Question

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

• A. $(9,12)$
• B. $(6,9)$
• C. $(3,6)$
• D. $(0,3)$

Answer 1

The correct option is B $(6,9)$

Given $f\left(x\right)f^{\prime \prime }\left(x\right)-\left(f^{\prime }\right(x){)}^2=0$
Let $h\left(x\right)=\frac{f\left(x\right)}{f^{\prime }\left(x\right)}$
Then $h^{\prime }\left(x\right)=0\Rightarrow h\left(x\right)=k$
$\Rightarrow \frac{f\left(x\right)}{f^{\prime }\left(x\right)}=k$
$\Rightarrow f\left(x\right)=k{f}^{\prime }\left(x\right)$
$\Rightarrow f\left(0\right)=k{f}^{\prime }\left(0\right)$
$\Rightarrow k=\frac{1}{2}$

Now, $f\left(x\right)=\frac{1}{2}{f}^{\prime }\left(x\right)$
$\Rightarrow \int 2dx=\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx$
$\Rightarrow 2x=\ln |f\left(x\right)|+C$
As $f\left(0\right)=1\Rightarrow C=0$
$\Rightarrow 2x=\ln |f\left(x\right)|$
$\Rightarrow f\left(x\right)=\pm e^{2x}$
As $f\left(0\right)=1\Rightarrow f\left(x\right)=e^{2x}$
$\therefore f\left(1\right)=e^2\approx 7.38$