2026 CSS P2 Q5

2026 CSS P2 Q5

Secondary 4
9 marks

The variables \(x\) and \(y\) are connected by the equation \(y=\dfrac12x^3+\dfrac1{5x}\).

Some corresponding values of \(x\) and \(y\), correct to 1 decimal place, are given in the table below.

\(x\)\(-2\)\(-1\)\(-0.5\)\(-0.1\)\(0.1\)\(0.5\)\(1\)\(2\)
\(y\)\(p\)\(-0.7\)\(-0.5\)\(-2.0\)\(2.0\)\(0.5\)\(0.7\)\(4.1\)
  1. Calculate the value of \(p\).[1]
  2. On the axes provided below, draw the graph of \(y=\dfrac12x^3+\dfrac1{5x}\).[2]
  1. Using your graph, find the value of \(y\) when \(x=-1.1\).[1]
  2. By drawing a tangent, find the gradient of the curve at \(x=1.5\).[2]
  3. By drawing a suitable straight line, find the solution of \(\dfrac12x^3+\dfrac1{5x}+2.5x+5=0\).[2]
  4. Explain why the curve does not touch the \(y\)-axis.[1]

Solution:

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Answer:(a) \(-4.1\) (b) See graph. (c) Approximately \(-0.85\) (d) Approximately \(3.29\) (e) \(x\approx-1.4\) and \(x\approx-0.04\) (f) Undefined at \(x=0\).

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