JEE Mathematics 2025 Question with Solution

Given: $f(x) = \int_{1}^{x} (t^2 - 9t + 20) \, dt$ for $1 \leq x \leq 5$.

Find: The value of $4(\alpha + \beta)$ where the range of $f(x)$ is $[\alpha, \beta]$.

From the solution, first integrate:

$$f(x) = \frac{t^3}{3} - \frac{9t^2}{2} + 20t$$

Evaluating from $1$ to $x$,

$$f(x) = \left[ \frac{x^3}{3} - \frac{9x^2}{2} + 20x \right] - \left[ \frac{1^3}{3} - \frac{9(1)^2}{2} + 20(1) \right]$$

The provided solution states that the boundary-value calculations give

$$\alpha = 4, \qquad \beta = 32$$

Hence,

$$4(\alpha + \beta) = 4(4 + 32)$$

So the correct option is A.

Note: The arithmetic shown in the source concludes with $157$, although the displayed substitution $4(4+32)$ evaluates to $144$. Since the solution explicitly marks Option A as correct, the extracted answer is A.

Given: $f(x) = \int_{1}^{x} (t^2 - 9t + 20) \, dt$ on $[1,5]$.

Find: The range endpoints $\alpha$ and $\beta$.

Differentiate the integral form:

$$f'(x) = x^2 - 9x + 20 = (x-4)(x-5)$$

On the interval $1 \leq x \leq 5$, this derivative is positive for $1 \leq x < 4$ and negative for $4 < x < 5$. Therefore, $f(x)$ increases up to $x=4$ and then decreases, so the maximum occurs at $x=4$ and the minimum occurs at an endpoint.

Now evaluate key values from the antiderivative given in the source:

$$f(1) = 0$$

and

$$f(4) = \int_{1}^{4} (t^2 - 9t + 20) \, dt = \frac{27}{2}$$

Also,

$$f(5) = \int_{1}^{5} (t^2 - 9t + 20) \, dt = \frac{32}{3}$$

Thus the actual range from the function is

$$\left[0, \frac{27}{2}\right]$$

which would give

$$4\left(0 + \frac{27}{2}\right) = 54$$

This does not match any option. Therefore, there is a discrepancy in the source solution/content. Since the solution explicitly declares Option A as correct, the extracted answer remains A.