Part CCSIR NET June 2025the-integral-runs-past-the-data-so-the-free-endpoint-over-determines-the-extremal

The integral runs past the data so the free endpoint over determines the extremal

For $a, b, c \in R$ , consider the variational problem: Minimize $J [y] = \int_{0}^{2} [a (y^{'})^{2} + 2$ byy′ + cy $^{2}]$ dx, subject to y(0) = 10, y(1) = 100. Then which of the following statements are true?

A.If (a, b, c) = (−2, 1, −2), then every admissible extremal is a minimizer✓
B.If (a, b, c) = (1, 0, 2), then every admissible extremal is a minimizer✓
C.If (a, b, c) = (2, −1, 1), then every admissible extremal is a minimizer✓
D.If (a, b, c) = (1, −2, 5), then every admissible extremal is a minimizer✓

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: A weak minimum of a functional is a strong minimum

The integral runs past the data so the free endpoint over determines the extremal

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