Question

Let f be a differentiable function such that $f\left(1\right)=2$ and $f^{\prime }\left(x\right)=f\left(x\right)$ for all $x\in R$. If $h\left(x\right)=f\left(f\right(x\left)\right)$, then $h^{\prime }\left(1\right)$ is equal to :

• A. 4e
• B. $4e^2$
• C. 2e
• D. $2e^2$

Answer 1

The correct option is A 4e

Given:

$f\left(1\right)=2$

$f^{\prime }\left(x\right)=f\left(x\right)$

$h\left(x\right)=f\left(f\right(x\left)\right)$

To find:

$h^{\prime }\left(1\right)$

Solution:

As given:

$f^{\prime }\left(x\right)=f\left(x\right)$

$only{e}^{x}isafuctionwhichcanfulfilltheabovecondition$

$\therefore f\left(x\right)=a{e}^{x-b}$

$\Rightarrow f^{\prime }\left(x\right)=a{e}^{x-b}$

Now,

$Asf\left(1\right)=2$

$\Rightarrow f\left(1\right)=a{e}^{1-b}$

$\Rightarrow 2=a{e}^{1-b}$

Comparing both the sides, we get power of e is 0,

$\therefore a=2and1-b=0$

$\Rightarrow a=2andb=1$

$\Rightarrow f\left(x\right)=2e^{x-1}$

Now,

$h\left(x\right)=f\left(f\right(x\left)\right)$

$\Rightarrow h^{\prime }\left(x\right)=f^{\prime }\left(f\right(x\left)\right)\cdot f^{\prime }\left(x\right)$

$\Rightarrow h^{\prime }\left(1\right)=f^{\prime }\left(f\right(1\left)\right)\cdot f^{\prime }\left(1\right)$

We know f'(x)=f(x) and f(1)=2,

$\Rightarrow h^{\prime }\left(1\right)=f^{\prime }\left(2\right)\cdot 2$

$\Rightarrow h^{\prime }\left(1\right)=2\cdot f^{\prime }\left(2\right).............\left(1\right)$

Here

$f\left(x\right)=f^{\prime }\left(x\right)=2e^{x-1}$

$\Rightarrow f^{\prime }\left(2\right)=2e^{2-1}$

$\Rightarrow f^{\prime }\left(2\right)=2e$

Putting in eq(1)

$\Rightarrow h^{\prime }\left(1\right)=2\cdot f^{\prime }\left(2\right)$

$\Rightarrow h^{\prime }\left(1\right)=2\cdot 2e$

$\Rightarrow h^{\prime }\left(1\right)=4e$