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CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Maths
Inverse Trigonometric Functions
Quiz 7
If $$\tan^{-1}x=\dfrac{\pi}{10}$$ for some $$x\in R$$, then the value of $$\cot^{-1}x$$ is
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$$\dfrac{\pi}{5}$$
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$$\dfrac{2\pi}{5}$$
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$$\dfrac{3\pi}{5}$$
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$$\dfrac{4\pi}{5}$$
Explanation
(B) is the correct answer.
We know that
$$\tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2}$$.
$$\Rightarrow \cot^{-1}x=\dfrac {\pi}{2}-\tan^{-1}x$$
$$\Rightarrow \cot^{-1}x=\dfrac{\pi}{2}-\dfrac{\pi}{10}=\dfrac{4\pi}{10}=\dfrac{2\pi}{5}$$
The value of $$ \sin^{-1}\left\{\cos \left( \dfrac{43\pi}{5}\right)\right\}$$ is
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$$\dfrac{3\pi}{5}$$
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$$\dfrac{-7\pi}{5}$$
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$$\dfrac{\pi}{10}$$
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$$-\dfrac{\pi}{10}$$
Explanation
(D) is the correct answer.
We have
$$\sin^{-1}\left( \cos \dfrac{40\pi +3\pi}{5}\right)$$
$$=\sin^{-1}\left\{\cos \left( 8\pi+\dfrac{3\pi}{5}\right\}\right\}$$
$$=\sin^{-1}\left( \cos \dfrac{3\pi}{5}\right)$$
$$=\sin^{-1}\left\{ \sin \left( \dfrac {\pi}{2}-\dfrac{3\pi}{5}\right)\right\}$$
$$=\sin^{-1}\left( \sin \left( -\dfrac{\pi}{10}\right)\right)$$
$$=-\dfrac{\pi}{10}$$
The value of $$\cot (\sin^{-1}x)$$ is
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$$\dfrac{\sqrt{1+x^2}}{x}$$
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$$\dfrac{x}{\sqrt{1+x^2}}$$
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$$\dfrac{1}{x}$$
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$$\dfrac{\sqrt{1-x^2}}{x}$$
Explanation
(D) is the correct answer.
Let $$\sin^{-1}x=\theta $$ ,
Then $$\sin \theta =x$$
$$\Rightarrow \csc \theta =\dfrac 1x $$
$$\Rightarrow \csc^2 \theta =\dfrac {1}{x^2}$$
$$\Rightarrow 1+\cot^2\theta =\dfrac {1}{x^2}$$
$$\Rightarrow \cot \theta =\dfrac {\sqrt{1-x^2}}{x}$$
The domain of $$y=\cos^{-1}(x^2-4)$$ is
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$$[3, 5]$$
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$$[0, \pi]$$
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$$[-\sqrt 5, -\sqrt 3] \cap [ -\sqrt 5, \sqrt 3]$$
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$$[ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, 5]$$
Explanation
(D) is the correct answer.
Given $$y=\cos^{-1}(x^2-4) \Rightarrow \cos y =x^2-4$$
$$\therefore \ -1 \le x^2 -4 \le 1 $$ ( since $$-1 \le \cos y \le 1)$$
$$\Rightarrow 3 \le x^2 \le 5$$
$$\Rightarrow \sqrt 3 \le |x| \le \sqrt 5$$
$$\Rightarrow x \in [ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, \sqrt 5]$$
Hence the domain of $$y $$ is $$[ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, 5]$$
The domain of the function $$y=\sin^{-1}(-x^2)$$ is
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$$[0, 1]$$
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$$(0, 1)$$
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$$[-1, 1]$$
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$$\phi$$
Explanation
(C) is the correct answer,
Given $$y=\sin^{-1}(-x^2) \Rightarrow \sin y =-x^2$$
$$\therefore\ -1 \le -x^2 \le 1$$ ( since $$-1 \le \sin y \le 1)$$
$$\Rightarrow 1 \ge x^2 \ge -1$$
$$\Rightarrow 0 \le x^2 \le 1$$
$$\Rightarrow |x| \le 1$$
$$\Rightarrow-1 \le x \le 1$$
Hence the domain is $$[-1, 1]$$
The value of $$\sin (2\sin^{-1}(0.6))$$ is
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$$0.48$$
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$$0.96$$
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$$1.2$$
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$$\sin 1.2$$
Explanation
(B) is the correct answer.
Let $$\sin^{-1}(0.6) =\theta \Rightarrow\sin \theta =0.6$$.
$$\therefore\ \sin (2\sin^{-1}(0.6))\\=\sin ( 2\theta )\\=2\sin \theta \cos \theta \\=2 \sin\theta\sqrt{1-\sin^2\theta}\\=2(0.6)(\sqrt{1-{0.6}^2})\\=2(0.6)(0.8)=0.96$$
The value of the expression $$\sin [ \cot^{-1}\{ \cos (\tan^{-1}1)\}]$$ is
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$$0$$
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$$1$$
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$$\dfrac{1}{\sqrt 3}$$
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$$\sqrt{\dfrac{2}{3}}$$
Explanation
(D) is the correct answer.
Let $$P=\sin [ \cot^{-1}\{ \cos (\tan^{-1}1)\}]$$
$$\\=\sin \left[ \cot^{-1}\left( \cos \dfrac {\pi}{4}\right) \right] \qquad\left[\because \tan^{-1}1=\dfrac{\pi}4\right]\\=\sin \left[ \cot^{-1}\dfrac{1}{\sqrt 2}\right] \qquad\left[\because \cos\dfrac{\pi}4=\dfrac1{\sqrt2}\right]\\$$
Let, $$\cot^{-1}\dfrac1{\sqrt2}=y\Rightarrow \cot y=\dfrac1{\sqrt2}$$
$$\\ \Rightarrow \sin y=\sqrt{\dfrac23}\Rightarrow y=\sin^{-1}\sqrt{\dfrac23}$$
$$\therefore P=\sin \left[ \sin^{-1}\sqrt{\dfrac 23}\right] =\sqrt{\dfrac 23}$$
The value of $$\tan^2 ( \sec^{-1}2)+\cot^2 (\csc^{-1}3)$$ is
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$$5$$
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$$11$$
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$$13$$
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$$15$$
Explanation
(B) is the correct answer.
We have
$$\tan^2 ( \sec^{-1}2) \cot^2 ( \csc^{-1}3) $$
$$=\sec^2 (\sec^{-1}2)-1+\csc^2 (\csc^{-1}3)-1\qquad[\because \tan^2\theta=\sec^2\theta-1,\ \ \ \cot^2\theta=\csc^2\theta-1]$$
$$=(2)^2-1+(3)^2-1=11$$
If $$\alpha \le 2\sin^{-1}x+\cos^{-1}x \le \beta$$, then
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$$\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{\pi}{2}$$
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$$\alpha =0, \beta =\pi$$
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$$\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{3\pi}{2}$$
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$$\alpha =0, \beta =2\pi$$
Explanation
(B) is the correct answer.
We know that
$$\dfrac {-\pi}{2}\le \sin^{-1}x \le \dfrac {\pi}{2}\\$$
$$\Rightarrow \dfrac {-\pi}{2}+\dfrac {\pi}{2} \le \sin^{-1}x + \dfrac {\pi}{2} \le \dfrac {\pi}{2}+\dfrac {\pi}{2}\qquad [$$Addidng $$\dfrac{\pi}2 ]$$
$$\\\Rightarrow 0 \le \sin^{-1}x+ (\sin^{-1}x+\cos^{-1}x) \le \pi\qquad\left[\because(\sin^{-1}x+\cos^{-1}x)= \dfrac{\pi}2 \right]\\$$
$$\Rightarrow 0 \le 2\sin^{-1}x+\cos^{-1}x \le \pi$$
$$\sin^{-1} (1 -x) - 2 \sin ^{-1} x = \dfrac{\pi }{2} , $$ then x
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$$0 , \dfrac{1 }{2}$$
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$$ 1, \dfrac{1}{2}$$
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$$ 0 $$
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$$ \dfrac{1 }{2}$$
Explanation
$$\sin^{-1} (1 -x) - 2 \sin ^{-1} x = \dfrac{\pi }{2} $$
$$ -2\sin ^{-1} x = \dfrac{\pi }{2} - \sin^{-1} (1 -x)$$
$$-2 \sin ^{-1} x = \cos^{-1} (1 -x)$$
$$\cos ( -2 \sin^{-1} x) = 1 -x$$
$$\cos ( 2 \sin^{-1} x) = 1 -x$$
$$\cos (2 sin ^{-1} x) = 1 -x $$
$$ 1 - 2 \sin^{2} (sin^{-1} x) = 1 - x $$
$$1 - 2x^{2} = 1- x\Rightarrow $$
$$2x^{2} -x = 0 \\x(2x -1) = 0\\ \Rightarrow x = 0 , x = \dfrac{1}{2}$$
$$\dfrac{1}{2}$$ does not satisfy the equation $$\therefore x = 0$$ is the only solution
Choose the correct answer
$$ \cos ^{-1} ( \cos \dfrac{7\pi }{6}) $$ 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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$$ \dfrac{\pi }{6} $$
Explanation
$$ \cos^{-1} \cos\left ( \dfrac{7\pi }{6} \right )$$
$$ \cos^{-1} \left ( \cos\left ( \pi + \dfrac{\pi }{6} \right ) \right ) = \cos^{-1} \left ( - \cos\dfrac{\pi}{6} \right ) $$
$$ =\pi- \cos^{-1} \left ( \cos\dfrac{\pi }{6} \right ) $$
$$=\pi-\dfrac{\pi}{6}$$
$$ = \dfrac{5\pi }{6} $$
Choose the correct answer :
$$ \sin \left ( \dfrac{\pi }{3} - \sin^{-1}\left ( -\dfrac{1}{2} \right ) \right ) $$ is equal to
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$$ \pi $$
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$$ - \dfrac{\pi }{2} $$
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$$ 1 $$
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$$ 2 \sqrt{3}$$
Explanation
$$ \sin \left [ \left ( \dfrac{\pi}{3} - \sin^{-1} \left ( \dfrac{-1}{2} \right ) \right ) \right ] $$
$$ \left [ \sin \left [ \left ( \dfrac{\pi}{3} - \sin^{-1} \left ( \dfrac{-1}{2} \right ) \right ) \right ] \right ] $$
$$ \sin \left ( \dfrac{\pi }{3} + \sin^{-1}\left ( \sin \dfrac{\pi }{6} \right ) \right ) $$
$$\left ( as \sin^{-1} \left ( \dfrac{-1}{2} \right ) = \sin^{-1}\dfrac{1}{2} and \sin \dfrac{\pi }{6} = \dfrac{1}{2} \right ) $$
$$ = \sin \left ( \dfrac{\pi }{3} + \dfrac{\pi }{6} \right ) $$
$$ = \sin\dfrac{\pi }{2} = 1 $$
If $$ \sin ^{-1} x = y $$ , then
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$$ 0 \leq y \leq \pi $$
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$$ - \dfrac{\pi}{2} \leq y \leq \dfrac{\pi }{2} $$
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$$ 0 < y < \pi $$
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$$ - \dfrac{\pi}{2} < y < \dfrac{\pi }{2} $$
Explanation
pv of $$ \sin^{-1} x$$ is $$\left [ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right ] $$
$$\therefore (B) - \dfrac{\pi}{2} \leq y \leq \dfrac{\pi }{2} $$ is correct
$$ \tan^{-1} \sqrt{3} - \sec^{-1} (-2) $$ is equal to
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$$ \pi $$
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$$ - \dfrac{\pi }{3} $$
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$$ \dfrac{\pi }{3} $$
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$$ - \dfrac{2\pi }{3} $$
Explanation
$$ \tan^{-1} \sqrt{3} = \dfrac{\pi }{3} \epsilon \left [ \dfrac{-\pi }{2} ,\dfrac{\pi }{2} \right ]$$
$$ \sec^{-1} (-2) = \pi - \dfrac{\pi }{3} = \dfrac{2 \pi}{3} \epsilon [0, \pi] $$
$$\therefore \tan^{-1} \sqrt{3} - sec^{-1} (-2) = \dfrac{\pi}{3} - \dfrac{2 \pi}{3} $$
$$ = -\dfrac{\pi}{3} $$
$$ \therefore $$ Option (B) is correct .
Multiple choice Questions :
$$ 2\sin \left ( \cos^{-1}\left ( \dfrac{-4}{5} \right ) \right )\times \cos \cos^{-1} \left ( \dfrac{-4}{5} \right )$$
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$$ \dfrac{24}{25}$$
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$$ - \dfrac{24}{25}$$
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$$ - \dfrac{6}{25}$$
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$$ \dfrac{- 6}{25}$$
Explanation
$$ 2\sin \left ( \cos^{-1}\left ( \dfrac{-4}{5} \right ) \right )\times \cos \cos^{-1} \left ( \dfrac{-4}{5} \right )$$
$$ = 2 \sqrt{1-\left ( \dfrac{-4}{5} \right )^{2}} \times\left ( \dfrac{-4}{5} \right )$$
$$= -\dfrac{8}{5} \sqrt{\dfrac{9}{25}} = \dfrac{-8}{5} \times\dfrac{3}{5} = \dfrac{-24}{25}$$
Multiple choice Questions :
$$ \sin^{-1}\left ( \dfrac{4}{5} \right ) + \cos^{-1}\left ( \dfrac{4}{5} \right )=$$
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$$ \dfrac{\pi}{2} $$
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$$ 0 $$
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$$ \pi $$
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$$ \dfrac{-\pi}{2} $$
Explanation
$$ \sin ^{1} x + \cos^{-1} x= \dfrac{\pi}{2} $$
$$ \sin^{-1}\left ( \dfrac{4}{5} \right ) + \cos^{-1}\left ( \dfrac{4}{5} \right )= \dfrac{\pi}{2} $$
Multiple choice Questions :
$$ \cos^{-1} \cos \left ( \dfrac{4 \pi}{3} \right ) = $$
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$$ \dfrac{4 \pi}{3} $$
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$$ \dfrac{2 \pi}{3} $$
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$$ \dfrac{- \pi}{3} $$
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$$ - \pi $$
Explanation
$$ \cos^{-1} \cos \left ( \dfrac{4 \pi}{3} \right ) = \cos^{1}\cos \left ( \pi + \dfrac{\pi}{3} \right ) $$
$$ = \cos^{-1} \left [ - cos \dfrac{\pi}{3} \right ] $$
$$ = \pi - cos^{-1} \left ( \cos\dfrac{\pi}{3} \right ) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$$
Multiple choice Questions :
The value of $$ \tan^{-1} (\tan 5)$$
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$$ 2 \pi $$
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$$ 5 - 2 \pi $$
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$$ 5 $$
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$$ \dfrac{2 \pi}{3} $$
Explanation
Let $$ 2\pi - 5 = \theta $$
$$ \Rightarrow 5 = 2 \pi - \theta $$
$$ \tan 5 = \tan ( 2 \pi - \theta )$$
$$ \tan^{-1} (\tan 5) = \tan^{-1} \tan ( 2 \pi - \theta)$$
$$ = \tan^{-1}[-\tan \theta]=-\theta$$
$$\tan ^{-1}\left ( \dfrac{x}{y} \right ) - \tan^{-1} \dfrac{x - y}{x + y}$$ is equal to
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$$\dfrac{\pi }{2}$$
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$$\dfrac{\pi }{3}$$
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$$\dfrac{\pi }{4}$$
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$$\dfrac{-3\pi }{4}$$
Explanation
$$\tan^{-1} \left ( \dfrac{x}{y} \right ) - \left ( \tan^{-1}\dfrac{\frac{x}{y}-1}{1+\dfrac{x}{y}} \right )$$
$$=\tan^{-1} \left ( \dfrac{x}{y} \right )-\tan ^{-1} \left ( \dfrac{x}{y} \right )+ \tan^{-1} (1)$$
$$\tan^{-1} 1 = \dfrac{\pi }{4}$$
Multiple choice Questions :
If a > b > c ,
$$ \cot^{-1} \left ( \dfrac{1 + ab}{a - b} \right ) + \cot^{-1}\left ( \dfrac{1 + bc}{b - c} \right ) + \cot^{-1} \left ( \dfrac{1 + ac}{c - a} \right ) $$
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$$ 0 $$
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$$ \pi $$
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$$ 2 \pi $$
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$$\dfrac{\pi}{2}$$
Explanation
$$ \tan ^{-1}\left ( \dfrac{a - b}{1 + ab} \right ) + \tan^{-1} \left ( \dfrac{b - c}{1 + bc} \right ) + \tan^{-1} \left ( \dfrac{c - a}{1 + ac} \right ) $$
$$ = \tan^{-1} a - \tan^{-1} b + \tan^{-1} b - \tan^{-1} c + \pi - \tan^{-1} a + \tan^{-1} c $$
$$ = \pi \ \ \ \ \ \ \ \ as \tan^{-1} \left ( \dfrac{c - a}{1 + ac} \right ) = \pi - \tan^{-1} \left ( \dfrac{a - c}{1 + ac} \right )$$
If $$\sin^{-1}\left(\dfrac{1}{2}\right)=x$$, then general value $x$$ is:
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$$2n\pi \pm \dfrac{\pi}{6}$$
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$$\dfrac{\pi}{6}$$
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$$n\pi \pm \dfrac{\pi}{6}$$
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$$n\pi +(-1)^n\dfrac{\pi}{6}$$
Explanation
Given,
$$\sin^{-1}\left(\dfrac{1}{2}\right)=x$$
$$\Rightarrow \sin x=\dfrac{1}{2}=\sin \dfrac{\pi}{6}$$
$$\Rightarrow x=\dfrac{\pi}{6}$$
$$\therefore$$ General value of $$x, \theta=n\pi +(-1)^{n}\dfrac{\pi}{6}$$
Hence, option $$(d)$$ is correct.
If $$\tan^{-1}(1)+\cos^{-1}(\dfrac{1}{\sqrt{2}})=\sin^{-1}x$$, then value of $$x$$ is
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$$-1$$
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$$0$$
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$$1$$
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$$-\dfrac{1}{2}$$
Explanation
$$\tan^{-1}(1)+\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right)=\sin^{-1}x$$
$$\Rightarrow \sin^{-1}x=\dfrac{\pi}{4}+\dfrac{\pi}{4}$$
$$\Rightarrow \sin^{-1}x=\dfrac{\pi}{2}$$
$$\Rightarrow x=\sin \dfrac{\pi}{2}$$
$$\Rightarrow x=1$$
Hence, option $$(c)$$ is correct.
$$2\tan(\tan^{-1}x+\tan^{-1}x^3)$$ is:
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$$\dfrac{2x}{1-x^2}$$
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$$1+x^2$$
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$$2x$$
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none of these
Explanation
$$2\tan (\tan^{-1}x+\tan^{-1}x^{3})$$
$$=2\tan \left\{\tan^{-1}\left(\dfrac{x+x^{3}}{1-x\times x^{3}}\right)\right\}$$
$$=2\tan \left\{\tan^{-1}\left(\dfrac{x(1+x^{2})}{1-x^{4}}\right)\right\}$$
$$=2\tan \left\{\tan^{-1}\dfrac{x(1+x^{2})}{(1-x^{2})(1+x^{2})}\right\}$$
$$=2\tan \dfrac{x}{1-x^{2}}=\dfrac{2x}{1-x^{2}}$$
Hence, option $$(a)$$ is correct.
Value of $$\sin^{-1}(\dfrac{\sqrt{3}}{2})+2\cos^{-1}(\dfrac{\sqrt{3}}{2})$$ is:
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$$\dfrac{\pi}{2}$$
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$$\dfrac{\pi}{3}$$
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$$\dfrac{2\pi}{3}$$
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$$\pi$$
Explanation
$$\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)+2\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$$
$$=\left[\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)+\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)\right]+\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$$
$$=\dfrac{\pi}{2}+\dfrac{\pi}{6}=\dfrac{2\pi}{3}\left[\because \sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}\right]$$
Hence, option $$(c)$$ is correct.
If $$\cot^{-1}x+\tan^{-1}\dfrac{1}{3}=\dfrac{\pi}{2}$$ then $$x$$ is:
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$$1$$
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$$3$$
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$$\dfrac{1}{3}$$
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none of these
Explanation
$$\cot^{-1}(x)+\tan^{-1}\left(\dfrac{1}{3}\right)=\dfrac{\pi}{2}$$
$$\Rightarrow \cot^{-1}x=\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{1}{3}\right)$$
$$\Rightarrow \cot^{-1}x=\cot^{-1}\dfrac{1}{3}$$
On comparing $$x=\dfrac{1}{3}$$
Hence, option $$(c)$$ is correct.
If $$\tan^{-1}(3x)+\tan^{-1} 2x=\dfrac{\pi}{4}$$, then $$x$$ is:
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$$\dfrac{1}{6}$$
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{10}$$
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$$\dfrac{1}{2}$$
Explanation
$$\tan^{-1}(3x)+\tan^{-1}(2x)=\dfrac{\pi}{4}$$
$$\Rightarrow \tan^{-1}\left\{\dfrac{3x+2x}{1-3x\times 2x}\right\}=\dfrac{\pi}{4}$$
$$\Rightarrow \left(\dfrac{5x}{1-6x^{2}}\right)=\tan \dfrac{\pi}{4}$$
$$\Rightarrow \dfrac{5x}{1-6x^{2}}=1$$
$$\Rightarrow 1-6x^{2}=5x$$
$$\Rightarrow 6x^{2}+5x-1=0$$
$$\Rightarrow 6x^{2}+6x-x-1=0$$
$$\Rightarrow 6x(x+1)-1(x+1)=0$$
$$\Rightarrow (x+1)(6x-1)=0$$
$$\Rightarrow x=-1, x=\dfrac{1}{6}$$
Hence, option $$(a)$$ is correct.
lf the equation $$\displaystyle \sin^{-1}(x^{2}+x+1)+\cos^{-1}(\lambda x+1)=\frac{\pi}{2}$$ has exactly two solutions, then $$\lambda$$ can not have the integral value(s)
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$$-1$$
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$$0$$
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$$1$$
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$$2$$
Explanation
$$\displaystyle \sin ^{-1}(x^{2}+x+1)+\cos ^{-1}(\lambda x+1)=\frac{\pi }{2}$$
$$\displaystyle \cos ^{-1}(\lambda x+1)=\frac{\pi }{2} -\sin ^{-1}(x^{2}+x+1)$$
Taking $$\cos$$ on both sides,
$$\lambda x+1=x^{2}+x+1$$
$$x^{2}+(1-\lambda )x=0$$
$$x(x+1-\lambda )=0$$
$$x=0$$ or $$\lambda =x+1$$
$$\sin ^{-1}(x^{2}+x+1)$$ is defined for $$-1 \leq x^{2}+x+1 \leq 1$$
$$x^{2}+x+2 \geq 0$$ is always true.
$$x(x+1) \leq 0 \Rightarrow x\in [-1,0]$$
$$\Rightarrow x+1 \in [0,1]$$
$$\lambda \in [0,1]$$
$$\lambda \neq -1,2$$
Also when $$\lambda =1$$, there is only one solution to the given equation i.e., $$x=0$$
So, $$\lambda\neq 1$$.
Assertion(A): $$\cos^{-1}x$$ and $$\tan^{-1}x$$ are positive for all positive real values of $$x$$ in their domain.
Reason(R): The domain of $$f(x)=\cos^{-1}x+\tan^{-1}x$$ is $$[-1, 1].$$
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Both A and R are true and R is the correct explanation of A
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Both A and R are true but R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
Assertion :
$$ \cos ^ {-1} x$$ range is $$ \left[0,\pi \right]$$ and domain is $$[-1,1]$$
$$\therefore \,\cos^{-1}x>0\,$$ for all $$x>0$$
$$ \tan ^ {-1} x $$ range is $$ \left(\dfrac{ -\pi }2 ,\dfrac \pi 2\right) $$ and range is $$R$$
$$ \forall x >0 \implies \tan ^ {-1} x >0$$
Reason :
The domain of $$ \cos ^ {-1} x $$ is $$ [-1,1]$$
The domain of $$ \tan ^ {-1} x $$ is $$R$$
So The domain of $$ \cos ^ {-1}x +\tan ^ {-1}x $$ is $$ R \cap [-1,1]$$
$$\implies [-1,1]$$
$$\sin^{-1}|\sin x|=\sqrt{\sin^{-1}|\sin x|}$$ then $$x=$$
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$$n\pi-1$$
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$$ n\pi$$
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$$n\pi+1$$
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$$n\displaystyle \frac{\pi}{2}+1$$
Explanation
Case - 1
$$|sin x|=sin x \ \ x\epsilon [2n\pi , (2n+1)\pi ]$$
$$sin ^{-1} sin x =\sqrt{sin ^{-1}sin x}$$
$$x=\sqrt{x}$$
$$\sqrt{x}=1\ \ or \ \ x=0$$
$$x=1$$
Case - 2
$$|sin x|=-sin x\ \ x\epsilon [(2n-1)\pi , 2n\pi ]$$
$$sin ^{-1}(-sin x)=\sqrt{sin ^{-1}(-sin x)}$$
$$-x=\sqrt{-x}$$
$$x=-1 , x=0$$
So,
$$x=n\pi -1 \ for\ n=0$$
$$x=n\pi\ for\ n=0$$
$$x=n\pi -1 \ for\ n=0$$
The number of solutions of:
$$\displaystyle \sin^{-1}(1+b+b^{2}+\ldots.\infty)+\cos^{-1}(a-\frac{a^{2}}{3}+\frac{a^{3}}{9}+\ldots\infty)=\frac{\pi}{2}$$
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$$1$$
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$$2$$
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$$3$$
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$$\infty$$
Explanation
As $$\displaystyle \sin ^{ -1 }{ \theta } $$ are defined for $$\displaystyle \theta \in \left[ -1,1 \right] $$
$$\therefore$$ For $$\displaystyle \sin ^{ -1 }{ \left( 1+b+{ b }^{ 3 }+...\infty \right) }$$
$$ \left| b \right| <1$$
Hence $$\displaystyle 1,b,{ b }^{ 2 },...\infty $$ is $$G.P.$$ of common ratio $$b$$
$$\displaystyle \therefore 1+b+{ b }^{ 2 }+...+\infty =\frac { 1 }{ 1-b } $$
Similarly for $$\displaystyle \cos ^{ -1 }{ \left( a-\frac { { a }^{ 2 } }{ 3 } +\frac { { a }^{ 3 } }{ 9 } +...\infty \right) } $$
$$\displaystyle\left| \frac { a }{ 3 } \right| <1$$
Hence $$\displaystyle 1,-\frac { { a }^{ 2 } }{ 3 } ,\frac { { a }^{ 3 } }{ 9 } ,...\infty $$ is $$G.P.$$ of common ration $$\displaystyle \left( \frac { -a }{ 3 } \right) $$
$$\displaystyle \therefore a-\frac { { a }^{ 2 } }{ 3 } +\frac { { a }^{ 3 } }{ 9 } +...\infty =\frac { a }{ 1+\frac { a }{ 3 } } =\frac { 3a }{ 3+a } $$
Now, as $$\displaystyle \sin ^{ -1 }{ \left( x \right) } +\cos ^{ -1 }{ \left( x \right) } =\frac { \pi }{ 2 } $$
$$\displaystyle \therefore \frac { 1 }{ 1-b } =\frac { 3a }{ 3+a } $$
$$\displaystyle \Rightarrow 3+a=3a-3ab$$
$$\displaystyle \Rightarrow 2a-3=3ab$$
This has infinite solution.
If $$(\tan^{-1}x)^{2}+(\cot^{-1}x)^{2} = \displaystyle \frac{5\pi^{2}}{8}$$, then $$x=$$
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$$-1$$
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$$1$$
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$$0$$
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$$2$$
Explanation
Let $$tan^{-1}x=y$$
Therefore
$$y^2+(\frac{\pi}{2}-y)^{2}=\frac{5\pi^2}{8}$$
$$2y^2-\frac{2\pi y}{2}+\frac{\pi^2}{4}=\frac{5\pi^2}{8}$$
$$2y^2-\pi y-\frac{3\pi^2}{8}=0$$
$$y=\frac{-\pi}{4}$$ and $$y=\frac{3\pi}{4}$$
Hence
$$tan^{-1}(x)=\frac{-\pi}{4}$$
$$x=-1$$ and
$$tan^{-1}(x)=\frac{3\pi}{4}$$
$$x=-1$$
However only $$x=-1$$ satisfies the above quadratic equation.
$$cos^{-1} \left (\sqrt{\dfrac{a-x}{a-b}} \right)$$ =$$sin^{-1} \left (\sqrt{\dfrac{x-b}{a-b}}\right)$$ is possible if
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$$a>x>b$$ or $$ a< x < b $$
0%
$$a=x=b$$
0%
$$a > b$$ and x takes any value
0%
$$a < b$$ and x takes any value
Explanation
In $$\cos ^{-1}x , x \in [-1,1]$$
In $$\sin ^{-1}x , x\in [-1,1]$$
$$\displaystyle -1 \leq \sqrt{\frac{a-x}{a-b}} \leq 1$$
$$\displaystyle 0 \leq \sqrt{\frac{a-x}{a-b}} \leq 1$$
$$0 \leq a-x \leq a-b$$
$$0\geq x-a \geq b-a$$ [on multiplying by (-ve) sing direction of equality changes]
$$a \geq x\geq b$$
and
$$\displaystyle 0\leq \sqrt{\frac{x-b}{a-b}}\leq 1$$
$$0\leq x-b \leq a-b$$
$$b \leq x\leq a$$
$$x\epsilon [b,a]$$
The solution set of the equation $$\tan^{-1}x -\cot^{-1}x =\cos^{-1}(2-x)$$ is
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0%
$$(0,1)$$
0%
$$(-1,1)$$
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$$[1,3)$$
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$$(1,3)$$
Explanation
Given, $$\tan ^{-1}x-\cot ^{-1}x=\cos ^{-1}(2-x) x\in [1,3]$$
$$2\tan ^{-1}x-\dfrac{\pi }{2}=\cos ^{-1}(2-x)$$ $$\left[\because \tan ^{-1}x+\cot ^{-1}x=\dfrac{\pi }{2}\right]$$
$$2 \tan ^{-1}x=\dfrac{\pi }{2}+\cos ^{-1}(2-x)$$
Take $$\sin$$ on both sides,
$$\Rightarrow 2\dfrac{x}{(1+x^{2})}=2-x \quad\quad........2\tan^{-1}x=sin^{-1}\dfrac{2x}{1+x^2}$$
$$\Rightarrow x^3-2x^{2}+3x-2=0$$
$$\Rightarrow x = 1$$
So, set containing solution is $$[1,3)$$.
If $$\sin^{-1}\alpha+\sin^{-1}\beta+\sin^{-1}\gamma =\displaystyle \frac{3\pi}{2}$$, then $$\alpha\beta+\alpha\gamma+\beta\gamma$$ is equal to :
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$$1$$
0%
$$0$$
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$$3$$
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$$-3$$
Explanation
The above expression is true for
$$\alpha=1$$ , $$\beta=1$$ and $$\gamma=1$$
Since $$\dfrac{-\pi}{2}\leq sin^{-1}(x) \leq \dfrac{\pi}{2}$$
Hence
$$\alpha\beta+\beta\gamma+\gamma\alpha$$
$$=(1)+(1)+(1)$$
$$=3$$
The number of positive integral solutions of the equation $$tan^{-1}x+cot^{-1}y =tan^{-1} 3$$ is :
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$$0$$
0%
$$1$$
0%
$$2$$
0%
$$3$$
Explanation
$$cot^{-1}y=tan ^{-1}3-tan ^{-1}x$$
$$\displaystyle cot^{-1}y=tan ^{-1}(\dfrac{3-x}{1+3x})$$
Take tan on both sides
$$\displaystyle \dfrac{1}{y}=\dfrac{3-x}{1+3x}$$
$$\displaystyle y=\dfrac{1+3x}{3-x} \Rightarrow y > 0$$
Put $$x=0 , y=\dfrac{1}{3}$$
$$\dfrac{1+3x}{3-x} > 0$$
$$x=1 , y=2$$
$$\dfrac{3x+1}{x=3} < 0$$
$$x=2 , y=7$$
$$x\epsilon (-\dfrac{1}{3} , 3)$$
The integral values of $$x=0,1,2$$
The domain of $$\displaystyle \mathrm{f}(\mathrm{x})=\cot^{-1}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}-[\mathrm{x}^{2}]}}\right)$$ is
( $$[\;.] $$ denotes the greatest integer function)
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$$(0,\infty)$$
0%
$$\mathrm{R}-\{0\}$$
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$$\mathrm{R}-\{\mathrm x:\mathrm{x}\in \mathrm{Z}\}$$
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$$(-\infty,0)$$
Explanation
$$f(x)=\cot^{-1}\left ( \displaystyle \frac{x}{\sqrt{x^2-[x^2]}} \right )$$
For $$f(x)$$ to be defined, $$x^2-[x^2]>0$$
$$\therefore x^2$$ should not be an integer
Hence, $$x$$ should not be an integer.
Hence, Option C is the answer
If $$(tan^{-1} x)^2 +(cot ^{-1}x)^2=\displaystyle \frac{5 \pi^2}{8}$$, then $$x$$ =
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0%
$$-1$$
0%
$$0$$
0%
$$1$$
0%
$$2$$
Explanation
We have, $$tan^{-1}x + cot^{-1}x =\displaystyle\frac{\pi}{2}$$
The given equation can be written as :
$$\displaystyle (tan^{-1}x + cot^{-1}x)^2 - 2tan^{-1}x\left ( \frac{\pi}{2}-tan^{-1}x \right )=\frac{5 \pi^2}{8}$$
$$\displaystyle 2(tan^{-1}x)^2 - 2\left ( \frac{\pi}{2} \right )tan^{-1}(x)- \frac{3 \pi ^2}{8}=0$$
$$\displaystyle tan^{-1}x=-\frac{\pi}{4} or \frac{3\pi}{4}$$
Hence, $$tan^{-1}x =1 $$
$$x=-1$$
The number of integral solutions of $$sin^{-1}\sqrt{4x-x^{2}-3}+tan^{-1}\sqrt{x^{2}-3x+2}=\frac{\pi }{2}$$ is
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0%
zero
0%
infinite
0%
four
0%
None of these
Explanation
Given equation is
$$\sin^{-1}\sqrt{4x-x^{2}-3}+\tan^{-1}\sqrt{x^{2}-3x+2}=\frac{\pi }{2}$$
$$\tan^{ -1 }\sqrt { x^{ 2 }-3x+2 }=\displaystyle\frac { \pi }{ 2 } -\sin^{ -1 }\sqrt { 4x-x^{ 2 }-3 } $$
$$\Rightarrow \tan^{ -1 }\sqrt { x^{ 2 }-3x+2 } =\cos^{ -1 }\sqrt { 4x-x^{ 2 }-3 } $$
Since, $$\sqrt { x^{ 2 }-3x+2 } \ge 0$$
$$\Rightarrow \tan ^{ -1 }{ \sqrt { x^{ 2 }-3x+2 } } <\dfrac { \pi }{ 2 } $$
Also, $$\sqrt { 4x-x^{ 2 }-3 } \ge 0$$
$$\Rightarrow 0<\cos^{ -1 }\sqrt { 4x-x^{ 2 }-3 } \le \dfrac { \pi }{ 2 } $$
The value of $$sin^{-1}(sin2010^{0})+cos^{-1}(cos2010^{0})+tan^{-1}(tan2010^{0})$$ is
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$$\frac{\pi }{6}$$
0%
$$\frac{\pi }{3}$$
0%
$$\frac{2\pi }{3}$$
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$$\frac{5\pi }{6}$$
Explanation
$$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}$$
The number of solutions of the equation $$1+x^{2}+2x\, sin\: (cos^{-1}y)=0$$ is
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0%
1
0%
2
0%
3
0%
4
Explanation
$$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$$ only one solution
The value of $$\displaystyle \sec^{-1}\left (\displaystyle \frac{1}{1-2x^{2}}\right)+4{\cos^{-1}}\sqrt{\displaystyle \frac{1+x}{2}}$$ is equal to
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$$\pi $$
0%
$$2\pi$$
0%
$$\dfrac{\pi}{2}$$
0%
None of these
Explanation
Given
$$\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 $$
The largest interval lying in $$\left ( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right )$$ for which the function $$\left [ f(x)=4^{-x^{2}}+\cos^{-1}\left ( \dfrac{x}{2}-1 \right )+\log (\cos x) \right ]$$ is defined, is-
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0%
$$[0,\pi ]$$
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$$\left ( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right )$$
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$$\left [- \dfrac{\pi }{4},\dfrac{\pi }{2} \right )$$
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$$\left [0,\dfrac{\pi }{2} \right )$$
Explanation
For given function to be defined $$-1 \leq \cfrac{x}{2}-1 \leq 1$$ and $$\cos x > 0$$
$$\Rightarrow 0 \leq \cfrac{x}{2} \leq 2$$ and $$ \cfrac{-\pi}{2} \leq x < \cfrac{\pi}{2}$$
$$\Rightarrow 0 \leq x \leq 4$$ and $$ \cfrac{-\pi}{2} \leq x < \cfrac{\pi}{2}$$
Taking common of both we get $$x \in \left[0,\cfrac{\pi}{2}\right)$$
The number of real solutions of $$tan^{-1} (\sqrt{x(x+1)}+sin^{-1} \displaystyle \sqrt{(x^{2}+x+1)}=\dfrac{\pi}{2}$$ is
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0%
$$0$$
0%
$$1$$
0%
$$2$$
0%
infinite
Explanation
From the above expression
$$-1\leq\sqrt{x^2+x+1}\leq 1$$ and $$x(x+1)\geq 0$$
$$\Rightarrow 0\le x^2+x+1\le1 \,and \,x^2+x+1\ge1 $$
$$\Rightarrow x^2+x+1=1$$
$$\Rightarrow x^2+x=0$$
$$\Rightarrow x(x+1)=0$$
Hence there will be two solutions. One at $$x=-1$$ and another at$$ x=0$$.
If $$x> 0\, $$ and $$\, cos^{-1}\left ( \dfrac{12}{x} \right )+cos^{-1}\left ( \dfrac{35}{x} \right )=\dfrac{\pi }{2},$$ then x is
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0%
$$7$$
0%
$$39$$
0%
$$37$$
0%
$$-37$$
Explanation
$$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$$
The value of $$\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
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$$0$$
0%
$$\dfrac{\pi }{2}$$
0%
$$-\dfrac{\pi }{2}$$
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$$\pi $$
Explanation
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,
$$\Rightarrow E= \sin^{-1} (\tan (15^0 + 30^0 - 45^0))$$
$$\Rightarrow E= \sin^{-1} (\tan0^0)$$
$$\Rightarrow E = \sin^{-1} 0 = 0$$
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
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$$\displaystyle\:\left [ -\frac{\pi }{2},-1\right ]\cup \left [ 1,\frac{\pi }{2} \right ]$$
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$$\displaystyle\:\left (-\frac{\pi }{2},-1\right )\cup \left (1,\frac{\pi }{2} \right )$$
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$$\displaystyle\:\left ( -\infty ,-1]\cup [1,\infty \right )$$
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$$\displaystyle\:\left ( -\infty ,-1)\cup (1,\infty \right )$$
Explanation
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
$$=\left[\dfrac{-\pi}{2},-1\right]\cup\left[1,\dfrac{\pi}{2}\right]$$
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
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$$[\tan { \sin { \cos { 1 } } } ,\sin { \tan { \cos { \sin { 1 } } } } ]$$
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$$[\sin { \tan { \cos { 1 } } } ,\tan { \sin { \cos { \sin { 1 } } } } ]$$
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$$[\tan { \sin { \cos { 1 } } } ,\tan { \sin { \cos { \sin { 1 } } } } ]$$
0%
None of these
Explanation
We have $$\left[ \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } } \right] =1$$ $$\Rightarrow 1\le \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } }\le \displaystyle\frac { \pi }{ 2 }$$
$$\Rightarrow \sin { 1 } \le \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } \le 1$$ $$\Rightarrow \cos { \sin { 1 } } \ge \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } \ge \cos { 1 }$$
$$\Rightarrow \sin { \cos { \sin { 1 } } } \ge \tan ^{ -1 }{ \theta } \ge \sin { \cos { 1 } }$$ $$\Rightarrow \tan { \sin { \cos { \sin { 1 } } } } \ge \theta \ge \tan { \sin { \cos { 1 } } } $$
$$\displaystyle \:\cos ^{-1}x+\cos ^{-1}\left ( \frac{x}{2}+\frac{1}{2}\sqrt{3-3x^{2}} \right )$$ is equal to
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$$\displaystyle \:\frac{\pi }{3} $$ for $$ x \epsilon \left [ \dfrac{1}{2},1 \right ]$$
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$$\displaystyle \:\frac{\pi }{3} $$ for $$ x\epsilon \left [ 0,\dfrac{1}{2} \right ]$$
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$$\displaystyle \:2\cos ^{-1}x-\cos ^{-1}\dfrac{1}{2} $$ for $$ x \epsilon \left [ \dfrac{1}{2},1 \right ]$$
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$$\displaystyle \:2\cos ^{-1}x-\cos ^{-1}\dfrac{1}{2}$$ for $$ x \epsilon \left [ 0,\dfrac{1}{2}\right ]$$
Explanation
$$cos^{ -1 }x+cos^{ -1 }\left( x.\cfrac { 1 }{ 2 } +\cfrac { \sqrt { 3 } }{ 2 } \sqrt { 1-x^{ 2 } } \right) =cos^{ -1 }x+cos^{ -1 }\cfrac { 1 }{ 3 } -cos^{ -1 }x$$
For $$\cfrac { 1 }{ 2 } \le x\le 1$$
$$cos^{ -1 }x+cos^{ -1 }\left( x.\cfrac { 1 }{ 2 } +\cfrac { \sqrt { 3 } }{ 2 } \sqrt { 1-x^{ 2 } } \right) =cos^{ -1 }x+\cfrac { \pi }{ 3 } -cos^{ -1 }x=\cfrac { \pi }{ 3 } $$
And for $$0\le x\le \cfrac { 1 }{ 2 } $$
$$os^{ -1 }x+cos^{ -1 }\left( x.\cfrac { 1 }{ 2 } +\cfrac { \sqrt { 3 } }{ 2 } \sqrt { 1-x^{ 2 } } \right) =cos^{ -1 }x+cos^{ -1 }x-cos^{ -1 }\cfrac { 1 }{ 2 } =2cos^{ -1 }x-cos^{ -1 }\cfrac { 1 }{ 2 } $$
The range of values of
p
for which the equation $$\sin \cos ^{-1}(\cos (\tan ^{-1}x))= p$$ has a solution is
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$$\left ( -\frac{1}{\sqrt{2}},\frac{2}{\sqrt{2}} \right )$$
0%
$$[0,1)$$
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$$\left ( \frac{1}{\sqrt{2}1} \right )$$
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$$(-1,1)$$
Explanation
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).
The solution set of the equation $$\displaystyle \sin^{-1} \sqrt{1-x} + \cos^{-1}x = \cot^{-1} \left ( \frac{\sqrt{1-x^{2}}}{x} \right )-\sin^{-1}x$$
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$$\displaystyle \left [ -1,\: 1 \right ] - \left \{ 0 \right \}$$
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$$\displaystyle \left ( 0, \: 1 \right ] \cup \left \{ -1 \right \}$$
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$$\displaystyle \left [ -1,\: 0 \right ) \cup \left \{ 1 \right \}$$
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$$\displaystyle \{ 1 \}$$
Explanation
$$\displaystyle \sin^{-1} \sqrt{1-x} + \cos^{-1}x = \cot^{-1} \left ( \frac{\sqrt{1-x^{2}}}{x} \right )-\sin^{-1}x$$
$$\Rightarrow \sin ^{ -1 } \sqrt { 1-x } +\dfrac { \pi }{ 2 } =\cot ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right) $$
$$\Rightarrow \dfrac { \pi }{ 2 } -\cot ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right) =-\sin ^{ -1 } \sqrt { 1-x } $$
$$\Rightarrow \tan ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right) =-\sin ^{ -1 } \sqrt { 1-x } $$
Thus is only true when, $$x = \{1\}$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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