2010-I-16

16-06-202121-06-2023 app.cch No Comments

Ans: (a) (i)

(36, 3)

(ii)

\frac{- 1}{144} (x - 32)^{2} + 8

(iii)

2^{\frac{- 1}{144} (x - 32)^{2} + 3} + 5

(b) (i)

36^{\circ} C

(ii)

v = 2^{\frac{- 1}{144} (t - 32)^{2} + 3} + 5

1. By the method of completing the square, we have
  $\begin{array}{rcl} f (x) & = & \frac{1}{2} x - \frac{1}{144} x^{2} - 6 \\ = & \frac{- 1}{144} (x^{2} - 72 x) - 6 \\ = & \frac{- 1}{144} (x^{2} - 72 x + 36^{2} 36^{2}) - 6 \\ = & \frac{- 1}{144} (x - 36)^{2} + 9 - 6 \\ = & \frac{- 1}{144} (x - 36)^{2} + 3 \end{array}$
  
  Therefore, the vertex of the graph $f (x)$ is $(36, 3)$ .
2. $\begin{array}{rcl} g (x) & = & f (x + 4) + 5 \\ = & \frac{- 1}{144} [(x + 4) - 36]^{2} + 3 + 5 \\ = & \frac{- 1}{144} (x - 32)^{2} + 8 \end{array}$
3. $\begin{array}{rcl} h (x) & = & 2^{f (x + 4)} + 5 \\ = & 2^{\frac{- 1}{144} [(x + 4) - 36]^{2} + 3} + 5 \\ = & 2^{\frac{- 1}{144} (x - 32)^{2} + 3} + 5 \end{array}$
1. Sub. $u = 8$ into $u = 2^{f (s)}$ , we have
  $\begin{array}{rcl} 8 & = & 2^{f (s)} \\ 2^{3} & = & 2^{f (s)} \\ 3 & = & f (s) \\ 3 & = & \frac{- 1}{144} (s - 36)^{2} + 3 \\ (s - 36)^{2} & = & 0 \\ s & = & 36 \end{array}$
  
  Therefore, the required temperature is $36^{\circ} C$ .
2. By observing Table 1(a) and Table 1(b), we have $s = t + 4$ and $u = v - 5$ . Hence, we have
  $\begin{array}{rcl} u & = & 2^{f (s)} \\ v - 5 & = & 2^{f (t + 4)} \\ v & = & 2^{\frac{- 1}{144} [(t + 4) - 36]^{2} + 3} + 5 \\ = & 2^{\frac{- 1}{144} (t - 32)^{2} + 3} + 5 \end{array}$

2010, HKCEE, Paper 1 Transformations

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