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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
Let $\mathrm{g}(\mathrm{x})$ 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
    $\log _e\left(\frac{4}{9}\right)-1$
  • ✓
    $\log _e\left(\frac{4}{9 e^{1 / 3}}\right)$
✓

Answer

Correct option: D.
$\log _e\left(\frac{4}{9 e^{1 / 3}}\right)$
d
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 $\mathrm{x}>0$

$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \cdot\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{x}-1} \cdot \frac{1}{(2+\mathrm{x})^2}$

$+\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{x}} \cdot \ln \left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right) \cdot\left(-\frac{1}{\mathrm{x}^2}\right)$

$\therefore \mathrm{f}^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right)$

$\text { And } \mathrm{f}(-1)=\mathrm{g}(-1)=-\mathrm{a}$

$\therefore \mathrm{a}=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9}$

$\therefore \mathrm{g}(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3}$

$=\ln \left(\frac{4}{9 \cdot \mathrm{e}^{1 / 3}}\right)$

 

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