Before You Begin

Question #40712

Difficulty: 8/10 | 🟢 Free

The Question

Let $Q(x) = x^2 - px + q$ be a quadratic polynomial with real roots $\alpha$ and $\beta$ ( $\alpha < \beta$ ). It is given that $p = \frac{1}{\pi} \int_{\alpha}^{\beta} \frac{x^2 - q}{\sqrt{-Q(x)}} \, dx$ Furthermore, the improper integral $I = \int_{0}^{\infty} \frac{dx}{x^2 + px + q}$ converges to a finite value, and satisfies the relation $q = e^{I \sqrt{p^2 - 4q}}$ . Find the value of $p^2 + q^2$ .

Options

A.
41
B.
61
C.
85
D.
113

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Quadratic Roots in Sequences|Fundamental Theorem of Calculus|Bounded vs Unbounded Regions

Why This Question Is Tricky

Common Mistakes

This question tests the 'Implicit Constraint' pattern by coupling algebraic fixed-point equations with the convergence criteria of improper integrals.

The primary trap lies in the Bounded vs Unbounded analysis: students easily derive the two possible parameter sets $(p, q) = (6, 5)$ and $(2/3, -1/3)$ from the integral evaluations.

However, they frequently overlook that the second integral's domain is $[0, \infty)$ .

For $(2/3, -1/3)$ , the quadratic $x^2+px+q$ has a root at $x=1/3$ , placing a non-integrable singularity directly inside the integration bounds and causing the integral to diverge.

Recognizing this forces the rejection of the extraneous solution, a classic JEE Advanced differentiator.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Quadratic Roots in Sequences
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free