2026 TPSS P2 Q10

2026 TPSS P2 Q10

Secondary 4
10 marks

The table below shows some values of \(x\) and its corresponding values of \(y\) for the graph of \(y=x^2+\dfrac{50}{x}-20\).

\(x\)\(0.5\)\(1\)\(1.5\)\(2\)\(3\)\(4\)\(6\)\(8\)
\(y\)\(80.3\)\(31\)\(15.6\)\(9\)\(5.7\)\(p\)\(24.3\)\(50.3\)
  1. Calculate the value of \(p\).[1]
  2. On the grid below, draw the graph of \(y=x^2+\dfrac{50}{x}-20\) for \(0.5\leq x\leq8\).[3]
  3. Draw a tangent and find the gradient of \(y=x^2+\dfrac{50}{x}-20\) at the point where \(x=4\).[2]
    1. Show that equation \(x^3-2x^2-48x+50=0\) can be solved by drawing the line \(y=2x+28\).[2]
    2. By drawing the line \(y=2x+28\), solve the equation of \(x^3-2x^2-48x+50=0\) for \(0.5\leq x\leq8\).[2]

Solution:

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Answer:(a) \(p=8.5\) (b) See graph. (c) Approximately \(4.9\) (d)(i) \(x^2+\dfrac{50}{x}-20=2x+28\) (ii) \(x\approx1.0\) or \(7.5\)

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