JEE Mathematics 2024 Question with Solution

Given:

$$f(x) = \int_{-1}^{x} (e^t - 1)^{11} (2t - 1)^5 (t - 2)^7 (t - 3)^{12} (2t - 10)^{61} \, dt$$

Find: The value of $p^2 + 2q$, where $p$ is the sum of squares of the $x$-values of local maxima and $q$ is the sum of the $x$-values of local minima.

By the Fundamental Theorem of Calculus,

$$f'(x) = (e^x - 1)^{11}(2x - 1)^5(x - 2)^7(x - 3)^{12}(2x - 10)^{61}$$

So the critical points are obtained from $f'(x)=0$:

$$x = 0, \quad x = \frac{1}{2}, \quad x = 2, \quad x = 3, \quad x = 5$$

Now determine the sign of $f'(x)$.

• $(e^x-1)^{11}$ changes sign at $x=0$ because the power is odd and $e^x-1$ changes sign there.
• $(2x-1)^5$ changes sign at $x=\frac{1}{2}$.
• $(x-2)^7$ changes sign at $x=2$.
• $(x-3)^{12}$ does not change sign at $x=3$ because the power is even.
• $(2x-10)^{61}$ changes sign at $x=5$.

Starting with $x \in (-1,0)$:

• $e^x-1<0$,
• $2x-1<0$,
• $x-2<0$,
• x-3<0$$ but its even power stays positive,
• $2x-10<0$.

Hence,

$$f'(x) = (-)(-)(-)(+)(-) = (+)$$

So the sign pattern changes as follows whenever an odd-power factor crosses zero:

$$(0, \tfrac12): (-), \quad (\tfrac12,2): (+), \quad (2,3): (-), \quad (3,5): (-), \quad (5,\infty): (+)$$

Including the initial interval, the full sign chart is:

$$(-1,0): (+), \quad (0,\tfrac12): (-), \quad (\tfrac12,2): (+), \quad (2,5): (-), \quad (5,\infty): (+)$$

Therefore:

• At $x=0$, sign changes $+ \to -$, so $x=0$ is a local maximum.
• At $x=\frac12$, sign changes $- \to +$, so $x=\frac12$ is a local minimum.
• At $x=2$, sign changes $+ \to -$, so $x=2$ is a local maximum.
• At $x=3$, no sign change, so no extremum.
• At $x=5$, sign changes $- \to +$, so $x=5$ is a local minimum.

Thus,

$$p = 0^2 + 2^2 = 4$$

and

$$q = \frac{1}{2} + 5 = \frac{11}{2}$$

Now compute:

$$p^2 + 2q = 4^2 + 2\left(\frac{11}{2}\right) = 16 + 11 = 27$$

Therefore, the correct option is B.

The first extracted approach contains inconsistent intermediate conclusions, but the second approach and the sign analysis above give the correct result $27$.

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