2007-I-14 – Solving Master

Ans: (a) (i)

39

(ii)

(x + 3) (x + 9) (4 x - 9)

(b) (i)

C = 16 x^{3} + 156 x^{2}

(ii)

\frac{9}{4} cm

1. Since $x + 3$ is a factor of $f (x)$ , then we have
  $\begin{array}{rcl} f (- 3) & = & 0 \\ 4 (- 3)^{3} + k (- 3)^{2} - 243 & = & 0 \\ 9 k - 351 & = & 0 \\ k & = & 39 \end{array}$
2. By long division, we have
  $\begin{aligned} 4 x^{2} + 27 x - 81 \\ x + 3 & 4 x^{3} + 39 x^{2} + 0 x - 243 \\ \underset{―}{4 x^{3} + 12 x^{2}} \\ 27 x^{2} + 0 x \\ \underset{―}{27 x^{2} + 81 x} \\ - 81 x - 243 \\ \underset{―}{- 81 x - 243} \end{aligned}$
  
  Hence, we have
  
  $\begin{array}{rcl} f (x) & = & (x + 3) (4 x^{2} + 27 x - 81) \\ = & (x + 3) (x + 9) (4 x - 9) \end{array}$
1. Let $C = k_{1} x^{3} + k_{2} x^{2}$ , where $k_{1}$ and $k_{2}$ are non-zero constants. When $x = 5.5$ , $C = 7381$ , we have
  $\begin{array}{rcl} k_{1} (5.5)^{3} + k_{2} (5.5)^{2} & = & 7381 \\ 5.5 k_{1} + k_{2} & = & 244 \dots ① \end{array}$
  
  When $x = 6$ , $C = 9072$ , we have
  
  $\begin{array}{rcl} k_{1} (6)^{3} + k_{2} (6)^{2} & = & 9072 \\ 6 k_{1} + k_{2} & = & 252 \dots ② \end{array}$
  
  $② - ①$ , we have
  
  $\begin{array}{rcl} 0.5 k_{1} & = & 8 \\ k_{1} & = & 16 \end{array}$
  
  Sub. $k_{1} = 16$ into $②$ , we have
  
  $\begin{array}{rcl} 6 (16) + k_{2} & = & 252 \\ k_{2} & = & 156 \end{array}$
  
  Therefore, $C = 16 x^{3} + 156 x^{2}$ .
2. Sub. $C = 972$ into $C = 16 x^{3} + 156 x^{2}$ , we have
  $\begin{array}{rcl} 16 x^{3} + 156 x^{2} & = & 972 \\ 4 x^{3} + 39 x^{2} - 243 & = & 0 \\ (x + 3) (x + 9) (4 x - 9) & = & 0 \end{array}$
  
  Therefore, $x = - 3$ , $x = - 9$ or $x = \frac{9}{4}$ .
  
  Since the length of a handicraft must be positive, the required length is $\frac{9}{4} cm$ .

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