Given that $f$ is an even function, show that $b = 0$.

Given that $g$ is an odd function, find the value of $r$.

The function $h$ is both odd and even, with domain $\mathbb{R}$.

Find $h(x)$.

EITHER

$f( - x) = f(x)$     M1

$\Rightarrow a{x^2} - bx + c = a{x^2} + bx + c \Rightarrow 2bx = 0,{\text{ }}(\forall x \in \mathbb{R})$     A1

OR

$y$-axis is eqn of symmetry     M1

so $\frac{{ - b}}{{2a}} = 0$     A1

THEN

$\Rightarrow b = 0$     AG

[2 marks]

$g( - x) =  - g(x) \Rightarrow p\sin ( - x) - qx + r =  - p\sin x - qx - r$

$\Rightarrow  - p\sin x - qx + r =  - p\sin x - qx - r$     M1

Note:     M1 is for knowing properties of sin.

$\Rightarrow 2r = 0 \Rightarrow r = 0$     A1

Note:     In (a) and (b) allow substitution of a particular value of $x$

[2 marks]

$h( - x) = h(x) =  - h(x) \Rightarrow 2h(x) = 0 \Rightarrow h(x) = 0,{\text{ }}(\forall x)$     M1A1

Note:     Accept geometrical explanations.

[2 marks]

Total [6 marks]

Sometimes backwards working but many correct approaches.

Some candidates did not know what odd and even functions were. Correct solutions from those who applied the definition.

Some realised: just apply the definitions. Some did very strange things involving $f$ and $g$.