Question
[Maximum mark: 6]
Let $f(x)=\sqrt{x+7}$, for $x \geq-7$.
(a) Find $f^{-1}(3)$ [3]
Let $g$ be a function such that $g^{-1}$ exists for all real numbers.
(b) Given that $g(9)=4$, find $\left(f \circ g^{-1}\right)(4)$. [3]
Answer/Explanation
(a) If we solve the equation $\left(f \circ f^{-1}\right)(3)=3$ for $y=f^{-1}(3)$, we get
$
\begin{array}{r}
f\left(f^{-1}(3)\right)=3 \\
f(y)=3 \\
\sqrt{y+7}=3 \\
y+7=9 \\
y=2 \\
f^{-1}(3)=2
\end{array}
$
(b) We have
$
\begin{array}{rlr}
4 & =g(9) \\
g^{-1}(4) & =g^{-1}(g(9)) \\
g^{-1}(4) & =9 \quad\left[\text { since }\left(g^{-1} \circ g\right)(9)=9\right]
\end{array}
$
Hence we obtain
$
\begin{aligned}
\left(f \circ g^{-1}\right)(4) & =f\left(g^{-1}(4)\right) \\
& =f(9) \\
& =\sqrt{9+7} \\
& =\sqrt{16} \\
& =4
\end{aligned}
$
Question
[Maximum mark: 4]
The points $\mathrm{A}$ and $\mathrm{B}$ have position vectors $\overrightarrow{\mathrm{OA}}=\left(\begin{array}{c}2 \\ 4 \\ 12\end{array}\right)$ and $\overrightarrow{\mathrm{OB}}=\left(\begin{array}{l}0 \\ 1 \\ 2\end{array}\right)$.
(a) Find $\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}$. [2]
(b) Hence find the area of the triangle OAB. [2]
Answer/Explanation
(a) Using the vector product formula, we get
$
\begin{aligned}
\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}} & =(2 \mathbf{i}+4 \mathbf{j}+12 \mathbf{k}) \times(\mathbf{j}+2 \mathbf{k}) \\
& =\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 4 & 12 \\
0 & 1 & 2
\end{array}\right| \\
& =[8-12] \mathbf{i}-[4-0] \mathbf{j}+[2-0] \mathbf{k} \\
& =-4 \mathbf{i}-4 \mathbf{j}+2 \mathbf{k} \\
& =\left(\begin{array}{r}
-4 \\
-4 \\
2
\end{array}\right)
\end{aligned}
$
(b) Using the area of a triangle formula, we obtain
$
\begin{aligned}
A_{\triangle \mathrm{OAB}} & =\frac{1}{2}|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}| \\
& =\frac{1}{2}|-4 \mathbf{i}-4 \mathbf{j}+2 \mathbf{k}| \\
& =\frac{1}{2} \sqrt{(-4)^2+(-4)^2+2^2} \\
& =\frac{1}{2} \sqrt{36} \\
& =3
\end{aligned}
$