SPM 2021 Add Maths Trial Paper (Selangor) – Paper 1 (Question 10)

Question 10:
(a) Given the gradient of the tangent to the curve y = x² (3 + px) is -3 when x = -1.
Find the value of p. [2 marks]

(b) Volume, V cm³, of a solid is given by $V=8\pi r^2+\frac{2}{3}\pi r^3$ , r is the radius. Find the approximate change in V, in terms of π, if r increases from 3 cm to 3.005 cm. [3 marks]


Solution:
(a)

$\begin{array}{l}\text{Given }\frac{dy}{dx}=-3,x=-1\\ y=x^2\left(3+px\right)\\ \frac{dy}{dx}=x^2\left(p\right)+\left(3+px\right)\left(2x\right)\\ -3=p{x}^2+6x+2p{x}^2\\ -3=p{\left(-1\right)}^2+6\left(-1\right)+2p{\left(-1\right)}^2\\ -3=p-6+2p\\ -3=3p-6\\ 3p=3\\ p=1\end{array}$


(b)

$\begin{array}{l}\text{Given }\delta r=3.005-3=0.005\\ \\ V=8\pi r^2+\frac{2}{3}\pi r^3\\ \frac{\delta V}{\delta r}=\frac{dV}{dr}\\ \delta V=\frac{dV}{dr}\times \delta r\\ \delta V=\left[2\left(8\right)\pi r+3\left(\frac{2}{3}\right)\pi r^2\right]\times 0.005\\ =\left(16\pi r+2\pi r^2\right)\times 0.005\\ =\left[16\pi \left(3\right)+2\pi {\left(3\right)}^2\right]\times 0.005\\ =66\pi \times 0.005\\ =0.33\pi \end{array}$