JEE Mathematics 2025 Question with Solution

Given: $f(x)$ is a polynomial of degree four, and it has extreme values at $x = 4$ and $x = 5$.

Find: $f(2)$

From the given limit,

$$\lim_{x \to 0} \frac{f(x)}{x^2} = 5$$

for a polynomial this requires the constant term and the coefficient of $x$ to be zero, and the coefficient of $x^2$ to be $5$. So we write

$$f(x) = ax^4 + bx^3 + 5x^2$$

Hence,

$$f'(x) = 4ax^3 + 3bx^2 + 10x$$

Since the function has extreme values at $x = 4$ and $x = 5$,

$$f'(4) = 0, \qquad f'(5) = 0$$

Using $f'(4)=0$,

$$4a(4)^3 + 3b(4)^2 + 10(4) = 0$$ $$256a + 48b + 40 = 0$$ $$64a + 12b + 10 = 0$$

Using $f'(5)=0$,

$$4a(5)^3 + 3b(5)^2 + 10(5) = 0$$ $$500a + 75b + 50 = 0$$ $$100a + 15b + 10 = 0$$

Subtracting,

$$(100a + 15b + 10) - (64a + 12b + 10) = 0$$ $$36a + 3b = 0$$ $$b = -12a$$

Substituting into $64a + 12b + 10 = 0$,

$$64a + 12(-12a) + 10 = 0$$ $$-80a + 10 = 0$$ $$a = \frac{1}{8}$$

Therefore,

$$b = -\frac{3}{2}$$

So,

$$f(x) = \frac{1}{8}x^4 - \frac{3}{2}x^3 + 5x^2$$

Now evaluate at $x = 2$:

$$f(2) = \frac{1}{8}(16) - \frac{3}{2}(8) + 5(4)$$ $$= 2 - 12 + 20$$ $$= 10$$

Therefore, $f(2) = 10$ and the correct option is B.

The limit

$$\lim_{x \to 0} \frac{f(x)}{x^2} = 5$$

means $f(0)=0$ and the coefficient of $x$ must also be zero; otherwise the quotient would either blow up or not approach a finite value. The coefficient of $x^2$ must be $5$.

Thus the quartic can be taken as

$$f(x) = ax^4 + bx^3 + 5x^2$$

Then use the extreme value condition, which gives derivative zero at $x=4$ and $x=5$, to determine $a$ and $b$. This yields

$$a = \frac{1}{8}, \qquad b = -\frac{3}{2}$$

and hence

$$f(2)=10$$

So the correct option is B.