Continuity And Differentiability Question 9
Question 9 - 2024 (31 Jan Shift 1)
Let $g(x)$ be a linear function and $f(x)=\begin{cases}g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x>0\end{cases}$, is continuous at $x=0$. If $f^{\prime}(1)=f(-1)$, then the value of $g(3)$ is
(1) $\frac{1}{3} \log _e\left(\frac{4}{9 e^{1 / 3}}\right)$
(2) $\frac{1}{3} \log _e\left(\frac{4}{9}\right)+1$
(3) $\log _{\circ}\left(\frac{4}{9}\right)-1$
(4) $\log _6\left(\frac{4}{9 e^{1 / 3}}\right)$
Show Answer
Answer (4)
Solution
Let $g(x)=a x+b$
Now function $f(x)$ in continuous at $x=0$
$\therefore \lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}=b$
$\Rightarrow 0=b$
$\therefore g(x)=a x$
Now, for $\mathbf{x}>0$
$f^{\prime}(x)=\frac{1}{x} \cdot\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}-1} \cdot \frac{1}{(2+x)^{2}}$
$+\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} \cdot \ln \left(\frac{1+x}{2+x}\right) \cdot\left(-\frac{1}{x^{2}}\right)$
$\therefore f^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right)$
And $f(-1)=g(-1)=-a$
$\therefore a=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9}$
$\therefore g(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3}$
$=\ln \left(\frac{4}{9 \cdot e^{1 / 3}}\right)$
