JEE Mathematics 2026 Question with Solution

Given: $f(x) = x^3 + x^2 f'(1) + 2x f''(2) + f'''(3)$, where $x \in \mathbb{R}$.

Find: $f'(5)$.

Treat $f'(1), f''(2), f'''(3)$ as constants. Let

$$a = f'(1), \qquad b = f''(2), \qquad c = f'''(3)$$

Then

$$f(x) = x^3 + ax^2 + 2bx + c$$

Differentiate repeatedly:

$$f'(x) = 3x^2 + 2ax + 2b$$ $$f''(x) = 6x + 2a$$ $$f'''(x) = 6$$

So,

$$c = f'''(3) = 6$$

Using $a = f'(1)$,

$$a = 3(1)^2 + 2a(1) + 2b$$ $$a + 2b = -3$$

Using $b = f''(2)$,

$$b = 6(2) + 2a$$ $$2a - b = -12$$

From the second equation,

$$b = 2a + 12$$

Substitute into the first equation:

$$a + 2(2a + 12) = -3$$ $$5a + 24 = -3$$ $$a = -\frac{27}{5}$$

Then

$$b = 2\left(-\frac{27}{5}\right) + 12 = \frac{6}{5}$$

Now evaluate $f'(5)$:

$$f'(5) = 3(5)^2 + 2a(5) + 2b$$ $$f'(5) = 75 + 10a + 2b$$ $$f'(5) = 75 + 10\left(-\frac{27}{5}\right) + 2\left(\frac{6}{5}\right)$$ $$f'(5) = 75 - 54 + \frac{12}{5}$$ $$f'(5) = 21 + \frac{12}{5} = \frac{117}{5}$$

Therefore, the correct option is D and $f'(5) = \frac{117}{5}$.

Given: The function is defined in terms of its own derivatives at fixed points.

Find: The value of $f'(5)$.

The key idea is that the quantities $f'(1)$, $f''(2)$ and $f'''(3)$ are fixed numbers, not functions of $x$. So rewrite the expression as a cubic polynomial with constant coefficients:

$$f(x) = x^3 + ax^2 + 2bx + c$$

where

$$a = f'(1), \qquad b = f''(2), \qquad c = f'''(3)$$

Now differentiate:

$$f'(x) = 3x^2 + 2ax + 2b$$ $$f''(x) = 6x + 2a$$ $$f'''(x) = 6$$

Since $c = f'''(3)$, we directly get

$$c = 6$$

Next, impose the self-consistency conditions. From $a = f'(1)$,

$$a = 3 + 2a + 2b$$

which gives

$$a + 2b = -3$$

From $b = f''(2)$,

$$b = 12 + 2a$$

which gives

$$2a - b = -12$$

Solve these two linear equations:

$$b = 2a + 12$$

Substitute into $a + 2b = -3$:

$$a + 2(2a + 12) = -3$$ $$5a = -27$$ $$a = -\frac{27}{5}$$

Then

$$b = 2\left(-\frac{27}{5}\right) + 12 = \frac{6}{5}$$

Now substitute into $f'(x)$ at $x = 5$:

$$f'(5) = 3 \cdot 25 + 2\left(-\frac{27}{5}\right) \cdot 5 + 2\left(\frac{6}{5}\right)$$ $$f'(5) = 75 - 54 + \frac{12}{5} = \frac{117}{5}$$

Hence, the correct option is D.