- Sub. $y=0$ into $y=f(x)$, we have$\begin{array}{rcl}
2x^2 -4kx + 3k^2 +5 & = & 0 \ldots (*)\\
\end{array}$Consider the discriminant of $(*)$, we have
$\begin{array}{rcl}
\Delta & = & (-4k)^2 – 4(2)(3k^2+5) \\
\Delta & = & 16k^2 -24k^2 -40\\
\Delta & = & -8k^2 -40 \\
\Delta & < & 0 \ \text{for all real constants $k$}
\end{array}$Therefore, the graph of $y=f(x)$ does not cut the $x$-axis.
-
$\begin{array}{rcl}
f(x) & = & 2x^2 -4kx +3k^2 +5 \\
f(x) & = & 2(x^2 -2kx) + 3k^2 +5 \\
f(x) & = & 2(x^2 -2kx + k^2 – k^2) +3k^2 +5 \\
f(x) & = & 2[(x-k)^2 -k^2] + 3k^2 +5 \\
f(x) & = & 2(x-k)^2 -2k^2 +3k^2 +5 \\
f(x) & = & 2(x-k)^2 +k^2 +5
\end{array}$Therefore, the coordinates of the vertex of the graph of $y=f(x)$ are $(k,k^2+5)$.
- Note that the graph of $y=2-f(x)$ is obtained by reflecting the graph of $y=f(x)$ about the $x$-axis and then translating upwards $2$ units. Therefore, the coordinates of the vertex of the graph of $y=2-f(x)$$\begin{array}{cl}
= & (k, -(k^2+5) +2) \\
= & (k, -k^2 -3)
\end{array}$When $S$ and $T$ are nearest to each other, $S$ and $T$ move to the vertices of the graph of $y=f(x)$ and $y=2-f(x)$ respectively. Therefore, the coordinates of $S$ and $T$ become $(k,k^2+5)$ and $(k,-k^2-3)$. Note that the circumcentre of $\Delta OST$ lies on the perpendicular bisector of $ST$. The mid-point of $ST$
$\begin{array}{cl}
= & \left(k, \dfrac{(k^2+5) + (-k^2 -3)}{2} \right) \\
= & (k, 1)
\end{array}$Note that $ST$ is a vertical line. Hence, the equation of the perpendicular bisector of $ST$ is $y=1$. Since the straight line $y=1$ is parallel to the $x$-axis and the circumcentre of $\Delta OST$ must lie on $y=1$, then the circumcentre of $\Delta OST$ does not lie on the $x$-axis. Hence, the claim is not correct.
2015-I-18
Ans: (a) No (b) $(k, k^2+5)$ (c) No
