Question
Part of the graph of a function, f , is shown in the following diagram. The graph of y = f (x) has a y-intercept at (0, 3), an x-intercept at (a, 0) and a horizontal asymptote y = -2.
Consider the function g(x) = |f (| x |)|.
(a) On the following grid, sketch the graph of y = g(x), labelling any axis intercepts and giving the equation of the asymptote
b) Find the possible values of k such that \({(g(x))}^2 = k\) has exactly two solutions .
Reflect f in x or y axis.
▶️Answer/Explanation
(a)
(b) To determine the possible values of \( k \) such that the equation \( g(x) = f'(|x|) = k \) has exactly two solutions, we need to analyze the graph of \( g(x) \) based on the given graph of \( f(x) \).
The graph of \( f(x) \) has a \( y \)-intercept at \( (0, 3) \), an \( x \)-intercept at \( (\alpha, 0) \), and a horizontal asymptote at \( y = -2 \).
Since \( g(x) \) involves the absolute value of \( f(x) \) and the absolute value of \( x \), the graph of \( g(x) \) will be the top half of the graph of \( f(x) \) reflected about the \( x \)-axis for \( x < 0 \).
The horizontal asymptote at \( y = -2 \) for \( f(x) \) becomes \( y = 2 \) for \( g(x) \), and the \( y \)-intercept remains at \( (0, 3) \). Therefore, the graph of \( g(x) \) will have a minimum point at \( (0, 3) \), and it will approach the horizontal line \( y = 2 \) from above.
For the equation \( g(x) = k \) to have exactly two solutions, the horizontal line \( y = k \) must intersect the graph of \( g(x) \) at exactly two points.
This can only occur if:
\(
4 \leq k < 9
\)
since the \( y \)-value of \( g(x) \) at \( x = 0 \) is 3, and the graph approaches \( y = 2 \) without ever touching it again.
Hence, the line \( y = k \) must be above \( y = 3 \) but below \( y = 9 \), which is the maximum value of \( f(x) \) before taking the absolute value.
Thus, the correct range for \( k \) is:
\(
4 \leq k < 9
\)
Question
The function f is defined by \(f (x) = \frac{7x+7}{2x-4} \) for x \(\epsilon\) R , x \(\neq\) 2.
(a) Find the zero of f (x).
(b) For the graph of y = f (x), write down the equation of
(i) the vertical asymptote;
(ii) the horizontal asymptote.
(c) Find \(f^{-1}\), the inverse function of f (x).
▶️Answer/Explanation
(a)
To find the zero of \( f(x) \), we need to solve for \( x \) when \( f(x) = 0 \).
Starting with the equation:
\(
0 = \frac{7x + 7}{2x – 4}
\)
We can multiply both sides of the equation by the denominator \( 2x – 4 \) to eliminate the fraction:
\(
0 \cdot (2x – 4) = 7x + 7
\)
Simplifying, we get:
\(
0 = 7x + 7
\)
Now, to find the value of \( x \), we subtract 7 from both sides of the equation:
\(
-7 = 7x
\)
Dividing both sides by 7 gives us:
\(
x = -1
\)
(b) (i) The vertical asymptote of a rational function occurs where the denominator is zero while the numerator is nonzero, leading to undefined values for the function. For the given function
\(
f(x) = \frac{7x + 7}{2x – 4}
\)
the denominator \( 2x – 4 \) is zero when \( x = 2 \). However, the numerator \( 7x + 7 \) is not zero at \( x = 2 \) (since \( 7(2) + 7 = 21 \), which is not equal to zero). Therefore, the equation of the vertical asymptote is \( x = 2 \).
It is important to note that the value \( x = -2 \) is excluded from the domain, but it does not affect the vertical asymptote since the denominator is not zero at \( x = -2 \).
(ii) The horizontal asymptote of the function \( f(x) = \frac{7x + 7}{2x – 4} \) is determined by examining the leading coefficients of the numerator and denominator since both are polynomials of the same degree. In this case, both the numerator and the denominator are first-degree polynomials.
To find the horizontal asymptote, we divide the leading coefficient of the numerator by the leading coefficient of the denominator. For the given function, the leading coefficient of the numerator (7) divided by the leading coefficient of the denominator (2) gives us:
\(
y = \frac{7}{2}
\)
Therefore, the equation of the horizontal asymptote is \( y = 3.5 \).
(c) To determine the inverse of the function \( f(x) = \frac{7x + 7}{2x – 4} \), we start by replacing \( f(x) \) with \( y \) and then interchange the roles of \( x \) and \( y \). This yields the equation:
\(
x = \frac{7y + 7}{2y – 4}
\)
By cross-multiplying, we obtain:
\(
2xy – 4x = 7y + 7
\)
We then group all terms containing \( y \) on one side to get:
\(
2xy – 7y = 4x + 7
\)
and factor out \( y \) as:
\(
y(2x – 7) = 4x + 7
\)
Finally, we isolate \( y \) to find the inverse function:
\(
f^{-1}(x) = \frac{4x + 7}{2x – 7}
\)
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
