If 3 ^{-1}≤ft({2 x}{1+x²})-4 ^{-1}≤ft({1-x²}{1+x²})+2 ^{-1}≤ft({2 x}{1-x²})=π/3 then x= ?

Correct Answer is : 13{1}{{3}}31

Correct Answer is : 13{1}{{3}}31

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MHTCET 20 Apr 2026 Shift 2 Paper

Question : 116 of 150

Marks: +1, -0

3 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}

x=

Solution:

Concept:

Use the substitution

x = \tan \theta

to convert the given inverse trigonometric expressions into standard angles.

Explanation:

Let

x = \tan \theta

. Then:

\frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta

\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta

\frac{2x}{1-x^2} = \frac{2\tan\theta}{1-\tan^2\theta} = \tan 2\theta

Test the option

x = \frac{1}{\sqrt{3}}

Since

\frac{1}{\sqrt{3}} = \tan\frac{\pi}{6}

, we get

2\theta = \frac{\pi}{3}

Therefore:

\sin^{-1}\left(\sin\frac{\pi}{3}\right) = \frac{\pi}{3}

\cos^{-1}\left(\cos\frac{\pi}{3}\right) = \frac{\pi}{3}

\tan^{-1}\left(\tan\frac{\pi}{3}\right) = \frac{\pi}{3}

Substitute into the given expression:

3\left(\frac{\pi}{3}\right) - 4\left(\frac{\pi}{3}\right) + 2\left(\frac{\pi}{3}\right)

= \pi - \frac{4\pi}{3} + \frac{2\pi}{3}

= \frac{\pi}{3}

This satisfies the equation.

Answer:

x = \frac{1}{\sqrt{3}}

Correct option: C.

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