Question 1
The diagram shows the graph of $y=a \cos b x+c$. Find the values of the constants $a, b$ and $c$. [3]
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Solution
Solution
The diagram shows the graph of $y = a \cos(bx) + c$, passing through $(-480, -6)$, $(0, 2)$, and $(480, -6)$. Find the values of the constants $a$, $b$, and $c$. Note that $x$ is measured in degrees.
Step 1: General Form of the Equation
The general form of the function is given as:
$ y = a \cos(bx) + c. $
Here, $a$ represents the amplitude, $b$ determines the frequency, and $c$ is the vertical shift.
Step 2: Analyzing Key Points
-
The graph passes through $(0, 2)$, so when $x = 0$, $y = 2$:
$ 2 = a \cos(0) + c. $
Since $\cos(0) = 1$, this simplifies to:
$ 2 = a + c. \tag{1} $
-
The graph also passes through $(-480, -6)$ and $(480, -6)$.
For these points, $y = -6$:
$ -6 = a \cos(-480b) + c $
and
$ -6 = a \cos(480b) + c. $
Using the property $\cos(-x) = \cos(x)$, these equations are equivalent:
$ -6 = a \cos(480b) + c. \tag{2} $
Step 3: Determine the Vertical Shift $c$
The maximum value of the function occurs at $y = 2$, and the minimum value occurs at $y = -6$. The vertical shift $c$ is the midpoint of these values:
$ c = \frac{\text{max} + \text{min}}{2} = \frac{2 + (-6)}{2} = -2. $
Substitute $c = -2$ into equation (1):
$ 2 = a - 2 \implies a = 4. \tag{3} $
Step 4: Determine the Frequency Constant $b$
The period of the cosine function is given by:
$ \text{Period} = \frac{360}{b}, $
where $b$ is in degrees. From the graph, the function completes one full cycle between $x = -480$ and $x = 480$. Therefore, the period is:
$ 480 - (-480) = 960. $
Equating this to the formula for the period:
$ 960 = \frac{360}{b} \implies b = \frac{360}{960} = \frac{3}{8}. \tag{4} $
Step 5: Final Equation
Substitute $a = 4$, $b = \frac{3}{8}$, and $c = -2$ into the general form:
$ y = 4 \cos\left(\frac{3}{8}x\right) - 2. $
Verification
-
At $x = 0$:
$ y = 4 \cos(0) - 2 = 4(1) - 2 = 2. $
-
At $x = 480$:
$ y = 4 \cos\left(\frac{3}{8} \cdot 480\right) - 2 = 4 \cos(180) - 2 = 4(-1) - 2 = -6. $
-
At $x = -480$:
$ y = 4 \cos\left(\frac{3}{8} \cdot (-480)\right) - 2 = 4 \cos(-180) - 2 = 4(-1) - 2 = -6. $
All points satisfy the equation. Thus, the solution is correct.
Question 2
DO NOT USE A CALCULATOR IN THIS QUESTION.
Solve the equation $(2+\sqrt{5}) x^2=4 x+3(2-\sqrt{5})$, giving your answers in the form $a+b \sqrt{5}$ where $a$ and $b$ are integers. [5]
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Solution
We are tasked to solve the quadratic equation
$ (2+\sqrt{5})x^2 = 4x + 3(2-\sqrt{5}), $
and express the solutions in the form $a + b\sqrt{5}$, where $a$ and $b$ are integers.
Step 1: Rearrange the equation
Rearrange all terms to one side of the equation:
$ (2+\sqrt{5})x^2 - 4x - 3(2-\sqrt{5}) = 0. $
Simplify the constant term $-3(2-\sqrt{5})$:
$ (2+\sqrt{5})x^2 - 4x - 6 + 3\sqrt{5} = 0. $
Step 2: Identify coefficients
The equation is now in standard quadratic form:
$ Ax^2 + Bx + C = 0, $
where:
- $A = 2 + \sqrt{5}$,
- $B = -4$,
- $C = -6 + 3\sqrt{5}$.
Step 3: Use the quadratic formula
The quadratic formula is:
$ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}. $
Substitute $A = 2 + \sqrt{5}$, $B = -4$, and $C = -6 + 3\sqrt{5}$:
$ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2+\sqrt{5})(-6+3\sqrt{5})}} {2(2+\sqrt{5})}. $
Step 4: Simplify the discriminant
Calculate $B^2$:
$ B^2 = (-4)^2 = 16. $
Calculate $4AC$:
$ \begin{aligned} 4AC &= 4(2+\sqrt{5})(-6+3\sqrt{5}) \\ &= 4 \cdot \big((-6)(2) + (-6)(\sqrt{5}) + (3\sqrt{5})(2) + (3\sqrt{5})(\sqrt{5})\big) \\ &= 4 \cdot \big(-12 - 6\sqrt{5} + 6\sqrt{5} + 15\big) \\ &= 4 \cdot (3) \\ &= 12. \end{aligned} $
Thus, the discriminant is:
$ B^2 - 4AC = 16 - 12 = 4. $
Step 5: Simplify the solution
We now substitute the discriminant back into the formula:
$ \begin{aligned} x &= \frac{4 \pm \sqrt{4}}{2(2+\sqrt{5})} \\ x &= \frac{4 \pm 2}{2(2+\sqrt{5})}. \end{aligned} $
Split into two cases:
- Case 1: $ x = \frac{4 + 2}{2(2+\sqrt{5})} = \frac{6}{4+2\sqrt{5}} $
- Case 2: $ x = \frac{4 - 2}{2(2+\sqrt{5})} = \frac{2}{4+2\sqrt{5}} $
Step 6: Rationalize the denominators
For both cases, multiply numerator and denominator by the conjugate of the denominator.
$ \begin{aligned} \text{Case 1: } x &= \frac{6}{4+2\sqrt{5}} \cdot \frac{4-2\sqrt{5}}{4-2\sqrt{5}} \\ &= \frac{6(4-2\sqrt{5})}{(4+2\sqrt{5})(4-2\sqrt{5})} \\ &= \frac{24 - 12\sqrt{5}}{16 - 20} \\ &= \frac{24 - 12\sqrt{5}}{-4} \\ &= -6 + 3\sqrt{5}. \end{aligned} $
$ \begin{aligned} \text{Case 2: } x &= \frac{2}{4+2\sqrt{5}} \cdot \frac{4-2\sqrt{5}}{4-2\sqrt{5}} \\ &= \frac{2(4-2\sqrt{5})}{(4+2\sqrt{5})(4-2\sqrt{5})} \\ &= \frac{8 - 4\sqrt{5}}{16 - 20} \\ &= \frac{8 - 4\sqrt{5}}{-4} \\ &= -2 + \sqrt{5}. \end{aligned} $
Final Answer
The solutions are:
$ \begin{aligned} x &= -6 + 3\sqrt{5}, \\ x &= -2 + \sqrt{5}. \end{aligned} $
Question 3
- (a) The diagram shows the graph of $y=|\mathrm{f}(x)|$, where $\mathrm{f}(x)$ is a cubic polynomial. Find, in factorised form, the possible expressions for $\mathrm{f}(x)$. [3]
- (b) Solve the inequality $|5 x-2| \leqslant |4 x+1|$. [4]
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Solution
Step 1: Analyze the graph and derive $\mathrm{f}(x)$
The graph of $y = |\mathrm{f}(x)|$ implies that $\mathrm{f}(x)$ is a cubic polynomial whose roots occur at $x = -2$, $x = 1$, and $x = 4$, as these are the points where $y = 0$ for $\mathrm{f}(x)$. Additionally, $\mathrm{f}(x)$ passes through the point $(0, 24)$.
Key Points
-
A cubic polynomial $\mathrm{f}(x)$ with roots
$x = -2$, $x = 1$, and $x = 4$
can be expressed as:
$ \mathrm{f}(x) = k(x + 2)(x - 1)(x - 4), $
where $k$ is a constant to be determined. - Substitute $(0, 24)$ into $\mathrm{f}(x)$ to find $k$.
Finding $k$
Substitute $x = 0$ and $\mathrm{f}(0) = 24$:
$ \begin{aligned} 24 &= k(0 + 2)(0 - 1)(0 - 4), \\ 24 &= k(2)(-1)(-4), \\ 24 &= k(8), \\ k &= 3. \end{aligned} $
Thus, the polynomial $\mathrm{f}(x)$ is:
$ \mathrm{f}(x) = 3(x + 2)(x - 1)(x - 4). $
Step 2: Solve the inequality $|5x - 2| \leqslant |4x + 1|$
Critical Values
The critical values are obtained by solving:
$ |5x - 2| = |4x + 1|. $
This leads to two cases:
$ \begin{aligned} 5x - 2 &= 4x + 1, \\ 5x - 2 &= -(4x + 1). \end{aligned} $
Case 1: $5x - 2 = 4x + 1$
$ \begin{aligned} 5x - 4x &= 1 + 2, \\ x &= 3. \end{aligned} $
Case 2: $5x - 2 = -(4x + 1)$
$ \begin{aligned} 5x - 2 &= -4x - 1, \\ 5x + 4x &= -1 + 2, \\ 9x &= 1, \\ x &= \frac{1}{9}. \end{aligned} $
The critical values are $x = \frac{1}{9}$ and $x = 3$. These values divide the number line into intervals for analysis.
Interval Analysis
We analyze the inequality $|5x - 2| \leqslant |4x + 1|$ over the intervals determined by the critical values:
- Interval 1: $x \lt \frac{1}{9}$
- Interval 2: $\frac{1}{9} \lt x \lt 3$
- Interval 3: $x \gt 3$
For each interval, take a valid number and check the statement $|5x - 2| \leqslant |4x + 1|$.
Interval 1: $x \lt \frac{1}{9}$
In this interval, take $x = 0$.
- $|5x - 2| = |5(0)-2| = 2$
- $|4x + 1| = |4(0)+1| = 1$
The inequality $|5x - 2| \leqslant |4x + 1|$ is not true.
Interval 2: $\frac{1}{9} \lt x \lt 3$
In this interval, take $x = 1$.
- $|5x - 2| = |5(1)-2| = 3$
- $|4x + 1| = |4(1)+1| = 5$
The inequality $|5x - 2| \leqslant |4x + 1|$ is true.
Interval 3: $x \gt 3$
In this interval, take $x = 4$.
- $|5x - 2| = |5(4)-2| = 18$
- $|4x + 1| = |4(4)+1| = 17$
The inequality $|5x - 2| \leqslant |4x + 1|$ is not true.
Solution
Combining the results from all intervals, the solution to $|5x - 2| \leqslant |4x + 1|$ is:
$ \frac{1}{9} \leqslant x \leqslant 3. $
Question 4
In this question all lengths are in centimetres and all angles are in radians. The diagram shows a circle with centre $O$ and radius $r$. The points $A$ and $B$ lie on the circumference of the circle such that the angle $AOB$ is $\theta$ and the length of the minor arc $AB$ is 12. The area of the minor sector $AOB$ is $57.6\mathrm{~cm}^2$. The point $C$ lies on the tangent to the circle at $A$ such that $OBC$ is a straight line.
- (a) Find the values of $r$ and $\theta$. [4]
- (b) Find the area of the shaded region. Give your answer correct to 1 decimal place. [3]
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Solution
Step 1: Understand the Problem
- The circle has center $O$ and radius $r$.
- The angle $\angle AOB=\theta$ radians.
- The length of the minor arc $AB$ is 12 cm.
- The area of the minor sector $AOB$ is $57.6\,\mathrm{cm}^2$.
- $C$ lies on the tangent to the circle at $A$ such that $OBC$ is a straight line.
-
Tasks:
- (a) Find $r$ and $\theta$.
- (b) Calculate the area of the shaded region to 1 decimal place.
Step 2: Relating Arc Length and Angle
The length of the minor arc $AB$ is given by
$$ \text{Arc Length}=r\theta. $$Substituting the given arc length of 12 cm:
$$ r\theta=12 \tag{1} $$Step 3: Relating Sector Area and Angle
The area of the minor sector $AOB$ is given by
$$ \text{Sector Area}=\frac12 r^2\theta. $$Substituting the given sector area of $57.6\,\mathrm{cm}^2$:
$$ \frac12 r^2\theta=57.6 \tag{2} $$Step 4: Solve for $r$ and $\theta$
From equation (1), express $\theta$ in terms of $r$:
$$ \theta=\frac{12}{r} \tag{3} $$Substitute equation (3) into equation (2):
$$ \begin{aligned} \frac12 r^2\left(\frac{12}{r}\right)&=57.6 \\ \frac12\cdot12r&=57.6 \\ 6r&=57.6 \\ r&=9.6 \end{aligned} $$Using equation (3) to find $\theta$:
$$ \begin{aligned} \theta&=\frac{12}{9.6} \\ \theta&=1.25\,\mathrm{radians} \end{aligned} $$Thus, $r=9.6\,\mathrm{cm}$ and $\theta=1.25\,\mathrm{radians}$.
Step 5: Area of the Shaded Region
The shaded region is the area outside the sector $AOB$ but within the triangle $OAC$.
- The triangle $\triangle OAC$ is a right triangle with $\angle OAC=\theta$ and $AC$ given by:
The area of $\triangle OAC$ is then:
$$ \text{Area of }\triangle OAC=\frac12 AO\cdot AC =\frac12 r\cdot(r\tan\theta) $$Substituting $r=9.6$ and $\theta=1.25$:
$$ \begin{aligned} \text{Area of }\triangle OAC &=\frac12\cdot9.6\cdot(9.6\tan1.25) \\ &=138.68\,\mathrm{cm}^2 \end{aligned} $$- The shaded area is the difference between $\triangle OAC$ and the sector $AOB$:
Thus, the area of the shaded region is $81.1\,\mathrm{cm}^2$ (to 1 decimal place).
Question 5
- (a) Find the exact solutions of the equation $6p^{\frac13}-5p^{-\frac13}-13=0$. [4]
- (b) Solve the equation $2\lg(2x+5)-\lg(x+2)=1$, giving your answers in exact form. [6]
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Solution
(a) Solve $6p^{\frac13}-5p^{-\frac13}-13=0$
Step 1: Introduce a substitution.
Let $x=p^{\frac13}$. Then $p^{-\frac13}=\frac1x$. Substituting these into the equation, we get:
$$ 6x-\frac5x-13=0 $$Step 2: Eliminate the fraction.
Multiply through by $x$ (where $x\neq0$):
$$ 6x^2-5-13x=0 $$Step 3: Rearrange into standard quadratic form.
$$ 6x^2-13x-5=0 $$Step 4: Solve the quadratic equation.
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=6$, $b=-13$, $c=-5$:
$$ x=\frac{-(-13)\pm\sqrt{(-13)^2-4(6)(-5)}}{2(6)} $$ $$ x=\frac{13\pm\sqrt{169+120}}{12} $$ $$ x=\frac{13\pm\sqrt{289}}{12} $$ $$ x=\frac{13\pm17}{12} $$ $$ x=\frac{30}{12}=\frac52, \qquad x=\frac{-4}{12}=-\frac13 $$Step 5: Back-substitute.
Recall $x=p^{\frac13}$, so:
$$ p^{\frac13}=\frac52 \implies p=\left(\frac52\right)^3=\frac{125}{8} $$ $$ p^{\frac13}=-\frac13 \implies p=\left(-\frac13\right)^3=-\frac1{27} $$Exact solutions: $p=\frac{125}{8}$ and $p=-\frac1{27}$.
(b) Solve $2\lg(2x+5)-\lg(x+2)=1$
Step 1: Apply logarithmic properties.
Using the property $a\lg b=\lg b^a$ on the first term:
$$ \lg\big((2x+5)^2\big)-\lg(x+2)=1 $$Using the property $\lg a-\lg b=\lg\frac{a}{b}$:
$$ \lg\frac{(2x+5)^2}{x+2}=1 $$Step 2: Rewrite in exponential form.
$$ \frac{(2x+5)^2}{x+2}=10^1=10 $$Step 3: Simplify the equation.
Multiply through by $(x+2)$ (where $x\neq-2$):
$$ (2x+5)^2=10(x+2) $$Expand both sides:
$$ 4x^2+20x+25=10x+20 $$ $$ 4x^2+20x+25-10x-20=0 $$ $$ 4x^2+10x+5=0 $$Step 4: Solve the quadratic equation.
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=4$, $b=10$, $c=5$:
$$ x=\frac{-(10)\pm\sqrt{(10)^2-4(4)(5)}}{2(4)} $$ $$ x=\frac{-10\pm\sqrt{100-80}}{8} $$ $$ x=\frac{-10\pm\sqrt{20}}{8} $$ $$ x=\frac{-10\pm2\sqrt5}{8} = \frac{-5\pm\sqrt5}{4} $$Step 5: Verify the domain.
The solutions must satisfy:
$$ 2x+5>0 \implies x>-\frac52 $$ $$ x+2>0 \implies x>-2 $$Both conditions are satisfied for $x=\frac{-5+\sqrt5}{4}$ and $x=\frac{-5-\sqrt5}{4}$.
Exact solutions: $x=\frac{-5+\sqrt5}{4}$ and $x=\frac{-5-\sqrt5}{4}$.
Question 6
- (a) Given that $\cot^2\theta=\frac{1}{y+2}$ and $\sec\theta=x-4$, find $y$ in terms of $x$. [2]
- (b) Solve the equation $\sqrt3\operatorname{cosec}\left(2\phi+\frac{3\pi}{4}\right)=2$, for $-\pi\lt\phi\lt\pi$, giving your answers in terms of $\pi$. [5]
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Solution
Step (a): Finding $y$ in terms of $x$
Given that $\cot^2\theta=\frac{1}{y+2}$ and $\sec\theta=x-4$, we want to find $y$ in terms of $x$.
- Recall the identity $\sec^2\theta=1+\tan^2\theta$.
- From the equation $\sec\theta=x-4$, we square both sides to get $\sec^2\theta=(x-4)^2$.
- Therefore, $1+\tan^2\theta=(x-4)^2$, which simplifies to
- Now, using the identity $\cot^2\theta=\frac{1}{\tan^2\theta}$, we substitute for $\tan^2\theta$:
- From the given equation $\cot^2\theta=\frac{1}{y+2}$, equate the two expressions for $\cot^2\theta$:
- Cross-multiply to solve for $y$:
- Finally, solve for $y$:
Thus, the solution is:
$$ y=(x-4)^2-3 $$Step (b): Solving the equation
We are asked to solve the equation
$$ \sqrt3\,\csc\left(2\phi+\frac{3\pi}{4}\right)=2 $$for $-\pi\lt\phi\lt\pi$, giving the answer in terms of $\pi$.
- Start by isolating $\csc\left(2\phi+\frac{3\pi}{4}\right)$:
- Recall that $\csc\theta=\frac{1}{\sin\theta}$, so we have:
- The general solutions to $\sin\theta=\frac{\sqrt3}{2}$ are $\theta=\frac{\pi}{3}+2k\pi$ or $\theta=\pi-\frac{\pi}{3}+2k\pi$, where $k$ is any integer.
- Applying this to $2\phi+\frac{3\pi}{4}$, we get two cases:
or
$$ 2\phi+\frac{3\pi}{4}=\pi-\frac{\pi}{3}+2k\pi $$First case
$$ 2\phi+\frac{3\pi}{4}=\frac{\pi}{3}+2k\pi $$Subtract $\frac{3\pi}{4}$ from both sides:
$$ 2\phi=\frac{\pi}{3}-\frac{3\pi}{4}+2k\pi $$Simplify the fractions:
$$ 2\phi=\frac{4\pi}{12}-\frac{9\pi}{12}+2k\pi =-\frac{5\pi}{12}+2k\pi $$Divide by 2:
$$ \phi=-\frac{5\pi}{24}+k\pi $$Second case
$$ 2\phi+\frac{3\pi}{4}=\pi-\frac{\pi}{3}+2k\pi $$Simplify the right side:
$$ 2\phi+\frac{3\pi}{4}=\frac{2\pi}{3}+2k\pi $$Subtract $\frac{3\pi}{4}$ from both sides:
$$ 2\phi=\frac{2\pi}{3}-\frac{3\pi}{4}+2k\pi $$Simplify the fractions:
$$ 2\phi=\frac{8\pi}{12}-\frac{9\pi}{12}+2k\pi =-\frac{\pi}{12}+2k\pi $$Divide by 2:
$$ \phi=-\frac{\pi}{24}+k\pi $$By combining the two cases,
$$ \phi=-\frac{5\pi}{24}+k\pi \quad\text{or}\quad \phi=-\frac{\pi}{24}+k\pi $$Finally, the solutions for $\phi$ in the interval $-\pi\lt \phi\lt\pi$ are:
$$ \boxed{ \phi= -\frac{5\pi}{24}, \quad -\frac{\pi}{24}, \quad \frac{19\pi}{24}, \quad \frac{23\pi}{24} } $$Question 7
- (a) Find the number of ways in which 14 people can be put into 4 groups containing 2, 3, 4 and 5 people. [3]
-
(b) 6-digit numbers are to be formed using the digits $0,1,2,3,4,5,6,7,8,9$. Each digit may be used only once in any 6-digit number. A 6-digit number must not start with 0. Find how many 6-digit numbers can be formed if
- (i) there are no further restrictions [1]
- (ii) the 6-digit number is divisible by 10 [1]
- (iii) the 6-digit number is greater than 500000 and even. [3]
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Solution
Step (a): Number of ways to divide 14 people into 4 groups
We need to find the number of ways to divide 14 people into 4 groups with sizes 2, 3, 4, and 5 people. This can be done by using the multinomial coefficient formula.
- First, we choose 2 people out of 14 to form the first group. This can be done in $ \binom{14}{2} $ ways.
- Next, we choose 3 people from the remaining 12 people to form the second group. This can be done in $ \binom{12}{3} $ ways.
- Then, we choose 4 people from the remaining 9 to form the third group. This can be done in $ \binom{9}{4} $ ways.
- Finally, the remaining 5 people automatically form the last group. There is only 1 way to do this.
The total number of ways to form the groups is:
$ \binom{14}{2} \binom{12}{3} \binom{9}{4} $
$ \frac{14!}{2!(14-2)!}\times \frac{12!}{3!(12-3)!} \times \frac{9!}{4!(9-4)!}=2522520 $
Step (b): Number of 6-digit numbers
We now turn to the problem of forming 6-digit numbers using the digits $0,1,2,3,4,5,6,7,8,9$ under various conditions.
(i) No further restrictions
A 6-digit number can be formed in the following way:
- The first digit can be any of the 9 non-zero digits (1 through 9), so there are 9 choices for the first digit.
- The second digit can be any of the remaining 9 digits (including 0), so there are 9 choices for the second digit.
- The third digit can be any of the remaining 8 digits, so there are 8 choices for the third digit.
- Similarly, the fourth, fifth, and sixth digits can be chosen from the remaining 7, 6, and 5 digits, respectively.
Thus, the total number of 6-digit numbers is:
$ 9 \times 9 \times 8 \times 7 \times 6 \times 5 = 136080 $
(ii) The 6-digit number is divisible by 10
For the number to be divisible by 10, the last digit must be 0. We can proceed as follows:
- The first digit can still be any of the 9 non-zero digits (1 through 9), so there are 9 choices for the first digit.
- The second, third, fourth, and fifth digits can be chosen from the remaining 8, 7, 6, and 5 digits, respectively.
- The last digit must be 0, so there is only 1 choice for the last digit.
Thus, the total number of 6-digit numbers divisible by 10 is:
$ 9 \times 8 \times 7 \times 6 \times 5 \times 1 = 15120 $
(iii) The 6-digit number is greater than 500000 and even
For the number to be greater than 500000 and even:
- The first digit must be 5, 6, 7, 8, or 9 (since the number must be greater than 500000), giving 5 choices for the first digit.
- The last digit must be even, so it must be 0, 2, 4, 6, or 8, giving 5 choices for the last digit.
- The second, third, fourth, and fifth digits can be chosen from the remaining 8, 7, 6, and 5 digits, respectively.
Moreover,
-
First digit odd, (5,7,9): 3 choices.
The last digit has 5 choices: (0,2,4,6,8).
Number of 6-digit numbers: $ 3 \times 5 \times 8 \times 7 \times 6 \times 5 = 25200 $ -
First digit even, (6,8): 2 choices.
The last digit has 4 choices from 0,2,4,6,8.
Number of 6-digit numbers: $ 2 \times 4 \times 8 \times 7 \times 6 \times 5 = 13440 $
Thus, the total number of 6-digit numbers greater than 500000 and even is:
$ \boxed{25200+13440=38640} $
Question 8
It is given that $f(x)=2 \ln(3x-4)$ for $x>a$.
- (a) Write down the least possible value of $a$. [1]
- (b) Write down the range of $f$. [1]
- (c) It is given that the equation $f(x)=f^{-1}(x)$ has two solutions. (You do not need to solve this equation). Using your answer to part (a), sketch the graphs of $y=f(x)$ and $y=f^{-1}(x)$ on the axes below, stating the coordinates of the points where the graphs meet the axes. [4]
It is given that $g(x)=2x-3$ for $x \ge 3$.
-
(d) (i) Find an expression for $g(g(x))$. [1]
(ii) Hence solve the equation $fg(g(x))=4$ giving your answer in exact form. [3]
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Solution
Step 1: Finding the least possible value of $a$
- The function $f(x)=2\ln(3x-4)$ is defined only when $3x-4>0$.
- Solving: $3x-4>0 \Rightarrow x>\frac{4}{3}$.
- Therefore, the least possible value of $a$ is: $a=\frac{4}{3}$.
Step 2: Range of $f$
- Domain: $x>\frac{4}{3}$.
- As $x \to \frac{4}{3}^+$, $3x-4 \to 0^+$ so $\ln(3x-4)\to -\infty$.
- As $x \to \infty$, $\ln(3x-4)\to \infty$.
- Hence multiplying by 2 does not restrict values, so the range is: $(-\infty,\infty)$.
Step 3: Sketch of $y=f(x)$ and $y=f^{-1}(x)$
- The graph $y=f^{-1}(x)$ is the reflection of $y=f(x)$ in the line $y=x$.
- The condition $f(x)=f^{-1}(x)$ means intersection points lie on $y=x$.
- Since there are two solutions, the curves meet the line $y=x$ at two points.
- $y=f(x)$ has vertical asymptote at $x=\frac{4}{3}$.
Step 4: Finding $g(g(x))$
- $g(x)=2x-3$
- $g(g(x)) = g(2x-3)$
- $=2(2x-3)-3=4x-6-3=4x-9$
Step 5: Solving $f(g(g(x)))=4$
- Substitute $g(g(x))=4x-9$: $f(g(g(x)))=2\ln(3(4x-9)-4)$
- Simplify: $3(4x-9)-4=12x-31$
- So: $2\ln(12x-31)=4$
- Divide by 2: $\ln(12x-31)=2$
- Exponentiate: $12x-31=e^2$
- Hence: $x=\frac{e^2+31}{12}$
Final Answers
- (a) $a=\frac{4}{3}$
- (b) Range: $(-\infty,\infty)$
- (c) Two intersections on $y=x$, with reflection symmetry between $f$ and $f^{-1}$
- (d)(i) $g(g(x))=4x-9$
- (d)(ii) $x=\frac{e^2+31}{12}$
Question 9
The diagram shows part of the curve $y=3+\frac{4}{2x+1}$ and the straight line $3y=2x+6$. Find the area of the shaded region, giving your answer in exact form. [10]
=================Click to view Solution
Solution
Step 1: Express the equations in suitable form
Straight line:
$3y=2x+6 \Rightarrow y=\frac{2}{3}x+2$
Curve:
$y=3+\frac{4}{2x+1}$
Step 2: Find points of intersection
Set the equations equal:
$3+\frac{4}{2x+1}=\frac{2}{3}x+2$
$\frac{4}{2x+1}=\frac{2}{3}x-1$
Multiply by $(2x+1)$:
$4=\left(\frac{2}{3}x-1\right)(2x+1)$
Expand:
$4=\frac{4}{3}x^2+\frac{2}{3}x-2x-1$
$4=\frac{4}{3}x^2-\frac{4}{3}x-1$
Multiply by 3:
$12=4x^2-4x-3$
$4x^2-4x-15=0$
Solve:
$x=\frac{4\pm\sqrt{16+240}}{8}=\frac{4\pm16}{8}$
$x=\frac{5}{2},\; -\frac{3}{2}$
Step 3: Set up the area integral
The required region is between the curve and line from $x=0$ to $x=\frac{5}{2}$:
$\text{Area}=\int_{0}^{\frac{5}{2}}\left[\left(3+\frac{4}{2x+1}\right)-\left(\frac{2}{3}x+2\right)\right]dx$
Simplify:
$\text{Area}=\int_{0}^{\frac{5}{2}}\left(1+\frac{4}{2x+1}-\frac{2}{3}x\right)dx$
Step 4: Evaluate the integral
Split the integral:
$\int_{0}^{\frac{5}{2}}1\,dx+\int_{0}^{\frac{5}{2}}\frac{4}{2x+1}\,dx-\int_{0}^{\frac{5}{2}}\frac{2}{3}x\,dx$
Part 1:
$\int_{0}^{\frac{5}{2}}1\,dx=\frac{5}{2}$
Part 2:
Let $u=2x+1$, then $du=2dx$
Limits: $x=0 \to u=1$, $x=\frac{5}{2} \to u=6$
$\int_{0}^{\frac{5}{2}}\frac{4}{2x+1}dx=2\int_{1}^{6}\frac{1}{u}du=2\ln 6$
Part 3:
$\int_{0}^{\frac{5}{2}}\frac{2}{3}x\,dx=\frac{2}{3}\cdot\frac{x^2}{2}\Big|_{0}^{\frac{5}{2}}=\frac{25}{12}$
Step 5: Combine results
$\text{Area}=\frac{5}{2}+2\ln 6-\frac{25}{12}$
$\text{Area}=\frac{30}{12}-\frac{25}{12}+2\ln 6$
$\text{Area}=\frac{5}{12}+2\ln 6$
Final Answer
$\boxed{\frac{5}{12}+2\ln 6}$
Question 10
-
(a) The first three terms of an arithmetic progression are $(2x+1), 4(2x+1)$ and $7(2x+1)$, where $x \ne -\frac{1}{2}$.
- (i) Show that the sum to $n$ terms can be written in the form $\frac{n}{2}(2x+1)(An+B)$, where $A$ and $B$ are integers. [2]
- (ii) Given that the sum to $n$ terms is $(54n+37)(2x+1)$, find $n$. [2]
- (iii) Given also that the sum to $n$ terms in part (ii) is equal to $1017.5$, find $x$. [2]
- (b) The first three terms of a geometric progression are $(2y+1), 3(2y+1)^2$ and $9(2y+1)^3$, where $y \ne -\frac{1}{2}$. Given that the $n$th term is equal to 4 times the $(n+2)$th term, find possible values of $y$. [4]
- (c) The first three terms of another geometric progression are $\sin\theta$, $2\sin^3\theta$ and $4\sin^5\theta$, for $0<\theta<\frac{\pi}{2}$. Find values of $\theta$ for which the progression has a sum to infinity. [3]
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Solution
Step (a)
(i) Sum to $n$ terms
First term: $a = (2x+1)$ Common ratio (difference):
$d = 4(2x+1) - (2x+1) = 3(2x+1)$
Sum of $n$ terms:
$S_n = \frac{n}{2}[2a + (n-1)d]$
$S_n = \frac{n}{2}\left[2(2x+1) + (n-1)3(2x+1)\right]$
$S_n = \frac{n}{2}(2x+1)\left[2 + 3(n-1)\right]$
$S_n = \frac{n}{2}(2x+1)(3n-1)$
Hence $A=3$, $B=-1$.
(ii) Find $n$
$\frac{n}{2}(2x+1)(3n-1) = (54n+37)(2x+1)$
Cancel $(2x+1)$:
$\frac{n}{2}(3n-1) = 54n+37$
$3n^2 - n = 108n + 74$
$3n^2 - 109n - 74 = 0$
$n = \frac{109 \pm \sqrt{12769}}{6} = \frac{109 \pm 113}{6}$
$n = 37$ (valid)
(iii) Find $x$
$S_n = (54n+37)(2x+1)$, with $n=37$
$1017.5 = (54\cdot 37 + 37)(2x+1)$
$1017.5 = 2035(2x+1)$
$2x+1 = 0.5$
$x = -0.25$
Step (b)
First term: $(2y+1)$ Common ratio:
$r = \frac{3(2y+1)^2}{2y+1} = 3(2y+1)$
For GP:
$T_n = ar^{n-1}, \quad T_{n+2} = ar^{n+1}$
$T_n = 4T_{n+2}$
$r^{n-1} = 4r^{n+1} \Rightarrow 1 = 4r^2$
$r^2 = \frac{1}{4} \Rightarrow r = \pm \frac{1}{2}$
So:
$3(2y+1) = \frac{1}{2}$ or $3(2y+1) = -\frac{1}{2}$
$2y+1 = \frac{1}{6}$ or $2y+1 = -\frac{1}{6}$
$y = -\frac{5}{12}, \; -\frac{7}{12}$
Step (c)
First three terms: $\sin\theta$, $2\sin^3\theta$, $4\sin^5\theta$
Common ratio:
$r = \frac{2\sin^3\theta}{\sin\theta} = 2\sin^2\theta$
For sum to infinity:
$|r| < 1$
$2\sin^2\theta < 1$
$\sin^2\theta < \frac{1}{2}$
For $0 < \theta < \frac{\pi}{2}$:
$\sin\theta < \frac{1}{\sqrt{2}} \Rightarrow \theta < \frac{\pi}{4}$
Final answer:
$\boxed{0 < \theta < \frac{\pi}{4}}$
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