Reciprocal, rational, exponential and logarithmic functions – IB Maths exam questions with mark scheme

1HardNon-calculator7 marks

Let $f(x)=\dfrac{2x+1}{x-3}$, $x\ne3$. The graph of $y=f(x)$ is shown, with its asymptotes drawn dashed. The graph crosses the y-axis at P and the x-axis at Q.

(a) Write down the equations of the vertical and horizontal asymptotes. [2]

(b) Find the coordinates of P and Q. [2]

Vertical asymptote $x=3$A1
Horizontal asymptote $y=2$A1
P is the y-intercept: $\left(0,-\tfrac13\right)$A1
Q is the x-intercept: $\left(-\tfrac12,0\right)$A1
$y(x-3)=2x+1$, collects $x$ terms: $x(y-2)=3y+1$ (or equivalent if $x$ and $y$ are interchanged first: $y(x-2)=3x+1$)M1
$f^{-1}(x)=\dfrac{3x+1}{x-2}$A1
Domain of $f^{-1}$ is $x\ne2$ (the range of $f$)A1
Answer: (a) $x=3$, $y=2$ (b) P$(0,-\tfrac13)$, Q$(-\tfrac12,0)$ (c) $\dfrac{3x+1}{x-2}$, $x\ne2$
Examiner tip: Dividing: $f(x)=2+\dfrac7{x-3}$ gives the horizontal asymptote immediately.

2CoreNon-calculator5 marks

Let $f(x)=\dfrac{8}{x}$, $x\neq 0$.

(a) Find $(f\circ f)(x)$ and hence write down $f^{-1}(x)$. [2]

(b) Write down the equations of the asymptotes of the graph of $y=f(x)$. [1]

(a) $(f\circ f)(x)=\dfrac{8}{8/x}=x$ shown (simplified to $x$)A1
$f^{-1}(x)=\dfrac{8}{x}$ (accept $f^{-1}=f$). Follow through from their $(f\circ f)(x)=x$A1
(b) $x=0$ and $y=0$ (both required; accept 'the $y$-axis and the $x$-axis')A1
(c) Sets $\dfrac{8}{x}=x$ leading to $x^{2}=8$M1
$(2\sqrt2,\,2\sqrt2)$ and $(-2\sqrt2,\,-2\sqrt2)$ (accept $\sqrt8$; both points needed, award A0 if only the positive point is given)A1
Answer: (a) $x$; $f^{-1}(x)=\dfrac{8}{x}$ (b) $x=0$, $y=0$ (c) $(2\sqrt2,2\sqrt2)$, $(-2\sqrt2,-2\sqrt2)$
Examiner tip: In (c) $x^2=8$ has two roots: the graph of $8/x$ has a branch in the third quadrant that also crosses $y=x$.

3CoreNon-calculator5 marks

Let $g(x)=2^{x+1}-8$, $x\in\mathbb{R}$.

(a) Find the $y$-intercept of the graph of $g$. [1]

(b) Find the $x$-intercept of the graph of $g$. [2]

(d) Write down the range of $g$. [1]

(a) $(0,-6)$ (accept $y=-6$)A1
(b) Sets $2^{x+1}=8$ and writes $8=2^{3}$ (or takes logs correctly)M1
$x=2$ (accept $(2,0)$)A1
(c) $y=-8$ (must be an equation)A1
(d) $g(x)>-8$ (accept $y>-8$, $]-8,\infty[$, $(-8,\infty)$); A0 for $g(x)\geq-8$A1
Answer: (a) $(0,-6)$ (b) $x=2$ (c) $y=-8$ (d) $g(x)>-8$
Examiner tip: The asymptote value is never reached, so the range needs a strict inequality: $g(x)\geq-8$ loses the mark.

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4StandardNon-calculator5 marks

The diagram shows part of the graph of $f(x)=\dfrac{px+q}{x+r}$, $x\neq -r$. The graph has a vertical asymptote $x=-2$ and a horizontal asymptote $y=3$, and it crosses the $y$-axis at $(0,-2)$.

(a) Write down the value of $r$ and the value of $p$. [2]

(b) Find the value of $q$. [2]

(a) $r=2$A1
$p=3$A1
(b) Substitutes $(0,-2)$: $\dfrac{q}{r}=-2$ or $\dfrac{q}{2}=-2$ (follow through their $r$)M1
$q=-4$A1
(c) $x=\dfrac{4}{3}$ (accept $\left(\frac43,0\right)$; follow through $-\frac{q}{p}$ from their values)A1
Answer: (a) $r=2$, $p=3$ (b) $q=-4$ (c) $x=\dfrac{4}{3}$
Examiner tip: The vertical asymptote is where the denominator is zero, so $x+r=0$ at $x=-2$ gives $r=+2$, not $-2$.

5StandardNon-calculator7 marks

Let $h(x)=\ln(2x-5)+1$, for $x>\dfrac{5}{2}$.

(a) Write down the equation of the vertical asymptote of the graph of $h$. [1]

(b) Find the exact $x$-intercept of the graph of $h$. [2]

(d) Write down the range of $h^{-1}$. [1]

(a) $x=\dfrac{5}{2}$ (accept $x=2.5$)A1
(b) $\ln(2x-5)=-1\Rightarrow 2x-5=e^{-1}$M1
$x=\dfrac{5+e^{-1}}{2}$ (accept $\dfrac{5}{2}+\dfrac{1}{2e}$ or equivalent exact form); A0 for $2.68$ onlyA1
(c) Interchanges $x$ and $y$ (may be seen at any stage) e.g. $x=\ln(2y-5)+1$M1
Removes the logarithm correctly: $2y-5=e^{x-1}$ (or $2x-5=e^{y-1}$)M1
$h^{-1}(x)=\dfrac{e^{x-1}+5}{2}$ (accept $\frac12e^{x-1}+\frac52$); must be in the form $h^{-1}(x)=$ or $y=$A1
(d) $h^{-1}(x)>\dfrac{5}{2}$ (accept $y>2.5$); range of $h^{-1}$ is the domain of $h$A1
Answer: (a) $x=\frac52$ (b) $x=\dfrac{5+e^{-1}}{2}$ (c) $h^{-1}(x)=\dfrac{e^{x-1}+5}{2}$ (d) $h^{-1}(x)>\frac52$
Examiner tip: Isolate the logarithm before exponentiating: $e^{\ln(2x-5)+1}$ is $e(2x-5)$, not $2x-5+e$.

6StandardCalculator7 marks

A cup of coffee is left to cool in a room. Its temperature, $T$ °C, $t$ minutes after it is poured is modelled by

$$T(t)=21+68e^{-0.045t},\quad t\geq 0.$$

(a) Write down the temperature of the coffee when it is poured. [1]

(b) Find the temperature of the coffee after 10 minutes. [2]

(d) Write down the equation of the horizontal asymptote of the graph of $T$ and interpret it in context. [2]

(a) $89$ °CA1
(b) Substitutes $t=10$: $21+68e^{-0.45}$M1
$64.4$ °C ($64.3587\ldots$)A1
(c) Sets $21+68e^{-0.045t}=55$ (or $e^{-0.045t}=0.5$), or a correct graph/solver approachM1
$t=15.4$ minutes ($15.4032\ldots$; accept $\dfrac{\ln2}{0.045}$)A1
(d) $T=21$ (must be an equation; accept $y=21$)A1
Interpretation: the temperature of the coffee approaches $21$ °C, the room temperature, (but never falls below it). Do not accept 'the coffee reaches 21 °C'R1
Answer: (a) $89$ °C (b) $64.4$ °C (c) $15.4$ min (d) $T=21$: coffee cools towards room temperature 21 °C
Examiner tip: Give answers to 3 s.f. unless told otherwise; and in (d) say the temperature tends to 21 °C, not that it reaches it.

7StandardNon-calculator6 marks

Let $f(x)=\dfrac{2x+3}{x-1}$, $x\neq1$.

(a) Show that $f(x)=2+\dfrac{5}{x-1}$. [2]

(b) Hence describe a sequence of transformations that maps the graph of $y=\dfrac{1}{x}$ onto the graph of $y=f(x)$. [3]

(a) Writes $2x+3=2(x-1)+5$, or combines $2+\frac{5}{x-1}=\frac{2(x-1)+5}{x-1}$ (or long division)M1
Fully correct working leading to the given result (AG). No marks for working backwards with no linkA1
(b) Stretch parallel to the $y$-axis (vertical), scale factor $5$ (accept stretch parallel to the $x$-axis, scale factor $5$, since $\frac{1}{x/5}=\frac5x$; do not accept 'enlargement')A1
Translation $1$ unit in the positive $x$-direction (if a horizontal stretch is used, this translation must come after the stretch)A1
Translation $2$ units in the positive $y$-direction (accept the single translation $\begin{pmatrix}1\\2\end{pmatrix}$). With a vertical stretch, award A0 if the vertical translation is applied before the stretchA1
(c) $f(x)\neq 2$ (accept $y\in\mathbb{R}$, $y\neq2$)A1
Answer: (a) shown (b) Vertical stretch, scale factor 5, then translation by $\begin{pmatrix}1\\2\end{pmatrix}$ (or horizontal stretch, scale factor 5, then the same translation) (c) $f(x)\neq2$
Examiner tip: If you translate up by 2 before stretching by 5, the asymptote moves to $y=10$: the vertical translation must come after the stretch.

8HardCalculator7 marks

The diagram shows the graph of $f(x)=\ln(x+a)+b$, $x>-a$. The graph has a vertical asymptote $x=-1$ and passes through the point $(0,2)$.

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

(b) Find the value of $b$. [1]

(d) The graphs of $f$ and $f^{-1}$ intersect at two points. Find the $x$-coordinates of these points. [2]

(e) Hence write down the set of values of $x$ for which $f(x)>f^{-1}(x)$. [1]

(a) $a=1$A1
(b) $\ln1+b=2\Rightarrow b=2$A1
(c) Interchanges $x$ and $y$ and exponentiates: $x-2=\ln(y+1)\Rightarrow y+1=e^{x-2}$ (or equivalent with their $a$, $b$)M1
$f^{-1}(x)=e^{x-2}-1$A1
(d) Uses the fact that the graphs meet on $y=x$, e.g. sets $\ln(x+1)+2=x$, or sketches/solves $f(x)=f^{-1}(x)$ on GDCM1
$x=-0.948$ and $x=3.51$ ($-0.94753\ldots$, $3.50524\ldots$; both needed)A1
(e) $-0.948A1
Answer: (a) $a=1$ (b) $b=2$ (c) $f^{-1}(x)=e^{x-2}-1$ (d) $x=-0.948,\ 3.51$ (e) $-0.948
Examiner tip: A common slip is exponentiating term by term, giving $e^{x}-e^{2}-1$: first subtract 2 so that $\ln(y+1)=x-2$, then exponentiate.

9StretchNon-calculator8 marks

Let $f(x)=\dfrac{3x+k}{x-3}$, $x\neq3$, where $k\in\mathbb{R}$ and $k\neq-9$.

(a) Explain why the value $k=-9$ is excluded. [1]

(b) Find $f^{-1}(x)$ and hence show that $f$ is its own inverse. [3]

(d) Find the set of values of $k$ for which the graph of $f$ meets the line $y=x$ at two distinct points. [3]

(a) When $k=-9$, $f(x)=\dfrac{3(x-3)}{x-3}=3$: the function is constant (so it has no inverse / the graph is a horizontal line, not a rational curve). Accept any equivalent explanation that identifies the common factorR1
(b) Interchanges $x$ and $y$ and multiplies out: $x(y-3)=3y+k$ (or $xy-3x=3y+k$), or equivalent starting from $y(x-3)=3x+k$M1
Collects the $y$ terms and factorises: $y(x-3)=3x+k$, giving $f^{-1}(x)=\dfrac{3x+k}{x-3}$A1
States $f^{-1}(x)=f(x)$, so $f$ is self-inverse (AG). Accept a valid alternative showing $(f\circ f)(x)=x$R1
(c) $x=3$ and $y=3$ (both needed)A1
(d) Sets $\dfrac{3x+k}{x-3}=x$ and forms a quadratic: $x^{2}-6x-k=0$M1
Uses discriminant $>0$: $36+4k>0$M1
$k>-9$. Accept with a comment that $x=3$ is never a root as $k\neq-9$; A0 for $k\geq-9$A1
Answer: (a) $f$ would be the constant $3$ (b) $f^{-1}(x)=\dfrac{3x+k}{x-3}=f(x)$ (c) $x=3$, $y=3$ (d) $k>-9$
Examiner tip: In (d) check the roots are not the excluded $x=3$: substituting gives $9-18-k=0$, i.e. $k=-9$, which is already excluded, so the answer is just $k>-9$.

10HardNon-calculatorHL only7 marks

Let $f(x)=\dfrac{x^{2}+x-2}{x-3}$, $x\neq3$.

(a) Find the coordinates of the points where the graph of $f$ meets the coordinate axes. [2]

(b) Write down the equation of the vertical asymptote. [1]

(d) Explain why the graph of $f$ never crosses its oblique asymptote. [1]

(a) $(-2,0)$ and $(1,0)$ (both needed; from $(x+2)(x-1)=0$)A1
$\left(0,\dfrac{2}{3}\right)$A1
(b) $x=3$A1
(c) Attempts polynomial division of $x^{2}+x-2$ by $x-3$ (or writes $x^{2}+x-2=(x-3)(Ax+B)+C$)M1
$f(x)=x+4+\dfrac{10}{x-3}$A1
$y=x+4$ (must be an equation)A1
(d) $f(x)-(x+4)=\dfrac{10}{x-3}$, which is never zero (numerator is a non-zero constant), so the curve and the line have no common pointR1
Answer: (a) $(-2,0)$, $(1,0)$, $\left(0,\frac23\right)$ (b) $x=3$ (c) $y=x+4$ (d) $\frac{10}{x-3}\neq0$ for all $x$
Examiner tip: The oblique asymptote is the quotient only: the remainder $\frac{10}{x-3}\to0$ as $x\to\pm\infty$, so it is not part of the line's equation.

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