2026 NASS P1 Q2

2026 NASS P1 Q2

Secondary 4
5 marks

Do not use a calculator in this question.

It is given that \(\sqrt{2a}=\sqrt b+12\), where \(a\) and \(b\) are positive real numbers.

  1. Show that \(2a-3b=216-2(\sqrt b-6)^2\).[4]
  2. Hence state the least possible positive value of \(\dfrac1{2a-3b}\).[1]

Solution:

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Answer:(a) Identity shown. (b) \(\dfrac1{216}\).

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