Expand $\left(2x – \dfrac{1}{x}\right)^{4}$ fully, and state the term that does not depend on $x$.
Question HKC78-AAHL-BIN-01
[Maximum mark: 5]
Expand $\left(2x – \dfrac{1}{x}\right)^{4}$ fully, and state the term that does not depend on $x$.
Mark Scheme
The general term is $dbinom{4}{k}(2x)^{4-k}left(-dfrac{1}{x}right)^{k}$.
$k=0$: $16x^{4}$. $k=1$: $dbinom{4}{1}(8x^{3})left(-dfrac{1}{x}right) = -32x^{2}$. $k=2$: $dbinom{4}{2}(4x^{2})left(dfrac{1}{x^{2}}right) = 24$. $k=3$: $dbinom{4}{3}(2x)left(-dfrac{1}{x^{3}}right) = -dfrac{8}{x^{2}}$. $k=4$: $dfrac{1}{x^{4}}$. [4]
So $left(2x – dfrac{1}{x}right)^{4} = 16x^{4} – 32x^{2} + 24 – dfrac{8}{x^{2}} + dfrac{1}{x^{4}}$. The term independent of $x$ is $24$. [1]
Total: 5 marks
An arithmetic sequence has a first term of 5 and a common difference of 3. Find the 10th term of the sequence.
Question Q1
An arithmetic sequence has a first term of 5 and a common difference of 3. Find the 10th term of the sequence.
Mark Scheme
Step 1: Identify given information.
$u_1 = 5$, $d = 3$, $n = 10$.
Step 2: Use formula $u_n = u_1 + (n-1)d$.
$u_{10} = 5 + (10-1)3$
Step 3: Calculate.
$u_{10} = 5 + (9)3 = 5 + 27 = 32$.
Answer: 32
A code has four characters. The first character is one of the letters A, B, C, D, E, or F. Each of the other three characters is a digit from $0$ to $9$, and a digit may be used more than once. Find the number of possible codes.
Question HKC78-AAHL-CNT-01
[Maximum mark: 4]
A code has four characters. The first character is one of the letters A, B, C, D, E, or F. Each of the other three characters is a digit from $0$ to $9$, and a digit may be used more than once. Find the number of possible codes.
Mark Scheme
There are $6$ choices for the letter. [1]
Each of the three digits has $10$ choices, and repetition is allowed, so those positions contribute $10^{3}$. [2]
Total: $6 times 10^{3} = 6000$. [1]
Total: 4 marks
Let $z = 1 – \sqrt{3}\, i$. Find $|z|$ and the principal argument of $z$, taking the argument in $(-\pi, \pi]$.
Question HKC78-AAHL-CPX-01
[Maximum mark: 5]
Let $z = 1 – \sqrt{3}\, i$. Find $|z|$ and the principal argument of $z$, taking the argument in $(-\pi, \pi]$.
Mark Scheme
$|z| = sqrt{1^{2} + (sqrt{3})^{2}} = sqrt{4} = 2$. [2]
The point $(1, -sqrt{3})$ is in the fourth quadrant. The reference angle is $dfrac{pi}{3}$. [2]
Therefore $arg z = -dfrac{pi}{3}$. [1]
Total: 5 marks
Prove that the square of any odd integer leaves a remainder of $1$ when it is divided by $8$.
Question HKC78-AAHL-PRF-01
[Maximum mark: 4]
Prove that the square of any odd integer leaves a remainder of $1$ when it is divided by $8$.
Mark Scheme
Let the odd integer be $n = 2k + 1$ for some integer $k$. [1]
Then $n^{2} = 4k^{2} + 4k + 1 = 4k(k+1) + 1$. [1]
$k$ and $k+1$ are consecutive integers, so $k(k+1)$ is even. Write $k(k+1) = 2m$. [1]
$n^{2} = 8m + 1$, so the remainder on division by $8$ is $1$. [1]
Total: 4 marks
Solve the system $2x + y = 7$ and $x – y = 2$.
Question HKC78-AAHL-SYS-01
[Maximum mark: 4]
Solve the system $2x + y = 7$ and $x – y = 2$.
Mark Scheme
Add the equations: $3x = 9$, so $x = 3$. [2]
From $x – y = 2$, $3 – y = 2$, so $y = 1$. Check: $2(3) + 1 = 7$. [2]
Answer: $x = 3$, $y = 1$.
Total: 4 marks
In the expansion of $(1 + 3x)^{n}$, where $n$ is an integer and $n > 2$, the coefficient of $x^{2}$ is $135$. Find $n$.
Question HKC78-AAHL-BIN-02
[Maximum mark: 5]
In the expansion of $(1 + 3x)^{n}$, where $n$ is an integer and $n > 2$, the coefficient of $x^{2}$ is $135$. Find $n$.
Mark Scheme
The coefficient of $x^{2}$ is $dbinom{n}{2} cdot 3^{2} = 135$. [1]
$dfrac{n(n-1)}{2} cdot 9 = 135$, so $n(n-1) = 30$. [2]
$n^{2} – n – 30 = 0$, hence $(n-6)(n+5) = 0$. [1]
$n = 6$, because $n$ is an integer greater than $2$. [1]
Total: 5 marks
For the sequence in Q1a, find the sum of the first 20 terms.
Question Q2
For the sequence in Q1a, find the sum of the first 20 terms.
Mark Scheme
Step 1: Identify given information.
$u_1 = 5$, $d = 3$, $n = 20$.
Step 2: Use formula $S_n = frac{n}{2}(2u_1 + (n-1)d)$.
$S_{20} = frac{20}{2}(2(5) + (20-1)3)$
Step 3: Calculate.
$S_{20} = 10(10 + (19)3) = 10(10 + 57) = 10(67) = 670$.
Answer: 670
A committee of $3$ people is chosen from $5$ teachers and $4$ students. Find the number of committees that include at least one student.
Question HKC78-AAHL-CNT-02
[Maximum mark: 6]
A committee of $3$ people is chosen from $5$ teachers and $4$ students. Find the number of committees that include at least one student.
Mark Scheme
The number of committees with no restriction is $dbinom{9}{3} = 84$. [1]
The number with no student is $dbinom{5}{3} = 10$. [1]
At least one student: $84 – 10 = 74$. [2]
Check by cases: exactly one student $dbinom{4}{1}dbinom{5}{2} = 40$, exactly two $dbinom{4}{2}dbinom{5}{1} = 30$, exactly three $dbinom{4}{3} = 4$. Sum $40+30+4 = 74$. [2]
Total: 6 marks
