JEE Main 12 Jan 2019 Evening Question 12
Question: Let f be a differentiable function such that $ f(1)=2 $ and $ f’(x)=f(x) $ for all $ x\in R. $ If $ h(x)=f(f(x)), $ then $ h’(1) $ is equal to [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ 2e^{2} $
B) 4e
C) 2e
D) $ 4e^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Given, $ \frac{f’(x)}{f(x)}=1\forall x\in R $
Integrating both sides,
we get $ \log f(x)=x+C\Rightarrow f(x)={e^{x+C}}=e^{x}.e^{C} $ $ \because $ $ f(1)=2\Rightarrow f(1)=e.e^{C}=2 $
$ \Rightarrow $ $ e^{C}=\frac{2}{e}\Rightarrow C=\log \frac{2}{e} $
$ \therefore $ $ f(x)=e^{x}.\frac{2}{e}=2{e^{x-1}} $
$ \Rightarrow $ $ f’(x)=2{e^{x-1}} $ Now, $ h(x)=(f(x)) $
$ \therefore $ $ h’(x)=f’(f(x)).f’(x) $
$ \therefore $ $ h’(1)=f’(f(1)).f’(1)=f’(2)f’(1) $ $ =2e.2=4e $
