Ho gaya!! arre i was makin a silly calc err....
Thnx all [1]
f'(x) = 2a (x^2 - 1) /(x^2 + ax + 1),
g'(x) = f'(e^x) . e ^ x /( 1 + e^2x)
It is B
When U derivate g(x)
g'(x)=exf'(ex)/(1+e2x)
f(x)= \frac{x^{2}+1-ax}{x^{2}+1+ax}
for x>0
As x increases , numerator decreases and denominator increases. Since 0<a<2 (Consider the term ax)
f'(ex)<0, since ex is always greater than 0
Clearly 1+e2x & ex are always +ve
Therefore g'(x) is always negative
ANSWER D. g'(x) doesnt change its sign[1][1]
[339]
f'(x) = 2a(x2-1)/(...)2
now, f'(ex) = 2a(e2x-1)/(...)2
for x<0, f'(ex) < 0
for x>0, f'(ex) >0
this determine the g'(x)'s sign .
so the ans is B.