Let f:R->R be continuous function which satisfies
f(x)=0∫x f(t)dt,
then the value of f(ln5) is:
(a) 0
(b) 2
(c) 4
(d) 6
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4 Answers
2305Applying Newton-Leibniz formula on the RHS,
d f(x)dx=f(x)
\int \frac{d f(x)}{f(x)}=\int dx
\ln(f(x))=x+c
f(x)=e^{x+c}
f(0)=\int^{0}_{0}f(t)dt=0=e^{C}
f(x)=e^{x+C}=e^{x}\cdot 0 = 0
Ans-A
486Is it (c)
- Shaswata Roy Did you get f(x)=e^x - 1.If so check whether it satisfies the 1st condition.Upvote·0· Reply ·2013-08-14 03:02:58
