Question

Let $g\left(x\right)$ be a polynomial of degree one and $f\left(x\right)$ be defined by
$f\left(x\right)=\{\begin{array}{cc}g\left(x\right),& x\le 0\\ {\left[\frac{1+x}{2+x}\right]}^{1/x},& x>0\end{array}$
Let $f\left(x\right)$ be a continuous function satisfying $f^{\prime }\left(1\right)=f(-1).$ Then $f(-2)$ is equal to

• A. $0$
• B. $-2$
• C. $\frac{2}{9}-\frac{4}{3}\log \left(\frac{2}{3}\right)$
• D. $\frac{1}{6}-\log \left(\frac{2}{3}\right)$

Answer 1

The correct option is C $\frac{2}{9}-\frac{4}{3}\log \left(\frac{2}{3}\right)$

Let $g\left(x\right)=ax+b$
$f\left(x\right)=\{\begin{array}{cc}ax+b,& x\le 0\\ {\left[\frac{1+x}{2+x}\right]}^{1/x},& x>0\end{array}$

Since, $f\left(x\right)$ be a continuous function.
$\Rightarrow \text{L.H.L}=\text{R.H.L}$
$\Rightarrow \underset{x\to 0^{-}}{lim}ax+b=\underset{x\to 0^{+}}{lim}{\left[\frac{1+x}{2+x}\right]}^{1/x}$
$\Rightarrow b={\left(\frac{1}{2}\right)}^{\infty }$
$\Rightarrow b=0$

For $x>0$
$\therefore f\left(x\right)={\left(\frac{1+x}{2+x}\right)}^{1/x},f\left(1\right)=\frac{2}{3}$
Taking $\log$ on both the sides, we get
$\ln f\left(x\right)=\frac{1}{x}\left[\ln \right(1+x)-\ln (2+x\left)\right]$

$\therefore \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=-\frac{1}{x^2}\ln \left(\frac{1+x}{2+x}\right)+\frac{1}{x(x+1)(x+2)}$
$\Rightarrow \frac{f^{\prime }\left(1\right)}{f\left(1\right)}=-\ln \frac{2}{3}+\frac{1}{6}$
$\Rightarrow f^{\prime }\left(1\right)=-\frac{2}{3}\ln \frac{2}{3}+\frac{1}{9}$

$\because f(-1)=b-a$
$\Rightarrow b-a=-\frac{2}{3}\ln \frac{2}{3}+\frac{1}{9}$
$\Rightarrow a=\frac{2}{3}\ln \frac{2}{3}-\frac{1}{9}(\because b=0)$

$\therefore f\left(x\right)=(\frac{2}{3}\ln \frac{2}{3}-\frac{1}{9})x$
$\Rightarrow f(-2)=\frac{2}{9}-\frac{4}{3}\log \left(\frac{2}{3}\right)$