Question
The functions \( f \) and \( g \) are defined by:
\( f(x) = \ln(2x – 9) \), where \( x > \frac{9}{2} \)
\( g(x) = 2\ln x – \ln d \), where \( x > 0 \), \( d \in \mathbb{R}^+ \).
(a) State the equation of the vertical asymptote to the graph of \( y = g(x) \). [1]
(b) (i) Show that, at the points of intersection, \( x^2 – 2dx + 9d = 0 \).
(ii) Hence show that \( d^2 – 9d > 0 \).
(iii) Find the range of possible values of \( d \). [9]
(c) In the case where \( d = 10 \), find the value of \( q – p \). Express your answer in the form \( a\sqrt{b} \), where \( a, b \in \mathbb{Z}^+ \). [5]
▶️Answer/Explanation
Detail Solution
(a)
To find the vertical asymptote of the graph of $ y=g(x)$
We can rewrite $g(x)$ using the logarithmic property $ ln{a}-ln{b}=ln (\frac{a}{b}) $
Next, we find the vertical asymptote. The logarithmic function
because the function is undefined for $x \leq 0 , g(x)=ln( \frac{x^2}{d})$
Vertical Asymptote is y axis and equation of y axis is x=0
(b) (i)
Step 1: Set the functions \( f(x) \) and \( g(x) \) equal to each other.
To find the points of intersection, we need to solve the equation \( f(x) = g(x) \).
This gives us:
\(\ln(2x – 9) = 2\ln x – \ln d\).
Step 2: Use properties of logarithms to simplify the equation.
We can rewrite \( 2\ln x – \ln d \) using the properties of logarithms:
\(\ln(2x – 9) = \ln\left(\frac{x^2}{d}\right)\).
This implies:
\(2x – 9 = \frac{x^2}{d}\).
Step 3: Multiply both sides by \( d \) to eliminate the fraction.
This gives us:
\(d(2x – 9) = x^2\).
Rearranging this equation results in:
\(x^2 – 2dx + 9d = 0\).
Thus, we have shown that at the points of intersection, the equation \( x^2 – 2dx + 9d = 0 \) holds true.
(ii)
Step 1: Express \( g(x) \) in terms of \( f(x) \).
We have \( g(x) = 2\ln x – \ln d \). Using the properties of logarithms, we can rewrite this as:
$$ g(x) = \ln(x^2) – \ln(d) = \ln\left(\frac{x^2}{d}\right) $$
Step 2: Set \( g(x) \) equal to \( f(x) \).
Since we need to show that \( d^2 – 9d > 0 \), we can set \( g(x) = f(x) \):
$$ \ln\left(\frac{x^2}{d}\right) = \ln(2x – 9) $$
Step 3: Exponentiate both sides to eliminate the logarithm.
This gives us:
$$ \frac{x^2}{d} = 2x – 9 $$
Multiplying both sides by \( d \) results in:
$$ x^2 = d(2x – 9) $$
Step 4: Rearrange the equation.
We can rewrite this as:
$$ x^2 – 2dx + 9d = 0 $$
Step 5: Analyze the discriminant of the quadratic equation.
For the quadratic \( x^2 – 2dx + 9d = 0 \) to have real solutions, the discriminant must be non-negative:
The discriminant \( D \) is given by:
$$ D = b^2 – 4ac = (-2d)^2 – 4(1)(9d) = 4d^2 – 36d $$
Setting the discriminant greater than zero for real solutions, we have:
$$ 4d^2 – 36d > 0 $$
Step 6: Factor the inequality.
Factoring out \( 4d \) gives:
$$ 4d(d – 9) > 0 $$
Dividing by 4 (which is positive) does not change the inequality:
$$ d(d – 9) > 0 $$
Step 7: Determine the intervals for \( d \).
The inequality \( d(d – 9) > 0 \) holds when \( d < 0 \) or \( d > 9 \). Since \( d \in \mathbb{R}^+ \), we only consider \( d > 9 \).
Step 8: Conclude the proof.
Thus, we have shown that for \( d > 9 \), the condition \( d^2 – 9d > 0 \) is satisfied.
(iii)
From the second part ,it’s clear \( d^2 – 9d > 0 \)
$$ d(d-9)>0$$
$$ d>0 ,(d-9)>0 \Rightarrow d>0 ,d>9 $$
solving above inequation $$d>9$$
(c)
Step 1: Define the functions with the given value of \( d \).
We have $$ g(x) = 2\ln x – \ln 10 $$
Step 2: Set the functions equal to each other to find \( x \) values.
We need to solve the equation \( f(x) = g(x) \):
$$ \ln(2x – 9) = 2\ln x – \ln 10$$
Step 3: Use properties of logarithms to simplify the equation.
We can rewrite \( g(x) \) as:
$$ \ln(2x – 9) = \ln\left(\frac{x^2}{10}\right)$$
This gives us the equation:
$$2x – 9 = \frac{x^2}{10}$$
Step 4: Multiply through by 10 to eliminate the fraction.
$$10(2x – 9) = x^2 $$ leads to:
$$20x – 90 = x^2$$
Step 5: Rearrange the equation into standard quadratic form.
This becomes:
$$ x^2 – 20x + 90 = 0$$
Step 6: Use the quadratic formula to find the roots.
The quadratic formula is given by:
$$ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
Here, \(a = 1\), \(b = -20\), and \(c = 90\).
Calculating the discriminant:
$$ b^2 – 4ac = (-20)^2 – 4(1)(90) = 400 – 360 = 40$$
Step 7: Substitute into the quadratic formula.
Thus, we have:
$$ x = \frac{20 \pm \sqrt{40}}{2}$$.
This simplifies to:
$$ x = \frac{20 \pm 2\sqrt{10}}{2} = 10 \pm \sqrt{10}$$
Step 8: Identify the values of \( p \) and \( q \).
Let \( p = 10 – \sqrt{10} \) and \( q = 10 + \sqrt{10} \).
Step 9: Calculate \( q – p \).
Thus,
$$ q – p = (10 + \sqrt{10}) – (10 – \sqrt{10}) = 2\sqrt{10}$$
Step 10: Express the answer in the required form \( a\sqrt{b} \).
Here, \( a = 2 \) and \( b = 10 \).
————Markscheme—————–
(a) The vertical asymptote is \( x = 0 \).
(b) (i) Setting \( f(x) = g(x) \):
\( \ln(2x – 9) = 2\ln x – \ln d \)
\( 2x – 9 = \frac{x^2}{d} \)
\( x^2 – 2dx + 9d = 0 \)
(ii) For two distinct real roots, the discriminant must be positive:
\( (2d)^2 – 4 \times 1 \times 9d > 0 \)
\( 4d^2 – 36d > 0 \)
\( d^2 – 9d > 0 \)
(iii) Solving \( d^2 – 9d > 0 \), we get \( d < 0 \) or \( d > 9 \). Since \( d \in \mathbb{R}^+ \), \( d > 9 \).
(c) For \( d = 10 \), solve \( x^2 – 20x + 90 = 0 \):
\( x = 10 \pm \sqrt{10} \)
\( q – p = 2\sqrt{10} \)
(a) The vertical asymptote is \( x = 0 \).
(b) (i) Setting \( f(x) = g(x) \):
\( \ln(2x – 9) = 2\ln x – \ln d \)
\( 2x – 9 = \frac{x^2}{d} \)
\( x^2 – 2dx + 9d = 0 \)
(ii) For two distinct real roots, the discriminant must be positive:
\( (2d)^2 – 4 \times 1 \times 9d > 0 \)
\( 4d^2 – 36d > 0 \)
\( d^2 – 9d > 0 \)
(iii) Solving \( d^2 – 9d > 0 \), we get \( d < 0 \) or \( d > 9 \). Since \( d \in \mathbb{R}^+ \), \( d > 9 \).
(c) For \( d = 10 \), solve \( x^2 – 20x + 90 = 0 \):
\( x = 10 \pm \sqrt{10} \)
\( q – p = 2\sqrt{10} \)
Question:
A function f is defined by \(f(x) = \frac{1}{x^{2}-2x-3},\) where x ∈ R, x ≠ -1, x ≠ 3 .
(a) Sketch the curve y = f (x), clearly indicating any asymptotes with their equations. State the coordinates of any local maximum or minimum points and any points of intersection with the coordinate axes. A function g is defined by \(g(x) = \frac{1}{x^{2}-2x-3},\) where x ∈ R, x >3.
▶️Answer/Explanation
Ans: 
Note: Accept an indication of \(-\frac{1}{3} on the y-axis.\)
vertical asymptotes x =−1 and x = 3
horizontal asymptote y = 0
uses a valid method to find the x-coordinate of the local maximum point
Note: For example, uses the axis of symmetry or attempts to solve f’ (x) = 0 .
local maximum point \(\left ( 1,-\frac{1}{4} \right )\)
Note: Award (M1)A0 for a local maximum point at x =1 and coordinates not given.
three correct branches with correct asymptotic behaviour and the key features in approximately correct relative positions to each other
(b) The inverse of g is g-1.
(i) Show that \(g^{-1}(x)=1+\frac{\sqrt{4x^{2}+x}}{x}.\)
(ii) State the domain of g-1.
A function h is defined by h(x) = arctan \(\frac{x}{2}\), where x ∈ R.
▶️Answer/Explanation
Ans: 
(c) Given that (h º g) (a) = \(\frac{\pi }{4},\) find the value of a.
Give your answer in the form \(p + \frac{q}{2}\sqrt{r},\) where p, q, r ∈ Z+ .
▶️Answer/Explanation
Ans: 
