Ans: (a) $1-4x+6x^2-4x^3+x^4$ (b) $\dfrac{1}{2}$
- By the binomial theorem, we have
$\begin{array}{cl}
& (1-x)^4 \\
= & C^4_0+C^4_1(-x)+C^4_2(-x)^2+C^4_3(-x)^3+C^4_4(-x)^4 \\
= & 1-4x+6x^2-4x^3+x^4
\end{array}$ -
$\begin{array}{cl}
& (1+kx)^9(1-x)^4 \\
= & (C^9_0+C^9_1(kx)+C^9_2(kx)^2+\ldots)(1-4x+6x^2-4x^3+x^4)
\end{array}$Consider the coefficient of $x^2$, we have
$\begin{array}{rcl}
C^9_0(6)+C^9_1(k)(-4)+C^9_2(k)^2(1) & = & -3 \\
6-36k+36k^2 & = & -3 \\
4k^2-4k+1 & = & 0 \\
(2k-1)^2 & = & 0
\end{array}$$\therefore k=\dfrac{1}{2}$ (repeated).
