Binomial for rational n; partial fractions – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

Find the binomial expansion of $$\sqrt{1+4x},\quad |x|<\tfrac14$$ in ascending powers of $x$, up to and including the term in $x^{3}$, giving each coefficient in its simplest form. [4]

$1+2x$ (first two terms correct)B1
Correct structure for the $x^{2}$ or $x^{3}$ term, e.g. $\dfrac{\frac12\left(-\frac12\right)}{2!}(4x)^{2}$ or $\dfrac{\frac12\left(-\frac12\right)\left(-\frac32\right)}{3!}(4x)^{3}$M1
$-2x^{2}$A1
$+4x^{3}$A1
Answer: $1+2x-2x^{2}+4x^{3}$
Examiner tip: Bracket the $4x$: $(4x)^{2}=16x^{2}$ – writing $4x^{2}$ is the commonest error.

2CoreCalculator3 marks

Express $\dfrac{5x+1}{x^{2}-1}$ in partial fractions. [3]

Factorises $x^{2}-1=(x+1)(x-1)$ and writes $5x+1\equiv A(x-1)+B(x+1)$, then substitutes a value of $x$ or compares coefficientsM1
One of $A=2$ (for $\frac{A}{x+1}$) or $B=3$ (for $\frac{B}{x-1}$)A1
$\dfrac{2}{x+1}+\dfrac{3}{x-1}$A1
Answer: $\dfrac{2}{x+1}+\dfrac{3}{x-1}$
Examiner tip: Always write the final answer as partial fractions – stopping at '$A=2$, $B=3$' can lose the last mark.

3StandardCalculator7 marks

(a) Find the first four terms, in ascending powers of $x$, of the binomial expansion of $$\frac{1}{\sqrt{4-x}},\quad |x|<4$$ giving each coefficient as an exact fraction in its simplest form. [4]

(b) By substituting $x=0.4$ into your expansion, find an approximation for $\sqrt{10}$, giving your answer to 5 decimal places. [3]

(a) Takes out the factor $4^{-\frac12}=\tfrac12$: $\tfrac12\left(1-\tfrac{x}{4}\right)^{-\frac12}$B1
Correct structure for the $x^{2}$ or $x^{3}$ term of $\left(1-\tfrac{x}{4}\right)^{-\frac12}$, e.g. $\dfrac{\left(-\frac12\right)\left(-\frac32\right)}{2!}\left(-\tfrac{x}{4}\right)^{2}$M1
Two of $\tfrac12,\ \tfrac{1}{16}x,\ \tfrac{3}{256}x^{2},\ \tfrac{5}{2048}x^{3}$ correctA1
$\tfrac12+\tfrac{1}{16}x+\tfrac{3}{256}x^{2}+\tfrac{5}{2048}x^{3}$A1
(b) Substitutes $x=0.4$ into their expansion: $0.52703\ldots$M1
Uses $\dfrac{1}{\sqrt{3.6}}=\dfrac{\sqrt{10}}{6}$ (or equivalent) so $\sqrt{10}\approx6\times$ their valueM1
$3.16219$ cao (premature rounding, e.g. $6\times0.52703=3.16218$, scores A0)A1
Answer: (a) $\tfrac12+\tfrac{1}{16}x+\tfrac{3}{256}x^{2}+\tfrac{5}{2048}x^{3}$ (b) $3.16219$
Examiner tip: Take the 4 out as $4^{-\frac12}$, not $4$ or $2$ – the power applies to the factor too.

4StandardCalculator9 marks

$$\mathrm{f}(x)=\frac{3x^{2}+3x+3}{(x+2)(x-1)^{2}}$$

(a) Express $\mathrm{f}(x)$ in partial fractions. [4]

(b) Hence, or otherwise, find the series expansion of $\mathrm{f}(x)$, in ascending powers of $x$, up to and including the term in $x^{2}$. Give each coefficient as a simplified fraction. [5]

(a) Correct form $\dfrac{A}{x+2}+\dfrac{B}{x-1}+\dfrac{C}{(x-1)^{2}}$B1
Uses $3x^{2}+3x+3\equiv A(x-1)^{2}+B(x+2)(x-1)+C(x+2)$ and substitutes values or compares coefficientsM1
Two of $A=1,\ B=2,\ C=3$A1
All three: $\dfrac{1}{x+2}+\dfrac{2}{x-1}+\dfrac{3}{(x-1)^{2}}$A1
(b) Attempts a binomial expansion of one term, e.g. $\tfrac12\left(1+\tfrac{x}{2}\right)^{-1}$ or $3(1-x)^{-2}$ or $-2(1-x)^{-1}$M1
Correct expansion of one term, e.g. $\tfrac12-\tfrac14x+\tfrac18x^{2}$A1
Correct expansion of a second term, e.g. $-2-2x-2x^{2}$ or $3+6x+9x^{2}$A1
Adds three expansions (of the right form) and collects like termsM1
$\tfrac32+\tfrac{15}{4}x+\tfrac{57}{8}x^{2}$A1
Answer: (a) $\dfrac{1}{x+2}+\dfrac{2}{x-1}+\dfrac{3}{(x-1)^{2}}$ (b) $\tfrac32+\tfrac{15}{4}x+\tfrac{57}{8}x^{2}$
Examiner tip: Rewrite $\dfrac{2}{x-1}$ as $-2(1-x)^{-1}$ before expanding – the sign change is where most marks are lost.

5HardCalculator7 marks

$$\mathrm{f}(x)=\frac{4x^{2}+7x}{2x^{2}+x-1},\quad x>1$$

(a) Given that $\mathrm{f}(x)$ can be written in the form $$A+\frac{B}{x+1}+\frac{C}{2x-1}$$ find the values of the constants $A$, $B$ and $C$. [4]

(b) Hence show that $\mathrm{f}$ is a decreasing function. [3]

(a) Divides or uses $4x^{2}+7x\equiv A(x+1)(2x-1)+B(2x-1)+C(x+1)$M1
$A=2$B1
Substitutes $x=-1$ or $x=\tfrac12$ (or compares coefficients) to find $B$ or $C$M1
$B=1$ and $C=3$A1
(b) Differentiates their partial fractions: $\mathrm{f}'(x)=-\dfrac{B}{(x+1)^{2}}-\dfrac{2C}{(2x-1)^{2}}$M1
$\mathrm{f}'(x)=-\dfrac{1}{(x+1)^{2}}-\dfrac{6}{(2x-1)^{2}}$A1
States both terms negative for $x>1$ (squares positive), so $\mathrm{f}'(x)<0$ and f is decreasingA1
Answer: (a) $A=2,\ B=1,\ C=3$ (b) $\mathrm{f}'(x)=-\dfrac{1}{(x+1)^{2}}-\dfrac{6}{(2x-1)^{2}}<0$
Examiner tip: The fraction is improper (degree 2 over degree 2) – you must include the constant $A$.

6StretchCalculator8 marks

In the binomial expansion of $(1+ax)^{n}$, where $a$ and $n$ are constants and $n$ is not a positive integer, the coefficient of $x$ is $-6$ and the coefficient of $x^{2}$ is $27$.

(a) Find the value of $a$ and the value of $n$. [5]

(b) Find the coefficient of $x^{3}$ in the expansion. [2]

(a) $na=-6$B1
$\dfrac{n(n-1)}{2}a^{2}=27$B1
Eliminates one variable, e.g. $\dfrac{(na)(na-a)}{2}=27\Rightarrow -6(-6-a)=54$M1
$a=3$A1
$n=-2$A1
(b) $\dfrac{n(n-1)(n-2)}{3!}a^{3}$ with their valuesM1
$-108$A1
(c) $|x|<\tfrac13$ (accept $-\tfrac13B1
Answer: (a) $a=3,\ n=-2$ (b) $-108$ (c) $|x|<\tfrac13$, i.e. $-\tfrac13
Examiner tip: Square the $a$ in the $x^{2}$ coefficient – $(ax)^{2}=a^{2}x^{2}$ – or the simultaneous equations will not give a sensible answer.

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