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Question #41604

Difficulty: 8/10 | 🟢 Free

The Question

Let Q(x)=∑k=03(3k)(−1)kx2k(sgn(x))kQ(x) = \sum_{k=0}^{3} \binom{3}{k} (-1)^k x^{2k} (\text{sgn}(x))^k, where sgn(x)\text{sgn}(x) denotes the signum function. Let y=y(x)y = y(x) be a differentiable function satisfying the differential equation (1−x∣x∣)dydx+6∣x∣y=x(Q(x))3(1 - x|x|) \frac{dy}{dx} + 6 |x| y = x (Q(x))^3 for all x∈Rx \in \mathbb{R}, with the initial condition y(0)=0y(0) = 0. Find the value of y(−1)y(-1).

Options

  • A.

    42

  • B.

    0

  • C.

    -42

  • D.

    84

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

Linear Differential Equation Structure|Coefficient of x^r in Expansion|Sign Flipping Rule

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize a piecewise binomial expansion and correctly compute the Integrating Factor (IF) across a sign-flip boundary.

The trap lies in the term x2k(sgn(x))kx^{2k}(\text{sgn}(x))^k, which simplifies to (x∣x∣)k(x|x|)^k, making Q(x)=(1−x∣x∣)3Q(x) = (1-x|x|)^3.

Students often ignore the absolute value or signum function, assuming x>0x>0 globally, which leads to Q(x)=(1−x2)3Q(x) = (1-x^2)^3 and an incorrect symmetric solution yielding y(−1)=0y(-1)=0.

Furthermore, integrating the piecewise LDE requires matching the constant of integration at x=0x=0, penalizing those who blindly apply standard formulas without respecting domain boundaries.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDifferential Equations - Class 12
ConceptLinear Differential Equation Structure
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree