jee-main 2024 Q73

jee-main · India · session1_31jan_shift1 Chain Rule Continuity Conditions via Composition
Let $g(x)$ be a linear function and $f(x) = \begin{cases} g(x), & x \leq 0 \\ \frac { 1 + x } { 2 + x } , & x > 0 \end{cases}$, is continuous at $x = 0$. If $f'(1) = f(-1)$, then the value of $g(3)$ is
(1) $\frac { 1 } { 3 } \log _ { e } \frac { 4 } { e^{1/3} }$
(2) $\frac { 1 } { 3 } \log _ { e } \frac { 4 } { 9 } + 1$
(3) $\log _ { e } \frac { 4 } { 9 } - 1$
(4) $\log _ { e } \frac { 4 } { 9 e ^ { 1/3 } }$
Let $g(x)$ be a linear function and $f(x) = \begin{cases} g(x), & x \leq 0 \\ \frac { 1 + x } { 2 + x } , & x > 0 \end{cases}$, is continuous at $x = 0$. If $f'(1) = f(-1)$, then the value of $g(3)$ is\\
(1) $\frac { 1 } { 3 } \log _ { e } \frac { 4 } { e^{1/3} }$\\
(2) $\frac { 1 } { 3 } \log _ { e } \frac { 4 } { 9 } + 1$\\
(3) $\log _ { e } \frac { 4 } { 9 } - 1$\\
(4) $\log _ { e } \frac { 4 } { 9 e ^ { 1/3 } }$