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JEE Main 20139 AprilOnline
Exam
JEE Main
Year
2013
Date
9 April
Mode
Online
Subject
Mathematics
Chapter
Differentiation

JEE Main 2013 · 9 April · Online · Mathematics · Differentiation

Let f(x)=(x^2-x)/(x^2+2 x) x ≠ 0,-2 . Then (d)/(d x)[f^ -1 (x)]...

Let f(x)=x2−xx2+2xx≠0,−2f(x)=\frac{x^2-x}{x^2+2 x} x \neq 0,-2. Then ddx[f−1(x)]\frac{d}{d x}\left[f^{-1}(x)\right] (wherever it is defined) is equal to:

Options

  1. A−1(1−x)2\frac{-1}{(1-x)^2}
  2. B3(1−x)2\frac{3}{(1-x)^2}
  3. C1(1−x)2\frac{1}{(1-x)^2}
  4. D−3(1−x)2\frac{-3}{(1-x)^2}

Correct answer

B. 3(1−x)2\frac{3}{(1-x)^2}

Step-by-step solution

Let y=x2−xx2+2xy=\frac{x^2-x}{x^2+2 x}⇒(x2+2x)y=x2−x⇒x(x+2)y=x(x−1)⇒x[(x+2)y−(x−1)]=0∵x≠0,∴(x+2)y−(x−1)=0⇒xy+2y−x+1=0⇒x(y−1)=−(2y+1)∴x=2y+11−y\Rightarrowf−1(x)=2x+11−xddx(f−1(x))=2(1−x)−(2x+1)(−1)(1−x)2=2−2x+2x+1(1−x)2=3(1−x)2\begin{aligned}& \Rightarrow \quad \left(x^2+2 x\right) y=x²-x \\ & \Rightarrow \quad x(x+2) y=x(x-1) \\ & \Rightarrow \quad x[(x+2) y-(x-1)]=0 \\ & \because \quad x ≠ 0, \quad \therefore \quad(x+2) y-(x-1)=0 \\ & \Rightarrow \quad x y+2 y-x+1=0 \\ & \Rightarrow \quad x(y-1)=-(2 y+1) \\ & \therefore \quad x=\frac{2 y+1}{1-y} \Rightarrowf^{-1}(x)=\frac{2 x+1}{1-x}\\ & \frac{d}{d x}\left(f^{-1}(x)\right)=\frac{2(1-x)-(2 x+1)(-1)}{(1-x)^2}\\ & \quad=\frac{2-2 x+2 x+1}{(1-x)^2}=\frac{3}{(1-x)^2} \end{aligned}$
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