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.

$f( - x) = 2\cos ( - x) + ( - x)\sin ( - x)$     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]