Ans: (a) $5$ (b) (i) $15m^3-63m^2+72m$ (ii) No
- Let $f(x)=kx^3-21x^2+24x-4$. If $x-2$ is a factor of $f(x)$, then we have
$\begin{array}{rcl}
f(2) & = & 0 \\
k(2)^3-21(2)^2+24(2)-4 & = & 0 \\
8k -40 & = & 0 \\
k & = & 5
\end{array}$ -
- The area of the rectangle $OPQR$
$\begin{array}{cl}
= & RQ\times PQ \\
= & m \times f(m) \\
= & m(15m^2-63m+72) \\
= & 15m^3 – 63m^2 +72m
\end{array}$ - For the area of the rectangle $OPQR$ is $12$,
$\begin{array}{rcl}
15m^3-63m^2+72m & = & 12 \\
5m^3-21m^2+24m-4 & = & 0 \\
(m-2)(5m^2-11m+2) & = & 0 \\
(m-2)(m-2)(5m+1) & = & 0
\end{array}$Therefore $m=2$ (repeated) or $m=\dfrac{-1}{5}$.
Hence, there are only two different points of $Q$ such that the area of the rectangle $OPQR$ is $12$.
- The area of the rectangle $OPQR$
