Mathematics 9709/11 — May/June 2019

Cambridge AS Level · Pure Mathematics 1 · worked solutions for every part, with the mark scheme

questions

marks

105

minutes

Topics Trigonometry · Series · Quadratics · Coordinate Geometry · Functions · Integration · +2 more

Q1SeriesFree sample

The term independent of $x$ in the expansion of $\left(2x + \frac{k}{x}\right)^6$ , where $k$ is a constant, is 540.

(i)

Find the value of $k$ .

DifficultyMedium-Easy

Worked solution

Approach

Identify the general term in the expansion of $\left(2x + \frac{k}{x}\right)^6$ , and isolate the term whose coefficient is given.

Working

The general term is

\text{T}_{r+1} = \binom{6}{r}(2x)^{6-r}\left(\frac{k}{x}\right)^r.

The power of $x$ in this term is

6-r-r = 6-2r.

For the term whose coefficient is given, set $6-2r=0$ , so $r=3$ . Therefore the required term is

\binom{6}{3}(2x)^3\left(\frac{k}{x}\right)^3 = 20 \cdot 8k^3 = 160k^3.

Equating the coefficient to $540$ :

160k^3 = 540

k^3 = \frac{540}{160} = \frac{27}{8}

k = \frac{3}{2}.

Answer

k = \frac{3}{2}

Final answer

k = 3/2

Detailed explanation

Walkthrough

In a binomial expansion, the term containing $(2x)^{6-r}\left(\frac{k}{x}\right)^r$ gives an $x$ -power of $6-2r$ . The term whose coefficient is given in the question is the one independent of $x$ , so we need this power to be $0$ ; this forces $r=3$ . Then $\binom{6}{3}=20$ and $(2x)^3\left(\frac{k}{x}\right)^3 = 8k^3$ , since the powers of $x$ cancel. This gives the isolated coefficient $160k^3$ . We set this equal to $540$ and solve for $k$ : $k^3=\frac{27}{8}$ , so $k=\frac{3}{2}$ .

Key Takeaways

The general term is the central tool in binomial coefficient questions. Separating the numerical part, the $k$ -part, and the $x$ -part lets you isolate the coefficient cleanly without carrying unnecessary powers of $x$ .

Common Mistakes

Not isolating the term: writing $20(2x)^3\left(\frac{k}{x}\right)^3$ instead of the coefficient $160k^3$ may lose the mark. The mark scheme requires the term to be isolated.
Choosing the wrong $r$ : any value of $r$ that does not make the power of $x$ equal to $0$ will not match the required term.
Arithmetic error when simplifying $\frac{540}{160}$ to $\frac{27}{8}$ .
Forgetting to take the cube root, leaving $k^3$ instead of $k$ .

Things to Be Careful About

The coefficient is just the number multiplying the $x^0$ term; do not include any power of $x$ in the coefficient. Work with exact fractions, so $k=\frac{3}{2}$ rather than an unsimplified decimal.

Techniques used

write the general binomial termisolate the required term in the expansionsolve the resulting equation for the constant

(ii)

For this value of $k$ , find the coefficient of $x^2$ in the expansion.

DifficultyMedium-Easy

Worked solution

Approach

Use the same general term. Determine which value of $r$ gives an $x^2$ term, isolate its coefficient, and substitute the value of $k$ found in part (i).

Working

The general term is

\text{T}_{r+1} = \binom{6}{r}(2x)^{6-r}\left(\frac{k}{x}\right)^r.

The power of $x$ is $6-2r$ . For an $x^2$ term, set

6-2r=2,

so $r=2$ . The $x^2$ term is

\binom{6}{2}(2x)^4\left(\frac{k}{x}\right)^2 = 15 \cdot 16k^2x^2 = 240k^2x^2.

Hence the coefficient of $x^2$ is $240k^2$ . With $k=\frac{3}{2}$ ,

240\left(\frac{3}{2}\right)^2 = 240 \cdot \frac{9}{4} = 540.

Answer

540

Final answer

540

Detailed explanation

Walkthrough

We repeat the general-term formula with $r$ chosen so that the power of $x$ is $2$ . Since the $x$ -power is $6-2r$ , setting $6-2r=2$ gives $r=2$ . The term is then $\binom{6}{2}(2x)^4\left(\frac{k}{x}\right)^2$ , which simplifies to $15 \times 16k^2x^2 = 240k^2x^2$ . The coefficient is therefore $240k^2$ . Substituting $k=\frac{3}{2}$ gives $240 \times \frac{9}{4}=540$ .

Key Takeaways

A coefficient is the numerical multiplier of a specific power of $x$ . Once the general term is written, identifying the correct $r$ is the key step. Keeping the $x$ -power separate from the coefficient prevents sign and arithmetic errors.

Common Mistakes

Using the previous part's $r=3$ ; that would give the wrong power of $x$ , not $x^2$ .
Forgetting to square $k$ when forming $k^2$ .
Writing $240k^2x^2$ as the coefficient; the coefficient is $240k^2$ , with the $x^2$ separate.
Arithmetically, $240 \times \frac{9}{4} = 540$ , not $480$ .

Things to Be Careful About

The mark scheme allows follow-through for the expression $240k^2$ even if the value of $k$ from part (i) were incorrect. When substituting, square the whole fraction $\frac{3}{2}$ , giving $\frac{9}{4}$ , and then multiply by $240$ .

Techniques used

use the general binomial term againfind the required power of xsubstitute the known value of k

The rest of this paper

10 more questions

Q2Quadratics5M
Q3Circular Measure4M
Q4Coordinate Geometry6M
Q5Quadratics · Functions6M
Q6Trigonometry7M
Q7Coordinate Geometry · Trigonometry7M
Q8Series8M
Q9Functions · Trigonometry7M
Q10Integration · Differentiation9M
Q11Differentiation · Integration11M

Loading the full paper…