Part CCSIR NET December 2025the-non-trivial-solvability-condition-pins-lambda-to-a-specific-value-first-then-the-boundary-data-pins-the-two-constants

The non trivial solvability condition pins lambda to a specific value first then the boundary data pins the two constants

Let $\lambda\in\mathbb{R}$ be such that the integral equation $y(x) = \lambda\int$ ₋ $_{1}^{1}(5$ xt $^{3}+4x^{2}t+3$ xt)y(t)dt admits a non-trivial solution y(x) such that y(1)=5/2. Then which of the following statements are true?

A.y(0)+y′(0) = 3/2✓
B.y(1/2)+y′(1/2) = 7/2✓
C.y(−1)+y′(−1) = −1✓
D.y(1/3)+y′(1/3) = 14/9

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

The non trivial solvability condition pins lambda to a specific value first then the boundary data pins the two constants

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