Straight lines, functions, composite and inverse functions – IB Maths exam questions with mark scheme

1CoreNon-calculator4 marks

Let $f(x)=2x+3$ and $g(x)=x^2$.

(a) Find $(f\circ g)(2)$. [2]

(b) Find $f^{-1}(x)$. [2]

$g(2)=4$ then $f(4)$M1
$11$A1
Sets $y=2x+3$ and rearranges for $x$ (or swaps and rearranges)M1
$f^{-1}(x)=\dfrac{x-3}2$A1
Answer: (a) $11$ (b) $\dfrac{x-3}{2}$
Examiner tip: $(f\circ g)(x)$ means apply $g$ first, then $f$.

2StandardNon-calculator5 marks

The function $f$ is defined by $f(x)=\sqrt{x-2}+1$ for $x\ge2$.

(a) Write down the range of $f$. [1]

(b) Find $f^{-1}(x)$. [3]

Range: $f(x)\ge1$ (accept $y\ge1$, $[1,\infty)$)A1
Interchanges $x$ and $y$ (seen anywhere): $x=\sqrt{y-2}+1$, or rearranges $y=\sqrt{x-2}+1$ for $x$M1
Isolates the root and squares correctly: $(x-1)^2=y-2$ (or $(y-1)^2=x-2$)A1
$f^{-1}(x)=(x-1)^2+2$ (accept $x^2-2x+3$)A1
$x\ge1$ (the range of $f$; accept $[1,\infty)$)A1
Answer: (a) $f(x)\ge1$ (b) $f^{-1}(x)=(x-1)^2+2$ (c) $x\ge1$
Examiner tip: The domain of the inverse is the range of the original; the formula alone is not enough.

3HardNon-calculator6 marks

(a) The function $f$ is defined by $f(x)=3x+a$, where $a$ is a constant. Given that $f(f(x))=9x+8$, find the value of $a$. [3]

(b) The function $h$ is defined by $h(x)=\dfrac{x+1}{x-1}$, $x\ne1$. Show that $h^{-1}(x)=h(x)$. [3]

$f(f(x))=3(3x+a)+a$M1
$9x+4a$A1
$4a=8$ so $a=2$A1
Sets $y=\dfrac{x+1}{x-1}$ and multiplies out: $xy-y=x+1$ (or, if $x$ and $y$ are interchanged first, $x=\dfrac{y+1}{y-1}$ and $xy-x=y+1$)M1
Collects terms: $x(y-1)=y+1$ so $x=\dfrac{y+1}{y-1}$ (or $y(x-1)=x+1$ so $y=\dfrac{x+1}{x-1}$)A1
$h^{-1}(x)=\dfrac{x+1}{x-1}=h(x)$ with conclusion stated (AG). Alternative method: M1 substitutes $h$ into itself, A1 simplifies $(h\circ h)(x)$ to $x$, A1 concludes $h^{-1}=h$A1
Answer: (a) $a=2$ (b) shown
Examiner tip: Finish with the explicit statement $h^{-1}(x)=h(x)$; just finding the inverse without comparing it to $h$ loses the last mark.

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4CoreNon-calculator6 marks

The points $A(-2,\,5)$ and $B(4,\,-1)$ lie on the line $L_1$.

(a) Find the gradient of $L_1$. [2]

The line $L_2$ is the perpendicular bisector of $[AB]$.

(b) Find the equation of $L_2$. Give your answer in the form $ax+by+d=0$, where $a,b,d\in\mathbb{Z}$. [3]

(a) Attempt at $\dfrac{\Delta y}{\Delta x}$ with the coordinates of $A$ and $B$, e.g. $\dfrac{-1-5}{4-(-2)}$M1
$m=-1$A1
(b) Midpoint of $[AB]$ is $(1,\,2)$A1
Uses perpendicular gradient $-\dfrac{1}{m}=1$ with their midpoint, e.g. $y-2=1(x-1)$ (follow through their $m$)M1
$x-y+1=0$ (accept any integer multiple, e.g. $-x+y-1=0$; must be in the required form)A1
(c) $x=-1$ (accept $(-1,\,0)$; follow through their line)A1
Answer: (a) $-1$ (b) $x-y+1=0$ (c) $-1$
Examiner tip: A perpendicular bisector needs BOTH the midpoint and the negative reciprocal gradient; using point $A$ or $B$ instead of the midpoint loses the last two marks in (b).

5CoreNon-calculator7 marks

The diagram shows the graph of $y=f(x)$ for $-2\le x\le 4$. The graph consists of two straight line segments joining the points $(-2,\,-3)$, $(0,\,1)$ and $(4,\,3)$.

(a) Write down the range of $f$. [1]

(b) Find $f^{-1}(2)$. [2]

(d) Find $(f\circ f)(0)$. [2]

(a) $-3\le y\le 3$ (accept $-3\le f(x)\le3$, $[-3,\,3]$; do not accept strict inequalities)A1
(b) Recognises that $f^{-1}(2)$ is the $x$-value where $f(x)=2$, e.g. reads across from $y=2$, or uses $f(x)=1+\tfrac12x=2$M1
$f^{-1}(2)=2$A1
(c) $(-3,\,-2)$A1
$(3,\,4)$ (A1 for each correct endpoint; A0A0 if the endpoints of $f$, $(-2,-3)$ and $(4,3)$, are given unchanged)A1
(d) $f(0)=1$ seenM1
$(f\circ f)(0)=f(1)=1.5$ (accept $\tfrac32$)A1
Answer: (a) $-3\le y\le3$ (b) $2$ (c) $(-3,-2)$ and $(3,4)$ (d) $1.5$
Examiner tip: The graph of $f^{-1}$ is the reflection of $y=f(x)$ in $y=x$, so every point $(a,b)$ becomes $(b,a)$; students who reflect in an axis instead lose both marks in (c).

6StandardNon-calculator6 marks

Let $f(x)=\dfrac{1}{x-2}$, $x\ne2$, and $g(x)=3x+1$, $x\in\mathbb{R}$.

(a) Find $(f\circ g)(x)$, simplifying your answer. [2]

(b) Write down the value of $x$ that must be excluded from the domain of $f\circ g$. [1]

(a) Substitutes $g$ into $f$: $\dfrac{1}{(3x+1)-2}$M1
$(f\circ g)(x)=\dfrac{1}{3x-1}$A1
(b) $x=\dfrac13$ (follow through their (a))A1
(c) Correct composite in the right order: $3\left(\dfrac{1}{x-2}\right)+1$M1
Sets equal to 7 and rearranges: $\dfrac{3}{x-2}=6$ or $x-2=\dfrac12$A1
$x=\dfrac52$ (accept $2.5$)A1
Answer: (a) $\dfrac{1}{3x-1}$ (b) $x=\dfrac13$ (c) $x=\dfrac52$
Examiner tip: $g\circ f$ means apply $f$ first; working with $f(g(x))$ in (c) gives $\frac{1}{3x-1}=7$ and loses all three marks.

7StandardCalculator7 marks

A venue charges a fixed booking fee plus a cost per guest. The total cost, in dollars, of an event with $n$ guests is modelled by $C(n)=150+12.5n$.

(a) Find $C(40)$. [1]

(b) Find $C^{-1}(x)$. [2]

A second venue models its total cost as $D(n)=90+15n$.

(d) Find the values of $n$ for which the second venue is cheaper. [2]

(a) $650$ (dollars)A1
(b) Interchanges variables or rearranges $x=150+12.5n$ for $n$, e.g. $n=\dfrac{x-150}{12.5}$M1
$C^{-1}(x)=\dfrac{x-150}{12.5}$ (accept $0.08x-12$)A1
(c) $C^{-1}(400)=20$A1
Interpretation: a total cost of \$400 corresponds to 20 guests (accept: with a \$400 budget, 20 guests can attend). Must refer to both cost and guestsR1
(d) Forms the inequality $90+15n<150+12.5n$ (or solves the equation $D(n)=C(n)$)M1
$n<24$ (accept $0\le n<24$, $n\le 23$, or 'fewer than 24 guests')A1
Answer: (a) $650$ (b) $\dfrac{x-150}{12.5}$ (c) $20$: a total cost of \$400 means 20 guests (d) $n<24$
Examiner tip: An interpretation must use the context: say what the input (\$400 total cost) and output (20 guests) of the inverse mean, not just 'the inverse of 400 is 20'.

8HardNon-calculator6 marks

The function $f$ is defined by $f(x)=\dfrac{2x-1}{x+3}$, $x\ne-3$.

(a) Find $f^{-1}(x)$, stating its domain. [4]

The function $h$ satisfies $(f\circ h)(x)=x+1$ for all $x$ in the domain of $h$.

(b) Find $h(x)$, giving your answer as a single fraction. [2]

(a) Interchanges $x$ and $y$ and multiplies out, e.g. $x(y+3)=2y-1$ (or equivalent starting from $y=\frac{2x-1}{x+3}$)M1
Collects the terms in the subject variable: $xy-2y=-1-3x$ (or equivalent)A1
$f^{-1}(x)=\dfrac{3x+1}{2-x}$ (accept $\dfrac{-3x-1}{x-2}$)A1
Domain $x\ne2$, $x\in\mathbb{R}$A1
(b) Recognises $h(x)=f^{-1}(x+1)$, e.g. applies $f^{-1}$ to both sides or substitutes $x+1$ into their $f^{-1}$M1
$h(x)=\dfrac{3x+4}{1-x}$ (accept $\dfrac{-3x-4}{x-1}$)A1
Answer: (a) $f^{-1}(x)=\dfrac{3x+1}{2-x}$, $x\ne2$ (b) $h(x)=\dfrac{3x+4}{1-x}$
Examiner tip: Since $f\circ f^{-1}$ is the identity function, applying $f^{-1}$ to both sides of $f(h(x))=x+1$ isolates $h$; trying to solve $f(h)=x+1$ from scratch without this is where time and marks go.

9StretchNon-calculatorHL only10 marks

Let $f(x)=x^2-6x+11$, $x\ge k$.

(a) Express $f(x)$ in the form $(x-h)^2+c$. [2]

(b) Write down the smallest value of $k$ for which $f^{-1}$ exists. [1]

For the rest of this question, use this value of $k$.

(d) Solve $f(x)=f^{-1}(x)$, giving your answer in the form $\dfrac{p+\sqrt{q}}{r}$ where $p,q,r\in\mathbb{Z}^+$. [3]

(a) Attempt to complete the square, e.g. $(x-3)^2-9+11$M1
$(x-3)^2+2$A1
(b) $k=3$ (follow through their $h$)A1
(c) Interchanges $x$ and $y$: $x=(y-3)^2+2$ (or rearranges for $x$)M1
$y-3=\pm\sqrt{x-2}$ (or $(y-3)^2=x-2$)A1
$f^{-1}(x)=3+\sqrt{x-2}$ (positive root only, justified by range of $f^{-1}$ being $\ge3$)A1
Domain $x\ge2$A1
(d) Uses $f(x)=x$ (since $f$ is increasing, the graphs of $f$ and $f^{-1}$ meet on $y=x$), or equates $f^{-1}(x)=x$M1
$x^2-7x+11=0$A1
$x=\dfrac{7+\sqrt5}{2}$ only (rejects $\frac{7-\sqrt5}{2}\approx2.38$ as it is less than 3). A0 if both roots givenA1
Answer: (a) $(x-3)^2+2$ (b) $k=3$ (c) $f^{-1}(x)=3+\sqrt{x-2}$, $x\ge2$ (d) $x=\dfrac{7+\sqrt5}{2}$
Examiner tip: In (d) setting $f(x)=f^{-1}(x)$ directly gives a quartic; use $f(x)=x$, and then check each root against the domain $x\ge3$.

10HardNon-calculator9 marks

The line $L_1$ passes through the point $P(1,\,9)$ and is parallel to the line $x+2y=4$.

(a) Find the equation of $L_1$ in the form $ax+by=c$. [3]

The line $L_2$ passes through the origin $O$ and is perpendicular to $L_1$.

(b) Write down the equation of $L_2$. [1]

(d) Find the exact area of triangle $OPQ$. [3]

(a) Gradient of $L_1$ is $-\dfrac12$A1
Substitutes $(1,9)$ into a line with their gradient, e.g. $y-9=-\frac12(x-1)$, or into $x+2y=c$M1
$x+2y=19$ (accept any integer multiple)A1
(b) $y=2x$A1
(c) Solves their equations simultaneously, e.g. $x+4x=19$M1
$Q\left(\dfrac{19}{5},\,\dfrac{38}{5}\right)$ (accept $(3.8,\,7.6)$)A1
(d) Valid method: recognises right angle at $Q$ and uses $\frac12\times OQ\times PQ$, or uses $\frac12|x_Py_Q-x_Qy_P|$, or another complete methodM1
Correct values, e.g. $OQ=\dfrac{19}{\sqrt5}$ and $PQ=\dfrac{7}{\sqrt5}$, or $\frac12\left|1(7.6)-9(3.8)\right|$A1
Area $=\dfrac{133}{10}$ (accept $13.3$)A1
Answer: (a) $x+2y=19$ (b) $y=2x$ (c) $Q(3.8,\,7.6)$ (d) $\dfrac{133}{10}$
Examiner tip: Spot that $L_2\perp L_1$ makes angle $OQP$ a right angle, so $OQ$ and $PQ$ are the base and height; using $OP$ as a base without its perpendicular height gives a wrong area.

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