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JEE Advanced 2008Paper 2
Exam
JEE Advanced
Year
2008
Paper
Paper 2
Subject
Mathematics
Chapter
Differentiation

JEE Advanced 2008 · Paper 2 · Mathematics · Differentiation

Paragraph: Consider the function f:(-∞, ∞) arrow(-∞, ∞) defined by...

Paragraph: Consider the function f:(−∞,∞)→(−∞,∞)f:(-\infty, \infty) \rightarrow(-\infty, \infty) defined by f(x)=x2−ax+1x2+ax+1;0<a<2f(x)=\frac{x^2-a x+1}{x^2+a x+1} ; 0 < a < 2Question: Which of the following is true ?

Options

  1. A(2+a)2f′′(1)+(2−a)2f′′(−1)=0(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0
  2. B(2−a)2f′′(1)−(2+a)2f′′(−1)=0(2-a)^2 f^{\prime \prime}(1)-(2+a)^2 f^{\prime \prime}(-1)=0
  3. Cf′(1)f′(−1)=(2−a)2f^{\prime}(1) f^{\prime}(-1)=(2-a)^2
  4. Df′(1)f′(−1)=−(2+a)2f^{\prime}(1) f^{\prime}(-1)=-(2+a)^2

Correct answer

A. (2+a)2f′′(1)+(2−a)2f′′(−1)=0(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0

Step-by-step solution

\because f(x)= & \frac{(x2+ax+1)\left(x^2+a x+1\right)-2 a x}{x^2+a x+1}=1-2axx2+ax+1\frac{2 a x}{x^2+a x+1} \\ \therefore \quad f^{\prime}(x) & =-\left[\frac{\left(x^2+a x+1\right) \cdot 2 a-2 a x(2 x+a)}{\left(x^2+a x+1\right)^2}\right] \\ & =-\left[\frac{-2 a x^2+2 a}{\left(x^2+a x+1\right)^2}\right]=2 a \left[\frac{\left(x^2-1\right)}{\left(x^2+a x+1\right)^2}\right] \end{aligned} \begin{aligned} & \text{\text{ and } } \begin{aligned} f^{\prime \prime}(x) & =2 a\left[\frac{\left(x^2+a x+1\right)^2(2 x)-2 \left(x^2-1\right)\left(x^2+a x+1\right)(2 x+a)}{\left(x^2+a x+1\right)^4}\right] \\ & =2 a \left[\frac{2 x\left(x^2+a x+1\right)-2\left(x^2-1\right)(2 x+a)}{\left(x^2+a x+1\right)^3}\right] \end{aligned} \\ &  Now, \text{ Now, } f^{\prime \prime}(1)=\frac{4 a(a+2)}{(a+2)^3}=\frac{4 a}{(a+2)^2} \end{aligned} and f(-1)=\frac{4 a(a-2)}{(2-a)^3}=-\frac{4 a}{(a-2)^2}\therefore(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=4 a-4 a=0
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