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Part CCSIR NET June 2024the-bound-on-lambda-kills-the-second-mode

The bound on lambda kills the second mode

For λ∈R\lambda \in \mathbb{R} such that |λ\lambda| < 5/32, let R(x,t,λ)R(x, t, \lambda) and u denote the resolvent kernel and the solution, respectively, of the Fredholm integral equation u(x)=x+(λ/2)∫u(x) = x + (\lambda/2)\int₋22(_{2}^{2} (xt +x2t2)u(t)+ x^{2}t^{2})u(t)dt. Then which of the following statements are true?

  1. A.R(x,t,λ)=3R(x, t, \lambda) = 3xt/(3−8λ)−5x2t2/(5−32λ)/(3 - 8\lambda) - 5x^{2}t^{2}/(5 - 32\lambda)
  2. B.R(x,t,λ)=3R(x, t, \lambda) = 3xt/(3−8λ)+5x2t2/(5−32λ)/(3 - 8\lambda) + 5x^{2}t^{2}/(5 - 32\lambda)✓ Correct
  3. C.u(1)=−5/(5−32λ)u(1) = -5/(5 - 32\lambda)
  4. D.u(1)=3/(3−8λ)u(1) = 3/(3 - 8\lambda)✓ Correct

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, and why this one tests boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

Related counterexample: Volterra equations have eigenvalues just like Fredholm equations

More on this topic

The chapter behind this: Separable kernels and resolvent kernels — free to read

From Linear Integral Equations › Separable kernels and resolvent kernels

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