| Date | May 2015 | Marks available | 2 | Reference code | 15M.1.hl.TZ1.5 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Show that | Question number | 5 | Adapted from | N/A |
Question
The functions \(f\) and \(g\) are defined by \(f(x) = a{x^2} + bx + c,{\text{ }}x \in \mathbb{R}\) and \(g(x) = p\sin x + qx + r,{\text{ }}x \in \mathbb{R}\) where \(a,{\text{ }}b,{\text{ }}c,{\text{ }}p,{\text{ }}q,{\text{ }}r\) are real constants.
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 \(h(x)\).
Markscheme
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]
Examiners report
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 \(g\).
