2}}{3}+\displaystyle \frac{a^{3}}{9}\cdots\infty$
(Infinite GP. series)

$\Rightarrow \displaystyle\frac{1}{1-b}=\frac{a}{1+\displaystyle \frac{a}{3}}$

$\Rightarrow \displaystyle\frac{1}{1-b}=\frac{3a}{a+3}$
which is true for infinite number of values of

$a$ and

$b$
So,

$\exists$ infinite solutions of the equation.

Express in terms of an inverse function the angle formed at the intesection of the diagonals of a cube.

0%
$sin^{1} 2/3$
0%
$cos^{1} 1/3$
0%
$tan^{1} 1/3$
0%
$sin^{1} 1/3$

$\tan^{-1}\left(\displaystyle\tan\frac{5\pi}{4}\right)=\alpha$ and

$\tan^{-1}\left(\displaystyle - \tan\frac{2\pi}{3}\right)=\beta$ then.

0%
$\displaystyle\alpha -\beta =\frac{7\pi}{12}$
0%
$\displaystyle \alpha +\beta =\frac{7\pi}{12}$
0%
$\displaystyle 2\alpha +3\beta =\frac{7\pi}{12}$
0%
$\displaystyle 4\alpha +3\beta =\frac{7\pi}{12}$

Explanation

$\tan^{ -1 }( \tan \dfrac { 5\pi }{ 4 } )=\alpha$

$\Rightarrow \dfrac { 5\pi }{ 4 } =\alpha$

Also, $\tan^{ -1 }( \tan -\dfrac { 2\pi }{ 3 } )=\beta$

$\Rightarrow -\dfrac { 2\pi }{ 3 } =\beta$

$\Rightarrow \alpha +\beta =\dfrac { 5\pi }{ 4 } -\dfrac { 2\pi }{ 3 }$

$\Rightarrow \alpha +\beta=\dfrac { 7\pi }{ 12 }$

$\displaystyle \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{4}{5}$ is equal to

0%
$\dfrac{\pi}{2}$
0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{6}$

Explanation

Given,

$\sin ^{ -1 } \cfrac { 3 }{ 5 } +sin^{ -1 }\cfrac { 4 }{ 5 }$

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

$\Rightarrow \sin ^{ -1 } \left(\cfrac { 3 }{ 5 } \sqrt { 1-\left(\cfrac { 4 }{ 5 } \right) } +\cfrac { 4 }{ 5 } \sqrt { 1-\left(\cfrac { 3 }{ 5 } \right)^{ 2 } } \right)$

$\Rightarrow \sin ^{ -1 } \left(\cfrac { 3 }{ 5 } \sqrt { \cfrac { 25-16 }{ 25 } ) } +\cfrac { 4 }{ 5 } \sqrt { \cfrac { 25-9 }{ 25 } } \right)$

$\Rightarrow \sin ^{ -1 } \left(\cfrac { 3 }{ 5 } \times \cfrac { 3 }{ 5 } +\cfrac { 4 }{ 5 } \times \cfrac { 4 }{ 5 } \right)$

$\Rightarrow \sin ^{ -1 } \left(\cfrac { 16 }{ 25 } +\cfrac { 9 }{ 25 } \right)$

$\Rightarrow \sin ^{ -1 } \left(\cfrac { 25 }{ 25 } \right)$

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

$\Rightarrow \cfrac { \pi }{ 2 }$

$0 < x_{1} < x_{2}$ which of following is true for

$y = \sec^{-1}x$ .

0%
$\sec^{-1}x_{1} + \sec^{-1}x_{2} > \sec^{-1} \left (\dfrac {x_{1} + x_{2}}{2}\right )$
0%
$\sec^{-1}x_{1} + \sec^{-1}x_{2} < 2\sec^{-1} \left (\dfrac {x_{1} + x_{2}}{2}\right )$
0%
$\sec^{-1}x_{1} > \sec^{-1}x_{2}$
0%
$\sec^{-1}x_{1} = \sec^{-1}x_{2}$

$3\cos ^{ -1 }{ x } +\sin ^{ -1 }{ x } =\pi$ , then

$x=.....$

0%
$\cfrac { \sqrt { 3 } }{ 2 }$
0%
$-\cfrac { 1 }{ \sqrt { 2 } }$
0%
$\cfrac { 1 }{ \sqrt { 2 } }$
0%
$\cfrac { 1 }{ 2 }$

Explanation

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

$3\cos^{-1}x+\sin^{-1}x=2\cos^{-1}x+\cos^{-1}x+\sin^{-1}x=\pi$

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

$\Rightarrow 2\cos^{-1}x=\cfrac{\pi}{2}$

$\cos^{-1}x=\cfrac{\pi}{4}$

$x=\cos (\cfrac{\pi}{4})=\cfrac{1}{\sqrt 2}$

The domain of the function

${\sin ^{ - 1}}2x$ is:

0%
$\left[ {0,\,1} \right]$
0%
$\left[ {-1,\,1} \right]$
0%
$\left[ {-2,\,2} \right]$
0%
$\left[ {\dfrac{{ - 1}}{2},\,\dfrac{1}{2}} \right]$

Explanation

$\sin^{-1} 2x$

$\sin^{-1} \theta$

$\theta \in [-1,1]$

$\Rightarrow 2x\in [-1,1]$

$\Rightarrow x \in [\cfrac{-1}{2},\cfrac{1}{2}]$

Domain of

$f(x)=\cot ^{ -1 }{ x } +\cos ^{ -1 }{ x } +co\sec ^{ -1 }{ x }$ is

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

Explanation

$f(x) = \cot^{-1}x + \cos^{-1}x + cosec^{-1}x$

Domain of

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

Domain of

$\cos^{-1}x = [-1,1]$

Domain of

$cosec^{-1}x = (-\infty, -1]\cup [1, \infty)$

These function are in addition.

So, we have to take the intersection of all domains.

So, answer is

$\{ -1, 1\}$

concept:

$f(x) = f_1(x) +f_2(x) + ...+f_n(x)$

domain of

$f(x) =$

Domain of

$f_1(x) \cap$ domain of

$f_2(x)$

$\cap$ domain of

$f_n(x)$

Let

$E_{1} = \left \{x \epsilon \mathbb {R} : x\neq 1\ and\ \dfrac {x}{x - 1} > 0\right \}$
and

$E_{2} = \left \{x \epsilon E_{1} : \sin^{-1} \left (\log_{e} \left (\dfrac {x}{x - 1}\right )\right )\text {is a real number}\right \}$ .
(Here, the inverse trigonometric function

$\sin^{-1}x$ assumes values in

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

$f : E_{1} \rightarrow \mathbb {R}$ be the function define by

$f(x) = \log_{e} \left (\dfrac {x}{x -1}\right )$ and

$g : E_{2}\rightarrow \mathbb{R}$ be the function defined by

$g(x) = \sin^{-1} \left (\log_{e} \left (\dfrac {x}{x - 1}\right )\right )$ .

LIST - I	LIST - II
P. The range of $f$ is	$\left (-\infty, \dfrac {1}{1 - e}\right ] \cup \left [\dfrac {e}{e - 1}, \infty \right )$
Q. The range of $g$ contins	$(0, 1)$
R. The domain of $f$ contains	$\left [-\dfrac {1}{2}, \dfrac {1}{2}\right ]$
S. The domain of $g$ is	$(-\infty, 0)\cup (0, \infty)$
	$\left (-\infty, \dfrac {e}{e - 1}\right ]$
	$(-\infty, 0)\cup \left (\dfrac {1}{2}, \dfrac {e}{e - 1}\right ]$

The correct option is

0%
$P\rightarrow 4; Q \rightarrow 2; R\rightarrow 1; S\rightarrow 1$
0%
$P\rightarrow 3; Q \rightarrow 3; R\rightarrow 6; S\rightarrow 5$
0%
$P\rightarrow 4; Q \rightarrow 2; R\rightarrow 1; S\rightarrow 6$
0%
$P\rightarrow 4; Q \rightarrow 3; R\rightarrow 6; S\rightarrow 5$

Explanation

$E_{1} : \dfrac {x}{x - 1} > 0$

$\Rightarrow E_{1} : x \epsilon (-\infty, 0) \cup (1, \infty)$

$E_{2} : -1\leq ln \left (\dfrac {x}{x - 1}\right )\leq 1$

$\dfrac {1}{e} \leq \dfrac {x}{x - 1}\leq e$

Now

$\dfrac {x}{x -1} - \dfrac {1}{e} \geq 0$

$\Rightarrow \dfrac {(e - 1)x + 1}{e(x - 1)}\geq 0$

$\Rightarrow x\epsilon \left (-\infty, \dfrac {1}{1 - e}\right ] \cup (1, \infty)$
also

$\dfrac {x}{x - 1} - e\leq 0$

$\dfrac {(e - 1)x - e}{x - 1} \geq 0$

$\Rightarrow x\epsilon (-\infty, 1) \cup \left [\dfrac {e}{e - 1}, \infty \right ]$

So,

$E_{2} : \left (-\infty, \dfrac {1}{1 - e}\right ]\cup \left [\dfrac {e}{e - 1}, \infty \right ]$

as Range of

$\dfrac {x}{x - 1}$ is

$R^{+} - \left \{1\right \}$

$\Rightarrow$ Range of

$f$ is

$R - \left \{0\right \}$ or

$(-\infty, 0)\cup (0, \infty)$

Range of

$g$ is

$\left [\dfrac {-\pi}{2}, \dfrac {\pi}{2}\right ]\cup \left \{0\right \}$ or

$\left [-\dfrac {\pi}{2}, 0\right )\cup \left (0, \dfrac {\pi}{2}\right ]$

Now

$P\rightarrow 4; Q \rightarrow 2; R\rightarrow 1; S\rightarrow 1$
Hence

$A$ is correct.

The value of

$sin^{-1} x + cos^{-1} x (|x| \geq 1)$ is

0%
1
0%
$\pi$
0%
$\pi / 2$
0%
$- \pi / 2$

Explanation

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

$\sin^{-1}x=\theta_{1}\,\,\Rightarrow \sin\theta_{1}=x$

$\cos^{-1}x=\theta_{2}\,\,\Rightarrow \cos\theta_{2}=x$

$\tan\theta_{1}=\cfrac{x}{\sqrt{1-x^{2}}}$

$\tan\theta_{2}=\cfrac{\sqrt{1-x^{2}}}{x}$

$\tan(\theta_{1}+\theta_{2})=\cfrac{\tan\theta_{1}+\tan\theta_{2}}{1-\tan\theta_{1}\tan\theta_{2}}$

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

$=\cfrac{x^{2}+1-x^{2}}{x\sqrt{1-x^{2}}(0)}$

$\tan(\theta_{1}+\theta_{2})=\infty$

$\theta_{1}+\theta_{2}=\cfrac{\pi}{2}$

$\therefore\,\sin^{-1}x+\cos^{-1}x=\cfrac{\pi}{2}$

Match the entries of Column - I and Column - II.

	Column - I		Column - II
a	If 4 $sin^{-1} x + cos^{-1} x = \pi$ , then x equals	1	ab
b	If $\angle C = 90^{0}$ , then the value of $tan^{-1}$ $\dfrac{a}{b + c}$ + $tan^{-1}$ $\dfrac{b}{c +a}$ is	2	$\pi$
c	$tan^{-1}$ 1 + $tan^{-1}$ 2 + $tan^{-1}$ 3 is	3	$\pi$ /4
d	If $sec^{-1}$ $\dfrac{x}{a}$ - $sec^{-1}$ $\dfrac{x}{b}$ = $sec^{-1}$ b - $sec^{-1}$ a, then x equals	4	1/2

0%
a-4, b-3, c-2, d-1
0%
a-1, b-3, c-2, d-4
0%
a-4, b-3, c-1, d-2
0%
a-3, b-4, c-2, d-1

Explanation

(a)

$\rightarrow$ (4).
4

$sin^{-1}$ x +

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

$\therefore$ 3

$sin^{-1}$ x =

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

$sin\dfrac{\pi}{6} = \dfrac{1}{2}$
(b)

$\rightarrow$ (3).
Use Rule 5 and

$a^{2}$ +

$b^{2}$ =

$c^{2}$ ,

$tan^{-1}$ 1 =

$\pi/4$
(c)

$\rightarrow$ (2).

$tan^{-1}$ 2 +

$tan^{-1}$ 3 =

$\pi$ +

$tan^{-1}$

$\dfrac{2 + 3}{1 - 2.3}$
=

$\pi$ -

$tan^{-1}$ 1

$\therefore$ L.H.S =

$tan^{-1}$ 1 +

$tan^{-1}$ 2 +

$tan^{-1}$ 3 =

$\pi$
(d)

$\rightarrow$ (1)

Let

$a={ (\sin ^{ -1 }{ x) } }^{ \sin ^{ -1 }{ x } },\quad b={ \left( \sin ^{ -1 }{ x } \right) }^{ \cos ^{ -1 }{ x } },\quad c={ \left( \cos ^{ -1 }{ x } \right) }^{ \sin ^{ -1 }{ x } },\quad d={ \left( \cos ^{ -1 }{ x } \right) }^{ \cos ^{ -1 }{ x } }$ and if

$x\in (0,1)$ then

0%
$a>b>d>c$
0%
$b>a>d>c$
0%
$d>c>a>b$
0%
none of these

$0< x < 1$ , then

$\tan^{-1}(\cfrac{\sqrt{1-x^{2}}}{1+x})$ is equal to

0%
$\cfrac{1}{2}\cos^{-1}x$
0%
${\cos ^{ - 1}}{{\sqrt {1 + x} } \over 2}$
0%
${\sin ^{ - 1}}\sqrt {{{1 - x} \over 2}}$
0%
${1 \over 2}\sqrt {{{1 + x} \over {1 - x}}}$

Explanation

$\tan^{-1}(\cfrac{\sqrt{1-x^{2}}}{1+x})$

Let

$x=\cos \theta$

$=\tan^{-1}(\cfrac{\sqrt{1-\cos^{2}\theta}}{1+\cos \theta})$

$\tan^{-1}\cfrac{\sin \theta}{1+\cos^{2}\cfrac{\theta}{2}-\sin^{2}\cfrac{\theta}{2}}$

$=\tan^{-1}\cfrac{2\sin \cfrac{\theta}{2} \cos \cfrac{\theta}{2}}{2\cos^{2}\cfrac{\theta}{2}}$

$=\tan^{-1}(\tan\cfrac{\theta}{2})=\cfrac{\theta}{2}$ =

$\cfrac{1}{2}\cos^{-1}x$

Let

$a, b, c$ be a positive real numbers

$\theta = \tan^{-1} \sqrt{\dfrac{a(a + b +c)}{bc}} + \tan^{-1} \sqrt{\dfrac{b(a + b+ c)}{ca}} + \tan^{-1} \sqrt{\dfrac{c(a + b + c)}{ab}}$ , then

$\tan \theta$

0%
$0$
0%
$3 \pi$
0%
$1$
0%
$4 \pi$

Explanation

$\theta =\tan^{-1} \sqrt{\dfrac{a\left ( a+b+c \right )}{bc}} +\tan^{-1} \sqrt{\dfrac{b\left ( a+b+c \right )}{ac}} +\tan^{-1} \sqrt{\dfrac{c\left ( a+b+c \right )}{ab}}$

$\alpha =\tan^{-1} \sqrt{\dfrac{a\left ( a+b+c \right )}{bc}}$

$\beta = \tan^{-1} \sqrt{\dfrac{b\left ( a+b+c \right )}{ac}}$

$\gamma = \tan^{-1} \sqrt{\dfrac{c\left ( a+b+c \right )}{ab}}$

$\Rightarrow \tan \alpha = \sqrt{\dfrac{a\left ( a+b+c \right )}{bc}}$

$\Rightarrow \tan \beta = \sqrt{\dfrac{b\left ( a+b+c \right )}{ac}}$

$\Rightarrow \tan \gamma = \sqrt{\dfrac{c\left ( a+b+c \right )}{ab}}$

$\tan \alpha+\tan \beta+\tan \gamma =\sqrt{\dfrac{a\left ( a+b+c \right )}{bc}}+\sqrt{\dfrac{b\left ( a+b+c \right )}{ac}}+\sqrt{\dfrac{c\left ( a+b+c \right )}{ab}}$

$=\dfrac{\left ( a+b+c \right )^{\frac{3}{2}}}{\sqrt{abc}}$

$=\tan \alpha\tan \beta\tan \gamma$

$\Rightarrow \tan \theta =\left [ \dfrac{(\tan \alpha+\tan \beta+\tan \gamma)-\tan \alpha\tan \beta\tan \gamma}{1-\tan \alpha\tan \beta-\tan \beta\tan \gamma-\tan \gamma\tan \alpha} \right ]$

$\Rightarrow \tan \theta =0$

$\\ \cos^{-1}\left[\dfrac {\sqrt {1+x}+\sqrt {1-x}}{2}\right]=\dfrac {\pi}{2}-\dfrac {1}{2}\cos^{-1}x$

0%
True
0%
False

$\cos ^{ -1 }{ \cfrac { x }{ 2 } } +\cos ^{ -1 }{ \cfrac { y }{ 3 } } =\theta$ , then

$9{x}^{2}-12xy\cos{\theta}+4{y}^{2}$ is equal to

0%
$36 \sin ^{ 2 }{ \theta }$
0%
$36 \cos ^{ 2 }{ \theta }$
0%
$36 \tan ^{ 2 }{ \theta }$
0%
None of these

The range of the function

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

0%
$\phi$
0%
$\left[ -\cfrac { \pi }{ 2 } ,\cfrac { \pi }{ 2 } \right]$
0%
$\cfrac { \pi }{ 2 }$
0%
none of these

Explanation

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

Domain of inverse sin function is

$[-1,1]$

$\Rightarrow -1\leq x^2-2x+2 \leq1$

On solving these inequalities separately, we get

$\Rightarrow x^2-2x+2\leq1$

$\Rightarrow x^2-2x+1\leq0$

$\Rightarrow (x-1)^2\leq0$

This inequality only holds good for

$x=1$

Substitute

$x=1$ in the above expression , we get

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

Therefore, Range of

$f(x)=\dfrac{\pi}{2}$

$2{\tan ^{ - 1}}\left[ {\sqrt {\dfrac{{a - b}}{{a + b}}} \tan \dfrac{\theta }{2}} \right] =$

0%
${\cos ^{ - 1}}\left( {\dfrac{{a\cos \theta + b}}{{a + b\cos \theta }}} \right)$
0%
${\cos ^{ - 1}}\left( {\dfrac{{a + b\cos \theta }}{{a\cos \theta + b}}} \right)$
0%
${\cos ^{ - 1}}\left( {\dfrac{{a\cos \theta }}{{a + b\cos \theta }}} \right)$
0%
${\cos ^{ - 1}}\left( {\dfrac{{b\cos \theta }}{{a\cos \theta + b}}} \right)$

Explanation

Let,

$t=2{{\tan }^{-1}}\left[ \sqrt{\dfrac{a-b}{a+b}}\tan \dfrac{\theta }{2} \right]$ ……. (1)

Let,

$b=a\cos \alpha$ ……. (2)

Then,

$\dfrac{a-b}{a+b}=\dfrac{a-a\cos \alpha }{a+a\cos \alpha }$

$\dfrac{a-b}{a+b}=\dfrac{1-\cos \alpha }{1+\cos \alpha }$

$\dfrac{a-b}{a+b}=\dfrac{1-\left( 1-2{{\sin }^{2}}\dfrac{\alpha }{2} \right)}{1+2{{\cos }^{2}}\dfrac{\alpha }{2}}$

$\dfrac{a-b}{a+b}=\dfrac{2{{\sin }^{2}}\dfrac{\alpha }{2}}{2{{\cos }^{2}}\dfrac{\alpha }{2}}$

$\dfrac{a-b}{a+b}={{\tan }^{2}}\dfrac{\alpha }{2}$

$\sqrt{\dfrac{a-b}{a+b}}=\tan \dfrac{\alpha }{2}$ …… (3)

From equation (1) and (3), we get

$\tan \dfrac{t}{2}=\tan \dfrac{\alpha }{2}\tan \dfrac{\theta }{2}$

$\Rightarrow \cos t=\dfrac{1-{{\tan }^{2}}\dfrac{t}{2}}{1+{{\tan }^{2}}\dfrac{t}{2}}$

$\cos t=\dfrac{1-{{\tan }^{2}}\dfrac{\alpha }{2}{{\tan }^{2}}\dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\alpha }{2}{{\tan }^{2}}\dfrac{\theta }{2}}$

$\cos t=\dfrac{1-\dfrac{{{\sin }^{2}}\dfrac{\alpha }{2}}{{{\cos }^{2}}\dfrac{\alpha }{2}}\dfrac{{{\sin }^{2}}\dfrac{\theta }{2}}{{{\cos }^{2}}\dfrac{\theta }{2}}}{1+\dfrac{{{\sin }^{2}}\dfrac{\alpha }{2}}{{{\cos }^{2}}\dfrac{\alpha }{2}}\dfrac{{{\sin }^{2}}\dfrac{\theta }{2}}{{{\cos }^{2}}\dfrac{\theta }{2}}}$

$\cos t=\dfrac{{{\cos }^{2}}\dfrac{\alpha }{2}{{\cos }^{2}}\dfrac{\theta }{2}-{{\sin }^{2}}\dfrac{\alpha }{2}{{\sin }^{2}}\dfrac{\theta }{2}}{{{\cos }^{2}}\dfrac{\alpha }{2}{{\cos }^{2}}\dfrac{\theta }{2}-{{\sin }^{2}}\dfrac{\alpha }{2}{{\sin }^{2}}\dfrac{\theta }{2}}$

$\cos t=\dfrac{\dfrac{1}{2}\left[ \cos \theta +\cos \alpha \right]}{\dfrac{1}{2}\left[ 1+\cos \alpha \cos \theta \right]}$

From equation (2), we get

$\cos t=\dfrac{\cos \theta +\dfrac{b}{a}}{1+\cos \theta \dfrac{b}{a}}=\dfrac{a\cos \theta +b}{a+b\cos x}$

$\Rightarrow t={{\cos }^{-1}}\left[ \dfrac{a\cos \theta +b}{a+b\cos \theta } \right]$

Hence, this is the required result.

$\sin { ^{ -1 }\dfrac { 3 }{ 5 } +\tan { ^{ -1 } } } \dfrac { 1 }{ 7 } =\dfrac { \pi }{ 2 }$

0%
True
0%
False

Explanation

$\sin ^{-1}\dfrac{3}{5}+\tan ^{-1}\dfrac{1}{7}=\dfrac{\pi}{2}$ True or false?

Let

$\sin ^{-1 \dfrac{3}{5}=\theta}\therefore \sin \theta=\dfrac{3}{5}$

$\therefore \tan \theta=\dfrac{3}{4}\Rightarrow \theta=\tan ^{-1}\dfrac{3}{4}$

$\tan ^{-1}\dfrac{3}{4}+\tan ^{-1}\dfrac{1}{7}=\tan ^{-1}\dfrac{\left(\dfrac{3}{4}+\dfrac{1}{7}\right)}{1-\dfrac{3}{7}\times \dfrac{1}{7}}$

$\boxed{\tan ^{-1}A+\tan ^{-1}B=\tan ^{-1}\dfrac{(A+B)}{1+AB}}$

$\tan ^{-1}\dfrac{3}{4}+\tan ^{-1}\dfrac{1}{7}=\tan ^{-1}=\tan ^{-1}\left(\dfrac{\dfrac{25}{28}}{\dfrac{25}{28}}\right)1$

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

2115556_1114616_ans_71cebb7feae24377b86f95f318e7cf4b.png

$\cos [\tan^{-1}\{ \sin(\cot^{-1}x)\}]$ is equal to:

0%
$\sqrt{\dfrac{x^2 + 2}{x^2 + 3}}$
0%
$\sqrt{\dfrac{x^2 + 2}{x^2 + 1}}$
0%
$\sqrt{\dfrac{x^2 + 1}{x^2 + 2}}$
0%
none of these

Explanation

Let

$\cot^{-1}x=\theta\Rightarrow \cot\theta=x$

Now,

$\sin\theta=\dfrac{1}{\sqrt{\cot^2\theta+1}}$

So,

$\sin\theta=\dfrac{1}{\sqrt{x^2+1}}\Rightarrow \theta=\sin^{-1}(\dfrac{1}{\sqrt{x^2+1}})$

Hence,

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

Again, Let

$\tan^{-1}(\dfrac{1}{\sqrt{x^2+1}})=\alpha\Rightarrow \tan\alpha=(\dfrac{1}{\sqrt{x^2+1}})$

Now,

$\cos\alpha=\dfrac{1}{\sqrt{\tan^2\alpha+1}}$

So,

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

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

$\sin^{-1}\left(a-\dfrac{a^2}{3}+\dfrac{a^3}{9}+...\right)+\cos^{-1}(1+b+b^2+...)=\dfrac{\pi}{2}$ when?

0%
$a=-3$ and $b=1$
0%
$a=1$ and $b=-\dfrac{1}{3}$
0%
$a=\dfrac{1}{6}$ and $b=\dfrac{1}{2}$
0%
None of these

Explanation

$\begin{matrix} { \sin ^{ -1 } }\left( { \frac { a }{ { 1-\left( { -\frac { a }{ 3 } } \right) } } } \right) +{ \cos ^{ -1 } }\left( { \frac { 1 }{ { 1-b } } } \right) =\frac { \pi }{ 2 } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \because Angle\, \, are\, \, in\, \, G.P\, \, then\, \, sum\, \, of\, inite\, G.P\, } \right] \\ { \sin ^{ -1 } }\left( { \frac { a }{ { 1+\frac { a }{ 3 } } } } \right) +{ \cos ^{ -1 } }\left( { \frac { 1 }{ { 1-b } } } \right) =\frac { \pi }{ 2 } \, is\, \, \left( { \frac { a }{ { 1-r } } } \right) \\ i{ n^{ -1 } }\left( { \frac { { 3a } }{ { a+3 } } } \right) +{ \cos ^{ -1 } }\left( { \frac { 1 }{ { 1-b } } } \right) =\frac { \pi }{ 4 } +\frac { \pi }{ 4 } .........\left( i \right) \\ by\, \, inverse\, \, property\, \, if\, angle\, \, of\, \sin \, \, \, and\, \cos ine\, \, is\, \, same\, \, then\, \, \\ { { Sin }^{ -1 } }x\, \, +{ \cos ^{ -1 } }x=\frac { \pi }{ 2 } \\ By\, \, compearing\, \, \\ { { Sin }^{ -1 } }\left( { \frac { { 3a } }{ { a+3 } } } \right) =\frac { \pi }{ 4 } \\ { Sin }\frac { \pi }{ 4 } =\frac { { 3a } }{ { a+3 } } \\ \frac { 1 }{ { \sqrt { 2 } } } =\frac { { 3a } }{ { a+3 } } \\ a+3=3\sqrt { 2a } \\ 3=a(3\sqrt { 2-1 } \\ a=\frac { 3 }{ { 3\sqrt { 2-1 } } } \\ { \cos ^{ -1 } }\left( { \frac { 1 }{ { 1-b } } } \right) =\frac { \pi }{ 4 } \\ \cos \frac { \pi }{ 4 } =\frac { 1 }{ { 1-b } } \\ \frac { 1 }{ { \sqrt { 2 } } } =\frac { 1 }{ { 1-b } } \\ \sqrt { 2 } =1-b \\ b=1-\sqrt { 2 } \\ \end{matrix}$

The range of

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

0%
$\left( 0,\pi \right)$
0%
$\left[ \dfrac { \pi }{ 4 } ,\dfrac { 3\pi }{ 4 } \right]$
0%
$\left[ \dfrac { -\pi }{ 4 } ,\dfrac { \pi }{ 4 } \right]$
0%
$\left[ 0,\dfrac { 3\pi }{ 4 } \right]$

${ x }_{ 1 },{ x }_{ 2},{ x }_{ 3}$ are positive roots of

$x^{ 3 }-6x^{ 2 }+3px-2p=0\quad (p\in R)$ , then the value of

$\sin ^{ -1 } \left( \frac { 1 }{ { x }_{ 1 } } +\frac { 1 }{ { x }_{ 2 } } \right) +\cos ^{ -1 }{ \left( \frac { 1 }{ { x }_{ 2 } } +\frac { 1 }{ { x }_{ 3 } } \right) } -\tan ^{ -1 }{ \left( \frac { 1 }{ { x }_{ 3 } } +\frac { 1 }{ { x }_{ 1 } } \right) }$ is equal to

0%
$\frac { \pi }{ 8 }$
0%
$\frac { \pi }{ 6 }$
0%
$\frac { \pi }{ 4 }$
0%
$\pi$

Explanation

$x^3-6x^2+3px-2p=0$

$x_1$

$x_2$

$x_3$

are the zeros

$x_1+x_2+x_3=6$

$x_1x_2+x_2x_3+x_3x_1=3p$

$\left[\because \dfrac{x_1+x_2+x_3}{3}\geq (2p)^{\dfrac{1}{3}}, 2\geq (2p)^{\dfrac{1}{3}}\Rightarrow p\leq 4\right]$

$x_1x_2x_3=2p$

$\dfrac{3p}{3}\geq ((x_1x_2x_3)^2)^{1/3}$

$p^3\geq 4p^2$

$p\geq 4$

$\Rightarrow p=4$

$\therefore x_1x_2+x_2x_3+x_3x_1=12$

$x_1x_2x_3=8$

$x_1+x_2+x_3=6$

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

$\dfrac{\pi}{2}+0-\dfrac{\pi}{4}=\dfrac{\pi}{4}$ .

The value of

$\sin ^{ -1 }{ \left[ \cot { \left( \sin ^{ -1 }{ \sqrt { \frac { 2-\sqrt { 3 } }{ 4 } } } +\cos ^{ -1 }{ \frac { \sqrt { 2 } }{ 4 } +\sec ^{ -1 }{ \sqrt { 2 } } } \right) } \right] }$ is equal to

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

$cos({ cos }^{ -1 }cos(\frac { 8\pi }{ 7 } )+{ tan }^{ -1 }tan(\frac { 8\pi }{ 7 } ))$ has the value equal to -

0%
$1$
0%
$-1$
0%
$cos\dfrac { \pi }{ 7 }$
0%
$0$

$\left( \tan ^ { - 1 } x \right) ^ { 2 } + \left( \cot ^ { - 1 } x \right) ^ { 2 } = \dfrac { 5 \pi ^ { 2 } } { 8 }$ , then

$x$ equals to

0%
$- 1$
0%
$1$
0%
$0$
0%
None Of These

The exhaustive set of values of

$'a'$ such that

$x^{2}+ax+sing^{-1}\ (x^{2}-4x+5)+\ cos^{-1}(x^{2}-4x+5)=0$ has at least one solution is

0%
$\left \{ -2-\frac{\pi }{4} \right \}$
0%
$\left ( -\propto,-2-\frac{\pi}{4} \right )$
0%
$(-\propto,-2-\frac{\pi}{4}]$
0%
$(-2-\frac{\pi}{4},+\propto]$

$\left |{\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\right |\,\, < \,\dfrac{\pi }{3},\,then\,:$

0%
$x \in \,\left[ { - \dfrac{1}{3},\dfrac{1}{{\sqrt 3 }}} \right]$
0%
$x \in \,\left[ { - \dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}} \right]$
0%
$x \in \,\left[ {0,\dfrac{1}{{\sqrt 3 }}} \right]$
0%
none of these

$\tanh^{1}\left(\dfrac {1}{3}\right)+\coth^{1}(3)=$ .....

0%
$\log 2$
0%
$\log 3$
0%
$\log \sqrt {3}$
0%
$\log \sqrt {2}$

$y=\dfrac{1}{2}\csc\ h^{-1}{\left(\dfrac{1}{2x\sqrt{1+x^{2}}}\right)}$ then

$x$ =

0%
coshy
0%
sinhy
0%
tanhy
0%
cothy

$3\cot^{-1}{\left(\dfrac{1}{2+\sqrt{3}}\right)}-\cot^{-1}\left(\dfrac{1}{x}\right)=\cot^{-1}\left(\dfrac{1}{3}\right)+\dfrac{\pi}{2}$ then

$x=$ ?

0%
$1$
0%
$2$
0%
$3$
0%
$\sqrt{3}$

$Sin h (cos h ^{-1} x) =$

0%
$\sqrt{x^2 +1}$
0%
$\dfrac{1}{\sqrt{x^2 +1}}$
0%
$\sqrt{x^2 -1}$
0%
$\dfrac{1}{\sqrt{x^2 -1}}$

The value of

$e^{sinh^{-1} (tan \theta)}$ is equal to

0%
$cosec \theta + cot \theta$
0%
$sec \theta + tan \theta$
0%
$cosec \theta + sec \theta$
0%
$tan \theta + cot \theta$

$f(x)={ sin }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } x-\frac { 1 }{ 2 } \sqrt { 1-{ x }^{ 2 } } \right) -\frac { 1 }{ 2 } \le x\le 1$ , then f(x) is equal to :

0%
${ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ sin }^{ -1 }(x)$
0%
${ sin }^{ -1 }x-\frac { \pi }{ 6 }$
0%
${ sin }^{ -1 }x+\frac { \pi }{ 6 }$
0%
None of these

The value of

$\tan{\left\{\dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}{(\dfrac{x}{y})}\right\}}+\tan{\left\{\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}{(\dfrac{x}{y})}\right\}}$

0%
$\dfrac{x}{y}$
0%
$\dfrac{y}{x}$
0%
$\dfrac{2y}{x}$
0%
$\dfrac{2x}{y}$

For which value of x,

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

0%
$\dfrac{1}{2}$
0%
$0$
0%
$1$
0%
$\dfrac{-1}{2}$

$cot^{-1} x - cot^{-1} (x+2) = 15^0$ then x is equal to

0%
$\sqrt{3}$
0%
$-\sqrt{3}$
0%
$\sqrt{3} +2$
0%
$-\sqrt{3} +2$

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 8 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 8 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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