| Date | None Specimen | Marks available | 3 | Reference code | SPNone.1.hl.TZ0.5 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Determine | Question number | 5 | Adapted from | N/A |
Question
The function f is defined, for \( - \frac{\pi }{2} \leqslant x \leqslant \frac{\pi }{2}\) , by \(f(x) = 2\cos x + x\sin x\) .
Determine whether f is even, odd or neither even nor odd.
Show that \(f''(0) = 0\) .
John states that, because \(f''(0) = 0\) , the graph of f has a point of inflexion at the point (0, 2) . Explain briefly whether John’s statement is correct or not.
Markscheme
\(f( - x) = 2\cos ( - x) + ( - x)\sin ( - x)\) M1
\( = 2\cos x + x\sin x\,\,\,\,\,\left( { = f(x)} \right)\) A1
therefore f is even A1
[3 marks]
\(f'(x) = - 2\sin x + \sin x + x\cos x\,\,\,\,\,( = - \sin x + x\cos x)\) A1
\(f''(x) = - \cos x + \cos x - x\sin x\,\,\,\,\,( = - x\sin x)\) A1
so \(f''(0) = 0\) AG
[2 marks]
John’s statement is incorrect because
either; there is a stationary point at (0, 2) and since f is an even function and therefore symmetrical about the y-axis it must be a maximum or a minimum
or; \(f''(x)\) is even and therefore has the same sign either side of (0, 2) R2
[2 marks]
