Q32 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
Let $f: \mathbf{R} \rightarrow(0, \infty)$ be a twice differentiable function such that $f(3)=18, f^{\prime}(3)=0$ and $f^{\prime \prime}(3)=4$. Then $\lim _{x \rightarrow 1}\left(\log _{\mathrm{e}}\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^{2}}}\right)$ is equal to :
- A1
- B18
- Correct answerC2
- D9
Solution
Let $T = \lim_{x \to 1} \left( \dfrac{f(x+2)}{f(3)} \right)^{\dfrac{18}{(x-1)^2}}$ ; $1^{\infty}$ form
$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{18}{(x-1)^2} \left( \dfrac{f(x+2) - f(3)}{f(3)} \right)}$
$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{18}{(x-1)^2} \left( \dfrac{f(x+2) - f(3)}{18} \right)}$
$\Rightarrow T = e^{\lim_{x \to 1} \left( \dfrac{f(x+2) - f(3)}{(x-1)^2} \right)}$ $\dfrac{0}{0}$ form, apply L'pital
$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{f'(x+2)}{2(x-1)}}$ ; $\dfrac{0}{0}$ form, apply L'pital
$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{f''(x+2)}{2}}$ $= e^{\dfrac{4}{2}} = e^2$
$\Rightarrow \log_e(T) = 2$
$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{18}{(x-1)^2} \left( \dfrac{f(x+2) - f(3)}{f(3)} \right)}$
$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{18}{(x-1)^2} \left( \dfrac{f(x+2) - f(3)}{18} \right)}$
$\Rightarrow T = e^{\lim_{x \to 1} \left( \dfrac{f(x+2) - f(3)}{(x-1)^2} \right)}$ $\dfrac{0}{0}$ form, apply L'pital
$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{f'(x+2)}{2(x-1)}}$ ; $\dfrac{0}{0}$ form, apply L'pital
$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{f''(x+2)}{2}}$ $= e^{\dfrac{4}{2}} = e^2$
$\Rightarrow \log_e(T) = 2$
Hence, the correct answer is option: 'C'.