JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18, f^{'} (3) = 0$ and $f^{''} (3) = 4$ . Then $x \to 1 lim (lo g_{e} \frac{f ( 2 + x )}{f ( 3 )})^{18/ (x - 1)^{2}}$ is equal to :

A
1
B
9
C
2
D
18

Solution

Near $x = 1$ , $lo g_{e} \frac{f ( x + 2 )}{f ( 3 )} = lo g_{e} (1 + \frac{f ( x + 2 ) - f ( 3 )}{f ( 3 )}) \approx \frac{f ( x + 2 ) - f ( 3 )}{18}$ , giving a $1^{\infty}$ form.

So the limit equals $e^{L}$ where $L = x \to 1 lim \frac{18}{( x - 1 ) ^{2}} \cdot \frac{f ( x + 2 ) - f ( 3 )}{18} = x \to 1 lim \frac{f ( x + 2 ) - f ( 3 )}{( x - 1 ) ^{2}}$ .

Two applications of L'Hôpital give $L = \frac{f ^{''} ( 3 )}{2} = 2$ .

The limit is $e^{2}$ , and $lo g_{e} (e^{2}) = 2$ .

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