SPM Additional Mathematics 2023, Paper 1 (Question 10 & 11)

Question 10:
Solutions using other then calculus are not accepted.

(a) On Sunday, Mus bought a number of spherical balls with radius 5 cm . Some of the balls will be arranged in a single line on a rack with length 2.5 m . On Monday, the volume of each ball decreased uniformly by 20π cm³.

Determine the maximum number of these balls which can be arranged on the rack on Monday.
[4 marks]

(b) Given that h(x) = 3x² + 7x – 8, determine the type of the turning point of h(x). Justify your answer.
[2 marks]

(c)
$$\text { The gradient function of a curve is }(2 x+1)^3 \text {. The curve passes through }\left(\frac{1}{2}, 5\right) \text {. }$$
Find the equation of the curve. Give your answer in the form y = a(2x + 1)^b + c such that a, b and c are constants.
[3 marks]

Solution:
(a)


$$\begin{aligned} & \text { Volume }_{\text {sphere }}=\frac{4}{3} \pi r^3, \quad \delta v=20 \pi \mathrm{~cm}^3 \\ & \begin{aligned} \frac{\delta v}{\delta r}& =\left(\frac{4}{3}\right)(3) \pi r^2 \\ & =4 \pi r^2 \end{aligned} \end{aligned}$$


$$\begin{aligned} &\text { When } r=5 \mathrm{~cm}, 4 \pi(5)^2=100 \pi\\ &\begin{aligned} & \frac{\delta r}{\delta v} \approx \frac{\mathrm{~d} r}{\mathrm{~d} v} \\ & \frac{\delta r}{-20 \pi} \approx \frac{1}{100 \pi} \\ & \delta r \approx-0.2 \\ & r_{\text {new }}=r_{\text {old }}+\delta r \\ &=5 \mathrm{~cm}+(-0.2) \\ &=4.8 \mathrm{~cm} \\ & \begin{aligned} d_{\text {new }} & =2(4.8 \mathrm{~cm}) \\ & =9.6 \mathrm{~cm} \end{aligned} \end{aligned}\\ &\begin{aligned} \text { Number of balls } & \approx \frac{250 \mathrm{~cm}}{9.6 \mathrm{~cm}} \\ & \approx 26.04 \\ & =26 \text { balls } \end{aligned} \end{aligned}$$


(b)
$$\begin{aligned} &\begin{aligned} & h(x)=3 x^2+7 x-8 \\ & h^{\prime}(x)=6 x+7 \\ & h^{\prime \prime}(x)=6>0 \end{aligned}\\ &\therefore h^{\prime \prime}(x)>0 \text {, minimum turning point. } \end{aligned}$$


(c)
$$\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x} & =(2 x+1)^3 \\ \int_{\mathrm{d} y} & =\int(2 x+1)^3 \mathrm{~d} x \\ y & =\frac{(2 x+1)^4}{(4)(2)}+c \\ y & =\frac{1}{8}(2 x+1)^4+c, \text a t\left(\frac{1}{2}, 5\right) \\ 5 & =\frac{1}{8}\left[2\left(\frac{1}{2}\right)+1\right]^4+c \\ 5 & =2+c \\ c & =3 \\ \therefore y & =\frac{1}{8}(2 x+1)^4+3 \end{aligned}$$

Question 11:
(a) It is found that the probability that a product from a company can be sold is 19/20.
If in a particular month, the company produces 820 units of the product, find the mean and the standard deviation of the number of products expected to be sold that month. [3 marks]

(b) There are 25 students in a class. A teacher wants to choose some students from the class to join a game.

(i) It is given that the number of different ways of choosing r students is equal to the number of different ways of choosing (r + 13) students. Find the value of r.

(ii) The teacher decides to choose 8 students to join the game. She wants to choose equal numbers of boys and girls. If these students are to be arranged in a row, find the number of different ways such that the boys cannot be next to each other.
[4 marks]

Solution:
(a)
$$\begin{aligned} n & =820 \\ p & =\frac{19}{20} \\ q & =1-\frac{19}{20} \\ & =\frac{1}{20} \end{aligned}$$
$$\text { Mean, } \begin{aligned} \mu & =n p \\ & =(820)\left(\frac{19}{20}\right) \\ & =779 \end{aligned}$$

$$\begin{aligned} &\text { Standard deviation, } \sigma=\sqrt{n p q}\\ &\begin{aligned} & =\sqrt{(820)\left(\frac{19}{20}\right)\left(\frac{1}{20}\right)} \\ & =\sqrt{\left(\frac{779}{20}\right)} \\ & =6.241 \end{aligned} \end{aligned}$$

(b)(i)
$$\begin{aligned} &\text { Formula }{ }^n C_r={ }^n C_{n-r}\\ &r+(n-r)=n \end{aligned}$$
$$\begin{aligned} n & =25 \\ { }^{25} C_r & ={ }^{25} C_{r+13} \\ r+(r+13) & =25 \\ 2 r+13 & =25 \\ 2 r & =12 \\ r & =6 \end{aligned}$$

(b)(ii)

4 girls to be chosen = 4!
Another 4 boys to be chosen whom do not next to each other = ⁵P₄
Number of ways = 4! × ⁵P₄
   = 2880