JEE Main 2026 — Jan 21, Shift 2 — Mathematics solution

JEE Main2026Jan 21, Shift 2Mathematics

Let $f (x) = x^{3} + x^{2} f^{'} (1) + 2 x f^{''} (2) + f^{'''} (3)$ , $x \in R$ . Then the value of $f^{'} (5)$ is :

A
$\frac{62}{5}$
B
$\frac{657}{5}$
C
$\frac{2}{5}$
D
$\frac{117}{5}$

Solution

Treat $f^{'} (1)$ , $f^{''} (2)$ , $f^{'''} (3)$ as constants. Differentiate:

$f^{'} (x) = 3 x^{2} + 2 x f^{'} (1) + 2 f^{''} (2)$ , $f^{''} (x) = 6 x + 2 f^{'} (1)$ , $f^{'''} (x) = 6$ .

Pin down the constants:

From $f^{''} (x)$ at $x = 2$ : $f^{''} (2) = 12 + 2 f^{'} (1)$ .

From $f^{'} (x)$ at $x = 1$ : $f^{'} (1) = 3 + 2 f^{'} (1) + 2 f^{''} (2) = 3 + 2 f^{'} (1) + 2 (12 + 2 f^{'} (1)) = 27 + 6 f^{'} (1)$ . So $- 5 f^{'} (1) = 27 \Rightarrow f^{'} (1) = - \frac{27}{5}$ , hence $f^{''} (2) = 12 - \frac{54}{5} = \frac{6}{5}$ .

Thus $f^{'} (x) = 3 x^{2} - \frac{54}{5} x + \frac{12}{5}$ .

$f^{'} (5) = 75 - 54 + \frac{12}{5} = 21 + \frac{12}{5} = \frac{117}{5}$ .

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