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CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 9 - MCQExams.com
CBSE
Class 12 Commerce Maths
Inverse Trigonometric Functions
Quiz 9
If $${ sin }^{ -1 }\left( \frac { x }{ 5 } \right) +{ cosec }^{ -1 }\left( \frac { 5 }{ 4 } \right) =\frac { \pi }{ 2 } $$, then value of x is :
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1
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3
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4
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5
The value of $${ tan }^{ -1 }\left( \frac { xcos\theta }{ 1-xsin\theta } \right) -{ cot }^{ -1 }\left( \frac { cos\theta }{ x-sin\theta } \right) $$ is :
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$$2\theta $$
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$$\theta $$
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$$\frac { \theta }{ 2 } $$
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Independent of $$\theta $$
Solve : $$sin h^{-1} (\dfrac{x}{\sqrt{1-x^2}}) =$$
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$$cos h^{-1} x$$
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$$-cos h^{-1} x$$
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$$tan h^{-1} x$$
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$$-tan h^{-1} x$$
The number of solution of the equation $${\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}$$ is are
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0
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1
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2
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More then Two
The value of $$\tan^{-1}\left[\dfrac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right],\left|x\right|<\dfrac{1}{2},x=0$$, is equal to:
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$$\dfrac{pi}{4}-\cos^{-1}x^{2}$$
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$$\dfrac{pi}{4}+\dfrac{1}{2}\cos^{-1}x^{2}$$
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$$\dfrac{pi}{4}+\cos^{-1}x^{2}$$
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$$\dfrac{pi}{4}-\dfrac{1}{2}\cos^{-1}x^{2}$$
If $${ sin }^{ -1 }\left( a-\frac { { a }^{ 2 } }{ 3 } +\frac { { a }^{ 3 } }{ 9 } -...........\infty \right) +{ cos }^{ -1 }\left( 1+b+{ b }^{ 2 }+.......\infty \right) =\frac { \pi }{ 2 } $$
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$$a=-3,b=1$$
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$$a=1,b=\dfrac { -1 }{ 3 } $$
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$$a=\dfrac { 1 }{ 6 } ,b=\dfrac { 1 }{ 2 } $$
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$$a=\dfrac { 1 }{ 6 } ,b=\dfrac { 1 }{ 3 } $$
If $$(cot^{-1}x)^{2}-3(cot^{-1}x)+2 > 0,$$ then x lies in
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(cot 2, cot 1)
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$$(-\infty , cot 2)\cup (cot 1,\infty )$$
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$$(cot 1 \infty)$$
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$$(-\infty , cot 1)\cup (cot 2,\infty )$$
Thee value of $$\cos^{-1}x+\cos^{-1}\left(\dfrac {x}{2}+\dfrac {\sqrt {3-x^{2}}}{2}\right)$$, where $$\dfrac {1}{2} \le x \le 1$$.
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$$\dfrac {\pi}{6}$$
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$$\dfrac {\pi}{3}$$
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$$\dfrac {\pi}{2}$$
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$$0$$
The solution of the equation $$sin^{-1}(\frac{dt}{dx})=x+y$$ is
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tan(x+y)+sec(x+y)=x+c
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tan(x+y)-sec(x+y)=x+c
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tan(x+y)+sec(x+y)+x+c=0
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None of these
$$cos^{-1}[cos(-\frac{17}{15}\pi )]$$ is equal to-
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$$-\frac{17\pi }{15}$$
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$$\frac{17\pi }{15}$$
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$$-\frac{2\pi }{15}$$
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$$\frac{13\pi }{15}$$
Let $$S_{n}=\cot^{-1}\left(3x+\dfrac{2}{x}\right)+ \cot^{-1}\left(6x+\dfrac{2}{x}\right)+ \cot^{-1}\left(10x+\dfrac{2}{x}\right)+.....+n$$ term where $$x>0$$. If $$\displaystyle \lim_{n\rightarrow \infty}S_{n}=$$ then $$x$$ equals
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$$\dfrac{\pi}{4}$$
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$$1$$
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$$\tan 1$$
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$$\cot 1$$
Let f:-$$\left\{ 0,4\pi \right\} $$$$\rightarrow \left[ 0,\pi \right] $$ be defined by f(x)=$${ cos }^{ -1 }\left( cosx \right) $$. The number of points x$$\in \left[ 0,4\pi \right] $$ satisfying the equation f(x)=$$\frac { 10-x }{ 10 } $$ is
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2
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1
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3
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4
Sum of maximum and minimum values of (sin$$^{-1}$$x)$$^4$$ + (cos$$^{-1}$$x)$$^4$$ is:
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$$\dfrac{137\pi^2}{128}$$
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$$\dfrac{\pi^2}{17}$$
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$$\dfrac{17\pi^4}{16}$$
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$$\dfrac{137\pi^4}{128}$$
Number of values of x for which $$(tan^{-1} x)^2 + (cot^{-1} x)^2 = \dfrac{\pi^2}{4}$$ is
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2
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4
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1
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3
if $${ tan }^{ -1 }\frac { \sqrt { 1+{ x }^{ 2 }-1 } }{ x } =4\quad then\quad x$$
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tan8
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tan4
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$$tan\frac { 1 }{ 4 } $$
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tan8
If $$f(x)=2tan^{-1}x+sin^{-1}\left (\dfrac {2x}{1+x^2}\right )$$ then for x>1, f(x)=
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$$sec^{-1}x$$
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$$sin^{-1}x$$
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$$\pi $$
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$$\cfrac {\pi }{2}$$
If $${ \tan \text{h} }^{ -1 }\left( \dfrac { 1 }{ 3 } \right) =\dfrac { 1 }{ 2 } \log t ,$$ then $$t$$ is equal to
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$$2$$
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$$3$$
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$$1$$
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$$4$$
$$\displaystyle \sum^{\infty}_{r=1}\tan^{-1}\left(\dfrac {1}{r^{2}+5r+7}\right)$$ equal to
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$$\tan^{-1}3$$
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$$\dfrac {3\pi}{4}$$
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$$\sin^{-1}\dfrac {1}{\sqrt {10}}$$
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$$\cot^{-1}2$$
$$tan^{-1}y=tan^{-1}x+tan^{-1}(\frac{2x}{1-x^{2}})$$ where $$|x| < \frac{1}{\sqrt{3}}$$. Then a value of y is:
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$$\dfrac{3x+x^{3}}{1+3x^{2}}$$
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$$\dfrac{3x-x^{3}}{1+3x^{2}}$$
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$$\dfrac{3x+x^{3}}{1-3x^{2}}$$
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$$\dfrac{3x-x^{3}}{1-3x^{2}}$$
$${ tanh }^{ -1 }\left( \frac { 1 }{ 3 } \right) +{ coth }^{ -1 }\left( 3 \right) =.....$$
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log 2
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log 3
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log$$\sqrt { 3 } $$
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$$log \sqrt { 2 } $$
The value of sin $$\left ( 3 sin^{-1}\left ( 0.8 \right ) \right )$$ is
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sin(2)
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sin(1.88)
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-sin(0.88)
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None of these
The value of $$3 tan^-1\left(\dfrac{1}{2}\right)+ 2tan^-1 \left(\dfrac{1}{5}\right)$$ is-
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$$\dfrac{\pi}{4}$$
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$$\dfrac{\pi}{2}$$
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$$\pi$$
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None
If $$f\left( x \right) =\cos ^{ -1 }{ x } +\cos ^{ -1 }{ \left\{ \frac { x }{ 2 } +\dfrac { 1 }{ 2 } \sqrt { 3-3{ x }^{ 2 } } \right\} } $$ then
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$$f\left( \dfrac { 2 }{ 3 } \right) =\dfrac { \pi }{ 3 } $$
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$$f\left( \dfrac { 2 }{ 3 } \right) =2\cos ^{ -1 }{ \dfrac { 2 }{ 3 } -\dfrac { \pi }{ 3 } } $$
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$$f\left( \dfrac { 1 }{ 3 } \right) =\dfrac { \pi }{ 3 }$$
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$$f\left( \dfrac { 1 }{ 3 } \right) =2\cos ^{ -1 }{ \dfrac { 1 }{ 3 } -\dfrac { \pi }{ 3 } } m$$
If $$(\cos^{-1}x)^{2}+(\cos^{-1}y)^{2}+2(\cos^{-1}x)(\cos^{-1}y)=4\pi^{2}$$ then $$x^{2}+y^{2}$$ is equal to
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$$1$$
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$$\dfrac{3}{2}$$
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$$2$$
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`Will depends on x and y
$$\cos ^{ -1 }{ \left( \cos { \dfrac { 7\pi }{ 6 } } \right) } $$ is equal to
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$$\dfrac {7\pi}{6}$$
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$$\dfrac {5\pi}{6}$$
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$$\dfrac {\pi}{3}$$
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$$\dfrac {\pi}{6}$$
The value of $$\sin^{-1}(\sin 3)+\cos^{-1}(\cos 7)-\tan^{-1}(\tan 5)$$ is
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$$\pi-1$$
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$$\pi$$
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$$3\pi-1$$
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$$2\pi-1$$
$$\sum^ { \infty }_ { n=1 } tan^{-1}\dfrac{4n}{n^4-2n^2+2}$$ is equal to:
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$$tan^{-1}2+tan^{-1}3$$
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$$4 tan^{-1}1$$
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$$\pi/2$$
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$$sec^{-1}(-\sqrt2)$$
The smallest and largest value of $$\tan^{-1}\left(\dfrac {1-x}{1+x}\right),0 \le x \le 1$$ are
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$$0, \pi$$
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$$0,\dfrac {\pi}{4}$$
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$$-\dfrac {\pi}{4}, \dfrac {\pi}{4}$$
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$$\dfrac {\pi}{4}, \dfrac {\pi}{2}$$
$${ cos }^{ -1 }(\frac { x }{ 3 } )+{ cos }^{ -1 }(\frac { y }{ 2 } )=(\frac { \theta }{ 2 } )$$ , then the value of $${ 4x }^{ 2 }$$-12xy cos$$(\frac { \theta }{ 2 } )$$+$${ 9y }^{ 2 }$$ is equal to
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$$18(1+cos\theta )$$
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$$18(1-cos\theta )$$
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$$36(1+cos\theta )$$
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$$36(1-cos\theta )$$
$$\cos ^{ -1 }{ \left\{ \dfrac { 1 }{ 2 } { x }^{ 2 }+\sqrt { { 1-x }^{ 2 } } .\sqrt { 1\dfrac { { x }^{ 2 } }{ 4 } } \right\} } =\cos ^{ -1 }{ \dfrac { x }{ 2 } } -\cos ^{ -1 }{ x } $$ holds for
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$$\left| x \right| \le 1$$
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$$x\in R$$
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$$0\le x\le 1$$
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$$-1\le x\le 0$$
If $$x=si{ n }^{ -1 }(sin10)$$ and $$y={ s }^{ -1 }(cos10)$$ then $$y-x$$ is equal to:
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0
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$$7\pi $$
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10
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$$\pi $$
Evaluate $$\cot ^{ -1 }{ \sum _{ n=1 }^{ 19 }{ \cot ^{ -1 }{ [1+\sum _{ p=1 }^{ n }{ 2p } ] } } } $$
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$$\frac { 23 }{ 22 } $$
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$$\frac { 19 }{ 23 } $$
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$$\frac { 23 }{ 19 } $$
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$$\frac { 22 }{ 23 } $$
If $$x={ sin }^{ -1 }(sin10)$$ and $$y={ cos }^{ -1 }(cos10)$$, then the value of (y - x) is
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$$\pi $$
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$$7\pi $$
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0
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10
Considering only the principal values of inverse functions, the set $$A=\left\{ x\quad \ge \quad 0\quad :tan^{ -1 }(2x)+tan^{ -1 }(3x)=\dfrac { \pi }{ 4 } \right\} $$
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Is an empty set
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contains more than two elements
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contains two elements
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is a singleton
The value of $$\sin ^{ -1 }{ (\cos { (\cos ^{ -1 }{ (\cos { x } ) } +\sin ^{ -1 }{ (\sin { x } ) } ) } ) } ,\quad where\quad x\in (\frac { \pi }{ 2 } ,\pi )$$, is equal to
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$$\frac { \pi }{ 2 } $$
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$$-\pi $$
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$$\pi $$
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$$-\frac { \pi }{ 2 } $$
If $$\theta ={ cot }^{ -1 }\sqrt { cosx } -{ tan }^{ -1 }\sqrt { cosx } $$, then $$sin\theta =$$
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$$tan\cfrac { 1 }{ 2 } x$$
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$${ tan }^{ 2 }(x/2)$$
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$$\cfrac { 1 }{ 2 } { tan }^{ -1 }(x2)$$
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None of these
The value of $$\sin^{-1}(\sin 12)-\cos^{-1}(\cos 12)=$$
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$$0$$
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$$\pi$$
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$$8\pi +24$$
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$$8\pi -24$$
Explanation
$${\sin}^{-1}{\left(\sin{12}\right)}-{\cos}^{-1}{\left(\cos{12}\right)}$$
$$={\cos}^{-1}{\sqrt{1-{\sin}^{2}{12}}}-{\cos}^{-1}{\left(\cos{12}\right)}$$ since $${\sin}^{-1}{x}={\cos}^{-1}{\sqrt{1-{x}^{2}}}$$
$$={\cos}^{-1}{\left(\cos{12}\right)}-{\cos}^{-1}{\left(\cos{12}\right)}=0$$
$$\tan ^{ -1 }{ (\frac { 5 }{ 12 } ) } +\sin ^{ -1 }{ (\frac { 24 }{ 25 } ) } =\cos ^{ -1 }{ (x) } \Rightarrow x=$$
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$$\frac { -31 }{ 325 } $$
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$$\frac { -33 }{ 325 } $$
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$$\frac { -36 }{ 325 } $$
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$$\frac { -39 }{ 325 } $$
If $$\cot \dfrac { 2{ x } }{ 3 } +\tan \dfrac { { x } }{ 3 } =\csc \dfrac { { kx } }{ 3 } ,$$ then the value of $$\tan ^{ { -1 } } (\tan { k } )$$ equals
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$$2$$
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$$2 - \pi$$
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$$\pi - 2$$
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$$2\pi - 2$$
$$ \sin ^{-1}\left(\dfrac{1}{\sqrt{2}}\right)+\sin ^{-1}\left(\dfrac{\sqrt{2}-1}{\sqrt{6}}\right)+\ldots+\sin ^{-1}\left(\dfrac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)+\ldots . \infty= $$
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$$\pi$$
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$$\cfrac {\pi}{2}$$
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$$\cfrac {\pi}{4}$$
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$$\cfrac {3\pi}{2}$$
$$4\tan ^{ -1 }{ \frac { 1 }{ 5 } } -\tan ^{ -1 }{ \frac { 1 }{ 70 } } +\tan ^{ -1 }{ \frac { 1 }{ 99 } } =$$
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$$\pi $$
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$${ \pi }/{ 2 }$$
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$${ \pi }/{ 4 }$$
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$${ 3\pi }/4$$
The value of the expression tan$$(\frac{1}{2} cos ^{-1}\frac{2}{\sqrt{5}})$$ is
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$$2-\sqrt{5}$$
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$$\sqrt{5}-2$$
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$$\frac{\sqrt{5}-2}{2}$$
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5-$$\sqrt{2}$$
If $$x = \sin ^ { - 1 } ( \sin 10 ) \text { and } y = \cos ^ { - 1 } ( \cos 10 )$$ then $$y - x$$ is equal to:
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$$\pi$$
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10
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$$7\pi$$
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0
If $$cos{ }^{ -1 }\frac { x }{ a } +cos{ }^{ -1 }\frac { y }{ b } =\alpha \quad then\quad \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { 2xy }{ ab } cos\alpha +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =$$
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$$sin^{2}\alpha$$
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$$cos^{2}\alpha$$
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$$tan^{2}\alpha$$
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$$cot^{2}\alpha$$
What is the value of $$\sin^{-1} \left\{ {\cos(\sin^{-1} x)} \right\} +\cos^{-1} \left\{ {\sin (\cos^{-1} x)} \right\} $$ ?
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$$2x$$
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$$2x+\pi$$
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$$\dfrac{\pi}{2}$$
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$$-\dfrac{\pi}{2}$$
$${\cot}^{-1}\left(\sqrt{\cos\alpha}\right) -{\tan}^{-1}\left(\sqrt{\cos\alpha}\right) =x$$, then $$\sin x$$ is equal to
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$$\displaystyle {\tan}^{2}\frac{\alpha}{2}$$
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$$\displaystyle {\cot}^{2}\frac{\alpha}{2}$$
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$$\tan\alpha$$
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$$\displaystyle \cot\frac{\alpha}{2}$$
The product of all values of x satisfying the equation.
$${ sin }^{ -1 }cos\left( \dfrac { { 2x }^{ 2 }+10\left| x \right| +4 }{ { x }^{ 2 }+5\left| x \right| +3 } \right) $$
$$=cot\left\{ { cot }^{ -1 }\left( \dfrac { 2-18\left| x \right| }{ 9\left| x \right| } \right) \right\} +\dfrac { \pi }{ 2 } $$ is
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9
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-9
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3
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-1
The value of $$\displaystyle sin^{1}\left ( sin\dfrac{5\pi}{3} \right )$$ is ......
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$$\dfrac{\pi}{3}$$
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$$\dfrac{5\pi}{3}$$
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$$\dfrac{\pi}{3}$$
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$$\dfrac{2\pi}{3}$$
The value of $$\tan \left[ \sin ^ { - 1 } \left( \cos \left( \sin ^ { - 1 } x \right) \right) \right]$$ $$\tan \left[ \cos ^ { - 1 } \left( \sin \left( \cos ^ { - 1 } x \right) \right) \right]$$ , $$( x \in ( 0,1 ) )$$ is equal to
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$$0$$
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$$1$$
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$$-1$$
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None
$$f ( x ) = \sin ^ { - 1 } \sqrt { \frac { \sqrt { 1 + x ^ { 2 } } - 1 } { 2 \sqrt { 1 + x ^ { 2 } } } } ,$$ then which of the following is (are) correct?
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$$f ( -1 ) = \frac { - 1 } { 4 }$$
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$$\operatorname { Ran } \mathscr { f } _ { f ( x ) }$ is $\left[ 0 , \frac { \pi } { 2 } \right]$$
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$$f ^ { \prime } ( x )$$ is an odd function
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$$\lim _ { x \rightarrow 0 } \frac { f ( x ) } { x } = \frac { 1 } { 2 }$$
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