Least value predicted by a linearised law

Least value predicted by a linearised law

IP 3
7 marks

For \(x>0\), the variables \(x\) and \(y\) satisfy \(y=ax+\dfrac{b}{x}\), where \(a\) and \(b\) are constants. The graph of \(xy\) against \(x^2\) is a straight line through \((1,5)\) and \((9,13)\).

  1. Explain why plotting \(xy\) against \(x^2\) gives a straight line, and find \(a\) and \(b\).[3]
  2. By writing \(y\) in the form \(\left(\sqrt{x}-\dfrac{p}{\sqrt{x}}\right)^2+q\), show that \(y\ge4\) for all \(x>0\), and find the value of \(x\) at which \(y=4\).[3]
  3. A recorded reading is \(y=3.5\) at \(x=1.5\). Explain what this suggests about the reading.[1]

Solution:

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Answer:(a) \(a=1,\ b=4\) (b) \(y=4\) at \(x=2\) (c) the reading is below the least possible value 4, so it is likely wrong

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