SPM Additional Mathematics 2019, Paper 2 (Question 4)

Question 4:
Diagram 2 shows the curve y = 4x – x² and tangent to the curve at point Q passes point P.

Diagram 2

(a) Show that h = 3. [4 marks]
(b) Calculate the area of the shaded region. [4 marks]

Solution:
(a)

\begin{array}{l} y = 4 x - x^{2} \\ \frac{d y}{d x} = 4 - 2 x \\ At point Q (h, 4 h - h^{2}) \\ \frac{d y}{d x} = 4 - 2 h \\ Equation of tangent at Q \\ y - y_{1} = \frac{d y}{d x} (x - x_{1}) \\ y - (4 h - h^{2}) = (4 - 2 h) (x - h) \end{array}

\begin{array}{l} At point P (2, 5), x = 2, y = 5 \\ 5 - 4 h + h^{2} = (4 - 2 h) (2 - h) \\ 5 - 4 h + h^{2} = 8 - 4 h - 4 h + 2 h^{2} \\ h^{2} - 4 h + 3 = 0 \\ (h - 1) (h - 3) = 0 \\ h = 1 (rejected), \\ h = 3 \end{array}

(b)

\begin{array}{l} Area of shaded region \\ = Area of trapezium - Area under the curve \\ = \frac{1}{2} (a + b) h - \int_{2}^{3} y d x \\ = \frac{1}{2} (5 + 3) 1 - \int_{2}^{3} 4 x - x^{2} d x \\ = 4 - {[\frac{4 x^{2}}{2} - \frac{x^{3}}{3}]}_{2}^{3} \\ = 4 - [18 - 9 - (8 - \frac{8}{3})] \\ = 4 - 9 + (8 - \frac{8}{3}) \\ = \frac{1}{3} {unit}^{2} \end{array}