# Let f be a twice differentiable function defined on

Question:

Let $f$ be a twice differentiable function defined on $\mathrm{R}$ such that $f(0)=1, f^{\prime}(0)=2$ and $f^{\prime}(x) \neq 0$ for all $\mathrm{X} \in \mathrm{R}$. If $\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0$, for all $\mathrm{x} \in \mathrm{R}$ then the value of $f(1)$ lies in the interval:

1. (1) $(9,12)$

2. (2) $(6,9)$

3. (3) $(3,6)$

4. (4) $(0,3)$

Correct Option: , 2

Solution:

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

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

$\Rightarrow h^{\prime}(x)=0 \quad \Rightarrow h(x)=k$

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

$\Rightarrow f(0)=k f^{\prime}(0) \quad \Rightarrow 1=k(2) \Rightarrow k=\frac{1}{2}$

Now $f(x)=\frac{1}{2} f^{\prime}(x) \Rightarrow \int 2 d x=\int \frac{f^{\prime}(x)}{f(x)} d x$

$\Rightarrow 2 x=\ln |f(x)|+C$

As $f(0)=1 \Rightarrow C=0$

$\Rightarrow 2 x=\ln |f(X)| \Rightarrow f(x)=\pm e^{2 x}$

As $f(0)=1 \Rightarrow f(x)=e^{2 x} \Rightarrow f(1)=e^{2}$