Aspire Faculty ID #17912 · Topic: JEE Main 2026 (21 January Evening Shift) · Just now
JEE Main 2026 (21 January Evening Shift)

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

Solution

$f'(x) = 3x^2 + 2x f'(1) + 2f''(2)$ 
$f''(x) = 6x + 2f'(1)$ $f'''(x) = 6$ 
$f(2) = 12 + 2f'(1)$ 
$f(x) = 3x^2 + 2x f'(1) + (12 + 2f'(1))$ 
Putting $x = 1$ 
$f(1) = 3 + 6f'(1) + 24$ $-5f'(1) = 27 \Rightarrow f'(1) = -\frac{27}{5}$ 
$\therefore f''(2) = 12 + 2\left(-\frac{27}{5}\right) = \frac{6}{5}$ 
$\therefore f'(x) = 3x^2 - \frac{54}{5}x + \frac{12}{5}$ 
$\therefore f'(5) = 75 - 54 + \frac{12}{5} = \frac{117}{5}$

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