June 2025 Paper 1 Q20
20
\[y = \frac{x^3}{3} - 5x^2 + 28x + 11\]Show that \(y\) is an increasing function for all values of \(x\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: completes the square | ||
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) x^2 - 10x + 28\) | M1 | oe 3 terms with at least two terms correct condone \(y = x^2 - 10x + 28\) |
| \((x - 5)^2 \ldots\ldots\) | M1 | must be ft from a quadratic derivative |
| \((x - 5)^2 + 3\) | M1dep | dep on both previous M marks must be from correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and not ft |
| \((x - 5)^2 \geqslant 0\) and so increasing function or the gradient or \((x - 5)^2 + 3\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is always positive | A1 | oe condone a statement that squares are always positive or \((x - 5)^2 \gt 0\) |
| Alternative method 2: uses the discriminant | ||
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) x^2 - 10x + 28\) | M1 | oe 3 terms with at least two terms correct condone \(y = x^2 - 10x + 28\) |
| \(10^2 - 4 \times 1 \times 28\) | M1 | oe must be ft from a quadratic derivative |
| No real roots stated or \(\sqrt{-12}\) or the gradient is never 0 | M1dep | dep on both previous M marks must be from correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and not ft |
| Statement to explain why this means the cubic is always an increasing function or the gradient is always positive | A1 | eg gradient can’t be 0 and when \(x = 1\) gradient is 19 so increasing function or graph of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) against \(x\) is always above the \(x\) axis so \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is always positive |