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If f(x)=sin⁡−1(2x+11+4x)f(x) = \sin^{-1}\left(\dfrac{2^{x+1}}{1+4^x}\right), then

Match List-I with List-II

List-IList-II
Value of xx12f(x)\dfrac{1}{2}f(x)
(A) x=0x = 0(I) tan⁡−12\tan^{-1}\sqrt{2}
(B) x=12x = \dfrac{1}{2}(II) tan⁡−14\tan^{-1}4
(C) x=1x = 1(III) π4\dfrac{\pi}{4}
(D) x=2x = 2(IV) tan⁡−12\tan^{-1}2

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

Put t=2xt=2^x. Then 2x+11+4x=2t1+t2=sin⁡(2tan⁡−1t)\frac{2^{x+1}}{1+4^x} = \frac{2t}{1+t^2} = \sin(2\tan^{-1}t), so f(x)=2tan⁡−1(2x)f(x) = 2\tan^{-1}(2^x) and 12f(x)=tan⁡−1(2x)\frac{1}{2}f(x) = \tan^{-1}(2^x).

Hence x=0→tan⁡−11=π4x=0\to\tan^{-1}1=\frac{\pi}{4}, x=12→tan⁡−12x=\frac{1}{2}\to\tan^{-1}\sqrt{2}, x=1→tan⁡−12x=1\to\tan^{-1}2, x=2→tan⁡−14x=2\to\tan^{-1}4.

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