Let f: R arrow(0, ∞) be a twice differentiable function such that f(3)=18, f^ (3)=0 and f^ (3)=4 . Then _ x arrow 1 ( _ e ((f(2+x))/(f(3)))^ 18 (x-1)^ 2 ) is equal to :

Correct answer: C. 2 Let T = _ x → 1 ( (f(x+2))/(f(3)) )^ (18)/((x-1)^2) ; 1^ ∞ form T = e^ _ x → 1 (18)/((x-1)^2) ( (f(x+2) - f(3))/(f(3)) ) T = e^ _ x → 1 (18)/((x-1)^2) ( (f(x+2) - f(3))/(18) ) T = e^ _ x → 1 ( (f(x+2) - f(3))/((x-1)^2) ) (0)/(0) form, apply L'pital T = e^ _ x → 1 (f'(x+2))/(2(x-1)) ; (0)/(0) form, apply L'pital T = e^ _ x → 1 (f''(x+2))/(2) = e^ (4)/(2) = e^2 _e(T) = 2

JEE Main 202621 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Limits

Let f: R arrow(0, ∞) be a twice differentiable function such that...

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 :

Options

Correct answer

C. 2

Step-by-step 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

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