2 Algebra and functions — printable coverage pack

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

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.

Tier 1 · Easy

1. Evaluate $81^{3/4}$ without using a calculator.[2 marks]
Answer
- $27$
Method: $81^{3/4}=(\sqrt[4]{81})^3=3^3=27$ .

Tier 2 · Standard

1. Given $x>0$ , simplify $\dfrac{x^{5/2}}{x^{-1/2}}$ to a single power of $x$ .[2 marks]
Answer
- $x^3$
Method: For division with a common base, subtract the exponents: $x^{5/2}/x^{-1/2}=x^{5/2-(-1/2)}=x^{6/2}=x^3$ .

Tier 3 · Hard

1. Solve $8^{2x-1}=4^{x+2}$ , giving the exact value of $x$ .[4 marks]
Answer
- $x=\dfrac{7}{4}$
Method: 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 $x=7/4$ .

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

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.

Tier 1 · Easy

1. Write $\sqrt{72}$ in the form $a\sqrt2$ , where $a$ is an integer.[2 marks]
Answer
- $6\sqrt2$
Method: Use the square factor $36$ : $\sqrt{72}=\sqrt{36\times2}=6\sqrt2$ .

Tier 2 · Standard

1. Rationalise and simplify $\dfrac{5}{2+\sqrt3}$ .[3 marks]
Answer
- $10-5\sqrt3$
Method: Multiply by the conjugate: $\dfrac{5}{2+\sqrt3}\times\dfrac{2-\sqrt3}{2-\sqrt3}=\dfrac{5(2-\sqrt3)}{4-3}=10-5\sqrt3$ .

Tier 3 · Hard

1. Solve $\sqrt{x+6}-\sqrt{x-3}=1$ .[5 marks]
Answer
- $x=19$
Method: 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.

2.3 · 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.

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.

Tier 1 · Easy

1. Write $x^2-8x+5$ in completed-square form.[2 marks]
Answer
- $(x-4)^2-11$
Method: Half the coefficient of $x$ is $-4$ . Since $(x-4)^2=x^2-8x+16$ , subtract $11$ to restore the constant $5$ : $x^2-8x+5=(x-4)^2-11$ .

Tier 2 · Standard

1. Find the two values of $k$ for which $2x^2+(k-1)x+8=0$ has a repeated root.[3 marks]
Answer
- $k=9$ or $k=-7$
Method: A repeated root requires discriminant zero: $(k-1)^2-4(2)(8)=0$ . Hence $(k-1)^2=64$ , so $k-1=\pm8$ and therefore $k=9$ or $k=-7$ .

Tier 3 · Hard

1. Solve $(x^2-5x)^2-5(x^2-5x)-14=0$ , giving exact answers.[6 marks]
Answer
- $x=\dfrac{5\pm\sqrt{53}}{2}$
- $x=\dfrac{5\pm\sqrt{17}}{2}$
Method: 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.

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

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.

Tier 1 · Easy

1. Solve $x+y=9$ and $x-y=1$ simultaneously.[2 marks]
Answer
- $x=5$ , $y=4$
Method: Add the equations to eliminate $y$ : $2x=10$ , so $x=5$ . Substitution into $x+y=9$ gives $y=4$ .

Tier 2 · Standard

1. Find every solution of $y=x+1$ and $x^2+y^2=25$ .[4 marks]
Answer
- $(x,y)=(3,4)$ or $(-4,-3)$
Method: Substitute $y=x+1$ : $x^2+(x+1)^2=25$ , so $2x^2+2x-24=0$ . Dividing by $2$ gives $(x+4)(x-3)=0$ , hence $x=3$ or $x=-4$ . Using $y=x+1$ gives the pairs $(3,4)$ and $(-4,-3)$ .

Tier 3 · Hard

1. Solve the system $x+2y=7$ and $x^2+y^2=13$ .[5 marks]
Answer
- $(x,y)=(3,2)$ or $\left(-\dfrac15,\dfrac{18}{5}\right)$
Method: 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.

2.5 · 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.

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.

Tier 1 · Easy

1. Solve $5-2x<11$ .[2 marks]
Answer
- $x>-3$
Method: Subtract $5$ to obtain $-2x<6$ . Dividing by the negative number $-2$ reverses the inequality, giving $x>-3$ .

Tier 2 · Standard

1. Solve $(x-4)(x+1)\leq0$ .[3 marks]
Answer
- $-1\leq x\leq4$
Method: The boundary values are $x=-1$ and $x=4$ . The upward-opening quadratic is non-positive between these roots, and equality includes both endpoints. Therefore $-1\leq x\leq4$ .

Tier 3 · Hard

1. Solve $\dfrac{x+2}{x-3}>2$ and give the result in set notation.[4 marks]
Answer
- $\{x\in\mathbb R:3<x<8\}$
Method: 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 $\{x\in\mathbb R:3<x<8\}$ .

2.6 · 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.

Polynomial manipulation uses expansion, collection and factorisation while preserving equality; choose the form that exposes the required structure.
The factor theorem states that $x-a$ is a factor of $f(x)$ exactly when $f(a)=0$ ; after finding a factor, divide 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.

Tier 1 · Easy

1. Factorise $x^2-9x+20$ completely.[2 marks]
Answer
- $(x-4)(x-5)$
Method: Find two numbers with product $20$ and sum $-9$ : they are $-4$ and $-5$ . Therefore $x^2-9x+20=(x-4)(x-5)$ .

Tier 2 · Standard

1. Use the factor theorem to factorise $x^3-4x^2+x+6$ fully.[4 marks]
Answer
- $(x-2)(x-3)(x+1)$
Method: Let $f(x)=x^3-4x^2+x+6$ . Since $f(2)=8-16+2+6=0$ , $x-2$ is a factor. Division gives $x^2-2x-3$ , which factorises as $(x-3)(x+1)$ . Hence $f(x)=(x-2)(x-3)(x+1)$ .

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 marks]
Answer
- $x-\dfrac{3}{x-2}$
- $x\ne2$
Method: 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 $x-3/(x-2)$ .

2.7 · 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.

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 $a$ .
An equation $f(x)=g(x)$ is solved graphically at the $x$ -coordinates where the two graphs intersect; modulus reflects negative outputs above the $x$ -axis.
Do not draw a reciprocal curve crossing an asymptote, and distinguish direct proportionality $y=kx$ from a relation that merely has positive correlation.

Tier 1 · Easy

1. For the curve $y=-\dfrac{6}{x}$ , state both asymptotes and the two quadrants containing its branches.[2 marks]
Answer
- Asymptotes $x=0$ and $y=0$ ; branches in quadrants II and IV.
Method: 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.

Tier 2 · Standard

1. Find the intersection values of $x$ for the graphs $y=|2x-3|$ and $y=x+4$ .[4 marks]
Answer
- $x=7$ or $x=-\dfrac13$
Method: For $x\geq3/2$ , solve $2x-3=x+4$ , giving $x=7$ . For $x<3/2$ , solve $-(2x-3)=x+4$ , so $-2x+3=x+4$ and $x=-1/3$ . Each value lies in its required branch, so both are intersections.

Tier 3 · Hard

1. The curves $y=\dfrac4x$ and $y=x+3$ meet twice. Determine the exact coordinates of both intersections.[4 marks]
Answer
- $(1,4)$ and $(-4,-1)$
Method: 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 $y=-1$ .

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

The composite $fg(x)$ means $f(g(x))$ : apply the right-hand function first and keep its whole output inside the outer function.
To find an inverse, write $y=f(x)$ , rearrange for $x$ , then exchange the variable labels; a restricted domain may be needed to make $f$ one-to-one.
A function and its inverse have graphs reflected in $y=x$ , with domain and range interchanged.
In general $fg\ne gf$ ; for inverse quadratics, use the stated domain to choose the correct square-root branch.

Tier 1 · Easy

1. Let $f(x)=3x-1$ and $g(x)=x^2+2$ . Calculate $fg(2)$ .[2 marks]
Answer
- $17$
Method: Apply $g$ first: $g(2)=2^2+2=6$ . Then $f(6)=3(6)-1=17$ .

Tier 2 · Standard

1. For $f(x)=\dfrac{3x+2}{x-1}$ , find $f^{-1}(x)$ and state its domain.[4 marks]
Answer
- $f^{-1}(x)=\dfrac{x+2}{x-3}$
- Domain: $x\ne3$
Method: Let $y=(3x+2)/(x-1)$ . Then $yx-y=3x+2$ , so $x(y-3)=y+2$ and $x=(y+2)/(y-3)$ . Exchanging the labels gives $f^{-1}(x)=(x+2)/(x-3)$ . The inverse denominator requires $x\ne3$ .

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 marks]
Answer
- $f^{-1}(x)=3+\sqrt{x-1}$ with domain $x\geq1$
- $x=0$ or $x=2$
Method: 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.

2.9 · 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.

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.
For example, a point $(u,v)$ on $y=f(x)$ maps to $(u/2,v+3)$ on $y=f(2x)+3$ .
Apply combined transformations to coordinates or rewrite the complete formula; describing $f(x+a)$ as a shift right is the standard sign error.

Tier 1 · Easy

1. Describe fully the transformation from $y=f(x)$ to $y=f(x-4)+2$ .[2 marks]
Answer
- Translation by vector $\begin{pmatrix}4\\2\end{pmatrix}$ .
Method: Replacing $x$ by $x-4$ moves the graph $4$ units right, and adding $2$ moves it $2$ units up. Together this is translation by $\begin{pmatrix}4\\2\end{pmatrix}$ .

Tier 2 · Standard

1. The point $(6,-1)$ lies on $y=f(x)$ . Find the corresponding point on $y=-2f(3x)+1$ .[3 marks]
Answer
- $(2,3)$
Method: For the transformed input to equal $6$ , set $3x=6$ , giving $x=2$ . The transformed output is $-2(-1)+1=3$ , so the corresponding point is $(2,3)$ .

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 marks]
Answer
- Turning point $(0,7)$ , a maximum
- $x$ -intercepts $\left(-\dfrac{\sqrt7}{2},0\right)$ and $\left(\dfrac{\sqrt7}{2},0\right)$
Method: 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 $x=\pm\sqrt7/2$ .

2.10 · 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).

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.

Tier 1 · Easy

1. Express $\dfrac7{(x+1)(x+2)}$ as partial fractions.[3 marks]
Answer
- $\dfrac7{x+1}-\dfrac7{x+2}$
Method: 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 $B=-7$ .

Tier 2 · Standard

1. Decompose $\dfrac{5x+1}{(x-1)(x+2)}$ into partial fractions.[4 marks]
Answer
- $\dfrac2{x-1}+\dfrac3{x+2}$
Method: Let $(5x+1)/((x-1)(x+2))=A/(x-1)+B/(x+2)$ . Hence $5x+1=A(x+2)+B(x-1)$ . Substituting $x=1$ gives $6=3A$ , so $A=2$ . Substituting $x=-2$ gives $-9=-3B$ , so $B=3$ .

Tier 3 · Hard

1. Express $\dfrac{5x+1}{(x-1)^2(x+2)}$ as three partial-fraction terms.[5 marks]
Answer
- $\dfrac1{x-1}+\dfrac2{(x-1)^2}-\dfrac1{x+2}$
Method: 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 $A=1$ .

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

A function model links clearly defined variables over a stated domain; parameters should have units or contextual meaning where possible.
Use the given data to determine parameters, then calculate with the model and interpret the result in context with appropriate rounding.
For example, in $P(t)=P_0r^t$ , $P_0$ is the value at $t=0$ and $r$ is the multiplier per time interval.
A model is an approximation: check whether extrapolation, a fixed growth rate, ignored constraints or measurement variation makes a prediction unreliable, and suggest a refinement tied to that weakness.

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 marks]
Answer
- $151$ organisms
Method: Substitute $t=3$ : $P(3)=120(1.08)^3=151.16544$ . Since the output counts organisms, round to the nearest whole number to obtain $151$ .

Tier 2 · Standard

1. 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.[5 marks]
Answer
- $11$ minutes
- For example, the model assumes a constant surrounding temperature of $18$ degrees Celsius.
Method: Solve $18+62e^{-0.15t}<30$ . This gives $e^{-0.15t}<6/31$ , so $t>-\ln(6/31)/0.15=10.95\ldots$ . The first whole minute is therefore $11$ . One limitation is that the fixed ambient temperature may not remain exactly $18$ degrees Celsius.

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 marks]
Answer
- $k=0.08$
- Maximum height $18$ m
- Width $10\sqrt3$ m
- For example, a real roof cross-section may not be exactly parabolic.
Method: 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.