Question 1
This question asks you to examine the number and nature of intersection points of the graph of $y = \log_a x$ where $a \in \mathbb{R}^+$, $a \neq 1$ and the line $y = x$ for particular sets of values of $a$.
In this question, you may either use the change of logarithm base formula
\[
\log_a x = \frac{\ln x}{\ln a}
\]
or a graphic display calculator “logarithm to any base feature”.
The function $f$ is defined by
\[
f(x) = \log_a x \quad \text{where } x \in \mathbb{R}^+ \text{ and } a \in \mathbb{R}^+, \quad a \neq 1.
\]
Topic – SL:2.6
(a) Consider the cases $a = 2$ and $a = 10$. On the same set of axes, sketch the following three graphs:
\[y = \log_2 x\]
\[y = \log_{10} x\]
\[y = x\]
Clearly label each graph with its equation and state the value of any non-zero $x$-axis intercepts.
In parts (b) and (c), consider the case where $a = e$. Note that $\ln x = \log_{e}x$.
Topic – SL:6.2
(b) Use calculus to find the minimum value of the expression $x – \ln x$, justifying that this value is a minimum.
Topic – SL:6.2
(c) Hence deduce that $x > \ln x$.
Topic – SL:2.6
(d) There exist values of $a$ for which the graph of $y = \log_{a}x$ and the line $y = x$ do have intersection points. The following table gives three intervals for the value of $a$.
By investigating the graph of $y = \log_{a}x$ for different values of $a$, write down the values of $p$, $q$ and $r$.
In parts (e) and (f), consider $a \in \mathbb{R}$, $a \neq 1$.
For $1.4 \leq a \leq 1.5$, a value of $a$ exists such that the line $y = x$ is a tangent to the graph of $y = \log_{a}x$ at a point $P$.
Topic – AHL:6.10
(e) Find the exact coordinates of $P$ and the exact value of $a$.
Topic – SL:2.6
(f) Write down the exact set of values for $a$ such that the graphs of $y = \log_{a}x$ and $y = x$ have
(i) two intersection points;
(ii) no intersection points.
▶️Answer/Explanation
Solution: –
(a) 
clearly labelled graphs of $y = log_{e} x$ and $y = log_{10} x$ with correct domain,
asymptotic behaviour and concavity evident
correct relative positions of the two log graphs both above and below the x-axis
(1,0) indicated (coordinates not required)
correct graph of $y = x$
(b)
$\frac{d}{dx}(x-ln~x)$
$=1-\frac{1}{x}$
attempts to solve their $\frac{dy}{dx}=0$ for x
$1-\frac{1}{x}=0\Rightarrow x=1$
(when $x=1$,) $x-ln~x=1$
EITHER
$\frac{d}{dx}\left(1-\frac{1}{x}\right)$
$=\frac{1}{x^{2}}$
$\frac{1}{x^{2}}>0$ (when $x=1$)
hence x-ln~x has a minimum value of 1
OR
for $(0<)x<1, 1-\frac{1}{x}<0$
for $x>1, 1-\frac{1}{x}>0$
hence $x-ln~x$ has a minimum value of 1
(c)
EITHER
\[x-ln~x\ge1(x\in\mathbb{R}^{+})\]
\[x-ln~x>0(x\in\mathbb{R}^{+})\]
THEN
so $x>ln~x$
(d)
Interval Number of intersection points
$0<a<1$ $p=1$
$1<a<1.4$ $q=2$
$1.5<a<2$ $r=0$
(e) METHOD 1
EITHER
$y = log_{a} x$
$\frac{dy}{dx}=\frac{1}{x~ln~a}$
attempts to solve $\frac{1}{x~ln~a}=1$ for x
OR
$y = x-log_{a} x$
$\frac{dy}{dx}=1-\frac{1}{x~ln~a}$
attempts to solve $1-\frac{1}{x~ln~a}=0$ for x
THEN
\[x=\frac{1}{ln~a}~OR~x~ln~a=1~OR~ln~a=\frac{1}{x}~OR~ln~a^{x}=1~OR~\frac{1}{a^{x}ln~a}=1\]
at
\[x=\frac{1}{ln~a},~log_{a}x=x\]
attempts to solve
\[\frac{ln~x}{ln~a}=\frac{1}{ln~a}~OR~ln~x=1~OR~\left(e^{\frac{1}{x}}\right)^{x}=x\]
for x
x=c
coordinates of P are (c,c) (accept x=c, y=c)
attempts to solve
\[\frac{1}{ln~a}=c~OR~log_{a}c=c\] for a analytically
$ln~a=\frac{1}{c}$ OR $a^{c}=c$
\[a=e^{\frac{1}{e}}\]
METHOD 2
EITHER
$y=log_{a} x$
$\frac{dy}{dx}=\frac{1}{x~ln~a}$
attempts to solve $\frac{1}{x~ln~a}=1$ for x
OR
$y=x-log_{a} x$
$\frac{dy}{dx}=1-\frac{1}{x~ln~a}$
attempts to solve $1-\frac{1}{x~ln~a}=0$ for x
THEN
$x=\frac{1}{ln~a}$ OR $x~ln~a=1$ OR $ln~a=\frac{1}{x}$ OR $ln~a^{x}=1$ OR $\frac{1}{a^{x}ln~a}=1$
at $x=\frac{1}{ln~a}$, $log_{a}x=x$
attempts to solve $log_{a}(\frac{1}{ln~a})=\frac{1}{ln~a}$ for a
EITHER
\[\frac{ln(\frac{1}{ln~a})}{ln~a}=\frac{1}{ln~a}\Rightarrow ln(\frac{1}{ln~a})=1\]
OR
for example, writes
\[a^{log_{a}(\frac{1}{ln~a})}=a^{\frac{1}{ln~a}}\]
and then attempts to apply appropriate
index$/log$ laws to both sides:
\[ln~a=\frac{log_{a}e}{log_{a}a}\]
and so
\[\frac{1}{ln~a}=log_{a}e\]
THEN
attempts to solve $\frac{1}{ln~a}=e$ OR $log_{a}e=e$ for a analytically
$ln~a=\frac{1}{e}$ OR $a^{e}=e$
$a=e^{\frac{1}{e}}$
$x=\frac{1}{ln~e^{\frac{1}{e}}}=\frac{1}{\frac{1}{e}}=e$
coordinates of P are (e,e) (accept x=e, y=e)
METHOD 3
$y=log_{a} x$
$\frac{dy}{dx}=\frac{1}{x~ln~a}$
(equation of the tangent at (x,y) is) $y=\frac{1}{x_{1}ln~a}(x-x_{1})+\frac{ln~x_{1}}{ln~a}$ (or equivalent)
compares this equation with y = x and attempts to form at least one of the following
\[\frac{1}{x_{1}ln~a}=1\]
OR
\[\frac{ln~x_{1}-1}{ln~a}=0\]
attempts to solve
\[\frac{1}{x_{1}ln~a}=1\]
OR
\[\frac{ln~x_{1}-1}{ln~a}=0\]
for x
$x_{1}=e$
coordinates of P are (e,e) (accept x=e, y=e)
attempts to solve $\frac{1}{e~ln~a}=1$ (or equivalent) for a analytically
$ln~a=\frac{1}{e}$
$a=e^{\frac{1}{e}}$
(f) (i) $1<a<e^{\frac{1}{e}}$
(ii) $a>e^{\frac{1}{e}}$
Question 2
This question asks you to examine linear and quadratic functions constructed in systematic ways using arithmetic sequences.
Consider the function $L(x) = mx + c$ for $x \in \mathbb{R}$ where $m, c \in \mathbb{R}$ and $m, c \neq 0$.
Let $r \in \mathbb{R}$ be the root of $L(x) = 0$.
If $m, r$, and $c$, in that order, are in arithmetic sequence then $L(x)$ is said to be an AS-linear function.
Topic – SL:1.3
(a) Show that $L(x) = 2x – 1$ is an AS-linear function.
Consider $L(x) = mx + c$.
Topic – SL:2.5
(b) (i) Show that $r = -\frac{c}{m}$.
Topic – SL:1.3
(ii) Given that $L(x)$ is an AS-linear function, show that $L(x) = mx – \frac{m^2}{m+2}$.
Topic – SL:1.3
(iii) State any further restrictions on the value of $m$.
There are only three integer sets of values of $m$, $r$, and $c$ that form an AS-linear function. One of these is $L(x) = -x – 1$.
Topic – SL:1.3
(c) Use part (b) to determine the other two AS-linear functions with integer values of $m$, $r$, and $c$.
Consider the function $Q(x) = ax^2 + bx + c$ for $x \in \mathbb{R}$ where $a \in \mathbb{R}$, $a \neq 0$, and $b, c \in \mathbb{R}$.
Let $r_1, r_2 \in \mathbb{R}$ be the roots of $Q(x) = 0$.
(d) Write down an expression for
Topic – SL:2.7
(i) the sum of roots, $r_1 + r_2$, in terms of $a$ and $b$.
Topic – SL:2.7
(ii) the product of roots, $r_1r_2$, in terms of $a$ and $c$.
If $a, r_{1}, b, r_{2}$, and $c$, in that order, are in arithmetic sequence, then $Q(x)$ is said to be an AS-quadratic function.
(e) Given that $Q(x)$ is an AS-quadratic function,
Topic – SL:2.7
(i) write down an expression for $r_{2}-r_{1}$ in terms of $a$ and $b$;
Topic – AHL:2.10
(ii) use your answers to parts (d)(i) and (e)(i) to show that
\[r_{1}=\frac{a^{2}-ab-b}{2a};\]
Topic – AHL:2.10
(iii) use the result from part (e)(ii) to show that $b=0$ or
\[a=-\frac{1}{2}.\]
Consider the case where $b=0$.
Topic – SL:2.7
(f) Determine the two AS-quadratic functions that satisfy this condition.
Now consider the case where
\[a=-\frac{1}{2}.\]
Topic – AHL:2.10
(g) (i) Find an expression for $r_{1}$ in terms of $b$.
Topic – AHL:2.10
(ii) Hence or otherwise, determine the exact values of $b$ and $c$ such that AS-quadratic functions are formed.
Give your answers in the form
\[\frac{-p\pm q\sqrt{s}}{2}\]
where $p,q,s\in\mathbb{Z}$.
▶️Answer/Explanation
Solution: –
(a) $m=2, c=-1$
$r=\frac{1}{2}$
$\frac{1}{2},-1$
EITHER
$d=\left(\frac{1}{2}-2=-1-\frac{1}{2}\right)=-\frac{3}{2}$
OR
this sequence has a common difference of $-\frac{3}{2}$
OR
the (arithmetic) mean of 2 and -1 is $\frac{1}{2}$
THEN
hence L(x)=2x-1 is an AS-linear function
(b) (i) $(L(r)=0\Rightarrow)mr+c=0$
$r=-\frac{c}{m}$
(ii) METHOD 1
EITHER
attempts to use $(d=)r-m=c-r$
$(d=)\frac{c}{m}-m=c-\left(-\frac{c}{m}\right)$
removes the denominator m from their expression involving m and c
$m^{2}+cm+2c=0$ (or equivalent)
OR
attempts to use
$\frac{m+c}{2}=r$
removes the denominator m from their expression involving and c
$m^{2}+cm+2c=0$ or equivalent)
OR
attempts to use $c=m+2d$
$c=m+2\left(-\frac{c}{m}-m\right)$
removes the denominator $m$ from their expression involving $m$ and $c$
$m^{2}+cm+2c=0$ (or equivalent)
OR
attempts to use $r=m+d$ and $c=m+2d~ (c=m+2(r-m))$
$m^{2}+dm+m+2d=0$ or equivalent)
substitutes $d=\frac{c-m}{2}$ into their expression involving m and d
$m^{2}+cm+2c=0$ (or equivalent)
THEN
$c(m+2)=-m^{2}\Rightarrow c=-\frac{m^{2}}{m+2}$
so \(L(x)=mx-\frac{m^{2}}{m+2}\)
not accept working backwards from the AG.
METHOD 2
considers $L(x)=mx-mr$
attempts to use $(d=)r-m=c-r$
(M1) for attempting to use $(d=)m-r=r-c_{-}$
$(d=)r-m=-mr-r$
attempts to express in terms of m
$2r+mr=m\Rightarrow r=\frac{m}{m+2}$
so \(L(x)=mx-\frac{m^{2}}{m+2}\)
(iii) $m\ne-2$ $(m\ne0)$
(c) Attempts to find an integer value of m
e.g. uses the result that $m+2$ exactly divides 2 OR uses a table OR uses a graph
and slider OR uses systematic trial and error
$m=-4$ OR $m=-3$
-4,2,8 OR -3,3,9
$L(x)=-4x+8$, $L(x)=-3x+9$
(d) (i)
$-\frac{b}{a}$
(ii)
$\frac{c}{a}$
(e) (i) b-a
(ii) attempts to eliminate $r_{2}$
$2r_{1}=-\frac{b}{a}-(b-a)\Rightarrow 2r_{1}=\frac{a^{2}-ab-b}{a}$ (or equivalent)
so $r_{1}=\frac{a^{2}-ab-b}{2a}$
(iii) METHOD 1
EITHER
$(r_{1}=)\frac{a+b}{2}$
attempts to equate two expressions for either $r_{1}$ or 2r_{1} in terms of a and b
\[\frac{a+b}{2}=\frac{a^{2}-ab-b}{2a}\] OR \[a+b=\frac{a^{2}-ab-b}{a}\]
OR
b-r_{1}=r_{1}-a
attempts to use b-r_{1}=r_{1}-a with $r_{1}=\frac{a^{2}-ab-b}{2a}$
\[b-\left(\frac{a^{2}-ab-b}{2a}\right)=\frac{a^{2}-ab-b}{2a}-a\]
OR
$(r_{1}=)a+d$
attempts to use $r_{1}=a+d$ with $r_{1}=\frac{a^{2}-ab-b}{2a}$ and $d=\frac{b-a}{2}$
\[\frac{a^{2}-ab-b}{2a}=a+\frac{b-a}{2}\]
THEN
$2a^{2}+2ab=2a^{2}-2ab-2b$ OR $a^{2}+ab=a^{2}-ab-b$
$4ab+2b=0$ OR $2ab+b=0$
$2b(2a+1)=0$ OR $b(2a+1)=0$
so $b=0$ or $a=-\frac{1}{2}$
METHOD 2
(b=)a+2d OR (r_{1}=)a+d
attempts to equate two expressions for either $r_{1}$ or $2r_{1}$ in terms of a and d
\[a+d=\frac{a^{2}-a(a+2d)-(a+2d)}{2a}\] OR \[2(a+d)=\frac{a^{2}-a(a+2d)-(a+2d)}{a}\]
\[2a^{2}+4ad+a+2d=0\]
\[(2a+1)(a+2d)=0\]
Do not accept numerical verification from the AG.
so $b=0$ or $a=-\frac{1}{2}$
(f) METHOD 1
$r_{1}=\frac{a}{2}$ $r_{2}=-\frac{a}{2}$ OR $d=-\frac{a}{2}$
$c=-a$
attempts to find the values of a
EITHER
the roots of $ax^{2}-a=0$ are ±1 and $\frac{a}{2}=\pm1$
OR
substitutes $x=\pm\frac{a}{2}$ into $ax^{2}-a=0$ giving $\frac{a^{3}}{4}-a=0$
OR
$(r_{1}r_{2}=)\frac{c}{a}=-\frac{a^{2}}{4}\Rightarrow c=-\frac{a^{3}}{4}$ and so $-a=-\frac{a^{3}}{4}\Rightarrow\frac{a^{3}}{4}-a=0$
OR
$c-r_{1}=r_{2}-b\Rightarrow\frac{a^{3}}{4}-\left(-\frac{a}{2}\right)=\left(\frac{a}{2}\right)-b\Rightarrow\frac{a^{3}}{4}-a=0$
THEN
$a=\pm2$
$(r_{1}=\pm1, b=0, r_{2}=\mp1, c=\mp2)$
so $Q(x)=2x^{2}-2$, $Q(x)=-2x^{2}+2$
METHOD 2
$r_{1}=-d$ OR $r_{2}=d$ OR $a=-2d$
$c=2d$
attempts to find the values of d
EITHER
the roots of $-2dx^{2}+2d=0$ are $\pm1$
OR
substitutes $x=\pm d$ into $-2dx^{2}+2d=0$ giving $-2d^{3}+2d=0$
OR
attempts to use $r_{1}r_{2}=\frac{c}{a}$ to form $-d^{2}=\frac{2d}{-2d}$
THEN
$d=\pm1$
$(a=\pm2, r_{1}=\pm1, b=0, r_{2}=\mp1, c=\mp2)$
so $Q(x)=2x^{2}-2$, $Q(x)=-2x^{2}+2$
METHOD 3
$a=2r_{1}$ OR $r_{2}=-r_{1}$ OR $d=-r_{1}$
$c=-2r_{1}$
attempts to find the values of $r_{1}$
EITHER
the roots of $2r_{1}x^{2}-2r_{1}=0$ are $\pm1$
OR
substitutes $x=\pm r_{1}$ into $2r_{1}x^{2}-2r_{1}=0$ giving $2r_{1}^{3}-2r_{1}=0$
OR
attempts to use $r_{1}r_{2}=\frac{c}{a}$ to form $-r_{1}^{2}=\frac{-2r_{1}}{2r_{1}}$
THEN
$r_{1}=\pm1$
$(a=\pm2, r_{1}=\pm1, b=0, r_{2}=\mp1, c=\mp2)$
so $Q(x)=2x^{2}-2$, $Q(x)=-2x^{2}+2$
(g) (i) attempts to express $r_{1}$ in terms of b with $a=-\frac{1}{2}$
Note: Do not award (M1) if $a=\frac{1}{2}$ is used.
EITHER
uses $r_{1}=\frac{a+b}{2}$
OR
uses $r_{1}=\frac{a^{2}-ab-b}{2a}$
OR
uses $r_{1}-a=b-r_{1}$
THEN
$r_{1}=\frac{2b-1}{4}=\frac{b}{2}-\frac{1}{4}=\frac{b-\frac{1}{2}}{2}$
(ii) METHOD 1
EITHER
substitutes their expression for $r_{1}$ with $a=-\frac{1}{2}$ into $Q(x)=0$
\[Q\left(\frac{2b-1}{4}\right)=0\Rightarrow-\frac{1}{2}\left(\frac{2b-1}{4}\right)^{2}+b\left(\frac{2b-1}{4}\right)+c=0\]
OR
$r_{2}=\frac{6b+1}{4}=\left(\frac{3b}{2}+\frac{1}{4}\right)$
substitutes their expression for $r_{2}$ with $a=-\frac{1}{2}$ into $Q(x)=0$
\[Q\left(\frac{6b+1}{4}\right)=0\Rightarrow-\frac{1}{2}\left(\frac{6b+1}{4}\right)^{2}+b\left(\frac{6b+1}{4}\right)+c=0\]
THEN
$c=\frac{4b+1}{2}=\left(2b+\frac{1}{2}\right)$ (seen anywhere)
\[4b^{2}+20b+5=0\]
attempts to solve their quadratic in b
\[b=\frac{-5\pm2\sqrt{5}}{2}\]
substitutes into $c=\frac{4b+1}{2}$
\[c=\frac{-9\pm4\sqrt{5}}{2}\]
METHOD 2
substitutes their expressions for $r_{1}$ and $r_{2}$ with $a=-\frac{1}{2}$ into $Q(x)$
\[-\frac{1}{2}\left(x-\left(\frac{2b-1}{4}\right)\right)\left(x-\left(\frac{6b+1}{4}\right)\right)\]
\[-\frac{1}{2}x^{2}+bx-\frac{3}{8}b^{2}+\frac{1}{8}b+\frac{1}{32}\]
$c=\frac{4b+1}{2}=\left(2b+\frac{1}{2}\right)$ (seen anywhere)
\[2b+\frac{1}{2}=-\frac{3}{8}b^{2}+\frac{1}{8}b+\frac{1}{32}\]
\[\frac{2b}{2}+\frac{1}{2}=-\frac{3}{8}b^{2}+\frac{1}{8}b+\frac{1}{32}\]
\[4b^{2}+20b+5=0\]
attempts to solve their quadratic in b
\[b=\frac{-5\pm2\sqrt{5}}{2}\]
substitutes into
\[c=\frac{4b+1}{2}\]
\[c=\frac{-9\pm4\sqrt{5}}{2}\]
METHOD 4
attempts to equate two expressions for $r_{1}$ with $a=-\frac{1}{2}$
\[\frac{-b\pm\sqrt{b^{2}+2c}}{-1}=\frac{2b-1}{4}(\pm\sqrt{b^{2}+2c}=\frac{2b+1}{4})\]
\[c=\frac{4b+1}{2}(=\left(2b+\frac{1}{2}\right))\]
(seen anywhere)
\[12b^{2}-4b-1+32\left(2b+\frac{1}{2}\right)=0\]
$(4b^{2}+20b+5=0)$
attempts to solve their quadratic in b
\[b=\frac{-5\pm2\sqrt{5}}{2}\]
substitutes into
\[c=\frac{4b+1}{2}\]
\[c=\frac{-9\pm4\sqrt{5}}{2}\]
METHOD 5
EITHER
$r_{1}=d-\frac{1}{2}$
substitutes their expression for $r_{1}$ in terms of d with $a=-\frac{1}{2}$ into $Q(x)=0$
\[Q\left(d-\frac{1}{2}\right)=0\Rightarrow-\frac{1}{2}\left(d-\frac{1}{2}\right)^{2}+b\left(d-\frac{1}{2}\right)+c(=0)\]
OR
$r_{2}=3d-\frac{1}{2}$
substitutes their expression for $r_{2}$ in terms of d with $a=-\frac{1}{2}$ into $Q(x)=0$
\[Q\left(3d-\frac{1}{2}\right)=0\Rightarrow-\frac{1}{2}\left(3d-\frac{1}{2}\right)^{2}+b\left(3d-\frac{1}{2}\right)+c(=0)\]
THEN
$b=2d-\frac{1}{2}$ and $c=4d-\frac{1}{2}$ (seen anywhere)
\[4d^{2}+8d-1=0\]
attempts to solve their quadratic in d
\[d=\frac{-2\pm\sqrt{5}}{2}\]
\[b=\frac{-5\pm2\sqrt{5}}{2}\]
substitutes into
\[c=\frac{4b+1}{2}\]
\[c=\frac{-9\pm4\sqrt{5}}{2}\]
METHOD 6
$r_{1}=d-\frac{1}{2}$ and $r_{2}=3d-\frac{1}{2}$
substitutes their expressions for $r_{1}$ and $r_{2}$ in terms of d with $a=-\frac{1}{2}$ into $r_{1}r_{2}=\frac{c}{a}$
\[\left(d-\frac{1}{2}\right)\left(3d-\frac{1}{2}\right)=\frac{c}{-\frac{1}{2}}\] (or equivalent)
$c=4d-\frac{1}{2}$ (seen anywhere)
\[4d^{2}+8d-1=0\]
attempts to solve their quadratic in d
\[d=\frac{-2\pm\sqrt{5}}{2}\]
\[b=\frac{-5\pm2\sqrt{5}}{2}\]
substitutes into
\[c=\frac{4b+1}{2}\]
\[c=\frac{-9\pm4\sqrt{5}}{2}\]