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

By KS Teng

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

Q3

Q4

Q5

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

(c)
Required probability = (0.1)(0.6) = 0.06

(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”.

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