JEE Mathematics 2025 Question with Solution

Given: $f(x) = x^3 + ax^2 + b \log|x| + 1$ and the critical points are $x = -1$ and $x = 2$.

Find: The value of $|M + m|$, where $m$ and $M$ are the absolute minimum and maximum values of $f$ on $\left[-2, -\frac{1}{2}\right]$.

Differentiate:

$$f'(x) = 3x^2 + 2ax + \frac{b}{x}$$

Since $x = -1$ and $x = 2$ are critical points,

$$f'(-1) = 0, \qquad f'(2) = 0$$

So,

$$3 - 2a - b = 0$$

and

$$12 + 4a + \frac{b}{2} = 0$$

which give

$$2a + b = 3, \qquad 8a + b = -24$$

Subtracting,

$$6a = -27 \Rightarrow a = -\frac{9}{2}$$

Then,

$$b = 3 - 2a = 12$$

Hence,

$$f(x) = x^3 - \frac{9}{2}x^2 + 12 \log|x| + 1$$

Now evaluate $f(x)$ at the endpoints and the critical point lying in the interval $\left[-2, -\frac{1}{2}\right]$, namely $x = -1$.

At $x = -2$,

$$f(-2) = -8 - 18 + 12\log 2 + 1 = -25 + 12\log 2 \approx -16.6$$

At $x = -\frac{1}{2}$,

$$f\left(-\frac{1}{2}\right) = -\frac{1}{8} - \frac{9}{8} + 12\log\left(\frac{1}{2}\right) + 1 \approx -8.6$$

At $x = -1$,

$$f(-1) = -1 - \frac{9}{2} + 1 = -\frac{9}{2} = -4.5$$

Therefore, on the interval $\left[-2, -\frac{1}{2}\right]$,

$$m = -16.6, \qquad M = -4.5$$

So,

$$|M + m| = |-4.5 + (-16.6)| = 21.1$$

Therefore, the correct option is A.

Given: $f'(x) = 3x^2 + 2ax + \frac{b}{x}$ with critical points at $x = -1$ and $x = 2$.

Find: $|M + m|$ on $\left[-2, -\frac{1}{2}\right]$.

From the critical point conditions,

$$2a + b = 3, \qquad 8a + b = -24$$

Thus,

$$a = -\frac{9}{2}, \qquad b = 12$$

Then

$$f'(x) = 3x^2 - 9x + \frac{12}{x} = \frac{3x^3 - 9x^2 + 12}{x}$$

and the numerator factors as

$$3x^3 - 9x^2 + 12 = 3(x+1)(x-2)^2$$

So,

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

This confirms that the only critical point in the interval $\left[-2, -\frac{1}{2}\right]$ is $x = -1$. Therefore, test $x = -2$, $x = -1$, and $x = -\frac{1}{2}$.

Using

$$f(x) = x^3 - \frac{9}{2}x^2 + 12 \log|x| + 1$$

we get

$$f(-2) = -25 + 12\log 2 \approx -16.6$$ $$f(-1) = -\frac{9}{2} = -4.5$$ $$f\left(-\frac{1}{2}\right) = -\frac{1}{8} - \frac{9}{8} - 12\log 2 + 1 \approx -8.6$$

Hence the absolute minimum is $m = -16.6$ and the absolute maximum is $M = -4.5$.

Now,

$$|M + m| = |-4.5 - 16.6| = 21.1$$

the solution states the correct option is A. Note that one part of the solution incorrectly evaluates $f(2)$ and treats it as relevant to the given interval, but $x = 2$ lies outside $\left[-2, -\frac{1}{2}\right]$, so it must not be used for determining $M$ and $m$ here.