Ans: (a) (ii) $a=2$, $b=3$ (b) $2x^2+3x-4$, $\dfrac{-3\pm\sqrt{41}}{4}$
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$\begin{array}{rcl}
f(1) & = & 1 \\
(1-a)(1-b)(1+1)-3 & = & 1 \\
2(1-a)(1-b) & = & 4 \\
(1-a)(1-b) & = & 2 \\
(a-1)(b-1) & = & 2
\end{array}$ - $a=2$, $b=3$.
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- By the result of (a)(ii), we have
$\begin{array}{cl}
& f(x) – g(x) \\
= & (x-2)(x-3)(x+1) – 3 – (x^3 – 6x^2 -2x +7) \\
= & x^3 -4x^2 +x +3 – x^3 + 6x^2 + 2x -7 \\
= & 2x^2 + 3x -4
\end{array}$Hence, we have
$\begin{array}{rcl}
f(x) & = & g(x) \\
f(x) – g(x) & = & 0 \\
2x^2 +3x – 4 & = & 0
\end{array}$Therefore, we have
$\begin{array}{rcl}
x & = & \dfrac{-(3) \pm \sqrt{(3)^2 – 4(2)(-4)}}{2(2)} \\
& = & \dfrac{-3 \pm \sqrt{41}}{4}
\end{array}$
