June 2024 Paper 2 Q10
10 The function \(\mathrm{f}\) is defined by
\[\mathrm{f}(x) = x^2 + 2\cos x \quad \text{for } -\pi \leqslant x \leqslant \pi\]Determine whether the curve with equation \(y = \mathrm{f}(x)\) has a point of inflection at the point where \(x = 0\)
Fully justify your answer. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Differentiates to obtain \(2x \pm 2\sin x\) | M1 | 1.1a |
| Differentiates again to obtain \(2 - 2\cos x\) | A1 | 1.1b |
| Concludes \(\mathrm{f}^{\prime\prime}(0) = 0\) and tests the sign of their \(\mathrm{f}^{\prime}(x)\) or \(\mathrm{f}^{\prime\prime}(x)\) either side of \(x\) = 0 Or Deduces that \(\mathrm{f}^{\prime\prime}(x) \geqslant 0\) | M1 | 2.1 |
| Completes a reasoned argument to conclude that \(y\) = f(\(x\)) does not have a point of inflection at \(x\) = 0 Or Completes a reasoned argument to conclude that \(y\) = f(\(x\)) does not have a point of inflection by consideration of the function | R1 | 2.4 |
| (4 marks) |
Typical solution
\[\mathrm{f}^{\prime}(x) = 2x - 2\sin x\]\[\mathrm{f}^{\prime\prime}(x) = 2 - 2\cos x\]\[\mathrm{f}^{\prime\prime}(0) = 2 - 2\cos(0) = 0\]\[\mathrm{f}^{\prime\prime}(-0.1) = 9.99 \times 10^{-3}\]\[\mathrm{f}^{\prime\prime}(0.1) = 9.99 \times 10^{-3}\]\(\mathrm{f}^{\prime\prime}(x)\) does not change sign either side of \(x\) = 0
Therefore, the curve does not have a point of inflection at \(x\) = 0