Consider the function $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ defined by \[ f(x)=\frac{x^{2}-a x+1}{x^{2}+a x+1}, 0<a<2 . \] Which of the following is true?

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

(B) $(2-a)^{2} f^{\prime \prime}(1)-(2+a)^{2} f^{\prime \prime}(-1)=0$

(C) $f^{\prime}(1) f^{\prime}(-1)=(2-a)^{2}$

(D) $f^{\prime}(1) f^{\prime}(-1)=-(2+a)^{2}$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Derivatives XI

Consider the function

f : (- \infty, \infty) \to (- \infty, \infty)

defined by

f (x) = \frac{x ^{2} - a x + 1}{x ^{2} + a x + 1}, 0 < a < 2.

Which of the following is true?

Answer: (A)

Computing

f^{''} (x)

and evaluating at

x = 1

and

x = - 1

f^{''} (1) = \frac{4 a}{( 2 + a ) ^{2}}

and

f^{''} (- 1) = \frac{- 4 a}{( 2 - a ) ^{2}}

. Then

(2 + a)^{2} f^{''} (1) + (2 - a)^{2} f^{''} (- 1) = 4 a - 4 a = 0

. So option (A) is correct.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option