The diagram shows the curve with equation $y = a\sin(bx) + c$ for $0 \le x \le 2\pi$, where $a$, $b$ and $c$ are positive constants. (a) State the values of $a$, $b$ and $c$. (b) For these values of $a$, $b$ and $c$, determine the number of solutions in the interval $0 \le x \le 2\pi$ for each of the following equations: (i) $a\sin(bx) + c = 7 - x$ (ii) $a\sin(bx) + c = 2\pi(x - 1)$.

(a) a = 4, b = 2, c = 3 · (b)(i) 5 · (b)(ii) 1

The equation of a circle is $x^2 + y^2 + px + 2y + q = 0$, where $p$ and $q$ are constants. (a) Express the equation in the form $(x - a)^2 + (y - b)^2 = r^2$, where $a$ is to be given in terms of $p$ and $r^2$ is to be given in terms of $p$ and $q$. (b) The line with equation $x + 2y = 10$ is the tangent to the circle at the point $A(4, 3)$. (i) Find the equation of the normal to the circle at the point $A$. (ii) Find the values of $p$ and $q$.

(a) a = -p/2, b = -1, r^2 = p^2/4 + 1 - q · (b)(i) y = 2x - 5 · (b)(ii) p = -4, q = -15

A function $f$ with domain $x > 0$ is such that $f'(x) = 8(2x - 3)^{-\frac{1}{3}} - 10x^{\frac{2}{3}}$. It is given that the curve with equation $y = f(x)$ passes through the point $(1, 0)$. (a) Find the equation of the normal to the curve at the point $(1, 0)$. (b) Find $f(x)$. (c) It is given that the equation $f'(x) = 0$ can be expressed in the form $$ 125x^2 - 128x + 192 = 0. $$ Determine, making your reasoning clear, whether $f$ is an increasing function, a decreasing…

(a) y = (x-1)/18 · (b) f(x) = 3(2x-3)^(4/3) - 6x^(5/3) + 3 · (c) decreasing function

9709/12 Oct/Nov 2024 — Mathematics Solved Past Paper

Q1MediumTrigonometry

Fig. 1

The diagram shows the curve with equation $y = a\sin(bx) + c$ for $0 \le x \le 2\pi$ , where $a$ , $b$ and $c$ are positive constants.

(a)

State the values of $a$ , $b$ and $c$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The equation is $y = a\sin(bx) + c$ . From the graph, we can determine $a$ (amplitude), $c$ (vertical shift/midline), and $b$ (related to the period).

Working

Find $c$ (vertical shift):
The maximum value is $y = 7$ and the minimum value is $y = -1$ . The midline is the average:

c = \frac{7 + (-1)}{2} = \frac{6}{2} = 3

Find $a$ (amplitude):
The amplitude is the distance from the midline to the maximum (or minimum):

a = 7 - 3 = 4

(Since $a$ is positive, $a = 4$ .)

Find $b$ :
The graph completes 2 full cycles between $x = 0$ and $x = 2\pi$ . Therefore, the period $T$ is:

T = \frac{2\pi}{2} = \pi

For $y = a\sin(bx)$ , the period is $\frac{2\pi}{b}$ . So:

\frac{2\pi}{b} = \pi \implies b = 2

Answer

$a = 4, b = 2, c = 3$

Final answer

a = 4, b = 2, c = 3

Detailed explanation

Walkthrough

The problem asks for the constants $a$ , $b$ , and $c$ in the equation $y = a\sin(bx) + c$ based on the provided graph.

Vertical shift ( $c$ ): The graph oscillates between a maximum of 7 and a minimum of -1. The midline (equilibrium position) is exactly halfway between these extremes: $(7 + (-1))/2 = 3$ . Thus, $c = 3$ .
Amplitude ( $a$ ): The amplitude is the distance from the midline to the peak. Peak is 7, midline is 3, so amplitude is $7 - 3 = 4$ . Since $a$ is positive, $a = 4$ .
Frequency/Period ( $b$ ): The graph shows two complete sine waves between $x=0$ and $x=2\pi$ . A standard sine wave $y=\sin(x)$ has period $2\pi$ . Here, the period is $\pi$ . The coefficient $b$ inside the sine function scales the period to $2\pi/b$ . Setting $2\pi/b = \pi$ gives $b=2$ .

Key Takeaways

The parameter $c$ in $y = a\sin(bx) + c$ represents the vertical translation (midline).
The parameter $a$ represents the amplitude (max displacement from midline).
The parameter $b$ determines the period: Period $= 2\pi/b$ .

Common Mistakes

Confusing the period with the number of cycles. The graph has 2 cycles in $2\pi$ , so the period is $\pi$ , not $2\pi$ .
Forgetting that $a$ is given as positive, though amplitude is always positive anyway.

Things to Be Careful About

Ensure units are consistent. The x-axis is in radians (indicated by $\pi$ ), so the period formula uses $2\pi$ .
Reading the minimum value correctly from the y-axis (it is -1, not 0 or 1).

Techniques used

identify amplitude from graphidentify vertical shift from graphdetermine period from graph

(b)

For these values of $a$ , $b$ and $c$ , determine the number of solutions in the interval $0 \le x \le 2\pi$ for each of the following equations:

2M

(i)

$a\sin(bx) + c = 7 - x$

1M Similar

DifficultyMedium

Worked solution

Approach

We need to find the number of solutions to $4\sin(2x) + 3 = 7 - x$ for $0 \le x \le 2\pi$ . This is equivalent to finding the number of intersections between the curve $y = 4\sin(2x) + 3$ and the line $y = 7 - x$ .

Working

Let $f(x) = 4\sin(2x) + 3$ and $g(x) = 7 - x$ .

$f(x)$ oscillates between -1 and 7 with period $\pi$ .
$g(x)$ is a straight line with y-intercept 7 and x-intercept 7.

We check values at key points (multiples of $\pi/4$ ):

At $x = 0$ : $f(0) = 3$ , $g(0) = 7$ . ( $g > f$ )
At $x = \pi/4 \approx 0.79$ : $f(\pi/4) = 7$ , $g(\pi/4) = 7 - \pi/4 \approx 6.21$ . ( $f > g$ ) → Intersection 1 between 0 and $\pi/4$ .
At $x = \pi/2 \approx 1.57$ : $f(\pi/2) = 3$ , $g(\pi/2) = 7 - \pi/2 \approx 5.43$ . ( $g > f$ ) → Intersection 2 between $\pi/4$ and $\pi/2$ .
At $x = 3\pi/4 \approx 2.36$ : $f(3\pi/4) = -1$ , $g(3\pi/4) \approx 4.64$ . ( $g > f$ )
At $x = \pi \approx 3.14$ : $f(\pi) = 3$ , $g(\pi) = 7 - \pi \approx 3.86$ . ( $g > f$ )
At $x = 5\pi/4 \approx 3.93$ : $f(5\pi/4) = 7$ , $g(5\pi/4) = 7 - 5\pi/4 \approx 3.07$ . ( $f > g$ ) → Intersection 3 between $\pi$ and $5\pi/4$ .
At $x = 3\pi/2 \approx 4.71$ : $f(3\pi/2) = 3$ , $g(3\pi/2) = 7 - 3\pi/2 \approx 2.29$ . ( $f > g$ )
At $x = 7\pi/4 \approx 5.50$ : $f(7\pi/4) = -1$ , $g(7\pi/4) = 7 - 7\pi/4 \approx 1.51$ . ( $g > f$ ) → Intersection 4 between $3\pi/2$ and $7\pi/4$ .
At $x = 2\pi \approx 6.28$ : $f(2\pi) = 3$ , $g(2\pi) = 7 - 2\pi \approx 0.72$ . ( $f > g$ ) → Intersection 5 between $7\pi/4$ and $2\pi$ .

Since $f(x)$ is continuous and changes relative position to $g(x)$ at each of these intervals, there are 5 solutions.

Answer

5

Final answer

5

Detailed explanation

Walkthrough

We are solving $4\sin(2x) + 3 = 7 - x$ . Instead of solving algebraically (which is impossible here), we treat this as finding the intersection of $y_1 = 4\sin(2x) + 3$ and $y_2 = 7 - x$ .

The curve $y_1$ has max 7, min -1, midline 3, period $\pi$ .
The line $y_2$ starts at $(0, 7)$ and goes down with slope -1, ending at $(2\pi, 7-2\pi) \approx (6.28, 0.72)$ .
At $x=0$ , line (7) is above curve (3). At the first peak $x=\pi/4$ , curve (7) is above line ( $\approx 6.21$ ). They must cross once.
At $x=\pi/2$ , curve (3) is below line ( $\approx 5.43$ ). They cross again.
The curve goes down to -1 and back to 3 at $x=\pi$ . The line is at $\approx 3.86$ . Line is still above curve at $x=\pi$ .
At the second peak $x=5\pi/4$ , curve (7) is above line ( $\approx 3.07$ ). They cross.
At $x=3\pi/2$ , curve (3) is above line ( $\approx 2.29$ ).
At $x=7\pi/4$ , curve (-1) is below line ( $\approx 1.51$ ). They cross.
At $x=2\pi$ , curve (3) is above line ( $\approx 0.72$ ). They cross.
Total crossings: 5.

Key Takeaways

Equations mixing trig and linear functions are best solved graphically or by analyzing intersections.
Checking values at key points (peaks, troughs, midline crossings) helps determine the number of solutions.

Common Mistakes

Assuming there is only 1 or 2 solutions without checking the full range.
Miscalculating the value of the line at key x-values.

Things to Be Careful About

The interval is $0 \le x \le 2\pi$ . Ensure the line doesn't go below the curve's minimum (-1) before the end of the interval. At $x=2\pi$ , line is $\approx 0.72 > -1$ , so it's still within range.
The line $y=7-x$ has a y-intercept of 7, which is the maximum of the sine curve. This means at $x=0$ , the line is at the max height, but the sine curve is at the midline.

Techniques used

graphical intersection methodevaluate functions at key points

(ii)

$a\sin(bx) + c = 2\pi(x - 1)$ .

1M Similar

DifficultyMedium

Worked solution

Approach

We need to find the number of solutions to $4\sin(2x) + 3 = 2\pi(x - 1)$ for $0 \le x \le 2\pi$ . This is equivalent to finding the number of intersections between $y = 4\sin(2x) + 3$ and $y = 2\pi(x - 1)$ .

Working

Let $f(x) = 4\sin(2x) + 3$ and $h(x) = 2\pi(x - 1)$ .

$f(x)$ oscillates between -1 and 7.
$h(x)$ is a straight line passing through $(1, 0)$ with slope $2\pi \approx 6.28$ .

Check key values:

At $x = 0$ : $f(0) = 3$ , $h(0) = -2\pi \approx -6.28$ . ( $f > h$ )
At $x = 1$ : $f(1) = 4\sin(2) + 3 \approx 6.64$ , $h(1) = 0$ . ( $f > h$ )
At $x = \pi/2 \approx 1.57$ : $f(\pi/2) = 3$ , $h(\pi/2) = 2\pi(\pi/2 - 1) = \pi^2 - 2\pi \approx 3.59$ . ( $h > f$ ) → Intersection between 1 and $\pi/2$ .
For $x > \pi/2$ , $h(x)$ increases rapidly. At $x = \pi \approx 3.14$ , $h(\pi) = 2\pi(\pi - 1) \approx 13.46$ , which is well above the maximum of $f(x)$ (which is 7).
Since $h(x) > 7$ for $x > 1 + 7/(2\pi) \approx 2.11$ , and $f(x) \le 7$ , there can be no further intersections for $x > 2.11$ .
We verified that between $x=1$ and $x=\pi/2$ , the line crosses from below to above the curve. Before $x=1$ , $h(x)$ is negative (below -1), so no intersection.

Thus, there is only 1 solution.

Answer

1

Final answer

1

Detailed explanation

Walkthrough

We solve $4\sin(2x) + 3 = 2\pi(x - 1)$ .

Curve $f(x) = 4\sin(2x) + 3$ is bounded between -1 and 7.
Line $h(x) = 2\pi(x - 1)$ passes through $(1, 0)$ and has a steep positive slope ( $\approx 6.28$ ).
At $x=0$ , line is at $-2\pi \approx -6.28$ (below curve min -1). Curve is at 3. No intersection near 0.
At $x=1$ , line is at 0. Curve is at $4\sin(2)+3 \approx 6.6$ . Curve is above line.
At $x=\pi/2 \approx 1.57$ , line is at $\approx 3.59$ . Curve is at 3. Line is now above curve. They must have crossed once between $x=1$ and $x=1.57$ .
For $x > 1.57$ , the line continues to rise steeply. At $x=2$ , line is $2\pi(1) \approx 6.28$ . At $x=2.11$ , line is 7 (max of curve). For all $x > 2.11$ , the line is above 7, while the curve is at most 7. So no more intersections.
Total solutions: 1.

Key Takeaways

When one function is bounded (trig) and the other is unbounded linear, intersections are limited to where the linear function is within the range of the bounded function.
Checking the value where the linear function exceeds the maximum of the trig function gives an upper bound for solutions.

Common Mistakes

Assuming multiple intersections because the trig function oscillates. The linear function rises too fast.
Forgetting that $x-1$ makes the line pass through $(1,0)$ , not the origin.

Things to Be Careful About

The slope $2\pi$ is large, so the line rises quickly out of the range $[-1, 7]$ .
Ensure you check the value at $x=1$ carefully; $\sin(2)$ is in radians, $\sin(2 \text{ rad}) \approx 0.909$ .

Techniques used

graphical intersection methodevaluate functions at key points

Q2MediumSeriesQuadratics

The first term of an arithmetic progression is $-20$ and the common difference is $5$ .

(a)

Find the sum of the first 20 terms of the progression.

2M Similar

(b)

It is given that the sum of the first $2k$ terms is 10 times the sum of the first $k$ terms.

Find the value of $k$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the arithmetic progression sum formula $S_n = \frac{n}{2}\left(2a + (n-1)d\right)$ for $n = k$ and $n = 2k$ . Then use the condition that $S_{2k}$ is ten times $S_k$ , simplify to a quadratic equation, and solve for $k$ .

Working

For $n = k$ :

S_k = \frac{k}{2}\left(2(-20) + (k-1)(5)\right) = \frac{k}{2}(-40 + 5k - 5) = \frac{k}{2}(5k - 45)

For $n = 2k$ :

S_{2k} = \frac{2k}{2}\left(2(-20) + (2k-1)(5)\right) = k(-40 + 10k - 5) = k(10k - 45)

The condition is $S_{2k} = 10S_k$ . Therefore

k(10k - 45) = 10 \cdot \frac{k}{2}(5k - 45)

10k^2 - 45k = 5(5k^2 - 45k)

10k^2 - 45k = 25k^2 - 225k

Bring all terms to one side:

0 = 15k^2 - 180k

15k(k - 12) = 0

So $k = 0$ or $k = 12$ . Since $k$ represents a positive number of terms, $k = 12$ .

Answer

k = 12

Final answer

k = 12

Detailed explanation

Walkthrough

We need to express $S_k$ and $S_{2k}$ using the AP sum formula $S_n = \frac{n}{2}(2a + (n-1)d)$ with $a = -20$ and $d = 5$ . Substituting $n = k$ gives $S_k = \frac{k}{2}(5k - 45)$ . Substituting $n = 2k$ gives $S_{2k} = k(10k - 45)$ . The condition says the sum of the first $2k$ terms is ten times the sum of the first $k$ terms, so we set $S_{2k} = 10S_k$ . Expanding both sides gives $10k^2 - 45k = 25k^2 - 225k$ , which simplifies to $15k^2 - 180k = 0$ . Factoring gives $15k(k - 12) = 0$ , so $k = 0$ or $k = 12$ . The root $k = 0$ corresponds to summing zero terms and is not meaningful in this context, so $k = 12$ .

Key Takeaways

This question combines the arithmetic progression sum formula with algebraic equation solving. You must substitute $n = k$ and $n = 2k$ correctly, expand carefully, and solve the resulting quadratic.

Common Mistakes

Using the term formula $a + (n-1)d$ instead of the sum formula.
Forgetting the factor $\frac{1}{2}$ when computing $S_k$ or when multiplying $S_k$ by 10.
Expanding the brackets incorrectly, leading to sign errors.
Keeping the trivial root $k = 0$ instead of rejecting it.

Things to Be Careful About

Substitute $2k$ into both $\frac{n}{2}$ and $(2a + (n-1)d)$ . When multiplying $S_k$ by 10, remember that $S_k = \frac{k}{2}(5k - 45)$ , so $10S_k = 5k(5k - 45)$ . Finally, reject the trivial solution $k = 0$ and state $k = 12$ .

Techniques used

apply the arithmetic progression sum formula for two different numbers of termsset up the given multiple conditionexpand and simplify to a quadratic equationsolve the quadratic equationdiscard the trivial root

Q3Medium-EasyDifferentiation

The equation of a curve is $y = 2x^2 - 3$ . Two points $A$ and $B$ with $x$ -coordinates 2 and $(2 + h)$ respectively lie on the curve.

(a)

Find and simplify an expression for the gradient of the chord $AB$ in terms of $h$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The gradient of a chord joining two points on a curve is the change in $y$ divided by the change in $x$ . We substitute the $x$ -coordinates of $A$ and $B$ into the curve equation to find the corresponding $y$ -values, then form the difference quotient and simplify.

Working

The $y$ -coordinate of $A$ (where $x = 2$ ):

f(2) = 2(2)^2 - 3 = 8 - 3 = 5

The $y$ -coordinate of $B$ (where $x = 2 + h$ ):

f(2+h) = 2(2+h)^2 - 3 = 2(4 + 4h + h^2) - 3 = 8 + 8h + 2h^2 - 3 = 5 + 8h + 2h^2

The gradient of chord $AB$ is the difference in $y$ -coordinates divided by the difference in $x$ -coordinates:

\frac{f(2+h) - f(2)}{(2+h) - 2} = \frac{(5 + 8h + 2h^2) - 5}{h}

Simplify:

\frac{8h + 2h^2}{h} = \frac{h(8 + 2h)}{h} = 8 + 2h

Answer

8 + 2h \quad \text{or} \quad 2(h+4)

Final answer

8 + 2h or 2(h+4)

Detailed explanation

Walkthrough

We want the gradient of the chord $AB$ . A chord is a straight line joining two points on the curve, so its gradient is given by the change in $y$ divided by the change in $x$ . Point $A$ has $x$ -coordinate 2 and point $B$ has $x$ -coordinate $2 + h$ .

First, find the $y$ -coordinates. Substituting $x = 2$ into $y = 2x^2 - 3$ gives $2(2)^2 - 3 = 8 - 3 = 5$ . Substituting $x = 2 + h$ gives $2(2+h)^2 - 3$ . Expand $(2+h)^2 = 4 + 4h + h^2$ , so the $y$ -coordinate of $B$ is $2(4 + 4h + h^2) - 3 = 8 + 8h + 2h^2 - 3 = 5 + 8h + 2h^2$ .

The gradient of the chord is the difference in $y$ -values divided by the difference in $x$ -values:

\frac{(5 + 8h + 2h^2) - 5}{(2 + h) - 2} = \frac{8h + 2h^2}{h}

Factor out $h$ from the numerator: $\frac{h(8 + 2h)}{h}$ . Since $A$ and $B$ are distinct points, $h \neq 0$ , so the $h$ cancels, leaving $8 + 2h$ .

Key Takeaways

This question introduces the difference quotient, which is the foundation of differentiation from first principles. The gradient of a chord is the average rate of change of the function over the interval $[2, 2+h]$ .

Common Mistakes

Forgetting to subtract the $y$ -value of $A$ (which is 5) when forming the numerator.
Expanding $(2+h)^2$ incorrectly — it must be $4 + 4h + h^2$ , not $4 + h^2$ .
Failing to simplify the fraction fully to $8 + 2h$ .

Things to Be Careful About

The difference in $x$ -coordinates is $(2+h) - 2 = h$ , not $h - 2$ .
We may cancel the $h$ only because $h \neq 0$ (the two points are distinct).
The mark scheme accepts the simplified form $2h + 8$ or $2(h+4)$ .

Techniques used

substitute x-coordinates into the curve equationform the difference quotient for the chord gradientsimplify the algebraic fraction

(b)

Explain how the gradient of the curve at the point $A$ can be deduced from the answer to part (a), and state the value of this gradient.

2M Similar

Q4MediumSeries

Find the term independent of $x$ in the expansion of each of the following:

(a)

\left(x + \frac{3}{x^2}\right)^6

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the binomial theorem to write the general term of the expansion. The term independent of $x$ is the one in which the power of $x$ is $0$ , so set the exponent of $x$ equal to $0$ and solve for the binomial index $r$ . Then evaluate the coefficient.

Working

The general term in $\left(x+\frac{3}{x^2}\right)^6$ is

\binom{6}{r} x^{6-r}\left(\frac{3}{x^2}\right)^r = \binom{6}{r} 3^r x^{6-3r}

For the term independent of $x$ , the exponent of $x$ must be $0$ :

6-3r=0 \quad \Rightarrow \quad r=2

So the required term is

\binom{6}{2} x^4 \left(\frac{3}{x^2}\right)^2 = \binom{6}{2} \cdot 9 \cdot x^{0} = 15 \times 9 = 135

Answer

135

Final answer

135

Detailed explanation

Walkthrough

The binomial expansion of $\left(x+\frac{3}{x^2}\right)^6$ is made up of terms of the form $\binom{6}{r} x^{6-r}\left(\frac{3}{x^2}\right)^r$ . Each such term has a power of $x$ equal to $6-3r$ . To find the term independent of $x$ , we need the power of $x$ to be $0$ , so we solve $6-3r=0$ , giving $r=2$ . This tells us which term in the expansion contributes the constant. We then evaluate $\binom{6}{2}3^2 = 15 \times 9 = 135$ . The coefficient $15$ is the binomial coefficient, and $3^2$ comes from the constant factor in the term.

Key Takeaways

This question tests the ability to use the binomial theorem in general form and to identify a particular term by setting the exponent of $x$ to the required value. The key idea is that the term independent of $x$ is the one whose $x$ -power is $0$ .

Common Mistakes

A common mistake is to use the wrong value of $r$ , such as $r=3$ , which gives a term in $x^{-3}$ rather than a constant. Another is to forget the factor $3^r$ when evaluating the coefficient, giving $15$ instead of $135$ .

Things to Be Careful About

Make sure the exponent of $x$ is simplified correctly: $6-r-2r = 6-3r$ . Also remember that the binomial coefficient $\binom{6}{2}$ is $15$ , not $6 \times 5 = 30$ ; the division by $2!$ is essential.

Techniques used

use the binomial general termset the exponent of x to zeroevaluate the binomial coefficient and power

(b)

(4x^3 - 5)\left(x + \frac{3}{x^2}\right)^6

4M Similar

DifficultyMedium

Worked solution

Approach

Write the general term of $\left(x+\frac{3}{x^2}\right)^6$ . When the whole product is expanded, the constant term receives two contributions: the term in $x^{-3}$ from the bracket multiplied by $4x^3$ , and the constant term from the bracket multiplied by $-5$ . Find both terms, then combine them.

Working

The general term of $\left(x+\frac{3}{x^2}\right)^6$ is

\binom{6}{r} x^{6-r}\left(\frac{3}{x^2}\right)^r = \binom{6}{r} 3^r x^{6-3r}

For the term in $x^{-3}$ , set $6-3r=-3$ , so $r=3$ :

\binom{6}{3} x^3 \left(\frac{3}{x^2}\right)^3 = \binom{6}{3} \cdot 27 \cdot x^{-3} = 20 \times 27 x^{-3} = \frac{540}{x^3}

For the term independent of $x$ , set $6-3r=0$ , so $r=2$ :

\binom{6}{2} x^4 \left(\frac{3}{x^2}\right)^2 = 15 \times 9 = 135

Now expand the product. The only terms that can produce a constant are:

4x^3 \cdot \frac{540}{x^3} + (-5)(135) = 2160 - 675 = 1485

Answer

1485

Final answer

1485

Detailed explanation

Walkthrough

Start with the general term $\binom{6}{r}3^r x^{6-3r}$ of the bracket. Multiplying by $4x^3$ will lower the exponent of $x$ by $3$ ; multiplying by $-5$ leaves the exponent unchanged. Therefore, to obtain a constant after multiplying by $4x^3$ , we need the bracket term with exponent $-3$ ; to obtain a constant after multiplying by $-5$ , we need the bracket term with exponent $0$ .

For the $x^{-3}$ term, solve $6-3r=-3$ , so $r=3$ . Its coefficient is $\binom{6}{3}3^3=20\times27=540$ , so the term is $\frac{540}{x^3}$ . For the constant term, solve $6-3r=0$ , so $r=2$ . Its coefficient is $\binom{6}{2}3^2=15\times9=135$ .

Then combine the contributions: $4x^3\cdot\frac{540}{x^3}=2160$ and $(-5)(135)=-675$ . Adding gives $2160-675=1485$ .

Key Takeaways

This question extends the binomial expansion to a product of a linear expression and a binomial power. The important skill is identifying which terms of the bracket are needed to form a requested term after multiplication, rather than expanding the whole product.

Common Mistakes

A common mistake is to only find the constant term of the bracket, $135$ , and multiply by $-5$ , forgetting the contribution from the $x^{-3}$ term multiplied by $4x^3$ . Another is to use $r=2$ for the $x^{-3}$ term instead of $r=3$ , or to omit the factor $3^r$ when computing coefficients.

Things to Be Careful About

When multiplying by $4x^3$ , the bracket term must contain $x^{-3}$ so that the $x$ -powers cancel. Check the exponent calculation: $6-r-2r = 6-3r$ . Also be careful with signs: the second contribution is $-5\times135$ , not $+5\times135$ .

Techniques used

use the binomial general termidentify the required x-powers for the constant termcombine contributions from the product expansion

Q5MediumFunctions

The function $f$ is defined by $f(x) = \frac{2x + 1}{2x - 1}$ for $x < \frac{1}{2}$ .

(a)

7M

(i)

State the value of $f(-1)$ .

1M Similar

(ii)

Fig. 2

The diagram shows the graph of $y = f(x)$ . Sketch the graph of $y = f^{-1}(x)$ on this diagram. Show any relevant mirror line.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$ . Draw the mirror line first, then sketch the reflected curve on the same axes.

Working

The original curve $y = f(x)$ has:

a vertical asymptote at $x = \frac{1}{2}$
a horizontal asymptote at $y = 1$
passes through $(0, -1)$ and $(-\frac{1}{2}, 0)$

Reflecting in the line $y = x$ swaps the $x$ - and $y$ -coordinates, so the curve $y = f^{-1}(x)$ has:

a horizontal asymptote at $y = \frac{1}{2}$
a vertical asymptote at $x = 1$
passes through $(-1, 0)$ and $(0, -\frac{1}{2})$

The two curves must intersect on the line $y = x$ . Solving $f(x) = x$ :

\frac{2x+1}{2x-1} = x \;\Rightarrow\; 2x+1 = 2x^2 - x \;\Rightarrow\; 2x^2 - 3x - 1 = 0

x = \frac{3 \pm \sqrt{17}}{4}

Only the negative root $\frac{3 - \sqrt{17}}{4} \approx -0.28$ lies in the domain $x < \frac{1}{2}$ , so the curves intersect at approximately $(-0.28, -0.28)$ in the third quadrant.

Answer

The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ , intersecting the original curve in the third quadrant.

Final answer

Reflection of y = f(x) in the line y = x, intersecting the original curve on y = x in the third quadrant.

Detailed explanation

Walkthrough

The graph of the inverse function $y = f^{-1}(x)$ is the mirror image of the graph of $y = f(x)$ in the line $y = x$ . This follows from the definition of the inverse: if $(a, b)$ lies on the original graph (so $f(a) = b$ ), then $(b, a)$ lies on the inverse graph (since $f^{-1}(b) = a$ ). Reflecting $(a, b)$ in the line $y = x$ swaps the coordinates to give $(b, a)$ , confirming the reflection property.

To sketch the inverse, first draw the line $y = x$ as a dashed mirror line. The original curve has a vertical asymptote at $x = 1/2$ and a horizontal asymptote at $y = 1$ . Under reflection, the vertical asymptote becomes a horizontal one at $y = 1/2$ , and the horizontal asymptote becomes a vertical one at $x = 1$ . The point $(0, -1)$ on the original becomes $(-1, 0)$ on the inverse, and $(-\frac{1}{2}, 0)$ becomes $(0, -\frac{1}{2})$ .

The two curves must meet on the line $y = x$ . Solving $\frac{2x+1}{2x-1} = x$ gives $2x^2 - 3x - 1 = 0$ , so $x = \frac{3 \pm \sqrt{17}}{4}$ . The negative root is approximately $-0.28$ , which lies in the third quadrant.

Key Takeaways

The graph of $y = f^{-1}(x)$ is the mirror image of $y = f(x)$ in the line $y = x$ .
Horizontal and vertical asymptotes swap under reflection.
The original and inverse curves always meet on the line $y = x$ where they cross it.

Common Mistakes

Translating the graph instead of reflecting it.
Omitting the line $y = x$ (this loses a B1 mark).
Drawing the asymptotes incorrectly on the reflected curve.

Things to Be Careful About

The mark scheme requires the reflected curve's shape: to the left of the intersection, the reflection lies below the original and crosses the $x$ -axis; to the right of the intersection, the reflection lies to the right of the original.

Techniques used

reflect the graph in the line y = xdraw the mirror line y = xidentify the corresponding asymptotes and key points

(iii)

Find an expression for $f^{-1}(x)$ and state the domain of the function $f^{-1}$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Let $y = \frac{2x+1}{2x-1}$ , swap the variables, then solve for $y$ . The domain of $f^{-1}$ equals the range of $f$ .

Working

\begin{aligned} y &= \frac{2x+1}{2x-1} \\ y(2x-1) &= 2x+1 \\ 2xy - y &= 2x + 1 \\ 2xy - 2x &= y + 1 \\ 2x(y-1) &= y + 1 \\ x &= \frac{y+1}{2(y-1)} \end{aligned}

Swapping $x$ and $y$ gives:

f^{-1}(x) = \frac{x+1}{2(x-1)}

To find the domain of $f^{-1}$ , determine the range of $f$ :

As $x \to \frac{1}{2}^-$ , the denominator $2x - 1 \to 0^-$ while the numerator $\to 2$ , so $f(x) \to -\infty$ .
As $x \to -\infty$ , $\frac{2x+1}{2x-1} \to 1$ from above.

Hence $f(x) < 1$ for all $x < \frac{1}{2}$ , so the range of $f$ is $x < 1$ . This is the domain of $f^{-1}$ .

Answer

f^{-1}(x) = \frac{x+1}{2(x-1)}, \quad \text{domain } x < 1

Final answer

f^{-1}(x) = (x+1)/(2(x-1)), domain x < 1

Detailed explanation

Walkthrough

Finding the inverse of a function requires solving $y = f(x)$ for $x$ and then swapping the variable names. Start with $y = \frac{2x+1}{2x-1}$ . Multiply both sides by $2x-1$ to clear the fraction, giving $y(2x-1) = 2x+1$ . Expand the left side: $2xy - y = 2x + 1$ . Collect all terms in $x$ on the left by subtracting $2x$ from both sides: $2xy - 2x = y + 1$ . Factor out $2x$ : $2x(y-1) = y + 1$ . Divide by $2(y-1)$ : $x = \frac{y+1}{2(y-1)}$ . Finally, swap $x$ and $y$ to get $f^{-1}(x) = \frac{x+1}{2(x-1)}$ .

The domain of the inverse is the range of the original function. As $x \to \frac{1}{2}$ from below, the denominator $2x - 1$ approaches $0$ from below, while the numerator $2x + 1$ approaches $2$ (positive), so $f(x) \to -\infty$ . As $x \to -\infty$ , the fraction $\frac{2x+1}{2x-1}$ tends to the ratio of leading coefficients, namely $1$ , approached from above. Therefore $f$ takes all values less than $1$ , i.e. its range is $x < 1$ , which is the domain of $f^{-1}$ .

Key Takeaways

To find an inverse, set $y = f(x)$ , swap variables, then solve.
The domain of $f^{-1}$ equals the range of $f$ , and vice versa.
Vertical and horizontal asymptotes swap between $f$ and $f^{-1}$ .

Common Mistakes

Writing $f^{-1}(x) = x$ (no inversion performed).
Sign errors when collecting the $x$ -terms.
Stating the domain of $f^{-1}$ as $x < \frac{1}{2}$ (the domain of $f$ ) instead of the range of $f$ .
Writing the answer as $x = \frac{x+1}{2(x-1)}$ , which is meaningless.

Things to Be Careful About

The mark scheme accepts equivalent forms such as $\frac{-x-1}{2-2x}$ or $\frac{1}{x-1} + \frac{1}{2}$ , but NOT $x = \frac{x+1}{2(x-1)}$ or fractions within fractions.
Acceptable domain notations: $x < 1$ , $-\infty < x < 1$ , $(-\infty, 1)$ . The notation $[-\infty, 1)$ is condoned but unconventional.

Techniques used

set y equal to f(x)swap x and yrearrange algebraically to isolate yidentify the range of f as the domain of f^{-1}

(b)

The function $g$ is defined by $g(x) = 3x + 2$ for $x \in \mathbb{R}$ .

Solve the equation $f(x) = gf\left(\frac{1}{4}\right)$ .

3M Similar

DifficultyMedium

Worked solution

Approach

First evaluate the composite $\mathrm{gf}(\frac{1}{4}) = g(f(\frac{1}{4}))$ , then solve $f(x) = \mathrm{gf}(\frac{1}{4})$ .

Working

Step 1: Find $f\left(\frac{1}{4}\right)$ .

\begin{aligned} f\left(\frac{1}{4}\right) &= \frac{2 \cdot \frac{1}{4} + 1}{2 \cdot \frac{1}{4} - 1} \\ &= \frac{\frac{1}{2} + 1}{\frac{1}{2} - 1} \\ &= \frac{\frac{3}{2}}{-\frac{1}{2}} \\ &= -3 \end{aligned}

Step 2: Apply $g$ to this result.

\mathrm{gf}\left(\frac{1}{4}\right) = g(-3) = 3(-3) + 2 = -7

Step 3: Solve $f(x) = -7$ .

\frac{2x+1}{2x-1} = -7

2x + 1 = -7(2x - 1)

2x + 1 = -14x + 7

16x = 6

x = \frac{6}{16} = \frac{3}{8}

Since $\frac{3}{8} < \frac{1}{2}$ , this value is in the domain of $f$ .

Answer

x = \frac{3}{8}

Final answer

x = 3/8

Detailed explanation

Walkthrough

The notation $\mathrm{gf}(x)$ means $g(f(x))$ , the composition of $g$ after $f$ . We need to evaluate this at $x = \frac{1}{4}$ .

First compute the inner function: $f(\frac{1}{4}) = \frac{2 \cdot \frac{1}{4} + 1}{2 \cdot \frac{1}{4} - 1} = \frac{\frac{3}{2}}{-\frac{1}{2}} = \frac{3}{2} \div (-\frac{1}{2}) = -3$ .

Next apply the outer function: $g(-3) = 3(-3) + 2 = -9 + 2 = -7$ .

Now solve $f(x) = -7$ , i.e. $\frac{2x+1}{2x-1} = -7$ . Multiply both sides by $2x-1$ to get $2x + 1 = -7(2x-1) = -14x + 7$ . Collect $x$ -terms on the left and constants on the right: $2x + 14x = 7 - 1$ , so $16x = 6$ , giving $x = \frac{6}{16} = \frac{3}{8}$ .

Check the domain: $\frac{3}{8} = 0.375 < 0.5$ , so this value is valid. As a sanity check, $f(\frac{3}{8}) = \frac{\frac{3}{4} + 1}{\frac{3}{4} - 1} = \frac{\frac{7}{4}}{-\frac{1}{4}} = -7$ ✓

Key Takeaways

The composition $g \circ f$ is evaluated right-to-left: first apply $f$ , then $g$ .
Setting a rational function equal to a constant leads to a linear equation after clearing the denominator.
Always check that solutions lie in the original function's domain.

Common Mistakes

Confusing $g \circ f$ with $f \circ g$ (which would give a different value).
Sign errors when clearing the fraction or applying $g$ .
Not simplifying $\frac{6}{16}$ to $\frac{3}{8}$ .

Things to Be Careful About

The mark scheme offers an alternative using the inverse: $x = f^{-1}(-7) = \frac{-7+1}{2(-7-1)} = \frac{-6}{-16} = \frac{3}{8}$ . This is a useful cross-check.
The condition $x < \frac{1}{2}$ must be satisfied.

Techniques used

evaluate f at x = 1/4compose g with the resultset f(x) equal to the composite valuesolve the resulting rational equationcheck the solution lies in the domain

Q66MMediumCircular MeasureQuadratics

Fig. 3

The diagram shows a metal plate $OABCDEF$ consisting of sectors of two circles, each with centre $O$ . The radii of sectors $AOB$ and $EOF$ are $r\text{ cm}$ and the radius of sector $COD$ is $2r\text{ cm}$ . Angle $AOB = \text{angle } EOF = \theta\text{ radians}$ and angle $COD = 2\theta\text{ radians}$ .

It is given that the perimeter of the plate is $14\text{ cm}$ and the area of the plate is $10\text{ cm}^2$ .

Given that $r > \frac{3}{2}$ and $\theta < \frac{3}{4}$ , find the values of $r$ and $\theta$ .

DifficultyMedium

Worked solution

Approach

Use the formulas for arc length and sector area to express the perimeter and area of the plate in terms of $r$ and $\theta$ . Set up a system of two equations and solve for $r$ and $\theta$ , applying the given constraints to select the valid solution.

Working

The perimeter of the plate consists of the two straight radii $OA$ and $OF$ , the two arcs $AB$ and $EF$ , the two straight segments $BC$ and $DE$ , and the arc $CD$ .

The lengths are:

$OA = OF = r$
Arc $AB =$ Arc $EF = r\theta$
$BC = DE = 2r - r = r$
Arc $CD = (2r)(2\theta) = 4r\theta$

Summing these gives the perimeter:

\text{Perimeter} = r + r\theta + r + 4r\theta + r + r\theta + r = 4r + 6r\theta

Given that the perimeter is $14\text{ cm}$ :

4r + 6r\theta = 14

The area of the plate is the sum of the areas of the three sectors:

Sector $AOB = \frac{1}{2}r^2\theta$
Sector $COD = \frac{1}{2}(2r)^2(2\theta) = 4r^2\theta$
Sector $EOF = \frac{1}{2}r^2\theta$

Summing these gives the total area:

\text{Area} = \frac{1}{2}r^2\theta + 4r^2\theta + \frac{1}{2}r^2\theta = 5r^2\theta

Given that the area is $10\text{ cm}^2$ :

5r^2\theta = 10

From the area equation, solve for $\theta$ :

r^2\theta = 2 \implies \theta = \frac{2}{r^2}

Substitute this into the perimeter equation:

4r + 6r\left(\frac{2}{r^2}\right) = 14

4r + \frac{12}{r} = 14

Multiply through by $r$ (since $r > 0$ ):

4r^2 + 12 = 14r

4r^2 - 14r + 12 = 0

Divide by 2:

2r^2 - 7r + 6 = 0

Factorise the quadratic:

(2r - 3)(r - 2) = 0

So $r = \frac{3}{2}$ or $r = 2$ .

Given the constraint $r > \frac{3}{2}$ , we must have $r = 2$ .

Substitute $r = 2$ back into the expression for $\theta$ :

\theta = \frac{2}{2^2} = \frac{2}{4} = 0.5

Check the constraint $\theta < \frac{3}{4}$ : $0.5 < 0.75$ is true.

Answer

r = 2, \quad \theta = 0.5

Final answer

r = 2, theta = 0.5

Detailed explanation

Walkthrough

First, we break down the boundary of the metal plate into its constituent parts to calculate the perimeter. The plate is made of three circular sectors sharing a common centre $O$ . The outer boundary consists of the straight lines $OA$ and $OF$ (length $r$ each), the arcs $AB$ and $EF$ (length $r\theta$ each), the straight segments $BC$ and $DE$ which connect the inner and outer circles (length $2r - r = r$ each), and the outer arc $CD$ (length $2r \times 2\theta = 4r\theta$ ). Adding these gives $4r + 6r\theta = 14$ .

Next, we calculate the total area by summing the areas of the three sectors using the formula $\frac{1}{2}r^2\theta$ . The two smaller sectors have area $\frac{1}{2}r^2\theta$ each, and the larger sector has radius $2r$ and angle $2\theta$ , giving area $\frac{1}{2}(2r)^2(2\theta) = 4r^2\theta$ . The total area is $5r^2\theta = 10$ .

We now have a system of two equations. Solving the area equation for $\theta$ gives $\theta = 2/r^2$ . Substituting this into the perimeter equation eliminates $\theta$ and yields a quadratic in $r$ : $2r^2 - 7r + 6 = 0$ . Factorising gives $r = 3/2$ or $r = 2$ . The problem states $r > 3/2$ , so we select $r = 2$ . Finally, we find $\theta = 0.5$ and verify it satisfies $\theta < 3/4$ .

Key Takeaways

The arc length formula is $s = r\theta$ and the sector area formula is $A = \frac{1}{2}r^2\theta$ , where $\theta$ is in radians.
Complex geometric shapes can be decomposed into simpler parts (sectors and straight lines) to find total perimeter and area.
Simultaneous equations involving a linear and a quadratic relationship can be solved by substitution, often leading to a quadratic equation that must be solved and checked against given constraints.

Common Mistakes

Forgetting to include the straight line segments $BC$ and $DE$ in the perimeter calculation. These have length $r$ each, not $0$ .
Using the wrong radius or angle for the middle sector $COD$ . Its radius is $2r$ and its angle is $2\theta$ , so its arc length is $4r\theta$ and area is $4r^2\theta$ .
Failing to apply the given constraints ( $r > 3/2$ and $\theta < 3/4$ ) to reject the extraneous solution $r = 3/2$ and $\theta = 8/9$ .

Things to Be Careful About

Always ensure angles are in radians when using the arc length and sector area formulae.
Check both constraints given in the question. The quadratic yields two valid mathematical solutions, but only one satisfies the physical or stated conditions of the problem.
When substituting $\theta = 2/r^2$ into the perimeter equation, remember to multiply by $r$ to clear the denominator, being careful not to lose the $r=0$ case (which is invalid here since $r > 3/2$ ).

Techniques used

apply arc length and sector area formulaeform simultaneous equationseliminate a variable to form a quadraticsolve quadratic equation and check constraints

Approach

Complete the square for $-2x^2 + 8x + 11$ by factoring out $-2$ from the quadratic and linear terms, then completing the square inside the brackets. The vertex coordinates can be read directly from the form $-a(x - b)^2 + c$ .

Working

Factor out $-2$ from the first two terms:

-2x^2 + 8x + 11 = -2(x^2 - 4x) + 11

Complete the square inside the brackets. Half of $-4$ is $-2$ , so:

x^2 - 4x = (x - 2)^2 - 4

Substitute back:

-2\big((x - 2)^2 - 4\big) + 11

Expand and simplify:

-2(x - 2)^2 + 8 + 11 = -2(x - 2)^2 + 19

So $a = 2$ , $b = 2$ , $c = 19$ , all positive integers as required.

The vertex of the parabola $y = -2(x - 2)^2 + 19$ is at $(b, c) = (2, 19)$ .

Answer

The vertex is at $(2, 19)$ .

Final answer

(2, 19)

Detailed explanation

Walkthrough

We need to express $-2x^2 + 8x + 11$ in the form $-a(x - b)^2 + c$ . First, factor out $-2$ from the $x^2$ and $x$ terms, giving $-2(x^2 - 4x) + 11$ . Next, complete the square inside the brackets: $x^2 - 4x$ becomes $(x - 2)^2 - 4$ because half of $-4$ is $-2$ , and we must subtract $(-2)^2 = 4$ to keep the expression equivalent. Substituting back gives $-2((x - 2)^2 - 4) + 11 = -2(x - 2)^2 + 8 + 11 = -2(x - 2)^2 + 19$ . This is now in the required form with $a = 2$ , $b = 2$ , $c = 19$ . The vertex of a parabola $y = -a(x - b)^2 + c$ is always at $(b, c)$ , so the vertex is $(2, 19)$ .

Key Takeaways

Completing the square for a quadratic with a negative leading coefficient requires factoring out the negative first.
The vertex form $y = -a(x - b)^2 + c$ immediately gives the vertex coordinates as $(b, c)$ .
Always verify that the values of $a$ , $b$ , $c$ satisfy any given conditions (here, positive integers).

Common Mistakes

Forgetting to distribute the $-2$ across the constant term when expanding $-2((x-2)^2 - 4)$ , leading to an incorrect value of $c$ .
Writing the completed square as $-2(x - 2)^2 + 4 + 11$ instead of $-2(x - 2)^2 + 8 + 11$ .

Things to Be Careful About

The question requires $a$ , $b$ , $c$ to be positive integers. After completing the square, check that all three values satisfy this condition.
The vertex is $(b, c)$ , not $(b, -c)$ or $(-b, c)$ .

Techniques used

factor out leading coefficientcomplete the square inside bracketsread vertex coordinates from completed square form

(b)

Fig. 4

The diagram shows part of the curve with equation $y = -2x^2 + 8x + 11$ and the line with equation $y = 8x + 9$ .

Find the area of the shaded region.

5M Similar

DifficultyMedium

Worked solution

Approach

Find the $x$ -coordinates of the intersection points by equating the curve and the line. The shaded region is bounded above by the curve and below by the line between these two $x$ -values. Integrate the difference (curve $-$ line) from the lower to the upper intersection point.

Working

Find the points of intersection by setting the equations equal:

-2x^2 + 8x + 11 = 8x + 9

Simplify by subtracting $8x + 9$ from both sides:

-2x^2 + 2 = 0

2x^2 = 2

x^2 = 1

x = \pm 1

The region is bounded between $x = -1$ and $x = 1$ .

Set up the integral for the area. The curve $y = -2x^2 + 8x + 11$ is above the line $y = 8x + 9$ in this region:

\text{Area} = \int_{-1}^{1} \big[(-2x^2 + 8x + 11) - (8x + 9)\big] \, \mathrm{d}x

Simplify the integrand:

\text{Area} = \int_{-1}^{1} (-2x^2 + 2) \, \mathrm{d}x

Integrate:

\text{Area} = \left[-\frac{2}{3}x^3 + 2x\right]_{-1}^{1}

Apply the upper limit $x = 1$ :

-\frac{2}{3}(1)^3 + 2(1) = -\frac{2}{3} + 2 = \frac{4}{3}

Apply the lower limit $x = -1$ :

-\frac{2}{3}(-1)^3 + 2(-1) = \frac{2}{3} - 2 = -\frac{4}{3}

Subtract:

\text{Area} = \frac{4}{3} - \left(-\frac{4}{3}\right) = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}

Answer

The area of the shaded region is $\frac{8}{3}$ square units.

Final answer

8/3

Detailed explanation

Walkthrough

First, find where the curve and line intersect by equating their equations: $-2x^2 + 8x + 11 = 8x + 9$ . The $8x$ terms cancel, leaving $-2x^2 + 2 = 0$ , which gives $x^2 = 1$ and so $x = \pm 1$ . These are the limits of integration.

Next, set up the integral for the shaded area. From the diagram, the curve is above the line between $x = -1$ and $x = 1$ , so the area is $\int_{-1}^{1} (\text{curve} - \text{line}) \, dx$ . The integrand simplifies to $(-2x^2 + 8x + 11) - (8x + 9) = -2x^2 + 2$ .

Integrate $-2x^2 + 2$ to get $-\frac{2}{3}x^3 + 2x$ . Evaluate at the upper limit $x = 1$ : $-\frac{2}{3} + 2 = \frac{4}{3}$ . Evaluate at the lower limit $x = -1$ : $\frac{2}{3} - 2 = -\frac{4}{3}$ . Subtract the lower from the upper: $\frac{4}{3} - (-\frac{4}{3}) = \frac{8}{3}$ .

Key Takeaways

To find the area between a curve and a line, first find the intersection points to determine the limits of integration.
Always subtract the lower function from the upper function inside the integral to ensure a positive area.
The $8x$ terms cancel in this problem, making the integrand a simple quadratic.

Common Mistakes

Using incorrect limits of integration (e.g., forgetting the negative intersection point $x = -1$ ).
Integrating the curve and line separately and subtracting the results incorrectly, leading to sign errors.
Mistaking the triangular region for a trapezium and using the wrong area formula, which gives $\frac{11}{3}$ instead of $\frac{8}{3}$ .
Getting a negative answer like $-\frac{8}{3}$ due to subtracting in the wrong order.

Things to Be Careful About

Check which function is on top in the region between the intersection points. Here, the curve is above the line, so integrate (curve $-$ line).
The mark scheme condones a sign error in the final answer (accepting $-\frac{8}{3} \to \frac{8}{3}$ ), but the working must show the correct application of limits.
Ensure the integration is performed correctly: $\int -2x^2 \, dx = -\frac{2}{3}x^3$ , not $-\frac{2}{2}x^3$ .

Techniques used

equate curve and line to find intersection pointsset up definite integral of difference between curve and lineevaluate definite integral with correct limits

Q8MediumCoordinate Geometry

The equation of a circle is $x^2 + y^2 + px + 2y + q = 0$ , where $p$ and $q$ are constants.

(a)

Express the equation in the form $(x - a)^2 + (y - b)^2 = r^2$ , where $a$ is to be given in terms of $p$ and $r^2$ is to be given in terms of $p$ and $q$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Complete the square separately for $x$ and $y$ , then rearrange so the constant terms are on the right.

Working

For $x$ :

x^2 + px = \left(x + \frac{p}{2}\right)^2 - \frac{p^2}{4}

For $y$ :

y^2 + 2y = (y+1)^2 - 1

Substitute into the circle equation:

\left(x + \frac{p}{2}\right)^2 - \frac{p^2}{4} + (y+1)^2 - 1 + q = 0

So

\left(x + \frac{p}{2}\right)^2 + (y+1)^2 = \frac{p^2}{4} + 1 - q

In the required form:

\left(x - \left(-\frac{p}{2}\right)\right)^2 + (y - (-1))^2 = -q + 1 + \left(-\frac{p}{2}\right)^2

Answer

$a = -\frac{p}{2}$ , $b = -1$ , $r^2 = \frac{p^2}{4} + 1 - q$ .

Final answer

a = -p/2, b = -1, r^2 = p^2/4 + 1 - q

Detailed explanation

Walkthrough

We start with $x^2 + y^2 + px + 2y + q = 0$ . The goal is to rewrite it as $(x-a)^2+(y-b)^2 = r^2$ , which reveals the centre $(a,b)$ and radius $r$ .

For the $x$ -terms, complete the square:

x^2 + px = \left(x + \frac{p}{2}\right)^2 - \frac{p^2}{4}

This is because expanding $\left(x+\frac{p}{2}\right)^2$ gives $x^2 + px + \frac{p^2}{4}$ , so we subtract $\frac{p^2}{4}$ to keep the expression equal.

For the $y$ -terms:

y^2 + 2y = (y+1)^2 - 1

Substitute both into the original equation and move the constant terms to the right-hand side:

\left(x+\frac{p}{2}\right)^2 + (y+1)^2 = \frac{p^2}{4} + 1 - q

So $a = -\frac{p}{2}$ , $b = -1$ and $r^2 = \frac{p^2}{4} + 1 - q$ .

Key Takeaways

Completing the square is the standard way to convert a general circle equation into centre-radius form. The centre is found by reading the signs inside the brackets; the radius squared is the constant on the right.

Common Mistakes

Forgetting to subtract $\frac{p^2}{4}$ and $1$ after completing the square.
Writing the $x$ -coordinate of the centre as $\frac{p}{2}$ instead of $-\frac{p}{2}$ .
Mixing up the sign of $q$ when moving it to the right-hand side.

Things to Be Careful About

The form asked for is $(x-a)^2$ , so the centre's $x$ -coordinate is $-\frac{p}{2}$ , not $\frac{p}{2}$ . The mark scheme accepts either the completed-square expression or an equivalent form such as $\left(x-\left(-\frac{p}{2}\right)\right)^2 + (y-(-1))^2 = -q + 1 + \left(-\frac{p}{2}\right)^2$ .

Techniques used

complete the square in x and yidentify the centre and radius squared from the completed-square form

(b)

The line with equation $x + 2y = 10$ is the tangent to the circle at the point $A(4, 3)$ .

8M

(i)

Find the equation of the normal to the circle at the point $A$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The tangent line is $x+2y=10$ . Find its gradient, then use the perpendicular gradient rule to get the gradient of the normal. Finally write the normal through $A(4,3)$ .

Working

From the tangent line:

x+2y=10 \Rightarrow y = 5 - \frac{x}{2}

So the tangent gradient is

m_T = -\frac{1}{2}

For the normal, $m_N m_T = -1$ , so

m_N = 2

The normal passes through $A(4,3)$ :

\frac{y-3}{x-4} = 2

y-3 = 2x - 8

y = 2x - 5

Answer

y = 2x - 5

Final answer

y = 2x - 5

Detailed explanation

Walkthrough

The tangent line is $x+2y=10$ . Rearranging gives $y = 5 - \frac{x}{2}$ , so its gradient is $-\frac{1}{2}$ . A normal is perpendicular to the tangent, so the product of their gradients is $-1$ : $m_N \times \left(-\frac{1}{2}\right) = -1$ , hence $m_N = 2$ .

Since the normal passes through $A(4,3)$ , use the point-gradient form:

\frac{y-3}{x-4} = 2

which simplifies to $y = 2x - 5$ .

Key Takeaways

For a circle, the tangent and normal at a point are perpendicular. The gradient of a line in the form $ax+by=c$ can be read as $-\frac{a}{b}$ .

Common Mistakes

Using the tangent gradient $-\frac{1}{2}$ as the normal gradient.
Forgetting to take the negative reciprocal.
Substituting the wrong point into the line equation.

Things to Be Careful About

The normal equation is used again in part (b)(ii) to find the centre, so keep it as $y=2x-5$ . The mark scheme allows $y=2x+c$ with $c$ found by substituting $A$ .

Techniques used

find the gradient of the tangent from its line equationapply the perpendicular gradient conditionwrite the equation of a line through a given point

(ii)

Find the values of $p$ and $q$ .

5M Similar

DifficultyMedium

Worked solution

Approach

The normal at a point on a circle passes through the centre. Use the centre $\left(-\frac{p}{2}, -1\right)$ from part (a) in the normal equation to find $p$ . Then use the radius-squared expression from part (a) and the distance from the centre to $A$ to find $q$ .

Working

The centre is

C = \left(-\frac{p}{2}, -1\right)

Since $C$ lies on the normal $y = 2x - 5$ :

-1 = 2\left(-\frac{p}{2}\right) - 5

-1 = -p - 5

p = -4

So the centre is $C = (2, -1)$ .

From part (a),

r^2 = -q + 1 + \frac{p^2}{4} = -q + 1 + \frac{16}{4} = -q + 5

The distance squared from $C(2,-1)$ to $A(4,3)$ is

r^2 = (4-2)^2 + (3-(-1))^2 = 4 + 16 = 20

Equating:

-q + 5 = 20

q = -15

Answer

p = -4, \quad q = -15

Final answer

p = -4, q = -15

Detailed explanation

Walkthrough

The centre of the circle from part (a) is $\left(-\frac{p}{2}, -1\right)$ . A key fact for circles is that the normal at a point on the circle passes through the centre. Since the normal is $y=2x-5$ , substitute the centre coordinates into this line:

-1 = 2\left(-\frac{p}{2}\right) - 5

This gives $-1 = -p - 5$ , so $p = -4$ . The centre is therefore $(2,-1)$ .

To find $q$ , use the radius. From part (a),

r^2 = -q + 1 + \frac{p^2}{4}

With $p=-4$ , this is $r^2 = -q + 5$ . The radius is also the distance from the centre to $A(4,3)$ :

r^2 = (4-2)^2 + (3-(-1))^2 = 4+16 = 20

Equating gives $-q + 5 = 20$ , so $q = -15$ .

Key Takeaways

This part combines the completed-square form of a circle with the fact that the normal passes through the centre. It also uses the distance formula, since the radius is the distance from the centre to any point on the circle.

Common Mistakes

Using the tangent line instead of the normal line to find $p$ .
Forgetting that the normal passes through the centre.
Substituting $p$ incorrectly into $r^2$ ; with $p=-4$ , $p^2=16$ .
Sign error when solving $-q+5=20$ , giving $q=15$ instead of $q=-15$ .
Giving final values without showing a method; the mark scheme requires method marks.

Things to Be Careful About

The mark scheme permits several methods. One alternative is to substitute $A(4,3)$ directly into the original circle equation to get $4p+q+31=0$ , then use $p=-4$ to find $q=-15$ . Whichever method is used, the final values must be consistent with the centre and radius found earlier.

Techniques used

use the fact that the normal to a circle passes through its centresubstitute the centre into the normal equation to find pcompute the squared distance from the centre to the given pointequate this to the completed-square expression for r^2

Q9MediumQuadratics

The equation of a curve is $y = \frac{1}{2}k^2x^2 - 2kx + 2$ and the equation of a line is $y = kx + p$ , where $k$ and $p$ are constants with $0 < k < 1$ .

(a)

It is given that one of the points of intersection of the curve and the line has coordinates $\left(\frac{5}{2}, \frac{1}{2}\right)$ .

Find the values of $k$ and $p$ , and find the coordinates of the other point of intersection.

7M Similar

DifficultyMedium

Worked solution

Approach

Substitute the known point into the curve equation to obtain a quadratic in $k$ . Solve it and select the root satisfying $0 < k < 1$ . Then use the line equation to find $p$ . Finally equate the curve and line with these values and solve the resulting quadratic to find the second intersection.

Working

Using $\left(\frac{5}{2}, \frac{1}{2}\right)$ in the curve equation:

\frac{1}{2}k^2\left(\frac{5}{2}\right)^2 - 2k\left(\frac{5}{2}\right) + 2 = \frac{1}{2}

Simplify:

\frac{25}{8}k^2 - 5k + 2 = \frac{1}{2}

\frac{25}{8}k^2 - 5k + \frac{3}{2} = 0

Multiply by 8:

25k^2 - 40k + 12 = 0

Factorise:

(5k - 2)(5k - 6) = 0

So $k = \frac{2}{5}$ or $k = \frac{6}{5}$ . Since $0 < k < 1$ , we take $k = \frac{2}{5}$ .

Use the line equation at $\left(\frac{5}{2}, \frac{1}{2}\right)$ :

\frac{1}{2} = \frac{2}{5}\cdot\frac{5}{2} + p = 1 + p

Therefore

p = -\frac{1}{2}

Now equate the curve and the line:

\frac{1}{2}\left(\frac{2}{5}\right)^2 x^2 - 2\left(\frac{2}{5}\right)x + 2 = \frac{2}{5}x - \frac{1}{2}

Simplify:

\frac{2}{25}x^2 - \frac{4}{5}x + 2 = \frac{2}{5}x - \frac{1}{2}

\frac{2}{25}x^2 - \frac{6}{5}x + \frac{5}{2} = 0

Multiply by 50:

4x^2 - 60x + 125 = 0

Factorise:

(2x - 5)(2x - 25) = 0

So $x = \frac{5}{2}$ or $x = \frac{25}{2}$ . The other intersection has $x = \frac{25}{2}$ .

Substitute into the line:

y = \frac{2}{5}\left(\frac{25}{2}\right) - \frac{1}{2} = 5 - \frac{1}{2} = \frac{9}{2}

Answer

$k = \frac{2}{5}$ , $p = -\frac{1}{2}$ , and the other point of intersection is $\left(\frac{25}{2}, \frac{9}{2}\right)$ .

Final answer

k = 2/5, p = -1/2, other point of intersection = (25/2, 9/2)

Detailed explanation

Walkthrough

We are told that $\left(\frac{5}{2}, \frac{1}{2}\right)$ is a point of intersection. This means it satisfies both the curve equation and the line equation. The first step is to substitute this point into the curve equation. Doing so removes $x$ and $y$ and leaves a quadratic equation in $k$ . Solve this quadratic; it gives two possible values of $k$ . The condition $0 < k < 1$ tells us which value is valid, so we discard the other root.

Once $k$ is known, substitute the same point into the line equation $y = kx + p$ . This gives a simple linear equation in $p$ , which is solved immediately.

With both $k$ and $p$ known, equate the curve and the line. This produces a quadratic equation in $x$ whose solutions are the $x$ -coordinates of the two intersection points. One of these is the given $x = \frac{5}{2}$ ; factorising the quadratic gives the other, $x = \frac{25}{2}$ . Finally, substitute this value into the line equation to obtain the corresponding $y$ -coordinate.

Key Takeaways

A point of intersection satisfies every equation involved.
Conditions such as $0 < k < 1$ can be used to choose between multiple algebraic solutions.
Intersections of a line and a quadratic curve are found by solving the quadratic obtained when the two equations are equated.
Quadratic equations can be solved by factorisation or the quadratic formula.

Common Mistakes

Forgetting to reject $k = \frac{6}{5}$ using the condition $0 < k < 1$ .
Making arithmetic errors when working with fractions such as $\frac{25}{8}k^2$ and $\frac{5}{2}$ .
Not equating the curve and the line after finding $k$ and $p$ , so the second intersection is never found.
Sign errors when rearranging the quadratic in $x$ .

Things to Be Careful About

The condition $0 < k < 1$ is essential for choosing the correct value of $k$ .
When multiplying an equation by a constant, multiply every term.
The known point gives one root of the quadratic in $x$ ; the other root is the new intersection.
Use the line equation (or the curve equation) to find the $y$ -coordinate of the other point.

Techniques used

substitute a known point into the curve equationsolve a quadratic equation in kuse the line equation to determine pequate the curve and line to form a quadratic in xsolve the resulting quadratic to find the second intersection point

(b)

It is given instead that the line and the curve do not intersect.

Find the set of possible values of $p$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Equate the curve and the line to obtain a quadratic in $x$ . For the line and curve not to intersect, this quadratic must have no real roots, so its discriminant must be negative. Simplify the discriminant and solve the resulting inequality for $p$ .

Working

Equating the curve and the line:

\frac{1}{2}k^2x^2 - 2kx + 2 = kx + p

Collect like terms:

\frac{1}{2}k^2x^2 - 3kx + 2 - p = 0

For no intersection, the discriminant must be negative. Here $a = \frac{1}{2}k^2$ , $b = -3k$ and $c = 2 - p$ , so

b^2 - 4ac = (-3k)^2 - 4\left(\frac{1}{2}k^2\right)(2 - p)

Simplify:

9k^2 - 2k^2(2 - p) = 9k^2 - 4k^2 + 2k^2p = 5k^2 + 2k^2p = k^2(5 + 2p)

Require $b^2 - 4ac < 0$ :

k^2(5 + 2p) < 0

Since $0 < k < 1$ , $k^2 > 0$ , so

5 + 2p < 0

p < -\frac{5}{2}

Answer

p < -\frac{5}{2}

Final answer

p < -5/2

Detailed explanation

Walkthrough

For the line and the curve not to intersect, the two equations must have no common solution. Equate the curve and the line and collect all terms on one side. This gives a quadratic in $x$ with coefficients involving $k$ and $p$ .

A quadratic equation has no real roots exactly when its discriminant is negative. Identify $a = \frac{1}{2}k^2$ , $b = -3k$ and $c = 2 - p$ , then substitute them into $b^2 - 4ac$ . Simplify the resulting expression: the $k^2$ terms combine, and the discriminant becomes $k^2(5 + 2p)$ .

For no intersection, set this discriminant less than zero. Because $0 < k < 1$ , $k^2$ is positive, so it can be divided out without reversing the inequality. This leaves $5 + 2p < 0$ , giving $p < -\frac{5}{2}$ .

Key Takeaways

Two graphs intersect exactly when their equations have a common solution.
No intersection corresponds to a quadratic with no real roots, i.e. discriminant $< 0$ .
Parameters can be handled symbolically in the discriminant.
Dividing an inequality by a positive quantity preserves its direction.

Common Mistakes

Using discriminant $> 0$ instead of $< 0$ .
Forgetting that $-2kx - kx = -3kx$ when collecting like terms.
Expanding $4ac$ incorrectly: $4 \times \frac{1}{2}k^2(2-p) = 2k^2(2-p)$ .
Dividing by $k^2$ without noting that $k \neq 0$ ; here the condition $0 < k < 1$ guarantees this.

Things to Be Careful About

The inequality is strict: $p < -\frac{5}{2}$ , not $p \leq -\frac{5}{2}$ . If the discriminant is zero, the line is tangent to the curve, so they do intersect at one point.
Keep the sign of the discriminant correct: no real roots means $b^2 - 4ac < 0$ .
Since $k^2 > 0$ , the inequality sign does not reverse when dividing by $k^2$ .

Techniques used

equate the curve and the line to form a quadratic in xapply the discriminant condition for no real rootssimplify the discriminant expression in k and psolve the resulting inequality for p

Q10MediumDifferentiationIntegrationQuadratics

A function $f$ with domain $x > 0$ is such that $f'(x) = 8(2x - 3)^{-\frac{1}{3}} - 10x^{\frac{2}{3}}$ . It is given that the curve with equation $y = f(x)$ passes through the point $(1, 0)$ .

(a)

Find the equation of the normal to the curve at the point $(1, 0)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Evaluate the gradient of the tangent at $(1,0)$ using $f'(1)$ . The normal is perpendicular to the tangent, so use $m_{\text{tangent}}m_{\text{normal}}=-1$ , then write the point-slope form of the line.

Working

At $x=1$ :

f'(1)=8(2(1)-3)^{1/3}-10(1)^{2/3}=8(-1)-10=-18

So the tangent gradient is $-18$ . The normal gradient is

m_{\text{normal}}=-\frac{1}{-18}=\frac{1}{18}

The normal passes through $(1,0)$ , so

y-0=\frac{1}{18}(x-1)

Answer

y=\frac{x-1}{18}

Equivalently, $x-18y-1=0$ .

Final answer

y = (x-1)/18

Detailed explanation

Walkthrough

The tangent gradient at the point is the value of the derivative at $x=1$ . Substitute $x=1$ into $f'(x)$ : the term $2(1)-3=-1$ , and $(-1)^{1/3}=-1$ , so $f'(1)=-18$ . The normal is perpendicular to the tangent, so its gradient is the negative reciprocal: $m_n=-1/(-18)=1/18$ . Since the normal passes through $(1,0)$ , use $y-y_1=m(x-x_1)$ to obtain the equation.

Key Takeaways

The gradient of a normal is the negative reciprocal of the gradient of the tangent.
The derivative must be evaluated at the specified $x$ -coordinate before applying the reciprocal.
A line with a known gradient $m$ and a point $(x_1,y_1)$ has equation $y-y_1=m(x-x_1)$ .

Common Mistakes

Forgetting the negative sign in the reciprocal, or using $-18$ as the normal gradient.
Using the wrong point, such as $(0,1)$ instead of $(1,0)$ .
Misreading $(-1)^{1/3}$ as $1$ instead of $-1$ .

Things to Be Careful About

$(-1)^{1/3}=-1$ .
The domain $x>0$ includes $x=1$ .
Equivalent forms such as $y-0=\frac{1}{18}(x-1)$ , $y=\frac{x-1}{18}$ , and $x-18y-1=0$ are all valid.

Techniques used

evaluate the derivative at a given pointapply the perpendicular gradient condition for a normalwrite the equation of a line through a point with a known gradient

(b)

Find $f(x)$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Integrate $f'(x)$ term by term, using the composite power rule for the first term and the ordinary power rule for the second. Then use $f(1)=0$ to determine the constant of integration.

Working

\begin{aligned} f(x) &= \int \left[8(2x-3)^{1/3}-10x^{2/3}\right]\,dx \\ &= 8\cdot\frac{(2x-3)^{4/3}}{2\cdot\frac{4}{3}} - 10\cdot\frac{x^{5/3}}{\frac{5}{3}} + c \\ &= 3(2x-3)^{4/3}-6x^{5/3}+c \end{aligned}

Use $f(1)=0$ :

0 = 3(2(1)-3)^{4/3}-6(1)^{5/3}+c=3-6+c

So $c=3$ . Therefore

f(x)=3(2x-3)^{4/3}-6x^{5/3}+3

Answer

f(x)=3(2x-3)^{4/3}-6x^{5/3}+3

Final answer

f(x) = 3(2x-3)^(4/3) - 6x^(5/3) + 3

Detailed explanation

Walkthrough

Integrate the two terms separately. For $8(2x-3)^{1/3}$ , use the composite power rule: increase the power by $1$ , then divide by the new power and by the derivative of the inside. For $10x^{2/3}$ , use the ordinary power rule. This gives $3(2x-3)^{4/3}-6x^{5/3}+c$ . Then use $f(1)=0$ : substitute $x=1$ and $y=0$ . Since $(-1)^{4/3}=1$ , the equation becomes $3-6+c=0$ , so $c=3$ .

Key Takeaways

$\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)}+C$ for $n\neq -1$ .
A point on the curve fixes the arbitrary constant of integration.
Fractional powers obey the same integration rule as integer powers.

Common Mistakes

Omitting the factor $a$ in the denominator when integrating $(ax+b)^n$ .
Forgetting the constant of integration before substituting the point.
Incorrectly simplifying $(-1)^{4/3}$ .
Substituting the point into $f'(x)$ instead of $f(x)$ .

Things to Be Careful About

The mark scheme awards one mark for each unsimplified integral; show both before simplifying.
Use $f(1)=0$ , not $f(0)=1$ .
$(-1)^{4/3}=1$ because the fourth power of $-1$ is $1$ before taking the cube root.

Techniques used

integrate a composite power functionintegrate a power of xuse an initial condition to determine the constant of integration

(c)

It is given that the equation $f'(x) = 0$ can be expressed in the form

125x^2 - 128x + 192 = 0.

Determine, making your reasoning clear, whether $f$ is an increasing function, a decreasing function or neither.

3M Similar

DifficultyMedium

Worked solution

Approach

Interpret $f'(x)=0$ as the quadratic $125x^2-128x+192=0$ . Use its discriminant to show that it has no real roots, so $f'(x)$ has no zeros on $x>0$ . Since $f'$ is continuous, its sign cannot change; check the sign at $x=1$ .

Working

\Delta = (-128)^2 - 4(125)(192) = 16384 - 96000 = -79616 < 0

The quadratic has no real roots, so $f'(x)\neq 0$ for all $x>0$ ; there are no stationary points.

Now evaluate the derivative at a convenient point:

f'(1) = 8(2(1)-3)^{1/3}-10(1)^{2/3}=8(-1)-10=-18<0

Since $f'$ has no zeros and is continuous on $x>0$ , it cannot change sign. Therefore $f'(x)<0$ for every $x>0$ , so $f$ is decreasing.

Answer

Final answer

decreasing function

Detailed explanation

Walkthrough

The equation $f'(x)=0$ is equivalent to the quadratic $125x^2-128x+192=0$ . Compute the discriminant $b^2-4ac$ . Since it is negative, the quadratic has no real solutions, so $f'(x)$ never equals $0$ on $x>0$ . A continuous derivative with no real zeros cannot change sign; it is either always positive or always negative. To decide which, substitute any convenient value in the domain, say $x=1$ . Since $f'(1)=-18<0$ , the derivative is negative everywhere on $x>0$ , so $f$ is decreasing.

Key Takeaways

The discriminant tells whether the equation $f'(x)=0$ has solutions, so it tells whether stationary points exist.
If the derivative has no zeros and is continuous, its sign is constant; one test point determines the sign.
A negative derivative means the function is decreasing.

Common Mistakes

Concluding that $f$ is increasing because the quadratic is always positive; the quadratic only records where $f'(x)=0$ , not the sign of $f'(x)$ .
Forgetting to check the sign of $f'(x)$ at a point.
Ignoring the domain $x>0$ .

Things to Be Careful About

The discriminant is $-79616$ , so there are no real roots.
The quadratic has a positive leading coefficient, but this is only part of the reasoning.
The mark scheme requires both "no real roots" and a demonstration that $f'(x)<0$ for a positive value of $x$ .

Techniques used

compute the discriminant of a quadraticdeduce that there are no real rootsevaluate the derivative at a convenient pointconclude the monotonicity of the function

Mathematics 9709/12 — October/November 2024

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