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JEE Main 201815 AprilEvening ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Evening Shift
Mode
Online
Subject
Mathematics
Chapter
Differentiation

JEE Main 2018 · 15 April · Evening Shift · Online · Mathematics · Differentiation

If f(x)= ^ -1 ((2 × 3^x)/(1+9^x)) , then f^ (-(1)/(2)) equals.

If f(x)=sin⁡−1(2×3x1+9x)f(x)=\sin ^{-1}\left(\frac{2 \times 3^x}{1+9^x}\right), then f′(−12)f^{\prime}\left(-\frac{1}{2}\right) equals.

Options

  1. A3log⁡e3\sqrt{3} \log _e \sqrt{3}
  2. B−3log⁡e3-\sqrt{3} \log _e \sqrt{3}
  3. C−3log⁡e3-\sqrt{3} \log _e 3
  4. D3log⁡e3\sqrt{3} \log _e 3

Correct answer

A. 3log⁡e3\sqrt{3} \log _e \sqrt{3}

Step-by-step solution

Since f(x)=sin⁡f(x)=\sin\left(\frac{2 \times 3^x}{1+9^x}\right)SupposeSuppose3^x=\tan t⇒f(x)=sin⁡−1(2tan⁡t1+tan⁡2t)=sin⁡−1(sin⁡2t)=2t=2tan⁡−1(3x) So, f′(x)=21+(3x)2×3x⋅log⁡e3∴f′(−12)=21+(3−12)2×3−12⋅log⁡e3=12×3×log⁡e3=3×log⁡e3\begin{aligned} &\Rightarrow f(x)=\sin ^{-1}\left(\frac{2 \tan t}{1+\tan ^2 t}\right) \\ &=\sin ^{-1}(\sin 2 t)=2 t=2 \tan ^{-1}(3 x) \\ &\text{ So, } f^{\prime}(x)=\frac{2}{1+\left(3^x\right)^2} \times 3^x \cdot \log _e 3 \\ &\therefore \quad f^{\prime}\left(-\frac{1}{2}\right)=\frac{2}{1+\left(3^{-\frac{1}{2}}\right)^2} \times 3^{-\frac{1}{2}} \cdot \log _e 3 \\ &=\frac{1}{2} \times \sqrt{3} \times \log _e 3=\sqrt{3} \times \log _e \sqrt{3} \end{aligned}
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