JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 13)
Let f be a differentiable function such that f(1) = 2 and f '(x) = f(x) for all x $$ \in $$ R R. If h(x) = f(f(x)), then h'(1) is equal to :
4e
2e2
4e2
2e
Explanation
$${{f'(x)} \over {f(x)}} = 1\forall x \in R$$
Intergrate & use f(1) = 2
f(x) = 2ex-1 $$ \Rightarrow $$ f '(x) = 2ex$$-$$1
h(x) = f(f(x)) $$ \Rightarrow $$ h'(x) = f '(f(x)) f'(x)
h'(1) = f '(f(1)) f'(1)
= f '(2) f '(1)
= 2e . 2 = 4e
Intergrate & use f(1) = 2
f(x) = 2ex-1 $$ \Rightarrow $$ f '(x) = 2ex$$-$$1
h(x) = f(f(x)) $$ \Rightarrow $$ h'(x) = f '(f(x)) f'(x)
h'(1) = f '(f(1)) f'(1)
= f '(2) f '(1)
= 2e . 2 = 4e
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