2021 A-level H1 Mathematics (8865) Paper 1 Suggested Solutions

All solutions here are suggested. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions. I will not be doing much formatting in this post.

Q1 [Maximum marks: 4]

Discriminant $< 0$
$(k-1)^2 - 4(1)(k+2) < 0$
$k^2 - 2k + 1 - 4k - 8 < 0$
$k^2 - 6k - 7 < 0$
$(k-7)(k+1) < 0$
$-1 < k < 7$

Q2 [Maximum marks: 5]

(a)
$\frac{d}{dx} ( 2x^3 - 5x^{1/3} )^2$
$= 2 ( 2x^3 - 5x^{1/3} ) (6x^2 - \frac{5}{3}x^{-2/3})$
$= -\frac{200}{3} x^{7/3} + 24x^5 + \frac{50}{3} x^{-1/3}$

(b)
$\int \frac{1}{\sqrt{5-4x}}dx$
$= - \frac{1}{2} \sqrt{5-4x} + c$

Q6 [Maximum marks: 4]
Let $X$ be the height of a sunflower plant, in m. $X \sim \text{N}( \mu, \sigma^2)$
$\text{P}( X > 1.72) = 0.30$
$\Rightarrow \frac{1.72 - \mu}{\sigma} = 0.5244005$
$0.5244005 \sigma + \mu = 1.72$ — (1)
$\text{P}( X < 1.52) = 0.10$
$\Rightarrow \frac{1.52 - \mu}{\sigma} = -1.28155$
$-1.28155 \sigma + \mu = 1.52$ — (2)
Using GC, $\mu \approx 1.66, \sigma \approx 0.111$

Q7 [Maximum marks: 4]
Observe that all the letters are distinct. Aside from P and G, we need to $\binom{7}{5}$ letters to put them between P and G.
Number of ways to form the group “P X X X X X G” with arrangements $= \binom{7}{5} \times 5! \times 2!$
We then need to arrange the group formed with the remaining 2 letters.
Total number of ways $= \binom{7}{5} \times 5! \times 2! \times 3! = 30240$

Q8 [Maximum marks: 8]
(a)
$\text{P} ( A \vert B) = \frac{\text{P}(A \cap B)}{\text{P}(B)}$
$0.8 = \frac{0.2}{\text{P}(B)}$
$\text{P}(B) = 0.25$
$\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B)$
$0.9 = \text{P}(A) + 0.25 - 0.2$
$\text{P}(A) = 0.85$
(b) It is the probability that both events A and B do not occur.
Observe by drawing a venn diagram that $\text{P}(A' \cap B') = 1 - \text{P}(A \cup B)$ .
$\text{P}(A' \cap B') = 1 - 0.9 = 0.1$
(c)
$\text{P}(A \vert B) = 0.8$
$\text{P}(A) = 0.85$
Since $\text{P}(A \vert B) \neq \text{P}(A)$ , events A and B are not independent.
(d)

Q9 [Maximum marks ] – CLT Topic

Q10 [Maximum marks: 11]

(a)

(b)
Required probability $= (0.8)(0.9) + (0.8)(0.1)(0.4) + (0.2)(0.4)(0.9) = 0.824$

(d)
Required probability $= 3(0.824)^2(1 - 0.824) \approx 0.358$

11 [Maximum marks: 11]

(a) Let $X$ be the number of orders delivered within 24 hours, out of 10 orders. $X \sim \text{B}(10, 0.75)$

(i) Required probability $= \text{P}(X = 4) + \text{P}(X=5) \approx 0.0746$

(ii) Required probability $= \text{P}(X > 6) = 1 - \text{P}(X \le 6) \approx 0.0776$

(b)
$\bar{x} = \frac{708.6}{60} = 11.81$
$s^2 = \frac{1}{59} \big( 8408.5 - \frac{708.6^2}{60} \big) = 0.67684$
Let $\mu$ be population mean delivery time.
Test $H_0 : \mu = 12$
against $H_1: \mu < 12$ at 5% significance level.
Under $H_0$ , $\bar{X} \sim \text{N} \big( 12, \frac{0.67684}{60} \big)$ approximately by central limit theorem.
From GC, $p = 0.036815$
Since $p \approx 0.0368 < 0.05$ , we reject $H_0$ and conclude that there is sufficient evidence at 5% significance level that the mean delivery time is less than 12 hours. Thus, manager’s claim is supported by the data.

(c) Since the sample size $n=60$ is sufficiently large for the sample mean distribution of delivery times to be approximated to a normal distribution by central limit theorem, it is not necessary to assume the distribution of the deliver times.

12 [Maximum marks: 13]

(a)
$\mathbb{E}(A - B) = 3.6 - 4.2 = -0.6$
$\text{Var}(A-B) = 0.5^2 + 0.8^2 = 0.89$
$A - B \sim \text{N}( -0.6, 0.89)$
$\text{P}(A - B > 0) \approx 0.262$

(b)
$\mathbb{E}(A_1 + A_2 + B_1 + B_2 + B_3) = 2(3.6) + 3(4.2) = 19.8$
$\text{Var}(A_1 + A_2 + B_1 + B_2 + B_3) = 2(0.5^2) + 3(0.8^2) = 2.42$
Let $D = A_1 + A_2 + B_1 + B_2 + B_3 \sim \text{N}( 19.8, 2.42)$
$\text{P}( 18 < D < 22) \approx 0.798$

(c)
Let $I$ be the interval timings, in minutes. $I \sim \text{N}(1.5, 0.4^2)$
Let $F = A_1 + \ldots + A_4 + B_1 + \ldots + B_4 + I_1 + \ldots + I_7$ .
$\mathbb{E}(F) = 4(3.6) + 4(4.2) + 7(1.5) = 41.7$
$\text{Var}(F) = 4(0.5^2) + 4(0.8^2) + 7(0.4)^2 = 4.68$
$F \sim \text{N}(41.7, 4.68)$
$\text{P}(F > 40) \approx 0.784$

(d)
Let $S$ be the second half duration, in minutes. $S \sim \text{N}(41.7, 4.68)$
$\text{P}(F > 40) \times \text{P}(S > 40) \approx 0.615$ .

(e)
The events found in (d) is a subset of the events found in that of “the two halves of the concert last for a total of more than 80 minutes”.

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

2021 A-level H1 Mathematics (8865) Paper 1 Suggested Solutions

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