2 Algebra and functions — revision question pack

11 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section 2. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

2.1 · Understand and use the laws of indices for all rational exponents.

Explanation

For a non-zero base, $a^m a^n=a^{m+n}$ , $a^m/a^n=a^{m-n}$ and $(a^m)^n=a^{mn}$ . Interpret rational powers through roots: $a^{p/q}=\sqrt[q]{a^p}=(\sqrt[q]{a})^p$ whenever the real-valued expression is defined.
For example, $16^{3/4}=(\sqrt[4]{16})^3=2^3=8$ .
A negative exponent means reciprocal, not a negative value: $a^{-r}=1/a^r$ ; also avoid applying index laws across addition.
The base must be non-zero for division and negative powers.
When an equation contains different-looking powers, rewrite them with a common positive base before equating exponents; this makes the index law being used visible and avoids premature decimal approximations.

Worked example

Given $x>0$ , simplify $\dfrac{x^{5/2}}{x^{-1/2}}$ to a single power of $x$ .

1.For division with a common base, subtract the exponents: $x^{5/2}/x^{-1/2}=x^{5/2-(-1/2)}=x^{6/2}=x^3$ .

Answer: $x^3$

Common mistakes

Don't apply $(a^m)^n=a^{m+n}$ instead of multiplying the exponents.
Don't treat $a^{-r}$ as a negative number instead of the reciprocal $1/a^r$ .

Exam tip

When solving an exponential equation, rewrite both sides with the same base and show the equation formed by equating exponents.

Tier 1 · Easy

1.
Evaluate $81^{3/4}$ without using a calculator.
(2)
(Total for Question 1 is 2 marks)
2.
Given $x>0$ , solve $x^{3/2}=125$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Given $x>0$ , simplify $\dfrac{\left(16x^{12}\right)^{3/4}}{2x^{-2}}$ .
(3)
(Total for Question 1 is 3 marks)
2.
Given $x>0$ , simplify $\dfrac{(27x^{-9})^{2/3}}{9x^{-1}}$ as a single power of $x$ , and then with positive indices.
(3)
(Total for Question 2 is 3 marks)
3.
For $x>0$ , the identity $(x^{p-1/2})^4x^{-3}\equiv x^7$ is true for all values of $x$ . Find the rational number $p$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
Solve $8^{2x-1}=4^{x+2}$ , giving the exact value of $x$ .
(4)
(Total for Question 1 is 4 marks)
2.
Solve $(27^{2x-1})^{1/2}9^{-x/3}=81^{x/4-1}$ , giving the exact value of $x$ .
(5)
(Total for Question 2 is 5 marks)
3.
Given $x>0$ and $y>0$ , simplify $\left(\dfrac{(x^{-1/2}y^{2/3})^6}{x^{-5}y}\right)^{3/5}$ as a product of powers with positive indices.
(4)
(Total for Question 3 is 4 marks)
4.
Given $x>0$ and $y>0$ , the identity $\left((x^py^q)^{-2}x^{1/2}y^3\right)^{1/3}\equiv x^{-3/2}y^{5/3}$ holds for all such $x$ and $y$ . Find the rational numbers $p$ and $q$ .
(5)
(Total for Question 4 is 5 marks)
5.
Given $x>0$ , solve $\dfrac{(x^{3/2})^{-2}(x^{-1/3})^6}{x^{1/4}}=32$ , giving the exact value of $x$ .
(4)
(Total for Question 5 is 4 marks)

2.2 · Use and manipulate surds, including rationalising the denominator.

Explanation

A surd is an exact irrational root; simplify it by extracting the largest square factor, as in $\sqrt{50}=5\sqrt2$ . Combine only like surds, and rationalise a binomial denominator by multiplying numerator and denominator by its conjugate.
For example, $1/(3+\sqrt2)=(3-\sqrt2)/(9-2)=(3-\sqrt2)/7$ .
Do not split roots across addition: $\sqrt{a+b}$ is not generally $\sqrt a+\sqrt b$ ; after squaring an equation, check every candidate in the original.
Keep surd answers exact unless the question requests a decimal.
In equations, record the domain before squaring and substitute every candidate back into the original radical equation because squaring can introduce an extraneous solution.

Worked example

Rationalise and simplify $\dfrac{5}{2+\sqrt3}$ .

1.Multiply by the conjugate: $\dfrac{5}{2+\sqrt3}\times\dfrac{2-\sqrt3}{2-\sqrt3}=\dfrac{5(2-\sqrt3)}{4-3}=10-5\sqrt3$ .

Answer: $10-5\sqrt3$

Common mistakes

Don't split $\sqrt{a+b}$ into $\sqrt a+\sqrt b$ , which is not a valid surd law.
Don't multiply a binomial denominator by itself instead of by its conjugate, so the denominator remains irrational.

Exam tip

For “exact” answers, simplify every surd and rationalise the denominator without converting to decimals.

Tier 1 · Easy

1.
Write $\sqrt{72}$ in the form $a\sqrt2$ , where $a$ is an integer.
(2)
(Total for Question 1 is 2 marks)
2.
Simplify $2\sqrt{45}-\sqrt{20}+\sqrt5$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Show that $\dfrac{2+\sqrt3}{2-\sqrt3}-\dfrac{2-\sqrt3}{2+\sqrt3}=8\sqrt3$ .
(4)
(Total for Question 1 is 4 marks)
2.
Rationalise the denominator of $\dfrac{4+\sqrt7}{\sqrt7-2}$ , giving your answer in exact simplified form.
(3)
(Total for Question 2 is 3 marks)
3.
The rational numbers $a$ and $b$ satisfy $(a+b\sqrt3)(2-\sqrt3)=7+\sqrt3$ . Find $a$ and $b$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Solve $\sqrt{x+6}-\sqrt{x-3}=1$ .
(5)
(Total for Question 1 is 5 marks)
2.
Let $y=\sqrt{11+6\sqrt2}$ . Express $y$ in the form $a+b\sqrt2$ , where $a$ and $b$ are positive integers. Hence express $1/y$ with a rational denominator.
(5)
(Total for Question 2 is 5 marks)
3.
Let $u=2+\sqrt3$ . Show that $u^{-1}=2-\sqrt3$ and hence find the exact value of $u^3+u^{-3}$ without expanding both cubes separately.
(5)
(Total for Question 3 is 5 marks)
4.
Rationalise the denominator of $\dfrac1{\sqrt2+\sqrt3+\sqrt5}$ , giving your answer in exact simplified form.
(5)
(Total for Question 4 is 5 marks)
5.
Let $\alpha=\sqrt5+\sqrt2$ and $\beta=\sqrt5-\sqrt2$ . Find a monic polynomial $P(x)$ with integer coefficients whose four roots are $\alpha$ , $-\alpha$ , $\beta$ and $-\beta$ .
(6)
(Total for Question 5 is 6 marks)

2.3 · Work with quadratic functions and their graphs; the discriminant, including conditions for real and repeated roots; completing the square; solution of quadratic equations, including solving quadratic equations in a function of the unknown.

Explanation

For $ax^2+bx+c$ , the discriminant $b^2-4ac$ is positive for two distinct real roots, zero for a repeated root and negative for no real roots.
Complete the square to expose the turning point and range, or use factorisation and the quadratic formula to solve equations.
For example, $x^2-6x+11=(x-3)^2+2$ , so its minimum point is $(3,2)$ and it has no real roots.
When the quadratic is in a function of the unknown, substitute a new variable, solve the quadratic in that variable, then solve every resulting equation and reject invalid roots.
The substituted function may be a power, trigonometric function, exponential or logarithm, so enforce its range and domain when returning to $x$ .

Worked example

Find the two values of $k$ for which $2x^2+(k-1)x+8=0$ has a repeated root.

1.A repeated root requires discriminant zero: $(k-1)^2-4(2)(8)=0$ .
2.Hence $(k-1)^2=64$ , so $k-1=\pm8$ and therefore $k=9$ or $k=-7$ .

Answer: $k=9$ or $k=-7$

Common mistakes

Don't use $b^2-4ac=0$ for two distinct roots instead of for one repeated root.
Don't solve the quadratic in that variable and fail to return to solve for the original unknown after substituting a new variable.

Exam tip

For a repeated-root condition, write $b^2-4ac=0$ before substituting the coefficients.

Tier 1 · Easy

1.
Write $x^2-8x+5$ in completed-square form.
(2)
(Total for Question 1 is 2 marks)
2.
State, with a reason, the number of real roots of $2x^2+3x+5=0$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The curve with equation $y=x^2-6x+k$ and the line with equation $y=2x-3$ meet at exactly one point. Find the value of $k$ and the coordinates of this point.
(4)
(Total for Question 1 is 4 marks)
2.
Find the set of real values of $k$ for which $x^2+(k-3)x+k=0$ has no real roots.
(4)
(Total for Question 2 is 4 marks)
3.
A monic quadratic $f(x)=x^2+px+q$ has roots $2+\sqrt3$ and $2-\sqrt3$ . Find $p$ and $q$ , and hence state the minimum value of $f(x)$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Solve $(x^2-5x)^2-5(x^2-5x)-14=0$ , giving exact answers.
(6)
(Total for Question 1 is 6 marks)
2.
Solve $25^x-26(5^x)+25=0$ , giving exact values.
(5)
(Total for Question 2 is 5 marks)
3.
Solve $\left(x+\dfrac1x\right)^2-5\left(x+\dfrac1x\right)+6=0$ for real $x$ , giving exact answers.
(6)
(Total for Question 3 is 6 marks)
4.
The roots of $2x^2-5x-3=0$ are $\alpha$ and $\beta$ . Without solving this equation, find an equation with integer coefficients whose roots are $\alpha^2$ and $\beta^2$ . Give your equation in the form $ax^2+bx+c=0$ , where $a>0$ and $a$ , $b$ , $c$ have greatest common divisor $1$ .
(5)
(Total for Question 4 is 5 marks)
5.
Find the exact range of $f(x)=\dfrac{x^2+4x+7}{x^2+1}$ for real $x$ .
(6)
(Total for Question 5 is 6 marks)

2.4 · Solve simultaneous equations in two variables by elimination and by substitution, including one linear and one quadratic equation.

Explanation

A simultaneous solution is an ordered pair satisfying every equation; graphically, solutions are intersection points. Use elimination when coefficients align conveniently, or substitute an expression from the linear equation into the nonlinear equation.
For example, substituting $y=5-x$ into $xy=6$ gives $x(5-x)=6$ , whose roots generate the two intersection pairs. After finding one coordinate, substitute each possible value back separately; losing the second quadratic root or mismatching coordinates are common errors.
If substitution produces a quadratic, its two real roots may give two intersection points.
Pair each value of the first coordinate with the value obtained from the same substitution, then verify both coordinates in both original equations.
Equations involving powers such as $y=2^x$ can similarly be reduced by substituting the exponential expression.

Worked example

Find every solution of $y=x+1$ and $x^2+y^2=25$ .

1.Substitute $y=x+1$ : $x^2+(x+1)^2=25$ , so $2x^2+2x-24=0$ .
2.Dividing by $2$ gives $(x+4)(x-3)=0$ , hence $x=3$ or $x=-4$ .
3.Using $y=x+1$ gives the pairs $(3,4)$ and $(-4,-3)$ .

Answer: $(x,y)=(3,4)$ or $(-4,-3)$

Common mistakes

Don't keep only one root of the quadratic produced by substitution and lose a valid intersection.
Don't pair an $x$ -value with the $y$ -value belonging to the other root.

Exam tip

State every ordered pair and check each pair in both original equations.

Tier 1 · Easy

1.
Solve $x+y=9$ and $x-y=1$ simultaneously.
(2)
(Total for Question 1 is 2 marks)
2.
Solve $3x+2y=19$ and $2x-y=1$ simultaneously.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Find the coordinates of the points of intersection of the curve $y=x^2-3x+1$ and the line $y=2x-5$ .
(4)
(Total for Question 1 is 4 marks)
2.
Find every solution of $y=2x+1$ and $xy=15$ .
(4)
(Total for Question 2 is 4 marks)
3.
A theatre sells $160$ tickets. An adult ticket costs £ $14$ and a student ticket costs £ $9$ . The total ticket income is £ $1860$ . Find the number of each type of ticket sold.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Solve the system $x+2y=7$ and $x^2+y^2=13$ .
(5)
(Total for Question 1 is 5 marks)
2.
Solve the simultaneous equations $x+y=7$ and $x^2-xy+y^2=13$ .
(5)
(Total for Question 2 is 5 marks)
3.
The simultaneous equations $(k+1)x+2y=7$ and $3x+(k-1)y=5$ do not have a unique solution for certain real values of $k$ . Find these values and determine whether each system has no solution or infinitely many solutions.
(6)
(Total for Question 3 is 6 marks)
4.
The line $y=2x+1$ meets the curve $y=x^2-4x-2$ at points $A$ and $B$ . Without finding the coordinates of $A$ and $B$ separately, find the exact coordinates of the midpoint of $AB$ and the exact length $AB$ .
(6)
(Total for Question 4 is 6 marks)
5.
The line $y=2x+k$ and the circle $x^2+y^2=20$ are given, where $k$ is a real constant. Find the values of $k$ for which the line is tangent to the circle and the corresponding point of contact in each case. State also the values of $k$ for which there are two distinct intersections and no intersections.
(6)
(Total for Question 5 is 6 marks)

2.5 · Solve linear and quadratic inequalities in one variable and interpret them graphically, including with brackets and fractions; express solutions using 'and'/'or' or set notation; represent linear and quadratic inequalities graphically.

Explanation

An inequality solution is a set of values; endpoints are included for $\leq$ or $\geq$ and excluded for $<$ or $>$ . For a quadratic or rational inequality, locate every zero and undefined value, then use a sign diagram or graph on the resulting intervals.
For example, $(x-2)(x+3)<0$ between its roots, so $-3<x<2$ .
Multiplying by an expression of unknown sign can reverse the inequality unpredictably; bring a fractional inequality to one side and analyse signs instead.
Mark open or closed endpoints carefully and exclude any value that makes an original denominator zero, even when it appears as a boundary on the sign diagram.
For a two-variable graphical inequality, draw a solid boundary for $\leq$ or $\geq$ , a dashed boundary for $<$ or $>$ , test a point and shade the correct region.

Boundary curves and a test point determine the shaded region of a two-variable inequality.

Worked example

Solve $(x-4)(x+1)\leq0$ .

1.The boundary values are $x=-1$ and $x=4$ .
2.The upward-opening quadratic is non-positive between these roots, and equality includes both endpoints.
3.Therefore $-1\leq x\leq4$ .

Answer: $-1\leq x\leq4$

Common mistakes

Don't divide an inequality by a negative quantity without reversing the inequality sign.
Don't cross-multiply a rational inequality by an expression of unknown sign, so the direction of the inequality is unjustified.

Exam tip

For a quadratic or rational inequality, show the critical values and a sign diagram before writing the final intervals.

Tier 1 · Easy

1.
Solve $5-2x<11$ .
(2)
(Total for Question 1 is 2 marks)
2.
Solve $4(x-3)\leq2x+6$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Solve the compound inequality $3(2x-1)\leq5-x<4x+10$ .
(4)
(Total for Question 1 is 4 marks)
2.
Solve $x^2-5x-14<0$ .
(3)
(Total for Question 2 is 3 marks)
3.
Sketch and shade the region satisfying $x\geq1$ , $y\geq0$ and $x+2y\leq7$ . State the coordinates of all vertices of the region.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Solve $\dfrac{x+2}{x-3}>2$ and give the result in set notation.
(4)
(Total for Question 1 is 4 marks)
2.
Solve $\dfrac{(x-4)(x+1)}{x-2}\leq0$ , giving your answer as inequalities.
(5)
(Total for Question 2 is 5 marks)
3.
Find the real values of $x$ that satisfy both $x^2-5x+4\leq0$ and $2x^2+x-6>0$ . Give your answer as inequalities.
(5)
(Total for Question 3 is 5 marks)
4.
Solve $\sqrt{2x+3}>x$ , giving your answer as an interval.
(5)
(Total for Question 4 is 5 marks)
5.
Solve $x^4-10x^2+9\leq0$ , giving your answer as inequalities.
(4)
(Total for Question 5 is 4 marks)

2.6 · Manipulate polynomials algebraically, including expanding brackets, collecting like terms, factorisation and simple algebraic division; use the factor theorem; simplify rational expressions by factorising, cancelling and algebraic division.

Explanation

Polynomial manipulation uses expansion, collection and factorisation while preserving equality; choose the form that exposes the required structure. The factor theorem states that $ax-b$ is a factor of $f(x)$ exactly when $f(b/a)=0$ ; after finding a factor, divide by the linear expression to obtain the remaining polynomial.
For example, $f(2)=0$ for $f(x)=x^3-3x^2-4x+12$ , and division by $x-2$ gives $x^2-x-6$ . Cancel factors, not separate terms, in rational expressions, and retain exclusions from the original denominator even when a factor cancels.
Only linear divisors $ax+b$ or $ax-b$ are required for algebraic division.
When simplifying a rational expression, factor the complete numerator and denominator before cancelling.
A cancelled factor still records an excluded input from the original expression, so state that restriction with the simplified answer.

Worked example

Use the factor theorem to factorise $x^3-4x^2+x+6$ fully.

1.Let $f(x)=x^3-4x^2+x+6$ .
2.Since $f(2)=8-16+2+6=0$ , $x-2$ is a factor.
3.Division gives $x^2-2x-3$ , which factorises as $(x-3)(x+1)$ .
4.Hence $f(x)=(x-2)(x-3)(x+1)$ .

Answer: $(x-2)(x-3)(x+1)$

Common mistakes

Don't cancel separate terms across addition, such as cancelling the $x$ in $(x+2)/x$ .
Don't forget the excluded value from the original denominator after a common factor has cancelled.

Exam tip

Use the factor theorem by showing $f(a)=0$ , then divide by $x-a$ and factorise the quotient fully.

Tier 1 · Easy

1.
Factorise $x^2-9x+20$ completely.
(2)
(Total for Question 1 is 2 marks)
2.
Expand and simplify $(2x-3)(x^2+x-4)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The polynomial $p(x)=x^3+ax^2-4x-12$ , where $a$ is a constant, has a factor $x+2$ . Find the value of $a$ and hence factorise $p(x)$ fully.
(4)
(Total for Question 1 is 4 marks)
2.
Divide $2x^3-3x^2+4x-5$ by $x-2$ , giving the quotient and remainder.
(4)
(Total for Question 2 is 4 marks)
3.
Find the remainder when $p(x)=4x^3-3x+7$ is divided by $2x+1$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
Simplify $\dfrac{x^3-4x^2+x+6}{x^2-4x+4}$ as far as possible, then use algebraic division. State any excluded value.
(5)
(Total for Question 1 is 5 marks)
2.
Both $x-1$ and $x+2$ divide $p(x)=x^3+ax^2+bx+6$ . Determine $a$ and $b$ , and hence factorise $p(x)$ fully.
(5)
(Total for Question 2 is 5 marks)
3.
For $x\ne-1,4$ , the rational function $F(x)=\dfrac{2x^3+ax^2+bx-4}{(x-4)(x+1)}$ simplifies exactly to a linear polynomial. Determine $a$ and $b$ , and find the simplified polynomial.
(6)
(Total for Question 3 is 6 marks)
4.
The polynomial $p(x)=2x^3+ax+b$ leaves the same remainder when divided by $x-2$ and $x+1$ , and $x+3$ is a factor. Determine $a$ and $b$ . Then divide $p(x)$ by $x-1$ , giving the quotient and remainder.
(6)
(Total for Question 4 is 6 marks)
5.
A monic quartic polynomial $p(x)$ has roots $-2$ , $1$ and a repeated root $r$ . Its coefficient of $x^3$ is $-5$ . Find $r$ , factorise the polynomial fully and state its constant term.
(5)
(Total for Question 5 is 5 marks)

2.7 · Understand and use graphs of functions; sketch curves including polynomials, the modulus of a linear function, y = a/x and y = a/x² with asymptotes; use graph intersections to solve equations; understand proportional relationships.

Explanation

A useful sketch records intercepts, turning points, end behaviour, symmetry and asymptotes rather than relying on a table of isolated points.
For $y=a/x$ the coordinate axes are asymptotes and the relation is inverse proportionality; $y=a/x^2$ is inverse-square, symmetric about the $y$ -axis and has the same sign as $a$ .
Translations such as $y=a/(x-p)+q$ move the asymptotes to $x=p$ and $y=q$ .
The graph of $y=|f(x)|$ reflects negative outputs above the $x$ -axis; use its intersections with another graph to solve equations or inequalities.
Do not draw a reciprocal curve crossing an asymptote, and distinguish direct proportionality $y=kx$ from a relation that merely has positive correlation.

The graph of

y=4/x

: the branches approach, but never cross, the coordinate-axis asymptotes.

Worked example

Find the intersection values of $x$ for the graphs $y=|2x-3|$ and $y=x+4$ .

1.For $x\geq3/2$ , solve $2x-3=x+4$ , giving $x=7$ .
2.For $x<3/2$ , solve $-(2x-3)=x+4$ , so $-2x+3=x+4$ and $x=-1/3$ .
3.Each value lies in its required branch, so both are intersections.

Answer: $x=7$ or $x=-\dfrac13$

Common mistakes

Don't draw a reciprocal branch crossing an axis even though the axes are asymptotes.
Don't call a relation directly proportional without a graph through the origin or an equation of the form $y=kx$ .

Exam tip

On a sketch, label intercepts, turning points and asymptotes; a table of unlabelled points is not enough.

Tier 1 · Easy

1.
For the curve $y=-\dfrac{6}{x}$ , state both asymptotes and the two quadrants containing its branches.
(2)
(Total for Question 1 is 2 marks)
2.
$y$ is directly proportional to $x^3$ . Given that $y=54$ when $x=3$ , find $y$ when $x=-2$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Sketch the graph of $y=|3x+6|-2$ , showing the coordinates of the vertex and all intercepts with the coordinate axes.
(4)
(Total for Question 1 is 4 marks)
2.
Sketch $y=(x+2)^2(x-1)(x-4)$ . Label every intercept and state the end behaviour and whether the graph crosses or touches the $x$ -axis at each root.
(4)
(Total for Question 2 is 4 marks)
3.
Sketch the curve $y=\dfrac{12}{x^2}-3$ . Label its asymptotes and intercepts with the coordinate axes, and state its range.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
The curves $y=\dfrac4x$ and $y=x+3$ meet twice. Determine the exact coordinates of both intersections.
(4)
(Total for Question 1 is 4 marks)
2.
The curve $y=k/x^2$ passes through $(2,9)$ . Find $k$ and hence determine the exact coordinates of every intersection of this curve with the line $y=x+1$ .
(5)
(Total for Question 2 is 5 marks)
3.
Find every real solution of $|x-1|=\dfrac4x$ , justifying the branches or domains that you reject.
(5)
(Total for Question 3 is 5 marks)
4.
A reciprocal curve has equation $y=\dfrac{a}{x-2}+3$ . It passes through $(4,1)$ . Find $a$ , state both asymptotes and find the intercepts with both coordinate axes. State the position of each branch relative to the asymptotes.
(6)
(Total for Question 4 is 6 marks)
5.
A polynomial $f$ has a graph that touches the $x$ -axis at $x=-1$ , crosses it at $x=3$ , and passes through $(0,6)$ . Find $f(x)$ of least possible degree, state its end behaviour and solve $f(x)\geq0$ .
(6)
(Total for Question 5 is 6 marks)

2.8 · Understand and use composite functions; inverse functions and their graphs.

Explanation

The composite $fg(x)$ means $f(g(x))$ : apply the right-hand function first and keep its whole output inside the outer function. A one-to-one mapping has no repeated outputs and can have an inverse; a many-to-one mapping needs a restricted domain first.
To find an inverse, write $y=f(x)$ , rearrange for $x$ , then exchange the variable labels and use the notation $f^{-1}$ . A function and its inverse have graphs reflected in $y=x$ , with domain and range interchanged.
In general $fg\ne gf$, whereas $f^{-1}f(x)=ff^{-1}(x)=x$ on the appropriate domains.
For inverse quadratics, use the stated domain to choose the correct square-root branch.
Check a proposed inverse by composing the functions on the permitted domain.

A function and its inverse are mirror images in the line

y=x

, so

(a,b)

becomes

(b,a)

Worked example

For $f(x)=\dfrac{3x+2}{x-1}$ , find $f^{-1}(x)$ and state its domain.

1.Let $y=(3x+2)/(x-1)$ .
2.Then $yx-y=3x+2$ , so $x(y-3)=y+2$ and $x=(y+2)/(y-3)$ .
3.Exchanging the labels gives $f^{-1}(x)=(x+2)/(x-3)$ .
4.The inverse denominator requires $x\ne3$ .

Answer: $f^{-1}(x)=\dfrac{x+2}{x-3}$ Domain: $x\ne3$

Common mistakes

Don't evaluate $fg(x)$ by applying $f$ first, even though $fg(x)=f(g(x))$ .
Don't use both square-root branches for the inverse of a restricted quadratic instead of choosing the branch fixed by the stated domain.

Exam tip

To find an inverse, rearrange $y=f(x)$ for $x$ and state the inverse domain obtained from the original range.

Tier 1 · Easy

1.
Let $f(x)=3x-1$ and $g(x)=x^2+2$ . Calculate $fg(2)$ .
(2)
(Total for Question 1 is 2 marks)
2.
Given $f(x)=5x+4$ , find $f^{-1}(x)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Let $f(x)=2x-3$ and let $g(x)=\dfrac1{x+1}$ . Find $fg(x)$ , state its domain and solve $fg(x)=5$ .
(4)
(Total for Question 1 is 4 marks)
2.
Let $f(x)=4x+1$ and $g(x)=\sqrt{x+2}$ . Find $f^{-1}g(7)$ . Then find $gf^{-1}(x)$ and state its domain.
(4)
(Total for Question 2 is 4 marks)
3.
Let $f(x)=3x-4$ and $g(x)=ax+b$ . Given that $fg(x)=x$ for every real $x$ , find $a$ and $b$ , and verify that $gf(x)=x$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
The function $f(x)=(x-3)^2+1$ has domain $x\geq3$ . Find $f^{-1}(x)$ , state its domain, and solve $f^{-1}(2x+1)=x+3$ .
(6)
(Total for Question 1 is 6 marks)
2.
Let $g(x)=x^2$ and let $f$ be a linear function. Given that $fg(x)=6x^2-5$ , find $f(x)$ . Hence solve $fg(x)=gf(x)$ .
(5)
(Total for Question 2 is 5 marks)
3.
The function $f(x)=x+\dfrac4x$ has domain $x\geq2$ . Find $f^{-1}(x)$ and state its domain. Hence find the exact value of $f^{-1}\left(\dfrac{13}{3}\right)$ .
(6)
(Total for Question 3 is 6 marks)
4.
Let $f(x)=x^3+x+1$ . Prove that the graphs of $y=f(x)$ and $y=f^{-1}(x)$ have exactly one point of intersection, and find its coordinates.
(6)
(Total for Question 4 is 6 marks)
5.
Let $f(x)=\sqrt{x-1}$ and $g(x)=\dfrac3{x-2}$ . Find $fg(x)$ and $gf(x)$ , stating the domain of each composite. Hence solve $fg(x)=1$ .
(6)
(Total for Question 5 is 6 marks)

2.9 · Understand the effect of simple transformations on the graph of y = f(x), including sketching associated graphs: y = af(x), y = f(x) + a, y = f(x + a), y = f(ax) and combinations of these transformations.

Explanation

Transformations outside $f$ act on output coordinates: $y=af(x)$ scales vertically by factor $|a|$ and reflects in the $x$ -axis when $a<0$ . Transformations inside $f$ act oppositely on inputs: $f(x+a)$ moves left by $a$ , while $f(ax)$ scales horizontally by factor $1/|a|$ and may reflect in the $y$ -axis.
The graph $y=|f(x)|$ reflects every negative part above the $x$ -axis.
For $y=|f(-x)|$ , first reflect $f$ in the $y$ -axis, then apply the output modulus.
For example, a point $(u,v)$ on $y=f(x)$ maps to $(u/2,v+3)$ on $y=f(2x)+3$ .
Map a general point through combined transformations so horizontal scale factors and shifts stay in the correct order.

Replacing

x

x-a

translates the graph of

f

right by

a

Worked example

The point $(6,-1)$ lies on $y=f(x)$ . Find the corresponding point on $y=-2f(3x)+1$ .

1.For the transformed input to equal $6$ , set $3x=6$ , giving $x=2$ .
2.The transformed output is $-2(-1)+1=3$ , so the corresponding point is $(2,3)$ .

Answer: $(2,3)$

Common mistakes

Don't describe $f(x+a)$ as a translation right by $a$ instead of left by $a$ .
Don't use horizontal scale factor $a$ for $f(ax)$ instead of the reciprocal factor $1/|a|$ .

Exam tip

Track a point $(u,v)$ through the transformation and give its new coordinates to verify every shift, stretch and reflection.

Tier 1 · Easy

1.
Describe fully the transformation from $y=f(x)$ to $y=f(x-4)+2$ .
(2)
(Total for Question 1 is 2 marks)
2.
Describe fully the two transformations that map the graph of $y=f(x)$ onto the graph of $y=f(-3x)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The graph of $y=2^x$ is transformed to the graph of $y=2^{x+1}-3$ . Describe the transformation, and state the equation of the horizontal asymptote and the $y$ -intercept of the transformed graph.
(4)
(Total for Question 1 is 4 marks)
2.
The curve $y=x^3$ is translated $2$ units left, stretched parallel to the $y$ -axis by scale factor $3$ , reflected in the $x$ -axis and then translated $1$ unit up. Find the equation of the transformed curve and the image of $(0,0)$ .
(4)
(Total for Question 2 is 4 marks)
3.
A function $f$ has domain $-1\leq x\leq4$ and range $-2\leq f(x)\leq5$ . Its minimum occurs at $(1,-2)$ and its maximum at $(3,5)$ . For $g(x)=3-2f(2x-1)$ , find the domain and range of $g$ and the images of these two turning points.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Let $f(x)=x^2-4x+1$ . For $g(x)=-f(2x+2)+4$ , find the turning point and both $x$ -intercepts, then state whether it is a maximum or minimum.
(5)
(Total for Question 1 is 5 marks)
2.
Let $f(x)=1/x$ and $g(x)=-2f(3x-6)+4$ . Give the coordinate mapping from $y=f(x)$ to $y=g(x)$ , state both asymptotes of $g$ , and find its $x$ -intercept.
(6)
(Total for Question 2 is 6 marks)
3.
A transformation sends $y=f(x)$ to $y=g(x)$ , where $g(x)=af(bx+c)+d$ and $a,b\ne0$ . The point $(-2,4)$ maps to $(3,-5)$ , and the corresponding point $(1,-1)$ maps to $\left(\dfrac32,5\right)$ . Find $a$ , $b$ , $c$ and $d$ , and hence write $g(x)$ in terms of $f$ .
(6)
(Total for Question 3 is 6 marks)
4.
A function $f$ has domain $-4\leq x\leq6$ . Within this domain, $f(x)\geq2$ exactly when $-3\leq x\leq-1$ or $4\leq x\leq6$ . For $g(x)=5-2f(3x+2)$ , find the domain of $g$ and solve $g(x)\leq1$ .
(5)
(Total for Question 4 is 5 marks)
5.
Starting with $y=f(x)$ , a translation $2$ units right is followed by a stretch parallel to the $x$ -axis with scale factor $3$ . Find the resulting equation and the image of the point $(-1,4)$ . Repeat when the same two transformations are applied in the reverse order, and explain why the results differ.
(5)
(Total for Question 5 is 5 marks)

2.10 · Decompose rational functions into partial fractions (denominators not more complicated than squared linear terms and with no more than 3 terms, numerators constant or linear).

Explanation

Factor the denominator first and assign one partial-fraction term to each linear factor. A repeated factor needs a term for every power: $(x-a)^2$ requires $A/(x-a)+B/(x-a)^2$ .
After multiplying through by the common denominator, substitute convenient roots or compare coefficients to determine the constants.
If the numerator degree is at least the denominator degree, divide first; omitting a repeated-factor term produces an identity that cannot hold.
Recombine the fractions to check the identity.
The simpler terms can then be differentiated, integrated or expanded as binomial/geometric series within the required domain.

Worked example

Decompose $\dfrac{5x+1}{(x-1)(x+2)}$ into partial fractions.

1.Let $(5x+1)/((x-1)(x+2))=A/(x-1)+B/(x+2)$ .
2.Hence $5x+1=A(x+2)+B(x-1)$ .
3.Substituting $x=1$ gives $6=3A$ , so $A=2$ .
4.Substituting $x=-2$ gives $-9=-3B$ , so $B=3$ .

Answer: $\dfrac2{x-1}+\dfrac3{x+2}$

Common mistakes

Don't write only one term for a repeated factor $(x-a)^2$ instead of both $A/(x-a)$ and $B/(x-a)^2$ .
Don't use an unnecessary linear numerator over a linear factor instead of a constant numerator.

Exam tip

Factor the denominator first, write every required fraction term, then multiply through before substituting convenient values.

Tier 1 · Easy

1.
Express $\dfrac7{(x+1)(x+2)}$ as partial fractions.
(3)
(Total for Question 1 is 3 marks)
2.
Given that $\dfrac{7x-1}{(x-3)^2}\equiv\dfrac{A}{x-3}+\dfrac{B}{(x-3)^2}$ , find $B$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Express $\dfrac{3x+5}{(x+1)^2}$ in the form $\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}$ , where $A$ and $B$ are constants.
(3)
(Total for Question 1 is 3 marks)
2.
Express $\dfrac{2x^2+6x-2}{x(x-1)(x+2)}$ as three partial-fraction terms.
(4)
(Total for Question 2 is 4 marks)
3.
Express $\dfrac{8x+3}{(2x-1)(x+3)}$ in partial fractions.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Express $\dfrac{5x+1}{(x-1)^2(x+2)}$ as three partial-fraction terms.
(5)
(Total for Question 1 is 5 marks)
2.
Decompose $\dfrac{2x^3-2x^2+x+3}{x(x-1)^2}$ fully into partial fractions, carrying out algebraic division first.
(6)
(Total for Question 2 is 6 marks)
3.
The identity $\dfrac{x^2+kx+10}{(x-1)(x+2)(x-3)}\equiv\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{x-3}$ has $A=C$ . Find $k$ and hence determine $A$ , $B$ and $C$ .
(6)
(Total for Question 3 is 6 marks)
4.
Express $\dfrac{10x^2-3x-4}{(2x-1)^2(x+1)}$ in partial fractions.
(5)
(Total for Question 4 is 5 marks)
5.
The decomposition of $\dfrac{2x^2+kx+5}{(x-1)(x+2)^2}$ has the form $\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$ . Given that $B=0$ , find $k$ , $A$ and $C$ .
(5)
(Total for Question 5 is 5 marks)

2.11 · Use of functions in modelling, including consideration of limitations and refinements of the models.

Explanation

A function model links clearly defined variables over a stated domain; it may use polynomial, exponential, logarithmic, trigonometric or reciprocal functions.
Use data to determine parameters, then calculate and interpret the result with units and appropriate rounding.
For example, $P(t)=P_0r^t$ models repeated growth, while a sine function can model periodic height and a reciprocal function can model inverse variation.
Evaluate assumptions such as fixed rates, exact periodicity, ignored constraints or extrapolation.
A refinement must answer the weakness: use a piecewise rate when growth changes, add a seasonal term when residuals are periodic, or restrict the domain when extrapolation is unreliable.

Worked example

A cooling model is $T(t)=18+62e^{-0.15t}$ , where $T$ is in degrees Celsius and $t$ is in minutes. Find the first whole minute for which the model gives $T<30$ , and state one limitation.

1.Solve $18+62e^{-0.15t}<30$ .
2.This gives $e^{-0.15t}<6/31$ , so $t>-\ln(6/31)/0.15=10.95\ldots$ .
3.The first whole minute is therefore $11$ .
4.One limitation is that the fixed ambient temperature may not remain exactly $18$ degrees Celsius.

Answer: $11$ minutes; for example, the model assumes a constant surrounding temperature of $18$ degrees Celsius.

Common mistakes

Don't report a decimal model output as an exact real-world count without appropriate contextual rounding.
Don't extrapolate beyond the stated domain without discussing whether the model assumptions remain plausible.

Exam tip

After interpreting the value, link each limitation to a concrete refinement that changes the model or restricts its domain.

Tier 1 · Easy

1.
A colony is modelled by $P(t)=120(1.08)^t$ , where $t$ is measured in days. Estimate the number of organisms after $3$ days.
(2)
(Total for Question 1 is 2 marks)
2.
A battery's charge percentage is modelled by $C(t)=96-7t$ for $0\leq t\leq10$ , where $t$ is measured in hours. Find $C(6)$ and explain why the model should not be used to predict $C(14)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A company uses $V=Ab^t$ to estimate the value of a car, where £ $V$ is its value $t$ years after purchase. The constants $A$ and $b$ satisfy $b>0$ . According to this model, the car is worth £ $18\,000$ initially and £ $14\,580$ after two years. Find $A$ and $b$ . Hence estimate the value of the car after $5$ years, giving your answer to the nearest pound. State one limitation of the model.
(5)
(Total for Question 1 is 5 marks)
2.
A concentration is modelled by $C(t)=\dfrac{a}{t+b}$ for $t\geq0$ , where $a$ and $b$ are positive constants. Given $C(0)=24$ and $C(4)=12$ , find $a$ and $b$ . Hence find the time when $C(t)=8$ .
(5)
(Total for Question 2 is 5 marks)
3.
A water level is modelled by $H(t)=a+b\cos(\pi t/6)$ for $0\leq t\leq12$ , where $H$ is measured in metres and $t$ in hours. The model gives $H(0)=8$ and $H(6)=2$ . Find $a$ and $b$ , then find the first time after $t=0$ when $H(t)=5$ . State the periodicity assumption made by the model.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The cross-section of a greenhouse roof is modelled by $h(x)=kx(30-x)$ for $0\leq x\leq30$ , with distances in metres. The point $(5,10)$ lies on the model. Find $k$ , the maximum modelled height, and the horizontal width over which $h(x)\geq12$ . Give one limitation of the model.
(7)
(Total for Question 1 is 7 marks)
2.
For $t\geq0$ , a storm collection model is $R(t)=\dfrac{at}{t+b}$ , with positive parameters $a$ and $b$ . Given $R(2)=18$ and $R(6)=36$ , find $a$ and $b$ and the limiting amount predicted by the model. A gauge later records $78$ cubic metres from this storm. Decide whether the model can represent the whole storm, and state a specific refinement.
(7)
(Total for Question 2 is 7 marks)
3.
A quantity is modelled at whole-number times $t\geq0$ by $P(t)=L-Ar^t$ , where $A>0$ , $L>0$ and $0<r<1$ . Given $P(0)=20$ , $P(1)=44$ and $P(2)=56$ , find $L$ , $A$ and $r$ . Hence find the first whole-number time for which $P(t)>65$ . A later measurement is lower than the model because the surrounding conditions changed at $t=3$ ; state a specific refinement that addresses this information.
(7)
(Total for Question 3 is 7 marks)
4.
A braking-distance model is $D=kv^n$ , where $D$ is measured in metres, $v$ in metres per second, and $k>0$ . The model gives $D=8$ when $v=10$ and $D=32$ when $v=20$ . Find $k$ and $n$ , then predict $D$ when $v=30$ . A measured braking distance at $30$ metres per second is $78$ metres. State the model error and give a specific refinement.
(6)
(Total for Question 4 is 6 marks)
5.
For $t\geq0$ , a population is modelled by $P(t)=50(2^{t/2})$ for $0\leq t\leq4$ and by $P(t)=A\left(\dfrac32\right)^{t-4}$ for $t>4$ . The model is continuous at $t=4$ . Find $A$ and the first whole-number time for which $P(t)>600$ . State one limitation of using the second branch indefinitely and give a specific refinement.
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

2.1 · Understand and use the laws of indices for all rational exponents.

Tier 1 · Easy

Mark scheme for 2.1 Tier 1 · Easy
Question	Scheme	Marks
1	$27$	2
(2 marks)		2
Notes
$81^{3/4}=(\sqrt[4]{81})^3=3^3=27$ .
2	$x=25$	2
(2 marks)		2
Notes
Since $x>0$ , $x^{3/2}=(\sqrt{x})^3$ . Therefore $\sqrt{x}=5$ , so $x=25$ .

Tier 2 · Standard

Mark scheme for 2.1 Tier 2 · Standard
Question	Scheme	Marks
1	$4x^{11}$	3
(3 marks)		3
Notes
Since $x>0$ , $\left(16x^{12}\right)^{3/4}=16^{3/4}x^9=8x^9$ . Therefore $\dfrac{8x^9}{2x^{-2}}=4x^{9-(-2)}=4x^{11}$ .
2	$x^{-5}=\dfrac1{x^5}$	3
(3 marks)		3
Notes
$(27x^{-9})^{2/3}=27^{2/3}x^{-6}=9x^{-6}$ . Dividing by $9x^{-1}$ gives $x^{-6-(-1)}=x^{-5}=1/x^5$ .
3	$p=3$	3
(3 marks)		3
Notes
The exponent on the left is $4(p-1/2)-3=4p-5$ . Equality for all positive $x$ requires $4p-5=7$ , so $4p=12$ and $p=3$ .

Tier 3 · Hard

Mark scheme for 2.1 Tier 3 · Hard
Question	Scheme	Marks
1	$x=\dfrac{7}{4}$	4
(4 marks)		4
Notes
Write both sides with base $2$ : $8^{2x-1}=2^{3(2x-1)}$ and $4^{x+2}=2^{2(x+2)}$ . Equal positive bases have equal exponents, so $6x-3=2x+4$ . Hence $4x=7$ and $x=7/4$ .
2	$x=-\dfrac{15}{8}$	5
(5 marks)		5
Notes
Write every term with base $3$ . The left-hand side is $3^{3(2x-1)/2}3^{-2x/3}=3^{7x/3-3/2}$ , while the right-hand side is $3^{x-4}$ . Equating exponents gives $7x/3-3/2=x-4$ . Multiplying by $6$ gives $14x-9=6x-24$ , so $8x=-15$ and $x=-15/8$ .
3	$x^{6/5}y^{9/5}$	4
(4 marks)		4
Notes
Inside the outer power, $(x^{-1/2}y^{2/3})^6=x^{-3}y^4$ . Dividing by $x^{-5}y$ gives $x^{-3-(-5)}y^{4-1}=x^2y^3$ . Raising this to the power $3/5$ gives $x^{6/5}y^{9/5}$ .
4	$p=\dfrac52$ and $q=-1$	5
(5 marks)		5
Notes
The exponent of $x$ on the left is $(-2p+1/2)/3$ , so $(-2p+1/2)/3=-3/2$ . Hence $-2p+1/2=-9/2$ , giving $p=5/2$ . The exponent of $y$ is $(-2q+3)/3$ , so $(-2q+3)/3=5/3$ . Hence $-2q+3=5$ , giving $q=-1$ .
5	$x=2^{-20/21}$	4
(4 marks)		4
Notes
Using the index laws, the exponent on the left is $-3-2-1/4=-21/4$ . Thus $x^{-21/4}=32=2^5$ . Raising both sides to the power $-4/21$ gives $x=(2^5)^{-4/21}=2^{-20/21}$ , which is positive as required.

2.2 · Use and manipulate surds, including rationalising the denominator.

Tier 1 · Easy

Mark scheme for 2.2 Tier 1 · Easy
Question	Scheme	Marks
1	$6\sqrt2$	2
(2 marks)		2
Notes
Use the square factor $36$ : $\sqrt{72}=\sqrt{36\times2}=6\sqrt2$ .
2	$5\sqrt5$	2
(2 marks)		2
Notes
$\sqrt{45}=3\sqrt5$ and $\sqrt{20}=2\sqrt5$ . Hence $2\sqrt{45}-\sqrt{20}+\sqrt5=6\sqrt5-2\sqrt5+\sqrt5=5\sqrt5$ .

Tier 2 · Standard

Mark scheme for 2.2 Tier 2 · Standard
Question	Scheme	Marks
1	$8\sqrt3$	4
(4 marks)		4
Notes
Rationalising each fraction gives $\dfrac{(2+\sqrt3)^2}{4-3}-\dfrac{(2-\sqrt3)^2}{4-3}=(7+4\sqrt3)-(7-4\sqrt3)=8\sqrt3$ , as required.
2	$5+2\sqrt7$	3
(3 marks)		3
Notes
Multiply numerator and denominator by $\sqrt7+2$ . The denominator becomes $7-4=3$ , while the numerator is $(4+\sqrt7)(\sqrt7+2)=15+6\sqrt7$ . Dividing by $3$ gives $5+2\sqrt7$ .
3	$a=17$ , $b=9$	4
(4 marks)		4
Notes
Expanding gives $(2a-3b)+(2b-a)\sqrt3=7+\sqrt3$ . Equating rational and irrational parts gives $2a-3b=7$ and $2b-a=1$ . The second equation gives $a=2b-1$ ; substituting into the first gives $b=9$ , and hence $a=17$ .

Tier 3 · Hard

Mark scheme for 2.2 Tier 3 · Hard
Question	Scheme	Marks
1	$x=19$	5
(5 marks)		5
Notes
The domain requires $x\geq3$ . Rearrange to $\sqrt{x+6}=1+\sqrt{x-3}$ and square: $x+6=x-2+2\sqrt{x-3}$ . Thus $8=2\sqrt{x-3}$ , so $\sqrt{x-3}=4$ and $x=19$ . Substitution gives $5-4=1$ , so the solution is valid.
2	$y=3+\sqrt2$ $\dfrac1y=\dfrac{3-\sqrt2}{7}$	5
(5 marks)		5
Notes
If $y=a+b\sqrt2$ , then $y^2=a^2+2b^2+2ab\sqrt2$ . Matching $11+6\sqrt2$ gives $a^2+2b^2=11$ and $2ab=6$ , so $ab=3$ and $(a,b)=(3,1)$ or $(1,3)$ ; only $(3,1)$ satisfies $a^2+2b^2=11$ , since $1+18=19\neq11$ . Thus $a=3$ , $b=1$ , and $(3+\sqrt2)^2=11+6\sqrt2$ , so the positive square root is $y=3+\sqrt2$ . Therefore $1/y=(3-\sqrt2)/(9-2)=(3-\sqrt2)/7$ .
3	$u^{-1}=2-\sqrt3$ $u^3+u^{-3}=52$	5
(5 marks)		5
Notes
Since $(2+\sqrt3)(2-\sqrt3)=1$ , it follows that $u^{-1}=2-\sqrt3$ . Therefore $u+u^{-1}=4$ . Using $a^3+b^3=(a+b)^3-3ab(a+b)$ with $a=u$ , $b=u^{-1}$ and $ab=1$ gives $u^3+u^{-3}=4^3-3(1)(4)=52$ .
4	$\dfrac{2\sqrt3+3\sqrt2-\sqrt{30}}{12}$	5
(5 marks)		5
Notes
First multiply numerator and denominator by $\sqrt2+\sqrt3-\sqrt5$ . The denominator becomes $(\sqrt2+\sqrt3)^2-5=2\sqrt6$ . The expression is therefore $(\sqrt2+\sqrt3-\sqrt5)/(2\sqrt6)$ . Multiplying by $\sqrt6/\sqrt6$ gives $(2\sqrt3+3\sqrt2-\sqrt{30})/12$ .
5	$P(x)=x^4-14x^2+9$	6
(6 marks)		6
Notes
The squared positive roots are $\alpha^2=7+2\sqrt{10}$ and $\beta^2=7-2\sqrt{10}$ . Hence $\alpha^2+\beta^2=14$ and $\alpha^2\beta^2=(7+2\sqrt{10})(7-2\sqrt{10})=9$ . Therefore $(x^2-\alpha^2)(x^2-\beta^2)=x^4-(\alpha^2+\beta^2)x^2+\alpha^2\beta^2=x^4-14x^2+9$ . Its roots are exactly the four stated values.

2.3 · Work with quadratic functions and their graphs; the discriminant, including conditions for real and repeated roots; completing the square; solution of quadratic equations, including solving quadratic equations in a function of the unknown.

Tier 1 · Easy

Mark scheme for 2.3 Tier 1 · Easy
Question	Scheme	Marks
1	$(x-4)^2-11$	2
(2 marks)		2
Notes
Half the coefficient of $x$ is $-4$ . Since $(x-4)^2=x^2-8x+16$ , subtract $11$ to restore the constant $5$ : $x^2-8x+5=(x-4)^2-11$ .
2	No real roots, because the discriminant is $-31<0$ .	2
(2 marks)		2
Notes
The discriminant is $b^2-4ac=3^2-4(2)(5)=9-40=-31$ . Since it is negative, the equation has no real roots.

Tier 2 · Standard

Mark scheme for 2.3 Tier 2 · Standard
Question	Scheme	Marks
1	$k=13$ The point is $(4,5)$ .	4
(4 marks)		4
Notes
At an intersection, $x^2-6x+k=2x-3$ , so $x^2-8x+k+3=0$ . Exactly one intersection requires a repeated root, hence $(-8)^2-4(k+3)=0$ . Therefore $64-4k-12=0$ and $k=13$ . The repeated root is $x=4$ , and the line gives $y=2(4)-3=5$ .
2	$1<k<9$	4
(4 marks)		4
Notes
No real roots requires a negative discriminant. Thus $(k-3)^2-4k<0$ , which simplifies to $k^2-10k+9<0$ , or $(k-1)(k-9)<0$ . This upward-opening quadratic is negative between its roots, so $1<k<9$ .
3	$p=-4$ , $q=1$ The minimum value is $-3$ .	4
(4 marks)		4
Notes
The sum of the roots is $4$ , so $-p=4$ and $p=-4$ . Their product is $(2+\sqrt3)(2-\sqrt3)=1$ , so $q=1$ . Hence $f(x)=x^2-4x+1=(x-2)^2-3$ , whose minimum value is $-3$ .

Tier 3 · Hard

Mark scheme for 2.3 Tier 3 · Hard
Question	Scheme	Marks
1	$x=\dfrac{5\pm\sqrt{53}}{2}$ $x=\dfrac{5\pm\sqrt{17}}{2}$	6
(6 marks)		6
Notes
Let $u=x^2-5x$ . Then $u^2-5u-14=0$ , so $(u-7)(u+2)=0$ . If $u=7$ , then $x^2-5x-7=0$ , giving $x=(5\pm\sqrt{53})/2$ . If $u=-2$ , then $x^2-5x+2=0$ , giving $x=(5\pm\sqrt{17})/2$ . All four values satisfy the original equation.
2	$x=0$ or $x=2$	5
(5 marks)		5
Notes
Let $u=5^x$ , where $u>0$ . Since $25^x=(5^x)^2$ , the equation becomes $u^2-26u+25=0=(u-1)(u-25)$ . Hence $5^x=1$ or $5^x=25$ , giving $x=0$ or $x=2$ .
3	$x=1$ or $x=\dfrac{3+\sqrt5}{2}$ or $x=\dfrac{3-\sqrt5}{2}$	6
(6 marks)		6
Notes
The original expression requires $x\ne0$ . Let $u=x+1/x$ . Then $u^2-5u+6=0$ , so $u=2$ or $u=3$ . If $u=2$ , then $x^2-2x+1=0$ , giving $x=1$ . If $u=3$ , then $x^2-3x+1=0$ , giving $x=(3\pm\sqrt5)/2$ . All three values are non-zero and satisfy the original equation.
4	$4x^2-37x+9=0$	5
(5 marks)		5
Notes
The original roots satisfy $\alpha+\beta=5/2$ and $\alpha\beta=-3/2$ . Hence $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=25/4+3=37/4$ , while $\alpha^2\beta^2=(\alpha\beta)^2=9/4$ . The required equation is $x^2-(37/4)x+9/4=0$ . Multiplying by $4$ gives the pinned integer form $4x^2-37x+9=0$ .
5	$4-\sqrt{13}\leq f(x)\leq4+\sqrt{13}$	6
(6 marks)		6
Notes
Set $y=(x^2+4x+7)/(x^2+1)$ ; the denominator is positive for every real $x$ . Rearranging gives $(y-1)x^2-4x+(y-7)=0$ . For $y\ne1$ , this has a real solution exactly when its discriminant is non-negative: $16-4(y-1)(y-7)\geq0$ . This simplifies to $y^2-8y+3\leq0$ , so $4-\sqrt{13}\leq y\leq4+\sqrt{13}$ . The value $y=1$ is also attained, at $x=-3/2$ , and lies in this interval. At each endpoint the quadratic has a repeated real root, so both endpoints are included.

2.4 · Solve simultaneous equations in two variables by elimination and by substitution, including one linear and one quadratic equation.

Tier 1 · Easy

Mark scheme for 2.4 Tier 1 · Easy
Question	Scheme	Marks
1	$x=5$ , $y=4$	2
(2 marks)		2
Notes
Add the equations to eliminate $y$ : $2x=10$ , so $x=5$ . Substitution into $x+y=9$ gives $y=4$ .
2	$x=3$ , $y=5$	3
(3 marks)		3
Notes
Doubling the second equation gives $4x-2y=2$ . Adding this to $3x+2y=19$ gives $7x=21$ , so $x=3$ . Then $2(3)-y=1$ , giving $y=5$ .

Tier 2 · Standard

Mark scheme for 2.4 Tier 2 · Standard
Question	Scheme	Marks
1	$(2,-1)$ and $(3,1)$	4
(4 marks)		4
Notes
Equating the two expressions for $y$ gives $x^2-3x+1=2x-5$ , so $x^2-5x+6=0$ . Hence $(x-2)(x-3)=0$ , giving $x=2$ or $x=3$ . Substitution into $y=2x-5$ gives $y=-1$ or $y=1$ , so the intersections are $(2,-1)$ and $(3,1)$ .
2	$(x,y)=\left(\dfrac52,6\right)$ or $(-3,-5)$	4
(4 marks)		4
Notes
Substitute $y=2x+1$ into $xy=15$ : $x(2x+1)=15$ , so $2x^2+x-15=0=(2x-5)(x+3)$ . Therefore $x=5/2$ or $x=-3$ . The corresponding values from $y=2x+1$ are $6$ and $-5$ .
3	$84$ adult tickets and $76$ student tickets	4
(4 marks)		4
Notes
Let $a$ and $s$ be the adult and student ticket counts. Then $a+s=160$ and $14a+9s=1860$ . Substituting $s=160-a$ into the income equation gives $14a+9(160-a)=1860$ , so $5a=420$ and $a=84$ . Hence $s=160-84=76$ .

Tier 3 · Hard

Mark scheme for 2.4 Tier 3 · Hard
Question	Scheme	Marks
1	$(x,y)=(3,2)$ or $\left(-\dfrac15,\dfrac{18}{5}\right)$	5
(5 marks)		5
Notes
From the line, $x=7-2y$ . Substitution gives $(7-2y)^2+y^2=13$ , so $5y^2-28y+36=0$ . The discriminant is $64$ , hence $y=(28\pm8)/10$ , giving $y=2$ or $y=18/5$ . Then $x=7-2y$ gives $x=3$ or $x=-1/5$ respectively.
2	$(x,y)=(3,4)$ or $(4,3)$	5
(5 marks)		5
Notes
Use $y=7-x$ in the second equation. This gives $x^2-x(7-x)+(7-x)^2=13$ , so $3x^2-21x+36=0$ . Dividing by $3$ gives $(x-3)(x-4)=0$ . Thus $x=3$ or $x=4$ , and $y=7-x$ gives the ordered pairs $(3,4)$ and $(4,3)$ .
3	$k=\sqrt7$ or $k=-\sqrt7$ ; for each value the system has no solution.	6
(6 marks)		6
Notes
Eliminating $y$ gives a coefficient of $(k+1)(k-1)-6=k^2-7$ on $x$ . A unique solution fails when $k^2-7=0$ , so $k=\pm\sqrt7$ . For infinitely many solutions, the coefficient and constant ratios would also have to agree. Comparing the $x$ -coefficients and constants would require $3/(k+1)=5/7$ , giving $k=16/5$ , which is neither $\sqrt7$ nor $-\sqrt7$ . Therefore the equations are inconsistent for both exceptional values, so each system has no solution.
4	Midpoint $(3,7)$ $AB=4\sqrt{15}$	6
(6 marks)		6
Notes
Equating the two expressions for $y$ gives $x^2-4x-2=2x+1$ , so the intersection inputs are the roots of $x^2-6x-3=0$ . Their sum is $6$ , so the midpoint has $x$ -coordinate $3$ and hence $y$ -coordinate $2(3)+1=7$ . The discriminant is $48$ , so the two $x$ -coordinates differ by $\sqrt{48}=4\sqrt3$ . Since $y=2x+1$ , the corresponding $y$ -coordinates differ by $8\sqrt3$ . Therefore $AB=\sqrt{(4\sqrt3)^2+(8\sqrt3)^2}=4\sqrt{15}$ .
5	$k=10$ gives contact point $(-4,2)$ ; $k=-10$ gives contact point $(4,-2)$ . There are two distinct intersections for $\|k\|<10$ and no intersections for $\|k\|>10$ .	6
(6 marks)		6
Notes
Substitution of $y=2x+k$ into the circle gives $5x^2+4kx+k^2-20=0$ . Its discriminant is $(4k)^2-20(k^2-20)=400-4k^2$ . Tangency requires this to be zero, so $k=\pm10$ . At a repeated root, $x=-4k/10=-2k/5$ . Thus $k=10$ gives $x=-4$ , $y=2$ , while $k=-10$ gives $x=4$ , $y=-2$ . The discriminant is positive when $\|k\|<10$ , giving two distinct intersections, and negative when $\|k\|>10$ , giving no intersections.

2.5 · Solve linear and quadratic inequalities in one variable and interpret them graphically, including with brackets and fractions; express solutions using 'and'/'or' or set notation; represent linear and quadratic inequalities graphically.

Tier 1 · Easy

Mark scheme for 2.5 Tier 1 · Easy
Question	Scheme	Marks
1	$x>-3$	2
(2 marks)		2
Notes
Subtract $5$ to obtain $-2x<6$ . Dividing by the negative number $-2$ reverses the inequality, giving $x>-3$ .
2	$x\leq9$	2
(2 marks)		2
Notes
Expanding gives $4x-12\leq2x+6$ . Hence $2x\leq18$ , so $x\leq9$ .

Tier 2 · Standard

Mark scheme for 2.5 Tier 2 · Standard
Question	Scheme	Marks
1	$-1<x\leq\dfrac87$	4
(4 marks)		4
Notes
From $3(2x-1)\leq5-x$ , $6x-3\leq5-x$ , so $7x\leq8$ and $x\leq8/7$ . From $5-x<4x+10$ , $-5<5x$ , so $x>-1$ . Both conditions must hold, giving $-1<x\leq8/7$ .
2	$-2<x<7$	3
(3 marks)		3
Notes
Factorise to obtain $(x-7)(x+2)<0$ . The upward-opening quadratic is negative between its roots, so the solution is $-2<x<7$ .
3	The closed triangular region bounded by $x=1$ , $y=0$ and $x+2y=7$ , with vertices $(1,0)$ , $(7,0)$ and $(1,3)$ .	4
(4 marks)		4
Notes
All three boundaries are solid because equality is included. The lines $x=1$ and $y=0$ meet at $(1,0)$ . Setting $y=0$ in $x+2y=7$ gives $(7,0)$ , and setting $x=1$ gives $y=3$ , producing $(1,3)$ . Testing a point inside these boundaries identifies the closed triangular region containing, for example, $(2,1)$ .

Tier 3 · Hard

Mark scheme for 2.5 Tier 3 · Hard
Question	Scheme	Marks
1	$\{x\in\mathbb R:3<x<8\}$	4
(4 marks)		4
Notes
Move everything to one side: $\dfrac{x+2}{x-3}-2=\dfrac{8-x}{x-3}>0$ . The critical values are $3$ , where the expression is undefined, and $8$ , where it is zero. A sign check shows the fraction is positive only for $3<x<8$ . Both endpoints are excluded, so the set is $\{x\in\mathbb R:3<x<8\}$ .
2	$x\leq-1$ or $2<x\leq4$	5
(5 marks)		5
Notes
The critical values are $-1$ and $4$ , where the numerator is zero, and $2$ , where the expression is undefined. A sign check on the four intervals gives negative, positive, negative, positive respectively. Include the numerator zeros because equality is allowed, but exclude $x=2$ . Hence $x\leq-1$ or $2<x\leq4$ .
3	$\dfrac32<x\leq4$	5
(5 marks)		5
Notes
The first inequality is $(x-1)(x-4)\leq0$ , so $1\leq x\leq4$ . The second is $(2x-3)(x+2)>0$ , so $x<-2$ or $x>3/2$ . Intersecting these two solution sets gives $3/2<x\leq4$ .
4	$\left[-\dfrac32,3\right)$	5
(5 marks)		5
Notes
The square root requires $x\geq-3/2$ . If $-3/2\leq x<0$ , the left-hand side is non-negative and the right-hand side is negative, so every such $x$ works. If $x\geq0$ , both sides are non-negative and squaring preserves the inequality: $2x+3>x^2$ . Thus $(x-3)(x+1)<0$ , giving $-1<x<3$ ; intersecting with $x\geq0$ gives $0\leq x<3$ . Combining the cases yields $-3/2\leq x<3$ .
5	$-3\leq x\leq-1$ or $1\leq x\leq3$	4
(4 marks)		4
Notes
Let $u=x^2$ , where $u\geq0$ . The inequality becomes $u^2-10u+9\leq0$ , or $(u-1)(u-9)\leq0$ , so $1\leq u\leq9$ . Hence $1\leq x^2\leq9$ , which gives $-3\leq x\leq-1$ or $1\leq x\leq3$ .

2.6 · Manipulate polynomials algebraically, including expanding brackets, collecting like terms, factorisation and simple algebraic division; use the factor theorem; simplify rational expressions by factorising, cancelling and algebraic division.

Tier 1 · Easy

Mark scheme for 2.6 Tier 1 · Easy
Question	Scheme	Marks
1	$(x-4)(x-5)$	2
(2 marks)		2
Notes
Find two numbers with product $20$ and sum $-9$ : they are $-4$ and $-5$ . Therefore $x^2-9x+20=(x-4)(x-5)$ .
2	$2x^3-x^2-11x+12$	2
(2 marks)		2
Notes
Expanding gives $2x^3+2x^2-8x-3x^2-3x+12$ . Collecting like terms produces $2x^3-x^2-11x+12$ .

Tier 2 · Standard

Mark scheme for 2.6 Tier 2 · Standard
Question	Scheme	Marks
1	$a=3$ $p(x)=(x+2)(x-2)(x+3)$	4
(4 marks)		4
Notes
By the factor theorem, $p(-2)=0$ . Hence $-8+4a+8-12=0$ , so $4a-12=0$ and $a=3$ . Then $p(x)=x^3+3x^2-4x-12=x^2(x+3)-4(x+3)=(x+3)(x^2-4)=(x+3)(x-2)(x+2)$ .
2	Quotient $2x^2+x+6$ ; remainder $7$	4
(4 marks)		4
Notes
Algebraic division gives leading terms $2x^2$ , then $x$ , then $6$ . Equivalently, $(x-2)(2x^2+x+6)=2x^3-3x^2+4x-12$ . Therefore $2x^3-3x^2+4x-5=(x-2)(2x^2+x+6)+7$ , so the quotient is $2x^2+x+6$ and the remainder is $7$ .
3	Remainder $8$	3
(3 marks)		3
Notes
For a divisor $2x+1$ , the remainder is $p(-1/2)$ . Therefore $p(-1/2)=4(-1/2)^3-3(-1/2)+7=-1/2+3/2+7=8$ .

Tier 3 · Hard

Mark scheme for 2.6 Tier 3 · Hard
Question	Scheme	Marks
1	$x-\dfrac{3}{x-2}$ $x\ne2$	5
(5 marks)		5
Notes
Factor the numerator as $(x-2)(x-3)(x+1)$ and the denominator as $(x-2)^2$ . Cancelling one factor gives $(x^2-2x-3)/(x-2)$ , while the original expression still requires $x\ne2$ . Division gives $x^2-2x-3=x(x-2)-3$ , so the simplified form is $x-3/(x-2)$ .
2	$a=-2$ , $b=-5$ $p(x)=(x-1)(x+2)(x-3)$	5
(5 marks)		5
Notes
The factor theorem gives $p(1)=0$ , so $a+b=-7$ . Also $p(-2)=0$ , so $-8+4a-2b+6=0$ , or $2a-b=1$ . Solving the two linear equations gives $a=-2$ and $b=-5$ . The polynomial is then $x^3-2x^2-5x+6$ ; division by the known factors, or inspection of the remaining root, gives $p(x)=(x-1)(x+2)(x-3)$ .
3	$a=-5$ , $b=-11$ $F(x)=2x+1$ for $x\ne-1,4$	6
(6 marks)		6
Notes
The quotient must have leading term $2x$ . Also $F(0)=(-4)/(-4)=1$ , so the linear quotient is $2x+1$ . Hence the numerator must equal $(2x+1)(x-4)(x+1)=(2x+1)(x^2-3x-4)=2x^3-5x^2-11x-4$ . Comparing coefficients gives $a=-5$ and $b=-11$ . The original excluded values $x=-1$ and $x=4$ remain excluded.
4	$a=-6$ , $b=36$ Quotient $2x^2+2x-4$ ; remainder $32$	6
(6 marks)		6
Notes
Equal remainders give $p(2)=p(-1)$ , so $16+2a+b=-2-a+b$ and hence $a=-6$ . Since $x+3$ is a factor, $p(-3)=0$ . Thus $-54-3a+b=0$ ; using $a=-6$ gives $b=36$ . Therefore $p(x)=2x^3-6x+36$ . Algebraic division by $x-1$ gives $p(x)=(x-1)(2x^2+2x-4)+32$ , so the quotient is $2x^2+2x-4$ and the remainder is $32$ .
5	$r=3$ $p(x)=(x+2)(x-1)(x-3)^2$ The constant term is $-18$ .	5
(5 marks)		5
Notes
The factorised polynomial is $(x+2)(x-1)(x-r)^2=(x^2+x-2)(x^2-2rx+r^2)$ . In the product, the $x^3$ terms are $-2rx^3$ from $x^2(-2rx)$ and $x^3$ from $x(x^2)$ , so the $x^3$ coefficient is $1-2r$ . Hence $1-2r=-5$ , giving $r=3$ . Therefore $p(x)=(x+2)(x-1)(x-3)^2$ . Setting $x=0$ gives the constant term $2(-1)(9)=-18$ .

2.7 · Understand and use graphs of functions; sketch curves including polynomials, the modulus of a linear function, y = a/x and y = a/x² with asymptotes; use graph intersections to solve equations; understand proportional relationships.

Tier 1 · Easy

Mark scheme for 2.7 Tier 1 · Easy
Question	Scheme	Marks
1	Asymptotes $x=0$ and $y=0$ ; branches in quadrants II and IV.	2
(2 marks)		2
Notes
The reciprocal form has the coordinate axes as asymptotes. Since $-6/x$ is negative for $x>0$ and positive for $x<0$ , its branches lie in quadrants IV and II respectively.
2	$y=-16$	2
(2 marks)		2
Notes
Write $y=kx^3$ . Using $(x,y)=(3,54)$ gives $54=27k$ , so $k=2$ . Therefore, when $x=-2$ , $y=2(-2)^3=-16$ .

Tier 2 · Standard

Mark scheme for 2.7 Tier 2 · Standard
Question	Scheme	Marks
1	Vertex $(-2,-2)$ $x$ -intercepts $\left(-\dfrac83,0\right)$ and $\left(-\dfrac43,0\right)$ $y$ -intercept $(0,4)$	4
(4 marks)		4
Notes
The vertex occurs where $3x+6=0$ , so it is $(-2,-2)$ . For the $x$ -intercepts, $\|3x+6\|=2$ , giving $3x+6=2$ or $3x+6=-2$ . Hence $x=-4/3$ or $x=-8/3$ . When $x=0$ , $y=\|6\|-2=4$ . The graph consists of two straight-line rays meeting at the vertex.
2	The graph touches at $(-2,0)$ , crosses at $(1,0)$ and $(4,0)$ , has $y$ -intercept $(0,16)$ , and tends to $+\infty$ at both ends.	4
(4 marks)		4
Notes
The roots are $-2$ , with even multiplicity, and $1$ and $4$ , each with odd multiplicity. Hence the graph touches the axis at $(-2,0)$ and crosses at $(1,0)$ and $(4,0)$ . At $x=0$ , $y=2^2(-1)(-4)=16$ . The leading term is $x^4$ , so $y\to+\infty$ as $x\to\pm\infty$ . These features pin one configuration: it remains above the axis through the touch at $-2$ , is negative only between $1$ and $4$ , and rises at both ends.
3	Asymptotes $x=0$ and $y=-3$ ; $x$ -intercepts $(-2,0)$ and $(2,0)$ ; no $y$ -intercept; range $y>-3$ .	4
(4 marks)		4
Notes
The denominator is zero at $x=0$ , giving the vertical asymptote $x=0$ . As $\|x\|\to\infty$ , $12/x^2\to0$ from above, so the horizontal asymptote is $y=-3$ and the range is $y>-3$ . Setting $y=0$ gives $12/x^2=3$ , hence $x=\pm2$ . The curve is undefined at $x=0$ , so there is no $y$ -intercept, and its two branches are symmetric about the $y$ -axis.

Tier 3 · Hard

Mark scheme for 2.7 Tier 3 · Hard
Question	Scheme	Marks
1	$(1,4)$ and $(-4,-1)$	4
(4 marks)		4
Notes
At an intersection, $4/x=x+3$ . Since $x\ne0$ , multiply by $x$ to get $x^2+3x-4=0$ . Factorising gives $(x-1)(x+4)=0$ , so $x=1$ or $x=-4$ . Substitution into $y=x+3$ gives $y=4$ or $y=-1$ .
2	$k=36$ The only intersection is $(3,4)$ .	5
(5 marks)		5
Notes
Using $(2,9)$ gives $9=k/4$ , so $k=36$ . At an intersection, $36/x^2=x+1$ , with $x\ne0$ , so $x^3+x^2-36=0$ . Since $x=3$ is a root, this factorises as $(x-3)(x^2+4x+12)=0$ . The quadratic factor has discriminant $16-48=-32$ , so it has no real roots. Thus $x=3$ is the only real intersection value and $y=3+1=4$ .
3	$x=\dfrac{1+\sqrt{17}}{2}$	5
(5 marks)		5
Notes
The right-hand side is negative for $x<0$ , whereas the modulus is non-negative, and $x=0$ is undefined. For $0<x<1$ , the equation is $1-x=4/x$ , giving $x^2-x+4=0$ , whose discriminant is $-15$ , so there is no real root on this branch. For $x\geq1$ , the equation is $x-1=4/x$ , so $x^2-x-4=0$ . Its roots are $(1\pm\sqrt{17})/2$ , and only $(1+\sqrt{17})/2$ lies in the required branch.
4	$a=-4$ Asymptotes $x=2$ and $y=3$ $x$ -intercept $\left(\dfrac{10}{3},0\right)$ and $y$ -intercept $(0,5)$ For $x>2$ , $y<3$ ; for $x<2$ , $y>3$ .	6
(6 marks)		6
Notes
Using $(4,1)$ gives $1=a/(4-2)+3$ , so $a=-4$ . The translated reciprocal form has asymptotes $x=2$ and $y=3$ . For the $x$ -intercept, solve $-4/(x-2)+3=0$ , giving $3(x-2)=4$ and $x=10/3$ . At $x=0$ , $y=-4/(-2)+3=5$ . When $x>2$ , the fraction $-4/(x-2)$ is negative, so $y<3$ ; when $x<2$ , it is positive, so $y>3$ .
5	$f(x)=-2(x+1)^2(x-3)$ As $x\to-\infty$ , $f(x)\to+\infty$ ; as $x\to+\infty$ , $f(x)\to-\infty$ . $x\leq3$	6
(6 marks)		6
Notes
Touching at $-1$ requires an even multiplicity, and the least possible is $2$ ; crossing at $3$ requires odd multiplicity, with least possible $1$ . Hence $f(x)=a(x+1)^2(x-3)$ . Since $f(0)=6$ , $-3a=6$ and $a=-2$ . The leading term is $-2x^3$ , giving the stated end behaviour. The squared factor is non-negative and does not change sign at $-1$ ; the sign is therefore controlled by $-2(x-3)$ . Thus $f(x)\geq0$ exactly when $x\leq3$ .

2.8 · Understand and use composite functions; inverse functions and their graphs.

Tier 1 · Easy

Mark scheme for 2.8 Tier 1 · Easy
Question	Scheme	Marks
1	$17$	2
(2 marks)		2
Notes
Apply $g$ first: $g(2)=2^2+2=6$ . Then $f(6)=3(6)-1=17$ .
2	$f^{-1}(x)=\dfrac{x-4}{5}$	2
(2 marks)		2
Notes
Write $y=5x+4$ and rearrange to $x=(y-4)/5$ . Exchanging the variable labels gives $f^{-1}(x)=(x-4)/5$ .

Tier 2 · Standard

Mark scheme for 2.8 Tier 2 · Standard
Question	Scheme	Marks
1	$fg(x)=\dfrac2{x+1}-3$ Domain: $x\ne-1$ $x=-\dfrac34$	4
(4 marks)		4
Notes
Applying $g$ first gives $fg(x)=f(g(x))=2\left(1/(x+1)\right)-3=2/(x+1)-3$ , with $x\ne-1$ . Solving $2/(x+1)-3=5$ gives $2/(x+1)=8$ , so $x+1=1/4$ and $x=-3/4$ .
2	$f^{-1}g(7)=\dfrac12$ $gf^{-1}(x)=\dfrac12\sqrt{x+7}$ with domain $x\geq-7$	4
(4 marks)		4
Notes
Since $f^{-1}(x)=(x-1)/4$ and $g(7)=3$ , $f^{-1}g(7)=f^{-1}(3)=1/2$ . Also $gf^{-1}(x)=g((x-1)/4)=\sqrt{(x-1)/4+2}=\frac12\sqrt{x+7}$ . The square-root input requires $x+7\geq0$ , so the domain is $x\geq-7$ .
3	$a=\dfrac13$ , $b=\dfrac43$ $gf(x)=x$	4
(4 marks)		4
Notes
Since $fg(x)=3(ax+b)-4=3ax+3b-4$ , comparison with $x$ gives $3a=1$ and $3b-4=0$ . Thus $a=1/3$ and $b=4/3$ . Then $gf(x)=((3x-4)/3)+4/3=x$ , as required.

Tier 3 · Hard

Mark scheme for 2.8 Tier 3 · Hard
Question	Scheme	Marks
1	$f^{-1}(x)=3+\sqrt{x-1}$ with domain $x\geq1$ $x=0$ or $x=2$	6
(6 marks)		6
Notes
From $y=(x-3)^2+1$ and $x\geq3$ , take the positive root: $x=3+\sqrt{y-1}$ . Thus $f^{-1}(x)=3+\sqrt{x-1}$ , whose domain is $x\geq1$ . The equation becomes $3+\sqrt{2x}=x+3$ , so $\sqrt{2x}=x$ with $x\geq0$ . Squaring gives $x^2=2x$ , hence $x=0$ or $x=2$ ; both satisfy the unsquared equation.
2	$f(x)=6x-5$ $x=1$	5
(5 marks)		5
Notes
Because $fg(x)=f(x^2)=6x^2-5$ , the linear function is $f(u)=6u-5$ , so $f(x)=6x-5$ . Then $gf(x)=g(6x-5)=(6x-5)^2$ . Equating the composites gives $6x^2-5=(6x-5)^2$ , so $30x^2-60x+30=0$ , or $30(x-1)^2=0$ . Hence $x=1$ .
3	$f^{-1}(x)=\dfrac{x+\sqrt{x^2-16}}2$ with domain $x\geq4$ $f^{-1}\left(\dfrac{13}{3}\right)=3$	6
(6 marks)		6
Notes
For $x\geq2$ , $f(x)-4=(x-2)^2/x\geq0$ , so the range of $f$ , and hence the domain of its inverse, is $[4,\infty)$ . Write $y=x+4/x$ and rearrange to $x^2-yx+4=0$ . Thus $x=(y\pm\sqrt{y^2-16})/2$ . The two roots of this quadratic multiply to $4$ , so they are $x$ and $4/x$ ; since $x\geq2$ gives $x\geq4/x$ , the required value is the larger root, so $f^{-1}(x)=(x+\sqrt{x^2-16})/2$ . At $x=13/3$ , the square root is $5/3$ , giving $f^{-1}(13/3)=((13/3)+(5/3))/2=3$ .
4	The unique point of intersection is $(-1,-1)$ .	6
(6 marks)		6
Notes
For $v>u$ , $f(v)-f(u)=(v-u)(v^2+uv+u^2+1)>0$ , because $v^2+uv+u^2=(v+u/2)^2+3u^2/4\geq0$ . Thus $f$ is strictly increasing; its cubic end behaviour also gives range $\mathbb R$ , so it has an inverse on $\mathbb R$ . If $(a,b)$ lies on both graphs, then $f(a)=b$ and $f(b)=a$ . If $a<b$ , strict increase would give $b=f(a)<f(b)=a$ , a contradiction; the case $a>b$ is similar. Hence $a=b$ . An intersection must therefore satisfy $f(a)=a$ , so $a^3+a+1=a$ , giving $a^3=-1$ and $a=-1$ . Substitution gives the unique point $(-1,-1)$ .
5	$fg(x)=\sqrt{\dfrac{5-x}{x-2}}$ with domain $2<x\leq5$ $gf(x)=\dfrac3{\sqrt{x-1}-2}$ with domain $x\geq1$ , $x\ne5$ $x=\dfrac72$	6
(6 marks)		6
Notes
Applying $g$ first gives $fg(x)=\sqrt{3/(x-2)-1}=\sqrt{(5-x)/(x-2)}$ . Its radicand is non-negative and $x\ne2$ , giving $2<x\leq5$ . Applying $f$ first gives $gf(x)=3/(\sqrt{x-1}-2)$ . This requires $x\geq1$ and excludes $\sqrt{x-1}=2$ , so $x\ne5$ . Finally, $fg(x)=1$ gives $(5-x)/(x-2)=1$ , hence $5-x=x-2$ and $x=7/2$ , which lies in the domain.

2.9 · Understand the effect of simple transformations on the graph of y = f(x), including sketching associated graphs: y = af(x), y = f(x) + a, y = f(x + a), y = f(ax) and combinations of these transformations.

Tier 1 · Easy

Mark scheme for 2.9 Tier 1 · Easy
Question	Scheme	Marks
1	Translation by vector $\begin{pmatrix}4\\2\end{pmatrix}$ .	2
(2 marks)		2
Notes
Replacing $x$ by $x-4$ moves the graph $4$ units right, and adding $2$ moves it $2$ units up. Together this is translation by $\begin{pmatrix}4\\2\end{pmatrix}$ .
2	A reflection in the $y$ -axis and a stretch parallel to the $x$ -axis with scale factor $\dfrac13$ .	2
(2 marks)		2
Notes
The negative sign inside $f$ reflects the graph in the $y$ -axis. The factor $3$ multiplying $x$ produces a horizontal scale factor of $1/3$ .

Tier 2 · Standard

Mark scheme for 2.9 Tier 2 · Standard
Question	Scheme	Marks
1	Translation by vector $\begin{pmatrix}-1\\-3\end{pmatrix}$ , or equivalently a stretch parallel to the $y$ -axis of scale factor $2$ followed by a translation $\begin{pmatrix}0\\-3\end{pmatrix}$ Horizontal asymptote $y=-3$ $y$ -intercept $(0,-1)$	4
(4 marks)		4
Notes
Replacing $x$ by $x+1$ translates the graph one unit left, and subtracting $3$ translates it three units down, giving the vector $(-1,-3)$ . A second description earns the same marks: because $2^{x+1}=2\times2^{x}$ , the same curve is a vertical stretch of scale factor $2$ followed by a translation three units down. Either is correct — do not discard your answer if it takes the stretch route. The asymptote $y=0$ moves to $y=-3$ . At $x=0$ , $y=2^1-3=-1$ , so the $y$ -intercept is $(0,-1)$ .
2	$y=-3(x+2)^3+1$ The image of $(0,0)$ is $(-2,1)$ .	4
(4 marks)		4
Notes
The horizontal translation replaces $x$ by $x+2$ . The vertical stretch and reflection multiply the output by $-3$ , and the final translation adds $1$ , giving $y=-3(x+2)^3+1$ . Tracking $(0,0)$ through the same changes gives $(-2,0)$ after the horizontal translation; the vertical stretch and reflection leave its zero output unchanged, and the final translation gives $(-2,1)$ .
3	Domain $0\leq x\leq\dfrac52$ ; range $-7\leq g(x)\leq7$ $(1,-2)$ maps to $(1,7)$ and $(3,5)$ maps to $(2,-7)$ .	5
(5 marks)		5
Notes
Write the original input as $u=2x-1$ , so $x=(u+1)/2$ . Mapping the original domain $-1\leq u\leq4$ gives $0\leq x\leq5/2$ . The output mapping is $v\mapsto3-2v$ , so $-2\leq v\leq5$ maps to $-7\leq g(x)\leq7$ . The point $(1,-2)$ maps to $((1+1)/2,3-2(-2))=(1,7)$ , while $(3,5)$ maps to $((3+1)/2,3-2(5))=(2,-7)$ .

Tier 3 · Hard

Mark scheme for 2.9 Tier 3 · Hard
Question	Scheme	Marks
1	Turning point $(0,7)$ , a maximum $x$ -intercepts $\left(-\dfrac{\sqrt7}{2},0\right)$ and $\left(\dfrac{\sqrt7}{2},0\right)$	5
(5 marks)		5
Notes
First write $f(u)=(u-2)^2-3$ . With $u=2x+2$ , $f(2x+2)=(2x)^2-3=4x^2-3$ , so $g(x)=-(4x^2-3)+4=7-4x^2$ . Its turning point is $(0,7)$ and the negative $x^2$ coefficient makes it a maximum. Setting $7-4x^2=0$ gives $x=\pm\sqrt7/2$ .
2	$(u,v)\mapsto\left(\dfrac{u}{3}+2,-2v+4\right)$ Asymptotes $x=2$ and $y=4$ $x$ -intercept $\left(\dfrac{13}{6},0\right)$	6
(6 marks)		6
Notes
The input relation $u=3x-6=3(x-2)$ gives $x=u/3+2$ , while the output becomes $-2v+4$ . Hence $(u,v)$ maps to $(u/3+2,-2v+4)$ . The asymptotes $u=0$ and $v=0$ therefore map to $x=2$ and $y=4$ . Also $g(x)=-2/(3x-6)+4$ . Setting this equal to zero gives $3x-6=1/2$ , so $x=13/6$ .
3	$a=-2$ , $b=-2$ , $c=4$ , $d=3$ $g(x)=-2f(-2x+4)+3$	6
(6 marks)		6
Notes
For a transformed point with new $x$ -coordinate $X$ , the original input is $bX+c$ . The two point correspondences therefore give $3b+c=-2$ and $(3/2)b+c=1$ . Subtracting gives $b=-2$ , and then $c=4$ . The output relation is $Y=av+d$ , so $4a+d=-5$ and $-a+d=5$ . These give $a=-2$ and $d=3$ . Hence $g(x)=-2f(-2x+4)+3$ .
4	The domain of $g$ is $-2\leq x\leq\dfrac43$ . $-\dfrac53\leq x\leq-1$ or $\dfrac23\leq x\leq\dfrac43$	5
(5 marks)		5
Notes
The input $u=3x+2$ must satisfy $-4\leq u\leq6$ , giving $-2\leq x\leq4/3$ . Also, $5-2f(u)\leq1$ is equivalent to $f(u)\geq2$ . Mapping $-3\leq u\leq-1$ through $x=(u-2)/3$ gives $-5/3\leq x\leq-1$ . Mapping $4\leq u\leq6$ gives $2/3\leq x\leq4/3$ . These intervals already lie in the domain of $g$ .
5	Translation then stretch: $y=f\left(\dfrac{x-6}{3}\right)$ and $(-1,4)\mapsto(3,4)$ . Stretch then translation: $y=f\left(\dfrac{x-2}{3}\right)$ and $(-1,4)\mapsto(-1,4)$ . The horizontal transformations do not commute.	5
(5 marks)		5
Notes
For translation then stretch, an original input coordinate $u$ maps first to $u+2$ and then to $3(u+2)=3u+6$ . Thus $u=(x-6)/3$ , giving $y=f((x-6)/3)$ ; $u=-1$ maps to $x=3$ . In the reverse order, $u$ maps to $3u$ and then to $3u+2$ , so $u=(x-2)/3$ , giving $y=f((x-2)/3)$ ; $u=-1$ maps to $x=-1$ . The different coordinate maps show that changing the order changes the graph.

2.10 · Decompose rational functions into partial fractions (denominators not more complicated than squared linear terms and with no more than 3 terms, numerators constant or linear).

Tier 1 · Easy

Mark scheme for 2.10 Tier 1 · Easy
Question	Scheme	Marks
1	$\dfrac7{x+1}-\dfrac7{x+2}$	3
(3 marks)		3
Notes
Write $7/((x+1)(x+2))=A/(x+1)+B/(x+2)$ . Then $7=A(x+2)+B(x+1)$ . Setting $x=-1$ gives $A=7$ , while $x=-2$ gives $B=-7$ .
2	$B=20$	2
(2 marks)		2
Notes
Multiplying by $(x-3)^2$ gives $7x-1=A(x-3)+B$ . Substituting $x=3$ gives $B=21-1=20$ .

Tier 2 · Standard

Mark scheme for 2.10 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac3{x+1}+\dfrac2{(x+1)^2}$	3
(3 marks)		3
Notes
Multiplying by $(x+1)^2$ gives $3x+5=A(x+1)+B=Ax+(A+B)$ . Comparing coefficients gives $A=3$ and $A+B=5$ , so $B=2$ . Therefore the decomposition is $3/(x+1)+2/(x+1)^2$ .
2	$\dfrac1x+\dfrac2{x-1}-\dfrac1{x+2}$	4
(4 marks)		4
Notes
Set the expression equal to $A/x+B/(x-1)+C/(x+2)$ . Multiplying through gives $2x^2+6x-2=A(x-1)(x+2)+Bx(x+2)+Cx(x-1)$ . Substituting $x=0,1,-2$ gives $A=1$ , $B=2$ and $C=-1$ respectively.
3	$\dfrac2{2x-1}+\dfrac3{x+3}$	4
(4 marks)		4
Notes
Write the expression as $A/(2x-1)+B/(x+3)$ . Multiplying through gives $8x+3=A(x+3)+B(2x-1)$ . Setting $x=1/2$ gives $7=7A/2$ , so $A=2$ . Setting $x=-3$ gives $-21=-7B$ , so $B=3$ .

Tier 3 · Hard

Mark scheme for 2.10 Tier 3 · Hard
Question	Scheme	Marks
1	$\dfrac1{x-1}+\dfrac2{(x-1)^2}-\dfrac1{x+2}$	5
(5 marks)		5
Notes
Set the expression equal to $A/(x-1)+B/(x-1)^2+C/(x+2)$ . Multiplying through gives $5x+1=A(x-1)(x+2)+B(x+2)+C(x-1)^2$ . With $x=1$ , $6=3B$ , so $B=2$ ; with $x=-2$ , $-9=9C$ , so $C=-1$ . Comparing the $x^2$ coefficients gives $A+C=0$ , hence $A=1$ .
2	$2+\dfrac3x-\dfrac1{x-1}+\dfrac4{(x-1)^2}$	6
(6 marks)		6
Notes
Since $x(x-1)^2=x^3-2x^2+x$ , division gives quotient $2$ and remainder $2x^2-x+3$ . Write the proper fraction as $A/x+B/(x-1)+C/(x-1)^2$ . Then $2x^2-x+3=A(x-1)^2+Bx(x-1)+Cx$ . Setting $x=0$ gives $A=3$ , setting $x=1$ gives $C=4$ , and comparing $x^2$ coefficients gives $A+B=2$ , so $B=-1$ . Combining with the quotient gives the stated decomposition.
3	$k=-8$ $A=-\dfrac12$ , $B=2$ , $C=-\dfrac12$	6
(6 marks)		6
Notes
Multiplying through and substituting $x=1$ gives $A=-(k+11)/6$ . Substituting $x=3$ gives $C=(3k+19)/10$ . Since $A=C$ , $-5(k+11)=3(3k+19)$ , so $-14k=112$ and $k=-8$ . Then $A=C=-1/2$ . Substituting $x=-2$ gives $30=15B$ , so $B=2$ .
4	$\dfrac3{2x-1}-\dfrac2{(2x-1)^2}+\dfrac1{x+1}$	5
(5 marks)		5
Notes
Write the expression as $A/(2x-1)+B/(2x-1)^2+C/(x+1)$ . Multiplying through gives $10x^2-3x-4=A(2x-1)(x+1)+B(x+1)+C(2x-1)^2$ . Setting $x=1/2$ gives $-3=(3/2)B$ , so $B=-2$ . Setting $x=-1$ gives $9=9C$ , so $C=1$ . Comparing $x^2$ coefficients gives $10=2A+4C$ , hence $A=3$ .
5	$k=11$ , $A=2$ and $C=3$	5
(5 marks)		5
Notes
Multiplying through gives $2x^2+kx+5=A(x+2)^2+B(x-1)(x+2)+C(x-1)$ . Since $B=0$ , comparison with $A(x+2)^2+C(x-1)$ gives $A=2$ from the $x^2$ coefficient. The constant term then gives $5=4A-C$ , so $C=3$ . Comparing the $x$ coefficients gives $k=4A+C=11$ . Thus the decomposition is $2/(x-1)+3/(x+2)^2$ .

2.11 · Use of functions in modelling, including consideration of limitations and refinements of the models.

Tier 1 · Easy

Mark scheme for 2.11 Tier 1 · Easy
Question	Scheme	Marks
1	$151$ organisms	2
(2 marks)		2
Notes
Substitute $t=3$ : $P(3)=120(1.08)^3=151.16544$ . Since the output counts organisms, round to the nearest whole number to obtain $151$ .
2	$C(6)=54$ $t=14$ is outside the stated domain of the model.	2
(2 marks)		2
Notes
Substitution gives $C(6)=96-7(6)=54$ . The model is stated only for $0\leq t\leq10$ , so using $t=14$ would be an unsupported extrapolation.

Tier 2 · Standard

Mark scheme for 2.11 Tier 2 · Standard
Question	Scheme	Marks
1	$A=18\,000$ and $b=0.9$ £ $10\,629$ For example, the model does not allow for sudden changes in value caused by damage or repairs.	5
(5 marks)		5
Notes
At $t=0$ , $A=18\,000$ . Using $t=2$ gives $14\,580=18\,000b^2$ , so $b^2=0.81$ and, since $b>0$ , $b=0.9$ . Then $V(5)=18\,000(0.9)^5=10\,628.82$ , giving £ $10\,629$ to the nearest pound. A limitation is that a constant annual multiplier cannot represent unexpected damage, repairs or changes in the second-hand market.
2	$a=96$ , $b=4$ $t=8$	5
(5 marks)		5
Notes
From $C(0)=24$ , $a/b=24$ , so $a=24b$ . From $C(4)=12$ , $a=12(b+4)$ . Equating these gives $24b=12b+48$ , hence $b=4$ and $a=96$ . Finally, $96/(t+4)=8$ gives $t+4=12$ , so $t=8$ .
3	$a=5$ , $b=3$ $t=3$ hours The water-level pattern is assumed to repeat exactly every $12$ hours.	5
(5 marks)		5
Notes
Since $\cos0=1$ and $\cos\pi=-1$ , the two observations give $a+b=8$ and $a-b=2$ . Hence $a=5$ and $b=3$ . Solving $5+3\cos(\pi t/6)=5$ gives $\cos(\pi t/6)=0$ ; the first positive solution is $\pi t/6=\pi/2$ , so $t=3$ . The cosine model assumes an exactly repeating $12$ -hour cycle.

Tier 3 · Hard

Mark scheme for 2.11 Tier 3 · Hard
Question	Scheme	Marks
1	$k=0.08$ Maximum height $18$ m Width $10\sqrt3$ m For example, a real roof cross-section may not be exactly parabolic.	7
(7 marks)		7
Notes
Using $(5,10)$ gives $10=k(5)(25)$ , so $k=0.08$ . The quadratic is symmetric about $x=15$ , where $h(15)=0.08(15)(15)=18$ . For $h(x)\geq12$ , solve $0.08x(30-x)\geq12$ , equivalent to $x^2-30x+150\leq0$ . The roots are $15\pm5\sqrt3$ , so the allowed interval has width $(15+5\sqrt3)-(15-5\sqrt3)=10\sqrt3$ metres. The exact-parabola assumption is a limitation because manufactured or loaded roofs may have a different profile.
2	$a=72$ , $b=6$ The limiting amount is $72$ cubic metres. No; the model never reaches $78$ cubic metres. Refit the model with a limiting parameter greater than $78$ cubic metres using the later data.	7
(7 marks)		7
Notes
Using $R(2)=18$ gives $18=2a/(2+b)$ , so $a=18+9b$ . Using $R(6)=36$ gives $36=6a/(6+b)$ , so $a=36+6b$ . Equating these expressions gives $b=6$ and then $a=72$ . As $t\to\infty$ , $at/(t+b)\to a$ , so the predicted limiting amount is $72$ cubic metres. For every finite $t\geq0$ , the model gives $R(t)<72$ , so it cannot represent a recorded total of $78$ cubic metres. A specific refinement is to refit the asymptote parameter using the later observation, making the limiting value exceed $78$ cubic metres.
3	$L=68$ , $A=48$ , $r=\dfrac12$ The first whole-number time is $t=5$ . Use a piecewise model with new parameter values from $t=3$ onwards.	7
(7 marks)		7
Notes
The first two increases are $44-20=24$ and $56-44=12$ . For this model successive increases have ratio $r$ , so $r=12/24=1/2$ . Also $P(1)-P(0)=A(1-r)=24$ , giving $A=48$ . Then $P(0)=L-A=20$ gives $L=68$ . The inequality $68-48(1/2)^t>65$ is equivalent to $(1/2)^t<1/16$ , so $t>4$ and the first whole-number time is $5$ . A piecewise model, refitted with different parameters for $t\geq3$ , directly represents the changed conditions.
4	$n=2$ and $k=0.08$ The model predicts $72$ metres. It underestimates by $6$ metres; for example, fit a piecewise power model using additional high-speed data.	6
(6 marks)		6
Notes
Dividing the two model equations gives $32/8=(20/10)^n$ , so $4=2^n$ and $n=2$ . Then $8=100k$ , giving $k=0.08$ . At $v=30$ , the model gives $D=0.08(30)^2=72$ metres. Compared with the measured $78$ metres, this is an underestimate of $6$ metres. A specific refinement is to collect further high-speed observations and fit a second power law above a stated speed threshold.
5	$A=200$ The first whole-number time is $t=7$ . A constant $50\%$ increase per time interval cannot remain realistic indefinitely; for example, replace the second branch by a model with a finite limiting population.	7
(7 marks)		7
Notes
At $t=4$ , the first branch gives $P(4)=50(2^2)=200$ . Continuity therefore requires $A=200$ . The first branch never exceeds $200$ . On the second branch, $P(6)=200(3/2)^2=450$ , while $P(7)=200(3/2)^3=675$ . Since the branch is increasing, the first whole-number time with $P(t)>600$ is $t=7$ . Indefinite fixed-percentage growth ignores limited resources, so a model with a carrying-capacity parameter would be a specific refinement.