Solve this following
Question:

Let $f$ be a twice differentiable function defined

on $\mathrm{R}$ such that $\mathrm{f}(0)=1, \mathrm{f}^{\prime}(0)=2$ and $\mathrm{f}^{\prime}(\mathrm{x}) \neq 0$ for

all $x \in R .$ If $\left|\begin{array}{cc}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0$, for all $x \in R$, then

the value of $\mathrm{f}(1)$ lies in the interval:

1. $(9,12)$

2. $(6,9)$

3. $(0,3)$

4. $(3,6)$

Correct Option: , 2

Solution:

$\mathrm{f}(\mathrm{x}) \mathrm{f}^{\prime \prime}(\mathrm{x})-\left(\mathrm{f}^{\prime}(\mathrm{x})\right)^{2}=0$

$\frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=\frac{f^{\prime}(x)}{f(x)}$

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

$f^{\prime}(x)=\operatorname{cf}(x)$

$\frac{f^{\prime}(x)}{f(x)}=c$

$\operatorname{lnf}(x)=c x+k_{1}$

$f(x)=k e^{c x}$

$f(0)=1=k$

$f^{\prime}(0)=c=2$

$f(x)=e^{2 x}$

$f(1)=e^{2} \in(6,9)$