| Date | None Specimen | Marks available | 3 | Reference code | SPNone.1.hl.TZ0.13 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Show that | Question number | 13 | Adapted from | N/A |
Question
The function f is defined by
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
{2x - 1,}&{x \leqslant 2} \\
{a{x^2} + bx - 5,}&{2 < x < 3}
\end{array}} \right.\]
where a , \(b \in \mathbb{R}\) .
Given that f and its derivative, \(f'\) , are continuous for all values in the domain of f , find the values of a and b .
Show that f is a one-to-one function.
Obtain expressions for the inverse function \({f^{ - 1}}\) and state their domains.
Markscheme
f continuous \( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ \div }} f(x)\) M1
\(4a + 2b = 8\) A1
\(f'(x) = \left\{ {\begin{array}{*{20}{c}}
{2,}&{x < 2} \\
{2ax + b,}&{2 < x < 3}
\end{array}} \right.\) A1
\(f'{\text{ continuous}} \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f'(x) = \mathop {\lim }\limits_{x \to {2^ \div }} f'(x)\)
\(4a + b = 2\) A1
solve simultaneously M1
to obtain a = –1 and b = 6 A1
[6 marks]
for \(x \leqslant 2,{\text{ }}f'(x) = 2 > 0\) A1
for \(2 < x < 3,{\text{ }}f'(x) = - 2x + 6 > 0\) A1
since \(f'(x) > 0\) for all values in the domain of f , f is increasing R1
therefore one-to-one AG
[3 marks]
\(x = 2y - 1 \Rightarrow y = \frac{{x + 1}}{2}\) M1
\(x = - {y^2} + 6y - 5 \Rightarrow {y^2} - 6y + x + 5 = 0\) M1
\(y = 3 \pm \sqrt {4 - x} \)
therefore
\({f^{ - 1}}(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{x + 1}}{2},}&{x \leqslant 3} \\
{3 - \sqrt {4 - x} ,}&{3 < x < 4}
\end{array}} \right.\) A1A1A1
Note: Award A1 for the first line and A1A1 for the second line.
[5 marks]
