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Q.

Let f be a twice differentiable function defined on R such that  f0=1,f'0=2and  f1x≠0for all x∈R . If fxf'xf'xf"x=0 for all x∈R, then the value of  f1 lies in the interval

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a

6,9

b

0,3

c

3,6

d

9,12

answer is A.

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Detailed Solution

fxf"x=f1x2⇒    f1x2−fxf11xf1x2=0   ⇒   ddxfxf1x=0⇒  fxf1x=k    ⇒   k=12fx=±ke2x  f0=1⇒f1=e2=7.38
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