Calculus - 2008 : I M GETTIN DIFF ANSWER.......

1

f'(x) = 2a (x^2 - 1) /(x^2 + ax + 1),

g'(x) = f'(e^x) . e ^ x /( 1 + e^2x)
It is B

1

When U derivate g(x)

g'(x)=e^xf'(e^x)/(1+e^2x)

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'(e^x)<0, since e^x is always greater than 0

Clearly 1+e^2x & e^x are always +ve

Therefore g'(x) is always negative

ANSWER D. g'(x) doesnt change its sign[1][1]
[339]

1

f'(x) = 2a(x²-1)/(...)²

now, f'(e^x) = 2a(e^2x-1)/(...)²

for x<0, f'(e^x) < 0

for x>0, f'(e^x) >0

this determine the g'(x)'s sign .

so the ans is B.

21

Ho gaya!! arre i was makin a silly calc err....

Thnx all [1]

2008 : I M GETTIN DIFF ANSWER.......