Question
[Maximum mark: 4]
At a sprinting competition the mean time of the first three sprinters is $10.5$ seconds. The time for the fourth sprinter is then recorded and the mean time of the first four sprinters is $11.0$ seconds.
Find the time achieved by the fourth sprinter.
Answer/Explanation
Using the mean formula, we have
$
\begin{aligned}
\bar{x}_{\text {three }} & =\frac{x_1+x_2+x_3}{3} \\
10.5 & =\frac{x_1+x_2+x_3}{3} \\
x_1+x_2+x_3 & =(10.5)(3) \\
x_1+x_2+x_3 & =31.5 \text { seconds }
\end{aligned}
$
Hence we get
$
\begin{aligned}
\bar{x}_{\text {four }} & =\frac{x_1+x_2+x_3+x_4}{4} \\
11.0 & =\frac{31.5+x_4}{4} \\
31.5+x_4 & =(11.0)(4) \\
x_4 & =44.0-31.5 \\
x_4 & =12.5 \text { seconds }
\end{aligned}
$
Question
[Maximum mark: 6]
Let $f(x)=x^2+k x$ and $g(x)=x+k$, for $x \in \mathbb{R}$, where $k$ is a constant. The graphs of $y=f(x)$ and $y=g(x)$ intersect at two distinct points.
Find the possible values of $k$.
Answer/Explanation
If we rewrite the equation $f(x)=g(x)$ in the form $a x^2+b x+c=0$, we have
$
\begin{aligned}
x^2+k x & =x+k \\
x^2+(k-1) x-k & =0
\end{aligned}
$
Hence the quadratic function $q(x)=x^2+(k-1) x-k$ has two distinct real roots.
Therefore we get
$
\begin{aligned}
{[\text { discriminant of } q] } & >0 \\
(k-1)^2-4(1)(-k) & >0 \quad\left[\Delta=b^2-4 a c\right] \\
\left(k^2-2 k+1\right)+4 k & >0 \\
k^2+2 k+1 & >0 \\
(k+1)^2 & >0 \\
|k+1| & >0 \quad \quad\quad\quad\quad \text { (1) }
\end{aligned}
$
From the inequality (1), we get
$
\begin{array}{rll}
k+1<0 & \text { or } & k+1>0 \\
k<-1 & \text { or } & k>-1
\end{array}
$
Hence the set of possible values of $k$ is $\mathbb{R} \backslash\{-1\}$