Question

Let $f$ be a function satisfying $f\left(0\right)=2,f^{\prime }\left(0\right)=3$ and $f^{\prime \prime }\left(x\right)=f\left(x\right)$. Then

• A. $f\left(4\right)=\frac{5e^8-1}{2e^4}$
• B. $f\left(4\right)=\frac{5e^8+1}{2e^4}$
• C. $f^{\prime }\left(4\right)=\frac{5e^8+1}{2e^4}$
• D. $f^{\prime }\left(4\right)=\frac{5e^8-1}{2e^4}$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(4\right)=\frac{5e^8-1}{2e^4}$
C $f^{\prime }\left(4\right)=\frac{5e^8+1}{2e^4}$

Given $f^{\prime \prime }\left(x\right)=f\left(x\right)$
$\Rightarrow 2f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)=2f\left(x\right)f{(}^{\prime }x)\Rightarrow \frac{d}{dx}{\left(f^{\prime }\right(x\left)\right)}^2=\frac{d}{dx}{\left(f\right(x\left)\right)}^2\Rightarrow {\left(f^{\prime }\right(x\left)\right)}^2={\left(f\right(x\left)\right)}^2+c\cdots (1)$

Now, $f\left(0\right)=2$ and $f^{\prime }\left(0\right)=3$
From $\left(1\right)$,
$3^2=2^2+c\Rightarrow c=5\therefore f^{\prime }\left(x\right)=\sqrt{{\left(f\right(x\left)\right)}^2+5}\Rightarrow \int \frac{1}{\sqrt{(\sqrt{5}{)}^2+{\left(f\right(x\left)\right)}^2}}d\left[f\right(x\left)\right]=\int 1dx\Rightarrow \ln \left[f\right(x)+\sqrt{{\left(f\right(x\left)\right)}^2+5}]=x+c_1$

Since $f\left(0\right)=2\Rightarrow c_1=\ln 5$
$\therefore \ln \left[f\right(x)+\sqrt{{\left(f\right(x\left)\right)}^2+5}]=x+\ln 5\Rightarrow \ln \left[\frac{f\left(x\right)+\sqrt{{\left(f\right(x\left)\right)}^2+5}}{5}\right]=x\Rightarrow f\left(x\right)+\sqrt{{\left(f\right(x\left)\right)}^2+5}=5e^x$
$\Rightarrow {\left(f\right(x\left)\right)}^2+5=(5e^x-f(x){)}^2$
$\Rightarrow {\left(f\right(x\left)\right)}^2+5=25e^{2x}+{\left(f\right(x\left)\right)}^2-10e^{x}f\left(x\right)$
$\Rightarrow 1=5e^{2x}-2e^{x}f\left(x\right)$
$\Rightarrow f\left(x\right)=\frac{1}{2}(5e^x-e^{-x})\therefore f\left(4\right)=\frac{1}{2}(5e^4-e^{-4})$
and $f^{\prime }\left(4\right)=\frac{1}{2}(5e^4+e^{-4})$