Question
A function $f$ is defined by $f(x)=\frac{2(x+3)}{3(x+2)}$, where $x \in \mathbb{R}, x \neq -2$.
The graph $y = f(x)$ is shown below.
(a) Write down the equation of the horizontal asymptote.
Consider $g(x) = mx + 1$, where $m \in \mathbb{R}, m \neq 0$.
(b) (i) Write down the number of solutions to $f(x) = g(x)$ for $m > 0$.
(ii) Determine the value of $m$ such that $f(x) = g(x)$ has only one solution for $x$.
(iii) Determine the range of values for $m$, where $f(x) = g(x)$ has two solutions for $x \geq 0$.
▶️Answer/Explanation
Detailed solution
Part (a): Equation of the Horizontal Asymptote
To find the horizontal asymptote of \( f(x) = \frac{2(x+3)}{3(x+2)} \), we need to examine the behavior of the function as \( x \to \pm \infty \).
The function is a rational function of the form \( f(x) = \frac{2(x+3)}{3(x+2)} \). Let’s rewrite it by expanding the numerator and denominator:
– Numerator: \( 2(x+3) = 2x + 6 \)
– Denominator: \( 3(x+2) = 3x + 6 \)
So, \( f(x) = \frac{2x + 6}{3x + 6} \). Now, divide the numerator and denominator by the highest power of \( x \), which is \( x \):
\[
f(x) = \frac{\frac{2x}{x} + \frac{6}{x}}{\frac{3x}{x} + \frac{6}{x}} = \frac{2 + \frac{6}{x}}{3 + \frac{6}{x}}
\]
As \( x \to \pm \infty \), the terms \( \frac{6}{x} \to 0 \), so:
\[
f(x) \to \frac{2 + 0}{3 + 0} = \frac{2}{3}
\]
Thus, the horizontal asymptote is at \( y = \frac{2}{3} \).
Part (b): Analyzing \( f(x) = g(x) \)
We’re now dealing with \( g(x) = mx + 1 \), where \( m \in \mathbb{R} \), \( m \neq 0 \), and we need to solve \( f(x) = g(x) \), i.e., \( \frac{2(x+3)}{3(x+2)} = mx + 1 \), under different conditions.
(b)(i): Number of Solutions for \( m > 0 \)
First, let’s set up the equation:
\[
\frac{2(x+3)}{3(x+2)} = mx + 1
\]
Multiply through by the denominator \( 3(x+2) \) to clear the fraction:
\[
2(x+3) = (mx + 1) \cdot 3(x+2)
\]
\[
2x + 6 = 3mx^2 + 6mx + 3x + 6
\]
\[
0 = 3mx^2 + 6mx + 3x + 6 – 2x – 6
\]
\[
0 = 3mx^2 + 6mx + 3x + 2x = 3mx^2 + (6m + 1)x
\]
This is a quadratic equation in \( x \):
\[
3mx^2 + (6m + 1)x = 0
\]
Factor out \( x \):
\[
x (3mx + 6m + 1) = 0
\]
The solutions are:
1. \( x = 0 \)
2. \( 3mx + 6m + 1 = 0 \)
Solving the second equation:
\[
3mx + 6m + 1 = 0
\]
\[
3mx = -6m – 1
\]
\[
x = \frac{-6m – 1}{3m}
\]
Since \( m \neq 0 \), this gives two distinct solutions: \( x = 0 \) and \( x = \frac{-6m – 1}{3m} \).
Hence, the number of solutions is 2.
(b) (ii) Value of \( m \) such that \( f(x) = g(x) \) has only one solution for \( x \).
For one solution, the quadratic \( 3mx^2 + (6m + 1)x = 0 \) must have a double root (discriminant = 0):
\[
(6m + 1)^2 = 0
\]
\[
6m + 1 = 0
\]
\[
m = -\frac{1}{6}
\]
(b)(iii): Range of \( m \) for Two Solutions for \( x \geq 0 \)
From (b)(i), for \( m > 0 \), we have two solutions: \( x = 0 \) (which is \( \geq 0 \)) and \( x = \frac{-6m – 1}{3m} < 0 \). This gives only one solution for \( x \geq 0 \).
For \( m < 0 \), let’s compute the solutions again:
\[
x = 0 \quad \text{and} \quad x = \frac{-6m – 1}{3m}
\]
If \( m < 0 \), then \( 3m < 0 \), and the numerator \( -6m – 1 \) (where \( -6m > 0 \)) needs to be analyzed:
– If \( -6m – 1 > 0 \), i.e., \( -6m > 1 \), then \( m < -\frac{1}{6} \), the second \( x \) is positive.
– If \( -6m – 1 < 0 \), i.e., \( m > -\frac{1}{6} \), the second \( x \) is negative.
So, for \( m < -\frac{1}{6} \), the second solution is positive, giving two solutions at \( x \geq 0 \): \( x = 0 \) and \( x = \frac{-6m – 1}{3m} > 0 \).
At \( m = -\frac{1}{6} \), we have one solution. For \( m > -\frac{1}{6} \), the second solution becomes negative, giving only one solution at \( x \geq 0 \).
hence, range of \( m \) is \( m < -\frac{1}{6} \).
………………………Markscheme…………………..
Solution:-
(a) $y=\frac{2}{3}$ (must be written as equation with $y=$)
(b) (i) 2
(ii) EITHER
$\frac{2(x+3)}{3(x+2)}=mx+1$
attempt to expand to obtain a quadratic equation
$2x+6=3mx^2+6mx+3x+6$
$3mx^2+(6m+1)x=0$ OR $3mx^2+6mx+x=0$
recognition that discriminant $\Delta=0$ for one solution
$(6m+1)^2=0$
OR
$\frac{2(x+3)}{3(x+2)}=mx+1$
attempt to expand to obtain a quadratic equation
$2x+6=3mx^2+6mx+3x+6$
$3mx^2+(6m+1)x=0$ OR $3mx^2+6mx+x=0$
attempt to solve their quadratic for $x$ and equating their solutions
$x(3mx+6m+1)=0$
$x=0$ OR $x=-\frac{6m+1}{3m}(=0)$
$-\frac{6m+1}{3m}=0$
OR
attempt to find $f'(x)$ using the quotient rule
$f'(x)=\frac{2}{3}(\frac{(x+2)-(x+3)}{(x+2)^2})=\frac{-2}{3(x+2)^2}$ OR $\frac{2(3x+6)-3(2x+6)}{(3x+6)^2}$ or equivalent
recognition that $m$ is the derivative of $f(x)$ at $x=0$
THEN
$\Rightarrow m=-\frac{1}{6}$
(iii)
$-\frac{1}{6}<m<0$