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JEE Mains 28 July 2022 Shift 1 Paper

Section: Mathematics

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Question : 4 of 90

Marks: +1, -0

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation

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

[28-Jul-2022-Shift-1]

0
1
$\frac{1}{2}$
$-\frac{1}{2}$

Solution:

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

For Domain :

x \in \left[ \frac{-1}{2}, \frac{1}{2} \right]

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

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

\Rightarrow \cos(3\cos^{-1} x) = -\cos(\cos^{-1} 2x)

\Rightarrow 4x^3 = x

\Rightarrow x = 3, \pm \frac{1}{2}

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