JEE Main 2026 Apr 5 Shift 2, Mathematics Q20: Differentiability Of A Function

JEE Main2026Apr 5, Shift 2Mathematics

Let $f (x)$ and $g (x)$ be twice differentiable functions satisfying $f^{''} (x) = g^{''} (x)$ for all $x \in R$ , $f^{'} (1) = 2 g^{'} (1) = 4$ and $g (2) = 3 f (2) = 9$ . Then $f (25) - g (25)$ is equal to:

A
20
B
40
C
$- 20$
D
$- 40$

Solution

From $f^{''} = g^{''}$ , integrating gives $f^{'} (x) = g^{'} (x) + C_{1}$ . At $x = 1$ : $f^{'} (1) = 4$ and $g^{'} (1) = 2$ , so $C_{1} = 2$ .

Integrating again: $f (x) = g (x) + 2 x + C_{2}$ . At $x = 2$ : $f (2) = 3$ and $g (2) = 9$ , so $3 = 9 + 4 + C_{2} \Rightarrow C_{2} = - 10$ .

Thus $f (x) - g (x) = 2 x - 10$ . At $x = 25$ : $f (25) - g (25) = 50 - 10 = 40$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q20: Differentiability Of A Function

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