Let g ( x ) be a linear function and f ( x ) = { g ( x ) , x ? 0 ( 1 + x 2 + x ) 1 x , x > 0 , is continuous at x = 0 . If f ' ( 1 ) = f ( - 1 ) , then the value of g ( 3 ) is [2024]

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Solution

Q.
Let $g (x)$ be a linear function and $f (x)$ $= {\begin{matrix} g (x), & x \leq 0 \\ {(\frac{1 + x}{2 + x})}^{\frac{1}{x}}, & x > 0 \end{matrix},$ is continuous at $x = 0 .$ If $f' (1) = f (- 1),$ then the value of $g (3)$ is [2024]

1 $\log_{e} (\frac{4}{9}) - 1$

2 $\frac{1}{3} \log_{e} (\frac{4}{9}) + 1$

3 $\frac{1}{3} \log_{e} (\frac{4}{9 e^{1 / 3}})$

4 $\log_{e} (\frac{4}{9 e^{1 / 3}})$

Ans.
(4)

$f (x)$ $= {\begin{matrix} g (x), & x < 0 \\ {(\frac{1 + x}{2 + x})}^{\frac{1}{x}}, & x > 0 \end{matrix}$ is continuous at $x = 0$

Let $g (x) = p x + q$

Since, $f (x)$ is continuous at, $x = 0 .$

$\Rightarrow g (x) = p (x) (q = 0)$

$Now, f' (1) = f (- 1), y = {(\frac{1 + x}{2 + x})}^{1 / x}$

$\Rightarrow \log y = \log {(\frac{1 + x}{2 + x})}^{1 / x} \Rightarrow \log y = \frac{1}{x} \log (\frac{1 + x}{2 + x})$

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = - \frac{1}{x^{2}} \log (\frac{1 + x}{2 + x}) + \frac{1}{x} \times (\frac{1}{\frac{1 + x}{2 + x}}) \times \frac{(x + 2) - (x + 1)}{{(2 + x)}^{2}}$

at $x = 1$

$f' (1) = \frac{2}{3} [- \log (\frac{2}{3}) + \frac{3}{2} (\frac{1}{9})] \Rightarrow - \frac{2}{3} \log (\frac{2}{3}) + \frac{1}{9}$

and at $x = - 1$

$f (- 1) = - p = - \frac{2}{3} \log (\frac{2}{3}) + \frac{1}{9} \Rightarrow p = \frac{2}{3} \log (\frac{2}{3}) - \frac{1}{9}$

$g (3) = 3 p = 2 \log (\frac{2}{3}) - \frac{1}{3} = \log (\frac{4}{9}) + \log_{e}^{- 1 / 3}$

$= \log (\frac{4}{9 e^{1 / 3}})$

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