Aspire Faculty ID #17894 · Topic: JEE Main 2026 (21 January Morning Shift) · Just now
JEE Main 2026 (21 January Morning Shift)

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f(3) = 18$, $f'(3) = 0$ and $f''(3) = 4$. Then

$\lim_{x \to 3} \left( \log_e \left( \frac{f(2 + x)}{f(3)} \right) \right)^{\frac{18}{(x-3)^2}}$

is equal to:

Solution

Let $T = \lim_{x \to 3} \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-3)^2}}$ ; $1^\infty$ form

$\Rightarrow T = e^{\lim \frac{18}{(x-3)^2} \cdot \frac{f(x+2) - f(3)}{f(3)}}$

$\Rightarrow T = e^{\lim \frac{18}{(x-3)^2} \cdot \frac{f(x+2) - f(3)}{18}}$

$\Rightarrow T = e^{\lim \frac{f(x+2) - f(3)}{(x-3)^2}}$ ; $0/0$ form apply L’Hospital

$\Rightarrow T = e^{\lim \frac{f'(x+2)}{2(x-3)}}$ ; $0/0$ form apply L’Hospital

$\Rightarrow T = e^{\lim \frac{f''(x+2)}{2}} = e^2$

$\Rightarrow \log_e (T) = 2$

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