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:
Correct Option: , 2
$\mathrm{f}(\mathrm{x}) \mathrm{f}^{\prime \prime}(\mathrm{x})-\left(\mathrm{f}^{\prime}(\mathrm{x})\right)^{2}=0$
$\frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=\frac{f^{\prime}(x)}{f(x)}$
$\ln \left(f^{\prime}(x)\right)=\ln f(x)+\ln c$
$f^{\prime}(x)=\operatorname{cf}(x)$
$\frac{f^{\prime}(x)}{f(x)}=c$
$\operatorname{lnf}(x)=c x+k_{1}$
$f(x)=k e^{c x}$
$f(0)=1=k$
$f^{\prime}(0)=c=2$
$f(x)=e^{2 x}$
$f(1)=e^{2} \in(6,9)$
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