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 :

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 $\Rightarrow T = e^{\lim_{x \to 1} \dfrac{f''(x+2)}{2}}$ $= e^{\dfrac{4}{2}} = e^2$ $\Rightarrow \log_e(T) = 2$

JEE Main 2026 — Mathematics Calculus

2026mcqhard

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 $\displaystyle \lim _{x \rightarrow 1}\left(\log _{\mathrm{e}}\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^{2}}}\right)$ is equal to :

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

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B
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Solution

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

$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{18}{(x-1)^2} \left( \dfrac{f(x+2) - f(3)}{f(3)} \right)}$

$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{18}{(x-1)^2} \left( \dfrac{f(x+2) - f(3)}{18} \right)}$

$\displaystyle \Rightarrow T = e^{\lim_{x \to 1} \left( \dfrac{f(x+2) - f(3)}{(x-1)^2} \right)}$ $\dfrac{0}{0}$ form, apply L'pital

$\displaystyle \Rightarrow T = e^{\lim_{x \to 1} \dfrac{f'(x+2)}{2(x-1)}}$ ; $\dfrac{0}{0}$ form, apply L'pital

$\displaystyle \Rightarrow T = e^{\lim_{x \to 1} \dfrac{f''(x+2)}{2}}$ $= e^{\dfrac{4}{2}} = e^2$

$\Rightarrow \log_e(T) = 2$

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Solution

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

$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{18}{(x-1)^2} \left( \dfrac{f(x+2) - f(3)}{f(3)} \right)}$

$\Rightarrow T = e^{\lim_{x \to 1} \dfrac{18}{(x-1)^2} \left( \dfrac{f(x+2) - f(3)}{18} \right)}$

$\displaystyle \Rightarrow T = e^{\lim_{x \to 1} \left( \dfrac{f(x+2) - f(3)}{(x-1)^2} \right)}$ $\dfrac{0}{0}$ form, apply L'pital

$\displaystyle \Rightarrow T = e^{\lim_{x \to 1} \dfrac{f'(x+2)}{2(x-1)}}$ ; $\dfrac{0}{0}$ form, apply L'pital

$\displaystyle \Rightarrow T = e^{\lim_{x \to 1} \dfrac{f''(x+2)}{2}}$ $= e^{\dfrac{4}{2}} = e^2$

$\Rightarrow \log_e(T) = 2$

JEE Main 2026 — Mathematics Calculus

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JEE Main 2026 — Mathematics Calculus

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