JEE MATH APEX

Inverse Trigonometric Functions-JEE-Mains-PYQ’s

Mathematics Questions

Considering the principal values of inverse trigonometric functions, the value of the expression
\tan \left( 2 \sin^{-1} \left( \frac{2}{\sqrt{13}} \right) - 2 \cos^{-1} \left( \frac{3}{\sqrt{10}} \right) \right)
is equal to :
(A) \frac{33}{56}
(B) -\frac{33}{56}
(C) -\frac{16}{63}
(D) \frac{16}{63}
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If the domain of the function f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right) is (-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty), then \alpha + \beta + \gamma + \delta is equal to
(A) 4
(B) 2
(C) 5
(D) 3
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The number of solutions of \tan^{-1} 4x + \tan^{-1} 6x = \frac{\pi}{6}, where -\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}, is equal to :
(A) 2
(B) 0
(C) 3
(D) 1
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If the domain of the function f(x) = \cos^{-1} \left( \frac{2x-5}{11-3x} \right) + \sin^{-1} (2x^2 - 3x + 1) is the interval [\alpha, \beta], then \alpha + 2\beta is equal to :
(A) 5
(B) 2
(C) 3
(D) 1
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Given that the inverse trigonometric function assumes principal values only. Let x, y be any two real numbers in [-1,1] such that \cos ^{-1} x-\sin ^{-1} y=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi. Then, the minimum value of x^{2}+y^{2}+2 x y \sin \alpha is
(A) 0
(B) -1
(C) \frac{1}{2}
(D) \frac{-1}{2}
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If the domain of the function \sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _{\mathrm{e}}\left(\frac{3 x^{2}-8 x+5}{x^{2}-3 x-10}\right) is (\alpha, \beta], then 3 \alpha+10 \beta is equal to:
(A) 95
(B) 100
(C) 97
(D) 98
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If a=\sin ^{-1}(\sin (5)) and b=\cos ^{-1}(\cos (5)), then a^{2}+b^{2} is equal to
(A) 25
(B) 4 \pi^{2}+25
(C) 8 \pi^{2}-40 \pi+50
(D) 4 \pi^{2}-20 \pi+50
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For \alpha, \beta, \gamma \neq 0, if \sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi and (\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta, then \gamma equals
(A) \sqrt{3}
(B) \frac{\sqrt{3}}{2}
(C) \frac{1}{\sqrt{2}}
(D) \frac{\sqrt{3}-1}{2 \sqrt{2}}
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Let x=\frac{m}{n} (m, n are co-prime natural numbers) be a solution of the equation \cos \left(2 \sin ^{-1} x\right)=\frac{1}{9} and let \alpha, \beta(\alpha>\beta) be the roots of the equation m x^{2}-n x-m+n=0. Then the point (\alpha, \beta) lies on the line
(A) 3 x-2 y=-2
(B) 3 x+2 y=2
(C) 5 x+8 y=9
(D) 5 x-8 y=-9
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Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying \tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4} is :
(A) more than 2
(B) 2
(C) 0
(D) 1
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