Date | None Specimen | Marks available | 2 | Reference code | SPNone.1.hl.TZ0.5 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Show that | Question number | 5 | Adapted from | N/A |
Question
The function f is defined, for −π2⩽x⩽π2 , by f(x)=2cosx+xsinx .
Determine whether f is even, odd or neither even nor odd.
Show that f″ .
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]