Let g ( x ) be a linear function and f(x)= cl g(x) & , x 0 (1+x 2… - Vidyadip

MCQ

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

A
$\frac{1}{3} \log _{ e }\left(\frac{4}{9 e ^{1 / 3}}\right)$
B
$\frac{1}{3} \log _e\left(\frac{4}{9}\right)+1$
C
$\log _e\left(\frac{4}{9}\right)-1$
D
$\log _e\left(\frac{4}{9 e^{1 / 3}}\right)$

✓

Answer

Let $g(x)=a x+b$
Now function $f(x)$ in continuous at $x=0$
$\therefore \lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}=b$
$\Rightarrow 0=b$
$\therefore g(x)=ax$
Now, for $x >0$
$f^{\prime}(x)=\frac{1}{x} \cdot\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}-1} \cdot \frac{1}{(2+x)^2}$
$+\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} \cdot \ln \left(\frac{1+x}{2+x}\right) \cdot\left(-\frac{1}{x^2}\right)$
$\therefore f^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right)$
And $f(-1)=g(-1)=-a$
$\therefore a=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9}$
$\therefore g(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3}$
$=\ln \left(\frac{4}{9 \cdot e^{1 / 3}}\right)$

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