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
Solution:-
(a) Equation of the horizontal asymptote for \(f(x) = \frac{2(x + 3)}{3(x + 2)}\), \(x \neq -2\)
As \(x \to \infty\) or \(x \to -\infty\), divide numerator and denominator by \(x\):
\[ f(x) = \frac{2(1 + \frac{3}{x})}{3(1 + \frac{2}{x})} \to \frac{2(1 + 0)}{3(1 + 0)} = \frac{2}{3} \]
Horizontal asymptote: \(y = \frac{2}{3}\).
(b) For \(g(x) = mx + 1\), \(m \in \mathbb{R}, m \neq 0\)
(i) Number of solutions to \(f(x) = g(x)\) for \(m > 0\)
\[ \frac{2(x + 3)}{3(x + 2)} = mx + 1 \]
Cross-multiply (since \(x \neq -2\)):
\[ 2(x + 3) = 3(mx + 1)(x + 2) \]
\[ 2x + 6 = 3m(x^2 + 2x) + 3(x + 2) \]
\[ 2x + 6 = 3mx^2 + 6mx + 3x + 6 \]
\[ 0 = 3mx^2 + (6m + 1)x \]
\[ x(3mx + 6m + 1) = 0 \]
Solutions:
– \(x = 0\)
– \(3mx + 6m + 1 = 0\)
\[ x = -\frac{6m + 1}{3m} = -2 – \frac{1}{3m} \]
For \(m > 0\), \(-\frac{1}{3m} < 0\), so \(x = -2 – \frac{1}{3m} < -2\). Both \(x = 0\) and \(x = -2 – \frac{1}{3m}\) are real and distinct. Thus, there are 2 solutions.
(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):
\[ (6m + 1)^2 = 0 \]
\[ 6m + 1 = 0 \]
\[ m = -\frac{1}{6} \]
————Markscheme————
(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)
$-\frac{1}{6}<m<0$
Question
Consider the function \(f(x)=\frac{4x+2}{x-2}, x\neq 2\)
(a) Sketch the graph of \(y=f(x)\). On your sketch, indicate the values of any axis intercepts and label any asymptotes with their equations.
(b) Write down the range of f.
Consider the function \(g(x)=x^{2}+bx+c\). The graph of g has an axis of symmetry at \(x=2\).
The two roots of \(g(x)=0\) are \(-\frac{1}{2}\) and p, where \(p\in \mathbb{Q}\).
(c) Show that \(p=\frac{9}{2}\).
(d) Find the value of b and the value of c.
(e) Find the y-coordinate of the vertex of the graph of \(y=g(x)\).
(f)Find the product of the solutions of the equation \(f(x)=g(x)\).
▶️Answer/Explanation
Ans:
(a) Step 1] Domain of function
The denominator
, so:
Domain:
, i.e.,
Step 2] X and Y intercept
Y-intercept (when x = 0 ):
Y-intercept:
X-intercept (when f(x)=0):
X-intercept:
The denominator is zero at
As
(left) f(x)→−∞ and As
→ (right) f(x)→+∞ , hence x=2 is the vertical asymptote to the curve
(b) Range \(y\neq 4\) or y
(c) We know that,
Axis of symmetry at
= 2 (given) , for \(g(x)=x^{2}+bx+c\)
Axis of symmetry at
= 2, put a =1,we get b = -4,
Applying the product of roots, we get
\( \frac{-1}{2}\times p = \frac{c}{a}= \frac{c}{1}\) , put p = \( \frac{9}{2}\)
\(\frac{-1}{2}\times \frac{9}{2}\ = \frac{c}{a}= \frac{c}{1}\)
Therefore, c = \( \frac{-1}{2}\times \frac{9}{2} = \frac{-9}{4}\)therefore \(g(x)=x^{2}+(-4)x+\frac{-9}{4} =(x+\frac{1}{2})(x-\frac{9}{2}) \)
(e) Substituting the axis of symmetry \(x=2\) into the \(g(x)=x^{2}+(-4)x+\frac{-9}{4}\) we get the y coordinate of vertex as
\(y = 2^{2}+(-4)2+\frac({-9}{4})\)
\(y=-\frac{25}{4}\)
Solving \(f(x)=g(x)\) can be done by following ways
Question
Let \(f\) be a quadratic function. Part of the graph of \(f\) is shown below.
The vertex is at P(\(4\), \(2\)) and the y-intercept is at Q(\(0\), \(6\)) .
a.Write down the equation of the axis of symmetry.[1]
b.The function f can be written in the form \(f(x) = a{(x – h)^2} + k\) .
Write down the value of h and of k .[2]
c.The function f can be written in the form \(f(x) = a{(x – h)^2} + k\) .
Find a .[3]
▶️Answer/Explanation
Markscheme
a.\(x = 4\) (must be an equation) A1 N1
[1 mark]
\(h = 4\) , \(k = 2\) A1A1 N2
[2 marks]
attempt to substitute coordinates of any point on the graph into f (M1)
e.g. \(f(0) = 6\) , \(6 = a{(0 – 4)^2} + 2\) , \(f(4) = 2\)
correct equation (do not accept an equation that results from \(f(4) = 2\) ) (A1)
e.g. \(6 = a{( – 4)^2} + 2\) , \(6 = 16a + 2\)
\(a = \frac{4}{{16}}\left( { = \frac{1}{4}} \right)\) A1 N2
[3 marks]
Question
Let \(f(x) = \sin x + \frac{1}{2}{x^2} – 2x\) , for \(0 \le x \le \pi \) .
Let \(g\) be a quadratic function such that \(g(0) = 5\) . The line \(x = 2\) is the axis of symmetry of the graph of \(g\) .
The function \(g\) can be expressed in the form \(g(x) = a{(x – h)^2} + 3\) .
a.Find \(f'(x)\) .[3]
(ii) Find the value of \(a\) .[4]
d.Find the value of \(x\) for which the tangent to the graph of \(f\) is parallel to the tangent to the graph of \(g\) . [6]
▶️Answer/Explanation
Markscheme
a.\(f'(x) = \cos x + x – 2\) A1A1A1 N3
Note: Award A1 for each term.
[3 marks]
recognizing \(g(0) = 5\) gives the point (\(0\), \(5\)) (R1)
recognize symmetry (M1)
eg vertex, sketch 
\(g(4) = 5\) A1 N3
[3 marks]
(i) \(h = 2\) A1 N1
(ii) substituting into \(g(x) = a{(x – 2)^2} + 3\) (not the vertex) (M1)
eg \(5 = a{(0 – 2)^2} + 3\) , \(5 = a{(4 – 2)^2} + 3\)
working towards solution (A1)
eg \(5 = 4a + 3\) , \(4a = 2\)
\(a = \frac{1}{2}\) A1 N2
[4 marks]
\(g(x) = \frac{1}{2}{(x – 2)^2} + 3 = \frac{1}{2}{x^2} – 2x + 5\)
correct derivative of \(g\) A1A1
eg \(2 \times \frac{1}{2}(x – 2)\) , \(x – 2\)
evidence of equating both derivatives (M1)
eg \(f’ = g’\)
correct equation (A1)
eg \(\cos x + x – 2 = x – 2\)
working towards a solution (A1)
eg \(\cos x = 0\) , combining like terms
\(x = \frac{\pi }{2}\) A1 N0
Note: Do not award final A1 if additional values are given.
[6 marks]