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Question #40462

Difficulty: 8/10 | 🟢 Free

The Question

Let P(z)=z2+(i−57)z−57iP(z) = z^2 + (i-57)z - 57i and Q(z)=z2+(i−233)z−233iQ(z) = z^2 + (i-233)z - 233i. Let R(z)=z2+pz+qR(z) = z^2 + pz + q where p,q∈Rp, q \in \mathbb{R}. Given that the polynomials P(z),Q(z),P(z), Q(z), and R(z)R(z) are linearly dependent over C\mathbb{C}, let α\alpha be the common root of P(z)=0P(z) = 0 and Q(z)=0Q(z) = 0. Find the value of (p+q)+∑x(p + q) + \sum x, where the sum is taken over all integers xx satisfying the inequality x2+7Im⁡(α)x+10≤0x^2 + 7 \operatorname{Im}(\alpha) x + 10 \le 0.

Options

  • A.

    15

  • B.

    -13

  • C.

    13

  • D.

    -15

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Algebraic Manipulation of i|Condition for Common Roots|Sign Flipping Rule|Linear Dependency of Rows

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of linear algebra (linear dependency of polynomials) and complex quadratic equations.

The primary trap lies in the 'Sign Flipping Rule' and the 'Conjugate Pair Theorem'.

Students often blindly assume that complex roots occur in conjugate pairs even when polynomial coefficients are non-real, leading them to incorrectly identify the common root as +i+i instead of −i-i.

This sign error flips the rational inequality, yielding a completely different set of integers.

Furthermore, the linear dependency condition rigorously forces R(−i)=0R(-i)=0, which then correctly invokes the conjugate pair theorem since R(z)R(z) has real coefficients, bridging matrix algebra with theory of equations.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterComplex Numbers and Quadratic Equations - Class 11
ConceptAlgebraic Manipulation of i
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree