The range of x for which the equation ^{-1}≤ft(2 x/1+x²)=2 ^{-1} x holds true

Correct Answer is : ∣x∣≤1| x | ≤ 1∣x∣≤1

Correct Answer is : ∣x∣≤1| x | ≤ 1∣x∣≤1

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COMEDK 2025 Morning Solved Paper

Question : 160 of 180

Marks: +1, -0

The range of

x

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

holds true

Solution:

Concept:

The inverse trigonometric identity

\sin^{-1}(\sin (2\theta)) = 2\theta

holds only when

2\theta

lies in the principal range of

\sin^{-1}

, i.e.,

[-\pi/2, \pi/2]

Explanation:

Let

\theta = \tan^{-1} x

. Then

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

We know

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

Thus

\sin^{-1}\left(\frac{2x}{1+x^2}\right) = \sin^{-1}(\sin(2\theta))

For the equation to hold, we need

\sin^{-1}(\sin(2\theta)) = 2\theta

This is true only if

2\theta

lies in

[-\pi/2, \pi/2]

, i.e.,

|\theta| \le \pi/4

Since

\theta = \tan^{-1}x

, the condition

|\tan^{-1}x| \le \pi/4

is equivalent to

|x| \le 1

Answer:

Option A:

|x| \le 1

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