Mathematics 9709/12 — February/March 2021

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

questions

marks

110

minutes

Topics Quadratics · Series · Trigonometry · Functions · Differentiation · Integration · +2 more

Q1SeriesFree sample

(a)

Find the first three terms in the expansion, in ascending powers of $x$ , of $(1 + x)^5$ .

DifficultyEasy

Worked solution

Approach

Use the binomial theorem for $(1+x)^5$ and keep only the terms up to $x^2$ .

Working

(1+x)^5 = \binom{5}{0} + \binom{5}{1}x + \binom{5}{2}x^2 + \cdots

Evaluating the binomial coefficients:

(1+x)^5 = 1 + 5x + 10x^2 + \cdots

Answer

1 + 5x + 10x^2

Final answer

1 + 5x + 10x^2

Detailed explanation

Walkthrough

This part asks for the first three terms in ascending powers of $x$ , so we need the constant term, the $x$ term and the $x^2$ term only. The binomial theorem says that in $(1+x)^5$ the term containing $x^r$ has coefficient $\binom{5}{r}$ . For $r=0$ , $r=1$ and $r=2$ , these coefficients are $\binom{5}{0}=1$ , $\binom{5}{1}=5$ and $\binom{5}{2}=10$ . Since the second term in the bracket is just $x$ , no extra factor is needed. Adding these three terms gives $1 + 5x + 10x^2$ .

Key Takeaways

The binomial theorem for a positive integer $n$ gives the coefficients of $x^r$ as $\binom{n}{r}$ . For small powers you can also read the coefficients from the row of Pascal's triangle, but the formula is more reliable for larger powers. 'Ascending powers' means the constant term first, then $x$ , then $x^2$ .

Common Mistakes

Writing the expansion from the wrong row of Pascal's triangle, or starting with the $x$ term instead of the constant term. Also, expanding all six terms when only the first three are required does not lose marks here if the first three are correct, but the instruction 'first three terms' should be followed.

Things to Be Careful About

Check that the exponents start at 0, not 1. The constant term is $1$ , not $0$ , and the coefficient of $x$ is $5$ , not $10$ (that is the coefficient of $x^2$ ). The mark scheme gives full marks for the single correct expression.

Techniques used

expand using the binomial theoremselect the first three terms in ascending powers

(b)

Find the first three terms in the expansion, in ascending powers of $x$ , of $(1 - 2x)^6$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the binomial theorem on $(1 - 2x)^6$ , treating the second term as $-2x$ , and keep terms up to $x^2$ .

Working

(1 - 2x)^6 = \binom{6}{0} + \binom{6}{1}(-2x) + \binom{6}{2}(-2x)^2 + \cdots

Evaluate each term:

(1 - 2x)^6 = 1 + 6(-2x) + 15(-2x)^2 + \cdots

Since $(-2x)^2 = 4x^2$ :

(1 - 2x)^6 = 1 - 12x + 60x^2 + \cdots

Answer

1 - 12x + 60x^2

Final answer

1 - 12x + 60x^2

Detailed explanation

Walkthrough

In $(1-2x)^6$ the two terms inside the bracket are $1$ and $-2x$ . Using the binomial theorem, the first three terms are formed by choosing $0$ , $1$ and $2$ factors of $-2x$ respectively. For $r=0$ the coefficient is $\binom{6}{0}=1$ . For $r=1$ the term is $\binom{6}{1}(-2x)=6(-2x)=-12x$ . For $r=2$ the term is $\binom{6}{2}(-2x)^2=15(4x^2)=60x^2$ . The square removes the negative sign, so the $x^2$ term is positive. Therefore the first three terms in ascending powers of $x$ are $1-12x+60x^2$ .

Key Takeaways

When a binomial has a negative second term, odd powers change sign and even powers stay positive. Squaring $-2x$ gives $4x^2$ , not $-4x^2$ . The coefficients come from the sixth row of Pascal's triangle or from $\binom{6}{r}$ : $1,6,15$ .

Common Mistakes

The most common error is writing $-60x^2$ instead of $60x^2$ by forgetting that $(-2x)^2=+4x^2$ . Another error is using the coefficient $\binom{6}{2}=15$ but multiplying by $2x$ instead of $(2x)^2$ , giving $30x^2$ instead of $60x^2$ .

Things to Be Careful About

Keep the minus sign attached to $2x$ throughout. It affects the sign of the $x$ term but not the $x^2$ term. The mark scheme awards B2 for all three terms correct and B1 for two correct components.

Techniques used

expand using the binomial theoremhandle a negative second termextract terms up to x squared

(c)

Hence find the coefficient of $x^2$ in the expansion of $(1 + x)^5(1 - 2x)^6$ .

DifficultyMedium-Easy

Worked solution

Approach

Multiply only the first three terms of each expansion, then collect the terms that contribute to $x^2$ .

Working

(1 + x)^5(1 - 2x)^6 = (1 + 5x + 10x^2)(1 - 12x + 60x^2) + \cdots

The coefficient of $x^2$ comes from three product pairs:

constant from the first factor with $x^2$ from the second: $1(60x^2)$
$x$ from the first factor with $x$ from the second: $5x(-12x)$
$x^2$ from the first factor with constant from the second: $10x^2(1)$

Therefore:

\text{coefficient of } x^2 = 60 - 60 + 10 = 10

Answer

10

Final answer

Detailed explanation

Walkthrough

To find the coefficient of $x^2$ in the product, we need only the terms up to $x^2$ from each expansion. When two brackets are multiplied, an $x^2$ term can be formed in exactly three ways: a constant from the first bracket times the $x^2$ term from the second; an $x$ term from the first times an $x$ term from the second; and the $x^2$ term from the first times the constant from the second. These give $1 \times 60x^2=60x^2$ , $5x \times (-12x)=-60x^2$ and $10x^2 \times 1=10x^2$ . Adding the coefficients gives $60-60+10=10$ . No other product from the first three terms of each bracket can produce $x^2$ .

Key Takeaways

The coefficient of a power in a product is obtained by collecting all combinations whose exponents add to that power. With binomial products, only a small number of terms contribute, so truncating each expansion at the required power is enough.

Common Mistakes

Forgetting the cross product $5x \times (-12x)$ and only computing $60+10$ . Another mistake is using the $x^2$ coefficient $10$ from the first bracket with the constant $1$ , but forgetting to include the product of the constants and $x^2$ from the second bracket. The mark scheme's method mark requires all three products to be considered.

Things to Be Careful About

Watch the negative sign in $-12x$ : the cross product is $-60x^2$ , so it cancels against the $60x^2$ term. The final coefficient is $10$ , not $70$ or $-50$ . The $x^2$ coefficient is just the number multiplying $x^2$ , so the answer should be $10$ (or $10x^2$ is allowed by the mark scheme).

Techniques used

multiply the first three terms of two expansionscollect terms contributing to x squaredcombine coefficients from three product pairs

The rest of this paper

10 more questions

Q2Quadratics4M
Q3Trigonometry4M
Q4Quadratics5M
Q5Functions6M
Q6Differentiation · Integration7M
Q7Quadratics · Functions8M
Q8Coordinate Geometry7M
Q9Series9M
Q10Circular Measure · Trigonometry8M
Q11Quadratics · Differentiation · Integration12M

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