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Part CCSIR NET June 2025the-degree-2-condition-forces-a-squared-plus-b-squared-equals-6-and-hence-root-3

The degree 2 condition forces a squared plus b squared equals 6 and hence root 3

If α,β∈R\alpha, \beta \in \mathbb{R} are such that the equation ∫03f(x)\int_{0}^{3} f(x) dx =(3/2)[f(α)+f(α+β)]= (3/2)[f(\alpha) + f(\alpha + \beta)] holds for all polynomials f(x) of degree less than or equal to 2, then which of the following statements are true?

  1. A.(α,β)=((3−3)/2,3)(\alpha, \beta) = ((3 - \sqrt{3})/2, \sqrt{3}) or (α,β)=((3+3)/2,−3)(\alpha, \beta) = ((3 + \sqrt{3})/2, -\sqrt{3})✓ Correct
  2. B.(α,β)=((3−2)/2,2)(\alpha, \beta) = ((3 - \sqrt{2})/2, \sqrt{2}) or (α,β)=((3+2)/2,−2)(\alpha, \beta) = ((3 + \sqrt{2})/2, -\sqrt{2})
  3. C.(α,β)=((3−5)/2,5)(\alpha, \beta) = ((3 - \sqrt{5})/2, \sqrt{5}) or (α,β)=((3+5)/2,−5)(\alpha, \beta) = ((3 + \sqrt{5})/2, -\sqrt{5})
  4. D.(α,β)=((3−7)/2,7)(\alpha, \beta) = ((3 - \sqrt{7})/2, \sqrt{7}) or (α,β)=((3+7)/2,−7)(\alpha, \beta) = ((3 + \sqrt{7})/2, -\sqrt{7})

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 execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Increasing the number of equally spaced interpolation nodes improves the approximation

More on this topic

The chapter behind this: Interpolation and numerical integration — free to read

From Numerical Analysis › Interpolation and numerical integration with error terms

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