Functions & graphs (modulus, composite, inverse, transformations) – A-level Maths exam questions with mark scheme

1CoreCalculator5 marks

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $$\mathrm{f}(x)=3x-5,\ x\in\mathbb{R}\qquad \mathrm{g}(x)=\frac{2}{x+1},\ x\in\mathbb{R},\ x\neq -1$$

(a) Find $\mathrm{f}^{-1}(x)$. [2]

(b) Solve the equation $\mathrm{fg}(x)=1$. [3]

(a) Attempts to make $x$ the subject of $y=3x-5$ (or swaps $x$ and $y$ first), e.g. $x=\dfrac{y+5}{3}$M1
$\mathrm{f}^{-1}(x)=\dfrac{x+5}{3}$ (must be in terms of $x$; allow $\tfrac13x+\tfrac53$)A1
(b) Correct order for composite: $\mathrm{fg}(x)=3\left(\dfrac{2}{x+1}\right)-5$M1
Sets $\dfrac{6}{x+1}-5=1$ and proceeds to a value of $x$ (e.g. $\dfrac{6}{x+1}=6\Rightarrow x+1=1$)M1
$x=0$A1
Answer: (a) $\mathrm{f}^{-1}(x)=\dfrac{x+5}{3}$ (b) $x=0$
Examiner tip: $\mathrm{fg}(x)$ means apply g first – students who work out $\mathrm{g}(\mathrm{f}(x))$ lose both method marks in (b).

2StandardCalculator6 marks

(a) Sketch the graph of $y=|2x-3|$, stating the coordinates of any points where the graph meets or cuts the coordinate axes. [2]

(b) Solve the equation $$|2x-3|=5-x$$ [4]

(a) V-shaped graph with its vertex on the positive $x$-axis (both branches above the $x$-axis)B1
Both $\left(\tfrac32,0\right)$ and $(0,3)$ correctly stated or marked on the sketchB1
(b) Solves $2x-3=5-x$ to find a value of $x$M1
$x=\dfrac{8}{3}$A1
Solves $-(2x-3)=5-x$ (or $3-2x=5-x$) to find a value of $x$M1
$x=-2$ (both $x=\tfrac83$ and $x=-2$ given, and no other values)A1
Answer: (a) V-shape, vertex $(1.5,0)$, $y$-intercept $(0,3)$ (b) $x=\tfrac83$ or $x=-2$
Examiner tip: The second solution comes from the reflected branch, $-(2x-3)=5-x$ – drawing $y=5-x$ on your sketch shows there are two intersections.

3StandardCalculator7 marks

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $$\mathrm{f}(x)=\mathrm{e}^{2x}+3,\ x\in\mathbb{R}\qquad \mathrm{g}(x)=\ln(x-1),\ x\in\mathbb{R},\ x>1$$

(a) State the range of $\mathrm{f}$. [1]

(b) Find $\mathrm{f}^{-1}(x)$, stating its domain. [3]

(a) $\mathrm{f}(x)>3$ (accept $y>3$, $(3,\infty)$; not $x>3$)B1
(b) Attempts to change the subject: $\mathrm{e}^{2x}=y-3\Rightarrow 2x=\ln(y-3)$M1
$\mathrm{f}^{-1}(x)=\tfrac12\ln(x-3)$A1
Domain $x>3$ (follow through from their range in (a))B1
(c) Uses $\mathrm{e}^{2\ln(x-1)}=(x-1)^{2}$ to obtain $\mathrm{fg}(x)=(x-1)^{2}+3$M1
Solves $(x-1)^{2}+3=12$ to obtain at least one value of $x$M1
$x=4$ only, with $x=-2$ rejected (as $x>1$ in the domain of g)A1
Answer: (a) $\mathrm{f}(x)>3$ (b) $\mathrm{f}^{-1}(x)=\tfrac12\ln(x-3),\ x>3$ (c) $x=4$
Examiner tip: Always check solutions against the domain – leaving $x=-2$ in (c) costs the final A mark.

4HardCalculator7 marks

The diagram shows a sketch of the curve with equation $y=\mathrm{f}(x)$, $x\in\mathbb{R}$. The curve has a maximum turning point at $P(-2,6)$ and a minimum turning point at $Q(3,-4)$. The curve has no other turning points and continues indefinitely in both directions as shown.

(a) State the coordinates of the stationary points on the graph of $y=|\mathrm{f}(x)|$. [2]

(b) State the coordinates of the points to which $P$ and $Q$ are mapped on the curve with equation $y=2\mathrm{f}(x-1)+3$. [3]

(a) $(-2,6)$B1
$(3,4)$ (the cusps where the graph meets the $x$-axis are not stationary points and are not required; do not award if extra stationary points are given)B1
(b) Either $x$-coordinate correct: $-1$ or $4$ (horizontal translation of $+1$)B1
$P\mapsto(-1,15)$B1
$Q\mapsto(4,-5)$B1
(c) Uses the $y$-coordinates of the turning points as the critical values, $-4$ and $6$M1
$-4A1
Answer: (a) $(-2,6)$ and $(3,4)$ (b) $(-1,15)$ and $(4,-5)$ (c) $-4
Examiner tip: In $2\mathrm{f}(x-1)+3$ the 2 is applied before the $+3$: $y\mapsto 2y+3$, so $-4\mapsto-5$, not $-2$.

5HardCalculator7 marks

The function $\mathrm{f}$ is defined by $$\mathrm{f}(x)=\frac{2x+3}{x-2},\quad x\in\mathbb{R},\ x\neq 2$$

(a) Show that $\mathrm{ff}(x)=x$. [3]

(b) Hence write down $\mathrm{f}^{-1}(x)$. [1]

(d) Find the values of $x$ for which $\mathrm{f}(x)=x$, giving your answers in simplest surd form. [2]

(a) Substitutes f into itself: $\dfrac{2\left(\frac{2x+3}{x-2}\right)+3}{\frac{2x+3}{x-2}-2}$M1
Multiplies numerator and denominator by $(x-2)$ (or combines fractions): $\dfrac{4x+6+3(x-2)}{2x+3-2(x-2)}$M1
$=\dfrac{7x}{7}=x$ csoA1*
(b) $\mathrm{f}^{-1}(x)=\dfrac{2x+3}{x-2}$ (accept $\mathrm{f}^{-1}(x)=\mathrm{f}(x)$)B1
(c) $\mathrm{f}(x)\neq 2$ (accept $y\in\mathbb{R},\ y\neq2$)B1
(d) Sets $\dfrac{2x+3}{x-2}=x$ and obtains a 3-term quadratic $x^{2}-4x-3=0$, attempting to solveM1
$x=2\pm\sqrt7$A1
Answer: (b) $\mathrm{f}^{-1}(x)=\dfrac{2x+3}{x-2}$ (c) $\mathrm{f}(x)\neq2$ (d) $x=2\pm\sqrt7$
Examiner tip: In a 'show that' every line must appear – jumping from the compound fraction straight to $x$ loses the A1*.

6StretchCalculator8 marks

Given that $a$ is a positive constant, the function $\mathrm{f}$ is defined by $$\mathrm{f}(x)=2|x+a|-3a,\quad x\in\mathbb{R}$$

(a) Sketch the graph of $y=\mathrm{f}(x)$, showing the coordinates of the vertex and of any points where the graph meets the coordinate axes, in terms of $a$. [3]

(b) Solve, in terms of $a$, the equation $\mathrm{f}(x)=x+a$. [4]

(a) V-shape with vertex at $(-a,-3a)$ in the third quadrantB1
$y$-intercept $(0,-a)$B1
$x$-intercepts $\left(-\tfrac52a,0\right)$ and $\left(\tfrac12a,0\right)$B1
(b) Solves $2(x+a)-3a=x+a$ to find $x$M1
$x=2a$A1
Solves $-2(x+a)-3a=x+a$ to find $x$M1
$x=-2a$ (both correct and no others)A1
(c) $k<-3a$B1
Answer: (a) vertex $(-a,-3a)$; meets axes at $(0,-a)$, $\left(-\tfrac52a,0\right)$, $\left(\tfrac12a,0\right)$ (b) $x=2a$ or $x=-2a$ (c) $k<-3a$
Examiner tip: Check each case actually lies on the line – with a parameter it is easy to keep an extra root that does not satisfy the original modulus equation.

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