Let f ( x )= x 3 + x 2 f prime (1)+2 x f prime prime (2)+f prime prime prime , x in R . Then the value of f prime (5) is :

Mathematics · Methods of Differentiation

JEE Main 2026 — 21 January, Evening Shift — Question 3

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

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

Answer: D

Step-by-step solution

Given $f(x) = x^3 + x^2 f'(1) + 2x f''(2) + f'''(3)$ . Differentiate: $f'(x) = 3x^2 + 2x f'(1) + 2 f''(2)$ . Differentiate again: $f''(x) = 6x + 2 f'(1)$ . Then $f''(2) = 12 + 2 f'(1)$ . Substitute into $f'(x)$ : $f'(x) = 3x^2 + 2x f'(1) + 2(12 + 2 f'(1)) = 3x^2 + 2x f'(1) + 24 + 4 f'(1)$ . Put $x = 1$ : $f'(1) = 3 + 2 f'(1) + 24 + 4 f'(1) = 27 + 6 f'(1)$ . Thus $-5 f'(1) = 27 \Rightarrow f'(1) = -\frac{27}{5}$ . Then $f''(2) = 12 + 2\left(-\frac{27}{5}\right) = 12 - \frac{54}{5} = \frac{6}{5}$ . Hence $f'(x) = 3x^2 + 2x\left(-\frac{27}{5}\right) + 2\left(\frac{6}{5}\right) = 3x^2 - \frac{54}{5}x + \frac{12}{5}$ . Finally $f'(5) = 3(25) - \frac{54}{5}(5) + \frac{12}{5} = 75 - 54 + \frac{12}{5} = 21 + \frac{12}{5} = \frac{105 + 12}{5} = \frac{117}{5}$ .

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Mathematics
Chapter: Methods of Differentiation
Topic: Methods of Differentiation