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Question #41426

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the set of all real values of kk for which the quadratic equations x2−kx−5=0x^2 - kx - 5 = 0 and x2+5x+k=0x^2 + 5x + k = 0 have at least one common root. Consider the integral I(k)=∫−∞∞dxx2+2kx+k3−6k2−14k+120I(k) = \int_{-\infty}^{\infty} \frac{dx}{x^2 + 2kx + k^3 - 6k^2 - 14k + 120} If k∈Sk \in S is chosen such that I(k)I(k) converges to a finite real number, find the value of I(k)I(k).

Options

  • A.

    π4\frac{\pi}{4}

  • B.

    π2\frac{\pi}{2}

  • C.

    π8\frac{\pi}{8}

  • D.

    π6\frac{\pi}{6}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Completing the Square in Quadratics|Condition for Common Roots|Wavy Curve (Method of Intervals)

Why This Question Is Tricky

Common Mistakes

The question fuses the algebraic 'Condition for Common Roots' with calculus convergence criteria.

The primary trap lies in the Wavy Curve method applied to the discriminant of the integral's denominator.

Students typically find k∈{−5,4}k \in \{-5, 4\} and might arbitrarily pick one, or assume both are valid, failing to realize that the integral's convergence strictly demands the denominator's discriminant to be negative (meaning the completed square's constant must be positive).

By completing the square, a cubic inequality (k+4)(k−5)(k−6)>0(k+4)(k-5)(k-6) > 0 emerges.

The trap is sprung when k=−5k = -5 algebraically satisfies the common root condition but falls into the negative region of the Wavy Curve, causing the integral to diverge.

Only k=4k = 4 survives this nested domain restriction.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptCompleting the Square in Quadratics
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree