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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6,9

0,3

3,6

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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