Higher June 2017 Paper 3 Q27
27 Prove that \(\quad x^2 + x + 1 \quad\) is always positive. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – completing the square | ||
| \(\left(x + \dfrac{1}{2}\right)^2 + \ldots\) | M1 | |
| \(\left(x + \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2 + 1\) or \(\left(x + \dfrac{1}{2}\right)^2 - \dfrac{1}{4} + 1\) or \(\left(x + \dfrac{1}{2}\right)^2 + \dfrac{3}{4}\) | A1 | oe |
| \(\left(x + \dfrac{1}{2}\right)^2 \geqslant 0\) and \(\dfrac{3}{4} \gt 0\) and always positive | A1 | oe |
| Alternative method 2 – real roots | ||
| \(\dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times 1}}{2 \times 1}\) or a correct sketch showing a quadratic curve with turning point above the \(x\)-axis | M1 | oe |
| States no values on \(x\)-axis | A1 | oe |
| States no values on \(x\)-axis and (minimum value =) \(\dfrac{3}{4}\) | A1 | oe |
| Alternative method 3 – Calculus | ||
| \(2x + 1 = 0\) | M1 | |
| \(x = -\dfrac{1}{2}\) | A1 | |
| (minimum value =) \(\dfrac{3}{4}\) | A1 | |
| Alternative method 4 – Explanation method | ||
| If \(x \geqslant 0\), \(x^2 \geqslant 0\) and \(x \geqslant 0\) \((1 \gt 0)\) so \(x^2 + x + 1 \gt 0\) and If \(-1 \lt x \lt 0\) \(x^2 \gt 0\) and \(x + 1 \gt 0\) so \(x^2 + x + 1 \gt 0\) and If \(x \leqslant -1\) \(x^2 \gt x\) and \(x^2 + x \gt 0\) so \(x^2 + x + 1 \gt 0\) | B3 | Accept \(x \gt 0\) for \(x \geqslant 0\) B2 for two correct statements B1 for one correct statement |
Additional guidance
| Calculating pairs of coordinates alone | M0A0A0 |