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 $f'(1)=f(-1)$, then the value of $g(3)$ is

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

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

(B) $\frac{1}{3}\log_{e}\left(\frac{4}{9}\right)+1$

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

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Continuity and Differentiability

Let

g (x)

be a linear function and

f (x) = {g (x) (\frac{1 + x}{2 + x})^{\frac{1}{x}}, x \leq 0, x > 0

, is continuous at

x = 0

. If

f^{'} (1) = f (- 1)

, then the value of

g (3)

Answer: (D)

Continuity at

x = 0

requires

g (0) = x \to 0^{+} lim (\frac{1 + x}{2 + x})^{1/ x}

. As

x \to 0^{+}

, the base

\to \frac{1}{2} < 1

while the exponent

\to \infty

, so this limit is

0

. So

g (0) = 0

, and since

g

is linear,

g (x) = a x

for some constant

a

.

For

x > 0

, let

h (x) = ln f (x) = \frac{1}{x} [ln (1 + x) - ln (2 + x)]

. Then

f^{'} (x) = f (x) \cdot h^{'} (x), h^{'} (x) = \frac{N ^{'} ( x ) x - N ( x )}{x ^{2}}, N (x) = ln (1 + x) - ln (2 + x)

x = 1

f (1) = \frac{2}{3}

N (1) = ln \frac{2}{3}

N^{'} (1) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}

, so

h^{'} (1) = \frac{1}{6} - ln \frac{2}{3} = \frac{1}{6} + ln \frac{3}{2}, f^{'} (1) = \frac{2}{3} (\frac{1}{6} + ln \frac{3}{2}) = \frac{1}{9} + \frac{2}{3} ln \frac{3}{2}

Also

f (- 1) = g (- 1) = - a

. Setting

f^{'} (1) = f (- 1)

- a = \frac{1}{9} + \frac{2}{3} ln \frac{3}{2} ⟹ a = - \frac{1}{9} - \frac{2}{3} ln \frac{3}{2}

g (3) = 3 a = - \frac{1}{3} - 2 ln \frac{3}{2} = - \frac{1}{3} - ln \frac{9}{4} = ln \frac{4}{9} - \frac{1}{3} = ln (\frac{4}{9 e ^{1/3}})

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option