Continuity And Differentiability Question 9
Question 9 - 2024 (31 Jan Shift 1)
Let $g(x)$ be a linear function and $f(x)=\left{\begin{array}{cl}g(x) & , x \leq 0 \ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x>0\end{array}\right.$, is continuous at $x=0$. If $f^{\prime}(1)=f(-1)$, then the value of $g(3)$ is
(1) $\frac{1}{3} \log _{\mathrm{e}}\left(\frac{4}{9 \mathrm{e}^{1 / 3}}\right)$
(2) $\frac{1}{3} \log _{\mathrm{e}}\left(\frac{4}{9}\right)+1$
(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 $\mathbf{x}>0$
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \cdot\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{x}-1} \cdot \frac{1}{(2+\mathrm{x})^{2}}$
$+\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{x}} \cdot \ln \left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right) \cdot\left(-\frac{1}{\mathrm{x}^{2}}\right)$
$\therefore \mathrm{f}^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right)$
And $f(-1)=g(-1)=-a$
$\therefore \mathrm{a}=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9}$
$\therefore \mathrm{g}(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3}$
$=\ln \left(\frac{4}{9 \cdot \mathrm{e}^{1 / 3}}\right)$
