33but f(x) is not given differentiable?
let f(X) be a continous funtion such that f(x)>0 and f(x)^101 =1+∫ f(t)dt.[integrate from 0 to x] the value of
(f(100))^101
a) 99
b) 100
c) 101
d) 98
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3 Answers
66There seems to be some trouble. Differentiating the given relation, we get
101f(x)^{100}f'(x)=f(x)
i.e.
101f(x)^{99}f'(x)=1
Integrating we get
\dfrac{101}{100}f(x)^{100}=x+c
Since f(0)=1, we get
c=101100
Hence,
f(x)100 = 100101 x + 1
i.e.
f(x)=\left(\dfrac{100}{101}x+1\right)^{\frac{1}{100}}
You can see that none of the options are correct.
66Well, we have been given that f(x) is continuous. So \int_0^x f(t)\ \mathrm dt is differentiable. That means 1+ \int_0^x f(t)\ \mathrm dt is differentiable.
Hence, f101 is differentiable. Also, one can see that the derivative is f'(x)=1101f(x)99. Since f(x)>0, so the derivative always exists, and hence f(x) is differentiable.