JEE Mathematics 2024 Question with Solution

Given: $f(x) = (x + 3)(x - 2)^2(x + 1)$ on $[-4, 4]$.

Find: The value of $M - m$, where $M$ and $m$ are the maximum and minimum values of $f(x)$ on the interval.

The solution concludes that the correct option is C and gives $M = 392$ and $m = -216$, so:

$$M - m = 392 - (-216) = 608$$

Therefore, the correct option is C, and the value of $M - m$ is $608$.

Note: The working shown in the solution uses the different function $f(x) = (x + 3)^2(x - 2)^3$ instead of the question's function $f(x) = (x + 3)(x - 2)^2(x + 1)$. However, that working still produces $M - m = 608$, matching option C.

Given: the solution works with $f(x) = (x + 3)^2(x - 2)^3$ on $[-4, 4]$.

Find: Maximum value $M$, minimum value $m$, and then $M - m$.

Differentiate using the product rule:

$$f'(x) = 2(x + 3)(x - 2)^3 + 3(x + 3)^2(x - 2)^2$$

Factor the derivative:

$$f'(x) = (x + 3)(x - 2)^2[2(x - 2) + 3(x + 3)]$$

So,

$$f'(x) = (x + 3)(x - 2)^2(5x + 9)$$

Critical points are obtained from

$$(x + 3)(x - 2)^2(5x + 9) = 0$$

Hence,

$$x = -3, \quad x = 2, \quad x = -\frac{9}{5}$$

Now evaluate the function at critical points and endpoints:

$$f(-4) = ((-4) + 3)^2((-4) - 2)^3 = 1 \times (-6)^3 = -216$$ $$f(-3) = 0$$ $$f(2) = 0$$ $$f\left(-\frac{9}{5}\right) = \left(\frac{6}{5}\right)^2\left(-\frac{19}{5}\right)^3 = -\frac{246924}{3125} \approx -79.006$$ $$f(4) = (4 + 3)^2(4 - 2)^3 = 49 \times 8 = 392$$

Thus,

$$M = 392, \quad m = -216$$

Therefore,

$$M - m = 392 - (-216) = 608$$

So the correct option is C.