Proof, algebra & functions (indices, surds, quadratics, inequalities, polynomials & factor theorem) – A-level Maths exam questions with mark scheme

1CoreCalculator2 marks

Write $\dfrac{6x^{3}-5\sqrt{x}}{2x}$ in the form $ax^{p}+bx^{q}$, where $a$, $b$, $p$ and $q$ are constants to be found. [2]

for one correct term, e.g. $3x^{2}$ or $-\tfrac52x^{-\frac12}$B1
for $3x^{2}-\tfrac52x^{-\frac12}$ fully correct (accept $a=3,\ p=2,\ b=-2.5,\ q=-\tfrac12$)B1
Answer: $3x^{2}-\tfrac{5}{2}x^{-\frac{1}{2}}$
Examiner tip: Write $\sqrt{x}$ as $x^{\frac12}$ before dividing – $x^{\frac12}\div x = x^{-\frac12}$, not $x^{\frac32}$.

2CoreCalculator3 marks

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Express $\dfrac{5+\sqrt{3}}{2-\sqrt{3}}$ in the form $a+b\sqrt{3}$, where $a$ and $b$ are integers. [3]

for multiplying numerator and denominator by $2+\sqrt3$M1
for expanding $(5+\sqrt3)(2+\sqrt3)$ with at least 3 of the 4 terms correct: $10+5\sqrt3+2\sqrt3+3$ (denominator $4-3=1$)M1
$13+7\sqrt3$A1
Answer: $13+7\sqrt{3}$
Examiner tip: Show the denominator becomes $4-3=1$ explicitly – with 'no calculator technology' the working is where the marks are.

3StandardCalculator5 marks

The equation $kx^{2}+(k+3)x+1=0$, where $k$ is a non-zero constant, is a quadratic equation in $x$.

(a) Show that the discriminant of the equation is $k^{2}+2k+9$. [2]

(b) Prove that, for all non-zero values of $k$, the equation has two distinct real roots. [3]

(a) for attempting $b^{2}-4ac$ with $a=k,\ b=k+3,\ c=1$M1
(a) $(k+3)^{2}-4k=k^{2}+6k+9-4k=k^{2}+2k+9$ (no errors – given answer)A1*
(b) for completing the square: $k^{2}+2k+9=(k+1)^{2}+8$M1
(b) for stating $(k+1)^{2}\geqslant 0$ so $(k+1)^{2}+8\geqslant 8>0$A1
(b) for a full conclusion: discriminant $>0$ for all $k$, so two distinct real rootsA1
Answer: (a) $k^{2}+2k+9$ (b) $(k+1)^{2}+8\geqslant 8>0$ so $b^{2}-4ac>0$ for all $k$
Examiner tip: A proof needs a conclusion that links back: say 'discriminant $>0$, therefore two distinct real roots', not just '$>0$'.

4StandardCalculator5 marks

Find the set of values of $x$ for which both

$$3x-7>2-x \quad\text{and}\quad x^{2}-5x-14<0.$$

Give your answer in set notation. [5]

$x>\tfrac94$B1
for finding critical values of the quadratic: $(x-7)(x+2)=0 \Rightarrow x=7,\ -2$M1
$-2A1
for combining their two inequalities (e.g. on a number line) to find the overlapM1
$\{x: \tfrac94\tfrac94\}\cap\{x:x<7\}$A1
Answer: $\left\{x:\tfrac{9}{4}
Examiner tip: For '$<0$' you want the region between the roots; sketch the parabola if unsure.

5HardCalculator8 marks

$$f(x)=2x^{3}-3x^{2}-11x+6$$

(a) Use the factor theorem to show that $(x-3)$ is a factor of $f(x)$. [2]

(b) Hence fully factorise $f(x)$. [3]

(c) Hence solve the equation $$2(8^{y})-3(4^{y})-11(2^{y})+6=0$$ giving your answers to 3 significant figures where appropriate. [3]

(a) for attempting $f(3)$M1
(a) $f(3)=54-27-33+6=0$ and conclusion so $(x-3)$ is a factorA1
(b) for dividing by $(x-3)$ or comparing coefficients to obtain a quadratic factor $2x^{2}+\dots x\pm 2$M1
(b) $(x-3)(2x^{2}+3x-2)$A1
(b) $(x-3)(2x-1)(x+2)$A1
(c) for recognising $x=2^{y}$ and solving $2^{y}=$ one of their positive roots using logsM1
(c) $y=-1$A1
(c) $y=\log_{2}3=1.58$ (awrt), with $2^{y}=-2$ rejected or ignoredA1
Answer: (a) $f(3)=0$ (b) $(x-3)(2x-1)(x+2)$ (c) $y=-1,\ y=1.58$
Examiner tip: In (c) the link is $8^{y}=(2^{y})^{3}$; reject $2^{y}=-2$ because $2^{y}>0$ – don't try to take $\log$ of a negative.

6StandardCalculator5 marks

(a) Prove that, for all integers $n$ that are not multiples of 3, $n^{2}-1$ is a multiple of 3. [3]

(b) Show, by means of a counter-example, that the statement

'$n^{2}+n+41$ is prime for all positive integers $n$'

is not true. [2]

(a) for writing $n=3k+1$ and $n=3k+2$ (or $3k-1$) and expanding $n^{2}-1$ in at least one case. Alternative: $n^2-1=(n-1)(n+1)$ and $n-1,\ n,\ n+1$ are three consecutive integers, so one is a multiple of 3M1
(a) $(3k+1)^{2}-1=9k^{2}+6k=3(3k^{2}+2k)$, a multiple of 3 (alt: $n$ is not a multiple of 3, so $n-1$ or $n+1$ is)A1
(a) $(3k+2)^{2}-1=9k^{2}+12k+3=3(3k^{2}+4k+1)$, a multiple of 3, with conclusion covering all cases (alt: hence $(n-1)(n+1)=n^2-1$ is a multiple of 3, with conclusion)A1
(b) for substituting a suitable value, e.g. $n=40$ or $n=41$M1
(b) e.g. $n=40$: $1681=41^{2}$ (or $n=41$: $41\times43$), not prime, so statement is falseA1
Answer: (a) both cases give $3\times$ integer (b) e.g. $n=40$ gives $41^{2}$
Examiner tip: Proof by exhaustion must cover every case and end with a conclusion; testing a few numbers is not a proof.

7HardCalculator8 marks

The diagram shows a sketch of the curve $C$ with equation $y=(x+1)(x-3)^{2}$ and the straight line $l$ with equation $y=9-3x$.

(a) Write down the coordinates of the points where $C$ meets the coordinate axes. [2]

(b) Using algebra, find the coordinates of all the points where $l$ meets $C$. [4]

The curve with equation $y=f(x-a)$, where $f(x)=(x+1)(x-3)^{2}$ and $a$ is a constant, passes through the origin.

(a) $(-1,0)$ and $(3,0)$B1
(a) $(0,9)$B1
(b) for setting $(x+1)(x-3)^{2}=9-3x$M1
(b) for collecting into a cubic $=0$ and factorising, e.g. $x^{3}-5x^{2}+6x=0$ or $(x-3)\left[(x+1)(x-3)+3\right]=0$M1
(b) $x=0,\ 2,\ 3$A1
(b) $(0,9),\ (2,3),\ (3,0)$A1
(c) one correct value, $a=1$ or $a=-3$B1
(c) both $a=1$ and $a=-3$ (and no others)B1
Answer: (a) $(-1,0),(3,0),(0,9)$ (b) $(0,9),(2,3),(3,0)$ (c) $a=1$ or $a=-3$
Examiner tip: Don't divide both sides by $(x-3)$ in (b) without recording $x=3$ as a solution – you'd lose a point.

8HardCalculator6 marks

The line $y=mx+4$, where $m$ is a constant, is a tangent to the curve with equation $y=x^{2}-2x+5$.

Find the possible values of $m$ and, for each value, the coordinates of the point where the line touches the curve. [6]

for equating and forming a 3-term quadratic: $x^{2}-(2+m)x+1=0$M1
for using $b^{2}-4ac=0$ with their quadratic: $(2+m)^{2}-4=0$M1
$m=0$ and $m=-4$A1
for substituting one value of $m$ back and solving for $x$ (repeated root)M1
$(1,4)$ for $m=0$A1
$(-1,8)$ for $m=-4$A1
Answer: $m=0$, touching at $(1,4)$; $m=-4$, touching at $(-1,8)$
Examiner tip: 'Tangent' means a repeated root, so the discriminant is zero; also remember $(2+m)^{2}=4$ gives two values.

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