Circles $C_1$ and $C_2$ have equations $$ x^2 + y^2 + 6x - 10y + 18 = 0 \text{ and } (x - 9)^2 + (y + 4)^2 - 64 = 0 $$ respectively. (a) Find the distance between the centres of the circles. (b) $P$ and $Q$ are points on $C_1$ and $C_2$ respectively. The distance between $P$ and $Q$ is denoted by $d$. Find the greatest and least possible values of $d$.

(a) 15 · (b) Least = 3, greatest = 27

9709/11 Oct/Nov 2024 — Mathematics Solved Past Paper

Q14MMediumSeries

In the expansion of $\left(kx + \frac{2}{x}\right)^4$ , where $k$ is a positive constant, the term independent of $x$ is equal to 150.

Find the value of $k$ and hence determine the coefficient of $x^2$ in the expansion.

DifficultyMedium

Worked solution

Approach

Use the binomial expansion of $(kx + \frac{2}{x})^4$ . The term independent of $x$ is the one where the powers of $x$ cancel, so set the index of $x$ to zero and solve for $k$ . Then use the same expansion to find the coefficient of $x^2$ .

Working

For $(kx + \frac{2}{x})^4$ , the general term with $r$ factors of $kx$ is

\binom{4}{r}(kx)^r\left(\frac{2}{x}\right)^{4-r}

The power of $x$ in this term is

x^{r-(4-r)} = x^{2r-4}

The term independent of $x$ occurs when $2r-4=0$ , so $r=2$ . Its coefficient is

\binom{4}{2}(kx)^2\left(\frac{2}{x}\right)^2 = 6k^2x^2 \cdot \frac{4}{x^2} = 24k^2

Equating this to 150:

24k^2 = 150

k^2 = \frac{150}{24} = \frac{25}{4}

Since $k$ is positive,

k = \frac{5}{2}

Now the coefficient of $x^2$ occurs when $2r-4=2$ , so $r=3$ . The corresponding term is

\binom{4}{3}(kx)^3\left(\frac{2}{x}\right) = 4k^3x^3 \cdot \frac{2}{x} = 8k^3x^2

Using $k = \frac{5}{2}$ :

8\left(\frac{5}{2}\right)^3 = 8 \cdot \frac{125}{8} = 125

Answer

k = \frac{5}{2}, \quad \text{coefficient of } x^2 = 125

Final answer

k = 5/2, coefficient of x^2 = 125

Detailed explanation

Walkthrough

We need to find the term in the binomial expansion of $(kx + \frac{2}{x})^4$ whose powers of $x$ cancel. In a binomial expansion, the term with $r$ factors of the first term has the form $\binom{4}{r}(kx)^r(\frac{2}{x})^{4-r}$ . The power of $x$ is $r-(4-r)=2r-4$ .

For the constant term, set $2r-4=0$ , so $r=2$ . This tells us which term to extract. Its coefficient is $\binom{4}{2}k^2 2^2 = 24k^2$ . Since the question tells us this coefficient equals 150, we solve $24k^2=150$ to get $k^2=\frac{25}{4}$ . Because $k$ is a positive constant, $k=\frac{5}{2}$ .

For the coefficient of $x^2$ , we need $2r-4=2$ , so $r=3$ . The term is $\binom{4}{3}(kx)^3(\frac{2}{x}) = 8k^3x^2$ . Substituting $k=\frac{5}{2}$ gives $8 \cdot \frac{125}{8}=125$ .

Key Takeaways

This question tests the binomial expansion of $(a+b)^n$ and the idea of identifying a particular term by its power of $x$ . Instead of writing out the whole expansion, use the general term $\binom{n}{r}a^{n-r}b^r$ and set the exponent of $x$ to the required value. It also tests solving a simple quadratic and substituting back into an expression.

Common Mistakes

Using $^4C_2$ instead of the numerical coefficient $\frac{4!}{2!2!}$ or $6$ ; the mark scheme requires the numerical coefficient or factorial form.
Forgetting that $k$ is positive, and incorrectly giving $k = -\frac{5}{2}$ as the only answer. The negative root should be ignored.
Confusing the term independent of $x$ with the first or last term of the expansion.
Substituting $k$ into the wrong term when finding the coefficient of $x^2$ .

Things to Be Careful About

The power of $x$ in the general term is $x^{2r-4}$ ; be careful with the signs when subtracting exponents.
The term independent of $x$ is the $r=2$ term, not the $r=0$ term.
When the question says "coefficient of $x^2$ ", give the number $125$ , not the whole term $125x^2$ (though $125x^2$ is accepted as a final answer).
The mark scheme accepts the coefficient as $125x^2$ ; either form is acceptable.

Techniques used

apply the binomial general termset the exponent of x to zero to find the constant termsolve a quadratic equation for kextract the coefficient of x^2

Similar questions

Q24MMedium-EasyDifferentiation

The curve $y = x^2 - \frac{a}{x}$ has a stationary point at $(-3, b)$ .

Find the values of the constants $a$ and $b$ .

DifficultyMedium-Easy

Worked solution

Approach

Differentiate $y$ with respect to $x$ , then use the fact that the gradient is zero at a stationary point. Substitute $x = -3$ into the derivative to find $a$ , then substitute $x = -3$ into the original equation to find $b$ .

Working

Given

y = x^2 - \frac{a}{x} = x^2 - a x^{-1}

Differentiate:

\frac{dy}{dx} = 2x + a x^{-2}

At a stationary point, $\frac{dy}{dx} = 0$ . Since the stationary point is at $x = -3$ :

2(-3) + a(-3)^{-2} = 0

-6 + \frac{a}{9} = 0

\frac{a}{9} = 6

a = 54

Now substitute $x = -3$ into $y = x^2 - \frac{a}{x}$ :

b = (-3)^2 - \frac{54}{-3}

b = 9 + 18 = 27

Answer

a = 54, \quad b = 27

Final answer

a = 54, b = 27

Detailed explanation

Walkthrough

We need to find two constants. The key fact is that a stationary point occurs where the gradient of the curve is zero, so we start by differentiating. The term $x^2$ differentiates to $2x$ . The term $-\frac{a}{x}$ can be written as $-a x^{-1}$ , and differentiating $-a x^{-1}$ gives $a x^{-2}$ because the power decreases by 1 and the sign changes. Setting the derivative equal to zero and using $x = -3$ gives an equation involving only $a$ . Solving it gives $a = 54$ . Then substitute $x = -3$ into the original equation to find the corresponding $y$ -coordinate, which is $b = 27$ .

Key Takeaways

A stationary point has zero gradient, so $\frac{dy}{dx}=0$ .
To differentiate $\frac{a}{x}$ , rewrite it as $a x^{-1}$ and use the power rule.
Once a constant is found, substitute back into the original equation to obtain the corresponding coordinate.

Common Mistakes

Differentiating $-\frac{a}{x}$ incorrectly, especially getting the sign wrong.
Forgetting that at a stationary point the derivative equals zero, not the value of $y$ .
Substituting $x = -3$ into the derivative but not solving for $a$ ; the mark scheme requires an attempt at differentiation and equating to zero.
Confusing $b$ with the gradient; $b$ is the $y$ -coordinate of the stationary point.

Things to Be Careful About

At a stationary point, set $\frac{dy}{dx} = 0$ , not $y = 0$ .
When substituting $x = -3$ , remember $(-3)^{-2} = \frac{1}{9}$ .
Keep signs correct: $2(-3) = -6$ , and $-\frac{54}{-3} = +18$ .
The mark scheme requires showing the differentiation and the substitution into the derivative for the method marks.

Techniques used

differentiate a power functionset derivative equal to zero at stationary pointsolve for a constantevaluate the original function

Similar questions

Q35MMedium-EasyCircular Measure

Fig. 1

The diagram shows a sector of a circle, centre $O$ , where $OB = OC = 15\text{ cm}$ . The size of angle $BOC$ is $\frac{2}{5}\pi$ radians. Points $A$ and $D$ on the lines $OB$ and $OC$ respectively are joined by an arc $AD$ of a circle with centre $O$ . The shaded region is bounded by the arcs $AD$ and $BC$ and by the straight lines $AB$ and $DC$ . It is given that the area of the shaded region is $\frac{209}{5}\pi\text{ cm}^2$ .

Find the perimeter of the shaded region. Give your answer in terms of $\pi$ .

DifficultyMedium-Easy

Worked solution

Approach

The shaded region is the difference between two sectors: the larger sector $OBC$ and the smaller sector $OAD$ . We use the sector area formula $\frac{1}{2}r^2\theta$ to find the inner radius $OA = OD$ , and then use the arc length formula $s = r\theta$ and the straight line segments to find the perimeter.

Working

Let $OA = OD = x$ cm. The angle $\theta = \frac{2}{5}\pi$ radians and the outer radius $R = 15$ cm.

The area of the shaded region is the area of sector $OBC$ minus the area of sector $OAD$ :

\text{Area} = \frac{1}{2}R^2\theta - \frac{1}{2}x^2\theta

Substitute the known values:

\frac{1}{2}(15)^2\left(\frac{2}{5}\pi\right) - \frac{1}{2}x^2\left(\frac{2}{5}\pi\right) = \frac{209}{5}\pi

Simplify the first term:

\frac{1}{2} \times 225 \times \frac{2}{5}\pi = 45\pi

So the equation becomes:

45\pi - \frac{1}{5}\pi x^2 = \frac{209}{5}\pi

Divide through by $\pi$ :

45 - \frac{1}{5}x^2 = \frac{209}{5}

Multiply by 5:

225 - x^2 = 209

x^2 = 16

Since $x$ is a length, $x = 4$ . So $OA = OD = 4$ cm.

The perimeter of the shaded region consists of the outer arc $BC$ , the inner arc $AD$ , and the two straight line segments $AB$ and $DC$ .

Arc length of $BC$ :

s_{BC} = R\theta = 15 \times \frac{2}{5}\pi = 6\pi \text{ cm}

Arc length of $AD$ :

s_{AD} = x\theta = 4 \times \frac{2}{5}\pi = \frac{8}{5}\pi \text{ cm}

Length of straight segments:

AB = OB - OA = 15 - 4 = 11 \text{ cm}

DC = OC - OD = 15 - 4 = 11 \text{ cm}

Total perimeter:

P = s_{BC} + s_{AD} + AB + DC = 6\pi + \frac{8}{5}\pi + 11 + 11

P = 22 + \left(\frac{30}{5} + \frac{8}{5}\right)\pi = 22 + \frac{38}{5}\pi \text{ cm}

Answer

22 + \frac{38}{5}\pi

Final answer

22 + 38/5 pi

Detailed explanation

Walkthrough

First, recognise that the shaded region is an annular sector, which is the difference between a larger sector $OBC$ and a smaller sector $OAD$ sharing the same centre $O$ and angle $\theta$ . Let the unknown inner radius be $x = OA = OD$ . The area of a sector is given by $\frac{1}{2}r^2\theta$ . Set up the equation $\frac{1}{2}(15)^2(\frac{2}{5}\pi) - \frac{1}{2}x^2(\frac{2}{5}\pi) = \frac{209}{5}\pi$ . Solve this algebraically to find $x = 4$ .

Next, calculate the perimeter. The boundary of the shaded region has four parts: the outer arc $BC$ , the inner arc $AD$ , and the two straight radial segments $AB$ and $DC$ . Use the arc length formula $s = r\theta$ to find the lengths of the arcs: $15 \times \frac{2}{5}\pi = 6\pi$ and $4 \times \frac{2}{5}\pi = \frac{8}{5}\pi$ . The straight segments are simply the difference in radii: $15 - 4 = 11$ cm each. Add all four components together and collect the $\pi$ terms to get the final answer.

Key Takeaways

The area of a sector is $\frac{1}{2}r^2\theta$ and the arc length is $s = r\theta$ , where $\theta$ is in radians.
A shaded region between two concentric sectors can be found by subtracting the smaller sector area from the larger one.
The perimeter of such a region includes both arcs and the straight radial segments connecting them.

Common Mistakes

Forgetting to include the straight line segments $AB$ and $DC$ in the perimeter calculation.
Using degrees instead of radians for the angle in the sector area and arc length formulas.
Rounding $x$ or $\pi$ too early, which leads to an incorrect final numerical answer.
Failing to collect like terms, so the answer is not fully simplified.

Things to Be Careful About

The angle must be in radians when using $s = r\theta$ and $\text{Area} = \frac{1}{2}r^2\theta$ . Here it is given as $\frac{2}{5}\pi$ radians, so no conversion is needed.
The final answer must be left in terms of $\pi$ as requested. Do not evaluate $\pi$ numerically.
Ensure all $\pi$ terms are combined into a single fraction or simplified expression.

Techniques used

use sector area formulaset up area equationsolve for inner radiususe arc length formulasum perimeter components

Similar questions

Q45MMediumQuadratics

Show that the curve with equation $x^2 - 3xy - 40 = 0$ and the line with equation $3x + y + k = 0$ meet for all values of the constant $k$ .

DifficultyMedium

Worked solution

Approach

Substitute $y$ from the line equation into the curve equation to obtain a quadratic in $x$ . Then use the discriminant to show that this quadratic has real roots for every real value of $k$ .

Working

From the line equation,

3x + y + k = 0 \quad \Rightarrow \quad y = -3x - k

Substitute $y = -3x - k$ into $x^2 - 3xy - 40 = 0$ :

\begin{aligned} x^2 - 3x(-3x - k) - 40 &= 0 \\ x^2 + 9x^2 + 3kx - 40 &= 0 \\ 10x^2 + 3kx - 40 &= 0 \end{aligned}

This is a quadratic in $x$ with $a = 10$ , $b = 3k$ , $c = -40$ . Its discriminant is

\Delta = b^2 - 4ac = (3k)^2 - 4(10)(-40) = 9k^2 + 1600

Since $k^2 \geq 0$ for all real $k$ , we have $9k^2 + 1600 \geq 1600 > 0$ . Therefore $\Delta > 0$ for every value of $k$ , so the quadratic always has two distinct real roots.

Answer

\Delta = 9k^2 + 1600 > 0 \quad \text{for all } k

Hence the line and the curve meet for all values of the constant $k$ .

Final answer

The discriminant is 9k^2 + 1600 > 0 for all k, so the line and curve meet for all values of k.

Detailed explanation

Walkthrough

We are given a curve and a line and asked to show they always meet. The key is to solve the two equations simultaneously. Because the line equation is linear, we can write $y$ in terms of $x$ : $y = -3x - k$ . Substituting this into the curve equation removes $y$ and leaves a quadratic in $x$ .

For the line and curve to meet, this quadratic must have at least one real solution for $x$ . A quadratic $ax^2 + bx + c = 0$ has real solutions exactly when its discriminant $b^2 - 4ac$ is greater than or equal to zero. So we compute the discriminant in terms of $k$ and check its sign.

Here $a = 10$ , $b = 3k$ and $c = -40$ , so

\Delta = (3k)^2 - 4(10)(-40) = 9k^2 + 1600

Because $k^2 \geq 0$ , the expression $9k^2 + 1600$ is always at least $1600$ , hence strictly positive. Therefore the quadratic always has two real roots, no matter what $k$ is. This means the line always intersects the curve.

Key Takeaways

Intersections of a line and a curve are found by substituting one equation into the other.
The discriminant of the resulting quadratic tells us how many intersection points exist.
A sum of a non-negative term and a positive constant is always positive, so such a discriminant guarantees real roots for all parameter values.

Common Mistakes

Forgetting to expand the bracket when substituting $y = -3x - k$ ; the term $-3xy$ becomes $+9x^2 + 3kx$ , not $-9x^2 - 3kx$ .
Making a sign error in the discriminant: since $c = -40$ , $-4ac = +1600$ .
Writing the discriminant as $9k^2 - 1600$ instead of $9k^2 + 1600$ .
Stating that the discriminant is positive without justifying that $k^2 \geq 0$ .
According to the mark scheme, the discriminant must be isolated; do not hide it inside the quadratic formula without explicitly showing $b^2 - 4ac$ .

Things to Be Careful About

The coefficient of $x^2$ is $10$ , not $1$ , after substitution.
$k$ is a real constant, so $k^2$ is never negative.
A positive discriminant gives two distinct real roots, so the line meets the curve in two points; the question only requires that they meet, so this is more than enough.
If you substitute for $x$ instead, you would obtain $10y^2 + 11ky + k^2 - 360 = 0$ and its discriminant $81k^2 + 14400$ , which is also always positive; both approaches are acceptable.

Techniques used

substitute the linear equation into the quadratic equationexpand and simplify to a single quadratic in xcompute the discriminant as a function of kargue that the discriminant is positive for all real k

Similar questions

Q5MediumDifferentiationQuadraticsIntegration

The equation of a curve is such that $\frac{dy}{dx} = 4x - 3\sqrt{x} + 1$ .

(a)

Find the $x$ -coordinate of the point on the curve at which the gradient is $\frac{11}{2}$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Set the derivative equal to the given gradient, rearrange into a quadratic in $\sqrt{x}$ , factorise, solve for $\sqrt{x}$ , then square to find $x$ . Since $\sqrt{x}$ is non-negative, discard the negative root.

Working

The gradient is $\frac{dy}{dx} = 4x - 3\sqrt{x} + 1$ . Set this equal to $\frac{11}{2}$ :

4x - ​3\sqrt{x} + 1 = \frac{11}{2}

Subtract $1$ from both sides:

4x - ​3\sqrt{x} = \frac{9}{2}

Let $y = \sqrt{x}$ , so $x = y^2$ . Then:

4y^2 - ​3y - \frac{9}{2} = ​0

Multiply by $2$ :

8y^2 - ​6y -​ ​9 = ​0

Factorise:

(2y - ​3)(4y +​ 3) = ​0

So $y = \frac{3}{2}$ or $y = -\frac{3}{4}$ . Since $y = \sqrt{x} \geq 0$ , discard $y = -\frac{3}{4}$ .

Hence:

\sqrt{x} = \frac{3}{2} \implies x = \left(\frac{3}{2}\right)^2 = \frac{9}{4}

Answer

x = \frac{9}{4}

Final answer

x = 9/4

Detailed explanation

Walkthrough

The derivative $\frac{dy}{dx}$ gives the gradient of the curve at any point. We are told this gradient equals $\frac{11}{2}$ . So we set

4x - ​3\sqrt{x} + 1 = \frac{11}{2}

and solve for $x$ . The equation contains both $x$ and $\sqrt{x}$ , so it is not a simple linear equation. Write $y = \sqrt{x}$ , so that $x = y^2$ . This converts the equation into a quadratic in $y$ :

8y^2 - ​6y -​ ​9 = ​0

Factorising gives two possible values for $y$ . Since $\sqrt{x}$ cannot be negative, the negative solution is discarded. The remaining solution $\sqrt{x} = \frac{3}{2}$ is squared to obtain $x = \frac{9}{4}$ .

Key Takeaways

A gradient condition is applied by setting $\frac{dy}{dx}$ equal to the given value.
An equation involving $x$ and $\sqrt{x}$ can be made quadratic by substituting $y = \sqrt{x}$ .
Always check whether a root is valid when the substituted variable has a restricted sign.

Common Mistakes

Forgetting to specify $y = \sqrt{x}$ before factorising; the mark scheme requires this substitution to be clear.
Keeping the negative root $\sqrt{x} = -\frac{3}{4}$ , which is impossible for a real square root.
Stopping at $\sqrt{x} = \frac{3}{2}$ instead of squaring to find $x$ .

Things to Be Careful About

The equation must be rearranged into a 3-term quadratic before factorising; a method with no working shown may only earn partial credit.
When squaring, only the positive root is used;the extraneous value $\frac{9}{16}$ must be ignored.
Use the substitution explicitly to avoid losing method marks.

Techniques used

set derivative equal to a given gradientsubstitute y = sqrt{x} to form a quadraticfactorise the quadraticdiscard the negative root and square

(b)

Given that the curve passes through the point $(4, 11)$ , find the equation of the curve.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Integrate the derivative term by term, add a constant of integration, then use the given point to determine the constant.

Working

Rewrite the derivative using powers:

\frac{dy}{dx} = 4x - ​3x^{1/2} + ​1.

Integrate:

y = \int \left(4x - ​3x^{1/2} + ​1\right)\,dx.

y = 2x^2 - ​2x^{3/2} + x + c.

Use $(4,11)$ :

11 = 2(4)^2 - ​2(4)^{3/2} +​ ​4 + c.

Since $(4)^{3/2} = (\sqrt{4})^3 = 8$ :

11 = 32 -​ ​16 +​ ​4 + c =​ ​20 + c.

So $c = -9$ . Therefore:

y = 2x^2 -​ ​2x^{3/2} + x -​ ​9.

Answer

y = 2x^2 -​ ​2x^{3/2} + x -​ ​9.

Final answer

y = 2x^2 - 2x^(3/2) + x - 9

Detailed explanation

Walkthrough

The derivative is given, so the equation of the curve is found by integrating:

y = \int \left(4x - ​3x^{1/2} + 1\right)\,dx.

Integrate each term using the reverse of differentiation: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + c$ . This gives

y = 2x^2 -​ ​2x^{3/2} + x + c.

The constant $c$ is unknown. Use the fact that the curve passes through $(4,11)$ : substitute $x=4$ and $y=11$ into this equation. Since $(4)^{3/2} = (\sqrt{4})^3 = 8$ , we get

11 = 32 -​ ​16 +​ ​4 + c =​ ​20 + c,

so $c = -9$ . Therefore the equation is

y = 2x^2 -​ ​2x^{3/2} + x -​ ​9.

Key Takeaways

Integration reverses differentiation and introduces an arbitrary constant.
The constant is determined by substituting a known point on the curve.
Powers like $\sqrt{x}$ must be written as $x^{1/2}$ before integrating.

Common Mistakes

Forgetting the constant of integration, then being unable to use the point condition.
Incorrectly integrating $-3\sqrt{x}$ ;the coefficient must be divided by $\frac{3}{2}$ , giving $-2$ .
Substituting $(4,11)$ into the derivative instead of the integrated function.

Things to Be Careful About

The final answer must be simplified;the mark scheme does not allow an unsimplified constant.
When computing $(4)^{3/2}$ , either take the square root first or cube the square root;both give $8$ .
Check the final equation by differentiating: $\frac{d}{dx}\left(2x^2 - 2x^{3/2} + x - 9\right) = 4x - 3\sqrt{x} + 1$ .

Techniques used

integrate term by termadd the constant of integrationsubstitute the given point to find the constantwrite the final simplified equation

Q6Medium-EasyCoordinate Geometry

Circles $C_1$ and $C_2$ have equations

x^2 + y^2 + 6x - 10y + 18 = 0 \text{ and } (x - 9)^2 + (y + 4)^2 - 64 = 0

respectively.

(a)

Find the distance between the centres of the circles.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Complete the square on the first circle equation to write it in centre-radius form. The second equation is already in centre-radius form. Then apply the distance formula between the two centres.

Working

For $C_1$ :

\begin{aligned} x^2 + y^2 + 6x - 10y + 18 &= 0 \\ (x+3)^2 - 9 + (y-5)^2 - 25 + 18 &= 0 \\ (x+3)^2 + (y-5)^2 &= 16 \end{aligned}

So the centre of $C_1$ is $(-3,5)$ and its radius is $4$ .

For $C_2$ :

(x-9)^2 + (y+4)^2 = 64

So the centre of $C_2$ is $(9,-4)$ and its radius is $8$ .

The distance between the centres is:

\sqrt{(9 - (-3))^2 + (-4 - 5)^2} = \sqrt{12^2 + (-9)^2} = \sqrt{144 + 81} = \sqrt{225} = 15

Answer

The distance between the centres is $15$ .

Final answer

15

Detailed explanation

Walkthrough

First rewrite $C_1$ in completed-square form. The $x$ -terms $x^2 + 6x$ become $(x+3)^2 - 9$ , and the $y$ -terms $y^2 - 10y$ become $(y-5)^2 - 25$ . Including the constant $18$ gives $(x+3)^2 - 9 + (y-5)^2 - 25 + 18 = 0$ , so $(x+3)^2 + (y-5)^2 = 16$ . Therefore the centre of $C_1$ is $(-3,5)$ and its radius is $\sqrt{16} = 4$ .

For $C_2$ , the equation $(x-9)^2 + (y+4)^2 = 64$ is already in centre-radius form, so its centre is $(9,-4)$ and its radius is $\sqrt{64} = 8$ .

The distance between the centres is found by applying the distance formula to $(-3,5)$ and $(9,-4)$ :

\sqrt{(9 - (-3))^2 + (-4 - 5)^2} = \sqrt{12^2 + (-9)^2} = 15

Key Takeaways

A circle equation in the form $(x-a)^2 + (y-b)^2 = r^2$ has centre $(a,b)$ and radius $r$ . Completing the square converts a general quadratic circle equation into this useful form. The distance formula measures the straight-line distance between two points.

Common Mistakes

Sign errors when reading the centre from completed-square form: $(x+3)^2 + (y-5)^2 = 16$ has centre $(-3,5)$ , not $(3,-5)$ .
Forgetting to subtract the squared constants when completing the square, leading to the wrong radius.
Confusing the radius with the distance between centres.

Things to Be Careful About

The mark scheme awards one mark for each centre and one mark for the distance calculation. Show the completed-square step clearly so the centres are evident. Also note that the radius is the square root of the constant on the right-hand side, not the constant itself.

Techniques used

complete the square to rewrite the circle equationread the centre and radius from centre-radius formapply the distance formula between two points

(b)

$P$ and $Q$ are points on $C_1$ and $C_2$ respectively. The distance between $P$ and $Q$ is denoted by $d$ .

Find the greatest and least possible values of $d$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The greatest distance between a point on one circle and a point on the other occurs when the two points lie on the line joining the centres, on opposite sides of the centres. The least distance occurs when the two points lie on the same side of the centres, between them. Therefore add or subtract the radii from the centre distance.

Working

From part (a), the distance between the centres is $15$ .

The radii are:

R_1 = 4, \quad R_2 = 8

So the sum of the radii is $4 + 8 = 12$ .

Since $15 > 12$ , the circles do not overlap.

Least distance:

d_{\min} = 15 - R_1 - R_2 = 15 - 4 - 8 = 3

Greatest distance:

d_{\max} = 15 + R_1 + R_2 = 15 + 4 + 8 = 27

Answer

The least possible value of $d$ is $3$ and the greatest possible value is $27$ .

Final answer

Least = 3, greatest = 27

Detailed explanation

Walkthrough

For two separate circles, the closest points lie on the straight line joining the centres, between the two circles. Starting from the centre of $C_1$ , move toward the centre of $C_2$ by the radius of $C_1$ to reach the nearest point on $C_1$ , then continue to the nearest point on $C_2$ by the radius of $C_2$ . The least distance is therefore the centre distance minus both radii.

The farthest points lie on the same line of centres but on opposite sides of the two circles. From the centre of $C_1$ , move away from $C_2$ by $R_1$ , and from the centre of $C_2$ , move away from $C_1$ by $R_2$ ; the distance between those two points is the centre distance plus both radii.

Using the values from part (a):

15 - 4 - 8 = 3

and

15 + 4 + 8 = 27

Key Takeaways

The line joining the centres of two circles is the line along which the closest and farthest points between the circles occur. For separate circles, the least distance is $\text{centre distance} - R_1 - R_2$ and the greatest distance is $\text{centre distance} + R_1 + R_2$ .

Common Mistakes

Giving the distance between the centres, $15$ , as the answer instead of adjusting for the radii.
Subtracting the radii incorrectly, e.g. using $15 - |R_1 - R_2|$ instead of $15 - (R_1 + R_2)$ .
Forgetting to consider both the least and greatest values when the question asks for both.

Things to Be Careful About

The mark scheme gives a mark for stating the radii $R_1 = 4$ and $R_2 = 8$ , a follow-through mark for one of the two distances, and a final follow-through mark for both. Always use the centre distance found in part (a), labelled as '15' in the mark scheme, and check that the circles are separate before assuming the least distance is positive. Here $15 > 12$ , so the least distance is indeed $3$ .

Techniques used

identify the line through the centres for closest and farthest pointssubtract the sum of the radii from the centre distanceadd the sum of the radii to the centre distance

Q7Medium-EasyDifferentiationCoordinate GeometryIntegration

Fig. 2

The diagram shows part of the curve with equation $y = \frac{12}{\sqrt[3]{2x + 1}}$ . The point $A$ on the curve has coordinates $\left(\frac{7}{2}, 6\right)$ .

(a)

Find the equation of the tangent to the curve at $A$ . Give your answer in the form $y = mx + c$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Rewrite the curve equation using a negative fractional exponent to make differentiation straightforward. Differentiate using the chain rule to find $\frac{dy}{dx}$ . Substitute $x = \frac{7}{2}$ to find the gradient of the tangent at $A$ . Use the point-slope form with point $A\left(\frac{7}{2}, 6\right)$ and the calculated gradient to find the equation of the tangent in the form $y = mx + c$ .

Working

Rewrite the equation:

y = 12(2x + 1)^{-\frac{1}{3}}

Differentiate using the chain rule:

\begin{aligned} \frac{dy}{dx} &= 12 \left(-\frac{1}{3}\right) (2x + 1)^{-\frac{4}{3}} \cdot 2 \\ &= -8(2x + 1)^{-\frac{4}{3}} \end{aligned}

At $x = \frac{7}{2}$ , the inner term is $2\left(\frac{7}{2}\right) + 1 = 8$ . Substitute this into the derivative:

\frac{dy}{dx} = -8(8)^{-\frac{4}{3}}

Evaluate $8^{-\frac{4}{3}}$ :

8^{-\frac{4}{3}} = \frac{1}{(\sqrt[3]{8})^4} = \frac{1}{2^4} = \frac{1}{16}

So the gradient $m$ is:

m = -8 \times \frac{1}{16} = -\frac{1}{2}

Use the point-slope form $y - y_1 = m(x - x_1)$ with $A\left(\frac{7}{2}, 6\right)$ and $m = -\frac{1}{2}$ :

y - 6 = -\frac{1}{2}\left(x - \frac{7}{2}\right)

Expand and rearrange into $y = mx + c$ form:

y - 6 = -\frac{1}{2}x + \frac{7}{4}

y = -\frac{1}{2}x + \frac{7}{4} + 6

y = -\frac{1}{2}x + \frac{31}{4}

Answer

y = -\frac{1}{2}x + \frac{31}{4}

Final answer

y = -1/2 x + 31/4

Detailed explanation

Walkthrough

First, rewrite the curve equation using a negative fractional exponent to make differentiation easier: $y = 12(2x+1)^{-1/3}$ . Then, apply the chain rule: bring down the power $-1/3$ , subtract 1 from the power to get $-4/3$ , and multiply by the derivative of the inner function $(2x+1)$ , which is 2. This gives $\frac{dy}{dx} = -8(2x+1)^{-4/3}$ . Next, substitute $x = 7/2$ into this derivative. Since $2(7/2)+1 = 8$ , we evaluate $-8 \times 8^{-4/3}$ . The cube root of 8 is 2, and $2^4 = 16$ , so $8^{-4/3} = 1/16$ . The gradient is $-8/16 = -1/2$ . Finally, use the point-slope formula $y - y_1 = m(x - x_1)$ with the coordinates of $A(7/2, 6)$ and the gradient $-1/2$ to find the tangent equation in the form $y = mx + c$ .

Key Takeaways

Rewriting fractional roots as negative fractional exponents simplifies differentiation.
The chain rule is essential for differentiating composite functions like $(ax+b)^n$ .
The point-slope form $y - y_1 = m(x - x_1)$ is a reliable way to find the equation of a tangent given a point and a gradient.

Common Mistakes

Forgetting to multiply by the derivative of the inner function $(2x+1)$ when using the chain rule.
Incorrectly evaluating fractional negative powers, such as $8^{-4/3}$ .
Algebra errors when rearranging the point-slope equation into $y = mx + c$ form.

Things to Be Careful About

Ensure the gradient is calculated from the differentiated expression, not by guessing or using a calculator if not allowed.
The final answer must be in the form $y = mx + c$ as requested.

Techniques used

rewrite curve using fractional exponentsapply chain rule for differentiationevaluate gradient at given x-coordinateuse point-slope form to find tangent equation

(b)

Find the area of the region bounded by the curve and the lines $x = 0$ , $x = \frac{7}{2}$ and $y = 0$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Set up the definite integral for the area under the curve from $x = 0$ to $x = \frac{7}{2}$ . Integrate $y = 12(2x+1)^{-1/3}$ using the reverse chain rule (or substitution). Evaluate the integrated expression at the upper limit $\frac{7}{2}$ and subtract the value at the lower limit $0$ .

Working

The area is given by the integral:

\text{Area} = \int_0^{\frac{7}{2}} 12(2x + 1)^{-\frac{1}{3}} \, dx

Integrate using the reverse chain rule. Bring up the power $-\frac{1}{3}$ to get $\frac{2}{3}$ , divide by the new power, and divide by the derivative of the inner function $(2)$ :

\begin{aligned} \int 12(2x + 1)^{-\frac{1}{3}} \, dx &= 12 \cdot \frac{(2x + 1)^{\frac{2}{3}}}{\frac{2}{3}} \cdot \frac{1}{2} \\ &= 12 \cdot \frac{3}{2} \cdot \frac{1}{2} (2x + 1)^{\frac{2}{3}} \\ &= 9(2x + 1)^{\frac{2}{3}} \end{aligned}

Now evaluate the definite integral from $0$ to $\frac{7}{2}$ :

\text{Area} = \left[ 9(2x + 1)^{\frac{2}{3}} \right]_0^{\frac{7}{2}}

Substitute the upper limit $x = \frac{7}{2}$ :

9\left(2\left(\frac{7}{2}\right) + 1\right)^{\frac{2}{3}} = 9(8)^{\frac{2}{3}} = 9(\sqrt[3]{8})^2 = 9(2)^2 = 9 \times 4 = 36

Substitute the lower limit $x = 0$ :

9(2(0) + 1)^{\frac{2}{3}} = 9(1)^{\frac{2}{3}} = 9 \times 1 = 9

Calculate the area:

\text{Area} = 36 - 9 = 27

Answer

27

Final answer

27

Detailed explanation

Walkthrough

Set up the definite integral for the area under the curve from $x = 0$ to $x = 7/2$ . The integrand is $y = 12(2x+1)^{-1/3}$ . To integrate, use the reverse chain rule: increase the power by 1 to get $2/3$ , divide by the new power $2/3$ , and divide by the derivative of the inner function $(2x+1)$ , which is 2. This simplifies to $9(2x+1)^{2/3}$ . Next, evaluate this antiderivative at the upper limit $x = 7/2$ : $9(8)^{2/3} = 9 \times 4 = 36$ . Then evaluate at the lower limit $x = 0$ : $9(1)^{2/3} = 9$ . Subtract the lower limit value from the upper limit value to get $36 - 9 = 27$ .

Key Takeaways

The reverse chain rule is used to integrate composite functions of the form $(ax+b)^n$ .
When integrating, remember to divide by the derivative of the inner function $a$ .
Definite integrals are evaluated by substituting the upper limit, subtracting the substitution of the lower limit.

Common Mistakes

Forgetting to divide by the derivative of the inner function $(2)$ after integrating.
Incorrectly evaluating fractional powers, such as $8^{2/3}$ .
Forgetting to subtract the lower limit evaluation from the upper limit evaluation.

Things to Be Careful About

Ensure the limits of integration are correctly identified from the problem statement ( $x = 0$ to $x = 7/2$ ).
Check that the integrated expression is correct before substituting the limits.

Techniques used

identify integrand and limitsapply reverse chain rule for integrationevaluate definite integral using upper and lower limits

Q8MediumTrigonometry

(a)

It is given that $\beta$ is an angle between $90^\circ$ and $180^\circ$ such that $\sin \beta = a$ .

Express $\tan^2 \beta - 3\sin \beta \cos \beta$ in terms of $a$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the identity $\tan \beta = \frac{\sin \beta}{\cos \beta}$ and the Pythagorean identity $\sin^2 \beta + \cos^2 \beta = 1$ to write everything in terms of $a$ . Since $\beta$ lies between $90^\circ$ and $180^\circ$ , $\cos \beta$ is negative, which fixes the sign of $\cos \beta$ .

Working

Given $\sin \beta = a$ .

Using $\tan^2 \beta = \frac{\sin^2 \beta}{\cos^2 \beta}$ :

\tan^2 \beta = \frac{a^2}{\cos^2 \beta}

From $\sin^2 \beta + \cos^2 \beta = 1$ :

\cos^2 \beta = 1 - \sin^2 \beta = 1 - a^2

Because $90^\circ < \beta < 180^\circ$ , $\cos \beta < 0$ , so:

\cos \beta = -\sqrt{1 - a^2}

Therefore:

\begin{aligned} \tan^2 \beta - 3\sin \beta \cos \beta &= \frac{a^2}{1 - a^2} - 3a\left(-\sqrt{1 - a^2}\right) \\ &= \frac{a^2}{1 - a^2} + 3a\sqrt{1 - a^2} \end{aligned}

Answer

\frac{a^2}{1 - a^2} + 3a\sqrt{1 - a^2}

Final answer

a^2/(1-a^2) + 3a sqrt(1-a^2)

Detailed explanation

Walkthrough

We are told $\sin \beta = a$ and $\beta$ is in the second quadrant. The expression contains $\tan^2 \beta$ and $\cos \beta$ , so we need to rewrite them using $a$ . The identity $\tan \beta = \frac{\sin \beta}{\cos \beta}$ gives $\tan^2 \beta = \frac{\sin^2 \beta}{\cos^2 \beta}$ . Since $\sin \beta = a$ , the numerator is $a^2$ . To get $\cos^2 \beta$ , use $\sin^2 \beta + \cos^2 \beta = 1$ , so $\cos^2 \beta = 1 - a^2$ . The quadrant matters: for $90^\circ < \beta < 180^\circ$ , $\cos \beta$ is negative, so $\cos \beta = -\sqrt{1 - a^2}$ . Substituting into $\tan^2 \beta - 3\sin \beta \cos \beta$ gives $\frac{a^2}{1 - a^2} - 3a(-\sqrt{1 - a^2}) = \frac{a^2}{1 - a^2} + 3a\sqrt{1 - a^2}$ . The two negatives cancel to produce the plus sign.

Key Takeaways

The Pythagorean identity $\sin^2 \beta + \cos^2 \beta = 1$ lets you replace one trigonometric function with another.
The identity $\tan \beta = \frac{\sin \beta}{\cos \beta}$ is essential for rewriting $\tan^2 \beta$ .
The quadrant of the angle determines the sign of $\cos \beta$ .

Common Mistakes

Forgetting that $\cos \beta$ is negative in the second quadrant, which changes the sign of the final term.
Writing $\tan^2 \beta = \frac{a^2}{\cos^2 \beta}$ without simplifying $\cos^2 \beta$ to $1 - a^2$ .
Dropping the square root or using $\cos \beta = \sqrt{1 - a^2}$ without considering the quadrant.

Things to Be Careful About

Since $90^\circ < \beta < 180^\circ$ , $\sin \beta$ is positive, so $a > 0$ .
$1 - a^2$ is positive, so the square root is defined.
The sign of $\cos \beta$ must be chosen before substituting into the expression.

Techniques used

use the Pythagorean identity to express cos^2 in terms of sin^2determine the sign of cos from the quadrantsubstitute and simplify the trigonometric expression

(b)

Solve the equation $\sin^2 \theta + 2\cos^2 \theta = 4\sin \theta + 3$ for $0^\circ < \theta < 360^\circ$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Replace $\cos^2 \theta$ using $\cos^2 \theta = 1 - \sin^2 \theta$ , then rearrange into a quadratic equation in $\sin \theta$ . Solve the quadratic, discard any root outside $[-1,1]$ , and use the quadrants where $\sin \theta$ is negative to find the required angles.

Working

Start with the equation:

\sin^2 \theta + 2\cos^2 \theta = 4\sin \theta + 3

Use $\cos^2 \theta = 1 - \sin^2 \theta$ :

\sin^2 \theta + 2(1 - \sin^2 \theta) = 4\sin \theta + 3

\sin^2 \theta + 2 - 2\sin^2 \theta = 4\sin \theta + 3

2 - \sin^2 \theta = 4\sin \theta + 3

Rearrange:

\sin^2 \theta + 4\sin \theta + 1 = 0

Solve the quadratic in $\sin \theta$ :

\sin \theta = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}

Since $\sqrt{3} \approx 1.732$ , the two values are:

-2 + \sqrt{3} \approx -0.2679, \qquad -2 - \sqrt{3} \approx -3.732

Only $\sin \theta = -2 + \sqrt{3} \approx -0.2679$ is valid.

Let $\alpha = \arcsin(2 - \sqrt{3}) \approx 15.5^\circ$ be the reference angle. Since $\sin \theta$ is negative, the solutions are in the third and fourth quadrants:

\theta = 180^\circ + \alpha = 180^\circ + 15.5^\circ \approx 195.5^\circ

\theta = 360^\circ - \alpha = 360^\circ - 15.5^\circ \approx 344.5^\circ

Answer

\theta \approx 195.5^\circ, \ 344.5^\circ

Final answer

195.5° and 344.5°

Detailed explanation

Walkthrough

Start by replacing $\cos^2 \theta$ with $1 - \sin^2 \theta$ so that the equation contains only $\sin \theta$ . Expanding gives $2 - \sin^2 \theta = 4\sin \theta + 3$ . Rearranging all terms to one side produces $\sin^2 \theta + 4\sin \theta + 1 = 0$ . This is a quadratic in $\sin \theta$ , so apply the quadratic formula: $\sin \theta = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$ . The value $-2 - \sqrt{3} \approx -3.732$ is outside the range of sine, so discard it. The valid value is $\sin \theta = -2 + \sqrt{3} \approx -0.2679$ . Since sine is negative, the solutions must be in the third and fourth quadrants. The reference angle is $\alpha = \arcsin(2 - \sqrt{3}) \approx 15.5^\circ$ . Therefore $\theta = 180^\circ + 15.5^\circ \approx 195.5^\circ$ and $\theta = 360^\circ - 15.5^\circ \approx 344.5^\circ$ .

Key Takeaways

The identity $\cos^2 \theta = 1 - \sin^2 \theta$ reduces a trigonometric equation to a quadratic.
A quadratic in $\sin \theta$ can have roots outside $[-1,1]$ , which must be rejected.
The sign of $\sin \theta$ tells you which quadrants contain the solutions.

Common Mistakes

Making a sign error when rearranging; the correct quadratic is $\sin^2 \theta + 4\sin \theta + 1 = 0$ .
Keeping the root $-2 - \sqrt{3}$ , which is less than $-1$ and therefore impossible.
Using $-15.5^\circ$ directly instead of using the positive reference angle $15.5^\circ$ .
Placing solutions in the wrong quadrants, such as using the first and second quadrants where sine is positive.

Things to Be Careful About

The interval is $0^\circ < \theta < 360^\circ$ , so $0^\circ$ and $360^\circ$ are not included, but neither appears as a solution.
If your calculator is in radian mode, convert the final answers: $195.5^\circ \approx 3.41$ radians and $344.5^\circ \approx 6.01$ radians.
Check that both final angles satisfy the original equation by substituting back if time permits.

Techniques used

use the Pythagorean identity to eliminate cos^2form and solve a quadratic in sin thetadiscard roots outside the range of sinefind angles in the correct quadrants

Q9MediumDifferentiationQuadratics

The equation of a curve is $y = 4 + 5x + 6x^2 - 3x^3$ .

(a)

Find the set of values of $x$ for which $y$ decreases as $x$ increases.

4M Similar

DifficultyMedium

Worked solution

Approach

The curve decreases as $x$ increases when its gradient is negative, so we require $\frac{dy}{dx} < 0$ . Differentiate $y$ , solve the resulting quadratic inequality, and state the set of values of $x$ .

Working

Differentiate the curve:

\frac{dy}{dx} = 5 + 12x - 9x^2

For the curve to be decreasing:

5 + 12x - 9x^2 < 0

Multiply by $-1$ and reverse the inequality:

9x^2 - 12x - 5 > 0

Factorise:

(3x - 5)(3x + 1) > 0

The critical values are:

x = -\frac{1}{3}, \quad x = \frac{5}{3}

Since the coefficient of $x^2$ is positive, the quadratic is positive outside the interval between the roots.

Answer

x < -\frac{1}{3} \quad \text{or} \quad x > \frac{5}{3}

Final answer

x < -1/3 or x > 5/3

Detailed explanation

Walkthrough

The phrase 'decreases as $x$ increases' means the curve is sloping downwards, so its gradient $\frac{dy}{dx}$ is negative. We differentiate $y = 4 + 5x + 6x^2 - 3x^3$ to obtain $\frac{dy}{dx} = 5 + 12x - 9x^2$ . Set this gradient less than zero and solve the quadratic inequality. Multiply by $-1$ to get $9x^2 - 12x - 5 > 0$ , and factorise as $(3x - 5)(3x + 1) > 0$ . The critical values are $x = -\frac{1}{3}$ and $x = \frac{5}{3}$ . Because the quadratic opens upwards, it is positive outside the interval between the roots, so the curve decreases for $x < -\frac{1}{3}$ or $x > \frac{5}{3}$ .

Key Takeaways

A curve decreases where its derivative is negative.
Differentiating a polynomial gives the gradient function.
Quadratic inequalities are solved by finding critical values and choosing the correct intervals.

Common Mistakes

Forgetting to reverse the inequality sign when multiplying by $-1$ .
Choosing the interval between the roots, which is where the curve increases.
Using $\le$ or $\ge$ instead of strict inequalities, since the curve is not decreasing exactly at the critical points.

Things to Be Careful About

The mark scheme requires a full method for solving the quadratic; an unsupported answer will not receive full marks. Also, write the final answer with $x$ on the correct side of each inequality: $x < -\frac{1}{3}$ or $x > \frac{5}{3}$ .

Techniques used

differentiate a polynomialset the gradient less than zerosolve a quadratic inequalityfactorise a quadratic

(b)

It is given that $y = 9x + k$ is a tangent to the curve.

Find the value of the constant $k$ .

4M Similar

DifficultyMedium

Worked solution

Approach

A line $y = 9x + k$ is tangent to the curve where its gradient, $9$ , equals the gradient of the curve. Set $\frac{dy}{dx} = 9$ , solve for $x$ , find the corresponding $y$ -value on the curve, and then use the tangent equation to find $k$ .

Working

The gradient of the curve is:

\frac{dy}{dx} = 5 + 12x - 9x^2

For the tangent, the gradient is $9$ , so:

5 + 12x - 9x^2 = 9

Simplify to a three-term quadratic:

-9x^2 + 12x - 4 = 0

Multiply by $-1$ :

9x^2 - 12x + 4 = 0

Factorise:

(3x - 2)^2 = 0

Therefore:

x = \frac{2}{3}

Find the corresponding $y$ -value on the curve:

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

Compute:

y = 4 + \frac{10}{3} + \frac{24}{9} - \frac{24}{27}

Using denominator $27$ :

y = \frac{108}{27} + \frac{90}{27} + \frac{72}{27} - \frac{24}{27} = \frac{246}{27} = \frac{82}{9}

The point of tangency is $\left(\frac{2}{3}, \frac{82}{9}\right)$ . Since it lies on $y = 9x + k$ :

\frac{82}{9} = 9\left(\frac{2}{3}\right) + k = 6 + k

Therefore:

k = \frac{82}{9} - 6 = \frac{82}{9} - \frac{54}{9} = \frac{28}{9}

Answer

k = \frac{28}{9}

Final answer

k = 28/9

Detailed explanation

Walkthrough

A line is tangent to a curve when its gradient equals the gradient of the curve at the point of contact. The line $y = 9x + k$ has gradient $9$ , so we set $\frac{dy}{dx} = 9$ . From part (a), $\frac{dy}{dx} = 5 + 12x - 9x^2$ , so we solve $5 + 12x - 9x^2 = 9$ . Rearranging gives $-9x^2 + 12x - 4 = 0$ , and multiplying by $-1$ gives $9x^2 - 12x + 4 = 0$ , which factorises as $(3x - 2)^2 = 0$ . Thus $x = \frac{2}{3}$ . Substitute this into the original curve equation to get $y = \frac{82}{9}$ . Since the point of tangency lies on the tangent line, $\frac{82}{9} = 9\left(\frac{2}{3}\right) + k = 6 + k$ , so $k = \frac{82}{9} - 6 = \frac{28}{9}$ .

Key Takeaways

A tangent line has the same gradient as the curve at the point of contact.
To find the point of contact, set $\frac{dy}{dx}$ equal to the gradient of the tangent line.
Once the point is known, the constant in the tangent line can be found using $y = mx + k$ .

Common Mistakes

Not simplifying $5 + 12x - 9x^2 = 9$ into a three-term quadratic; the mark scheme requires this simplification.
Using the wrong sign when rearranging the quadratic.
Forgetting to find the $y$ -coordinate of the point of tangency before finding $k$ .
Substituting the point into the curve equation again instead of into the tangent line equation.

Things to Be Careful About

There is a repeated root, so the tangent line touches the curve at exactly one point. Use the original curve equation to compute $y$ , not the tangent equation, because $k$ is unknown. Then subtract carefully: $k = \frac{82}{9} - 6 = \frac{82}{9} - \frac{54}{9} = \frac{28}{9}$ .

Techniques used

differentiate a polynomialequate the derivative to the tangent gradientsolve a quadratic equationfind the point of tangencyuse the tangent line equation to find a constant

Q10MediumSeriesQuadratics

An arithmetic progression has first term 5 and common difference $d$ , where $d > 0$ . The second, fifth and eleventh terms of the arithmetic progression, in that order, are the first three terms of a geometric progression.

(a)

Find the value of $d$ .

3M Similar

DifficultyMedium

Worked solution

Approach

The AP terms are written using the formula for the $n$ th term. Since the second, fifth and eleventh terms form the first three terms of a GP, the common ratio condition is used to form a quadratic equation in $d$ . The root is then selected using $d > 0$ .

Working

The AP has first term $a = 5$ and common difference $d$ . Its second, fifth and eleventh terms are:

u_2 = 5 + d

u_5 = 5 + 4d

u_{11} = 5 + 10d

These are the first three terms of a GP, so the ratios of consecutive terms are equal:

\frac{5 + 4d}{5 + d} = \frac{5 + 10d}{5 + 4d}

Cross-multiplying gives:

(5 + 4d)^2 = (5 + d)(5 + 10d)

Expand both sides:

25 + 40d + 16d^2 = 25 + 55d + 10d^2

Simplify:

6d^2 - 15d = 0

Factorise:

3d(2d - 5) = 0

So $d = 0$ or $d = 2.5$ . Since $d > 0$ , reject $d = 0$ .

Answer

d = 2.5

Final answer

d = 2.5

Detailed explanation

Walkthrough

The AP has first term $5$ and common difference $d$ , so its $n$ th term is $u_n = 5 + (n-1)d$ . To get the second, fifth and eleventh terms, substitute $n = 2, 5, 11$ :

u_2 = 5 + d, \quad u_5 = 5 + 4d, \quad u_{11} = 5 + 10d

These three numbers are the first three terms of a GP. In a GP the ratio between consecutive terms is constant, so:

\frac{5 + 4d}{5 + d} = \frac{5 + 10d}{5 + 4d}

Cross-multiplying gives:

(5 + 4d)^2 = (5 + d)(5 + 10d)

Expanding and simplifying gives $6d^2 - 15d = 0$ , which factorises as $3d(2d - 5) = 0$ . Hence $d = 0$ or $d = 2.5$ . Since the question states $d > 0$ , the solution is $d = 2.5$ .

Key Takeaways

This question links the definitions of arithmetic and geometric progressions. The key idea is that consecutive terms of a GP have a common ratio, so the square of the middle term equals the product of the outer terms. It also tests forming and solving a quadratic equation from a geometric condition.

Common Mistakes

Forgetting to reject $d = 0$ , even though the question specifies $d > 0$ .
Making expansion errors, especially with $(5 + 4d)^2$ or $(5 + d)(5 + 10d)$ .
Using the wrong AP terms, such as $5 + 2d$ for the second term instead of $5 + d$ .

Things to Be Careful About

The nth term of an AP with first term $a$ and common difference $d$ is $a + (n-1)d$ , not $a + nd$ .
Use the given order: second, fifth and eleventh terms are the first three GP terms.
The zero root must be rejected because $d > 0$ .

Techniques used

write the AP terms using the nth term formulaequate GP ratios to form an equationsolve the resulting quadratic equationselect the positive root using d > 0

(b)

The sum of the first 77 terms of the arithmetic progression is denoted by $S_{77}$ . The sum of the first 10 terms of the geometric progression is denoted by $G_{10}$ .

Find the value of $S_{77} - G_{10}$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use the AP sum formula with $a = 5$ , $d = 2.5$ and $n = 77$ . Then identify the GP first term and common ratio from the terms found in part (a), use the GP sum formula with $n = 10$ , and subtract.

Working

For the AP, $a = 5$ , $d = 2.5$ and $n = 77$ :

S_{77} = \frac{77}{2}\left[2(5) + (77 - 1)(2.5)\right]

Evaluate the bracket:

2(5) = 10, \quad (76)(2.5) = 190

Therefore:

S_{77} = \frac{77}{2}(200) = 7700

The GP has first three terms $7.5, 15, 30$ , so its first term is $7.5$ and its common ratio is $2$ .

Using the GP sum formula for $n = 10$ :

G_{10} = \frac{7.5(2^{10} - 1)}{2 - 1}

Since $2^{10} = 1024$ :

G_{10} = 7.5(1023) = 7672.5

Therefore:

S_{77} - G_{10} = 7700 - 7672.5 = 27.5

Answer

S_{77} - G_{10} = 27.5

Final answer

S_77 - G_10 = 27.5

Detailed explanation

Walkthrough

From part (a), $d = 2.5$ . The AP has first term $5$ and common difference $2.5$ . The sum of the first $77$ terms is found using the AP sum formula:

S_n = \frac{n}{2}\left(2a + (n-1)d\right)

Substituting $n = 77$ , $a = 5$ , $d = 2.5$ :

S_{77} = \frac{77}{2}\left(10 + 76 \times 2.5\right) = \frac{77}{2}(200) = 7700

The GP has first three terms equal to the second, fifth and eleventh AP terms:

7.5, \quad 15, \quad 30

So the GP first term is $7.5$ and the common ratio is $15/7.5 = 2$ . The sum of the first $10$ terms is:

G_{10} = \frac{7.5(2^{10} - 1)}{2 - 1} = 7.5(1023) = 7672.5

Finally:

S_{77} - G_{10} = 7700 - 7672.5 = 27.5

Key Takeaways

This part tests fluency with both the AP and GP sum formulae. It also shows that once the common difference is known, the GP is fully determined by the AP terms, so the two progressions are linked.

Common Mistakes

Using the wrong sum formula: AP uses $\frac{n}{2}(2a + (n-1)d)$ , while GP uses $\frac{a(r^n - 1)}{r - 1}$ .
Using $5$ as the first term of the GP; the GP first term is actually $7.5$ .
Arithmetic slips with $2^{10} = 1024$ or with decimal multiplication.

Things to Be Careful About

The GP common ratio is $2$ , not $1.5$ or $2.5$ .
$S_{77}$ is $7700$ and $G_{10}$ is $7672.5$ ; subtract carefully to get $27.5$ .
The mark scheme expects the correct sum formula to be used with the value of $d$ found in part (a).

Techniques used

use the AP sum formulaidentify the GP first term and common ratiouse the GP sum formulasubtract the two sums

Q11MediumQuadraticsFunctions

The function $f$ is defined by $f(x) = 3 + 6x - 2x^2$ for $x \in \mathbb{R}$ .

(a)

Express $f(x)$ in the form $a - b(x - c)^2$ , where $a$ , $b$ and $c$ are constants, and state the range of $f$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Factor out the coefficient of $x^2$ from the quadratic terms, complete the square, then expand to obtain the form $a - b(x-c)^2$ . Use the fact that a squared expression is always non-negative to determine the maximum value and hence the range.

Working

Factor out $-2$ from the $x$ -terms:

f(x) = -2(x^2 - 3x) + 3

Complete the square inside the bracket:

x^2 - 3x = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}

Substitute back:

f(x) = -2\left[\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right] + 3

f(x) = -2\left(x - \frac{3}{2}\right)^2 + \frac{9}{2} + 3

f(x) = \frac{15}{2} - 2\left(x - \frac{3}{2}\right)^2

So $a = \frac{15}{2}$ , $b = 2$ , $c = \frac{3}{2}$ .

Since $\left(x - \frac{3}{2}\right)^2 \geq 0$ for all $x \in \mathbb{R}$ , we have $-2\left(x - \frac{3}{2}\right)^2 \leq 0$ , and therefore:

f(x) \leq \frac{15}{2}

The range of $f$ is $y \leq \frac{15}{2}$ .

Answer

f(x) = \frac{15}{2} - 2\left(x - \frac{3}{2}\right)^2, \quad \text{range: } y \leq \frac{15}{2}

Final answer

f(x) = 15/2 - 2(x - 3/2)^2, range y ≤ 15/2

Detailed explanation

Walkthrough

We are asked to rewrite $f(x) = 3 + 6x - 2x^2$ in the form $a - b(x-c)^2$ . This is a completing-the-square problem.

Step 1: Factor out the coefficient of $x^2$ , which is $-2$ , from the $x$ -terms only:
$f(x) = -2(x^2 - 3x) + 3$

Step 2: Complete the square for $x^2 - 3x$ . Half of $-3$ is $-\frac{3}{2}$ , so:
$x^2 - 3x = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}$

Step 3: Substitute back and distribute the $-2$ :
$f(x) = -2\left(x - \frac{3}{2}\right)^2 + \frac{9}{2} + 3 = \frac{15}{2} - 2\left(x - \frac{3}{2}\right)^2$

Step 4: Determine the range. Since $\left(x - \frac{3}{2}\right)^2 \geq 0$ , multiplying by $-2$ gives $-2\left(x - \frac{3}{2}\right)^2 \leq 0$ . Adding $\frac{15}{2}$ means $f(x) \leq \frac{15}{2}$ , with equality when $x = \frac{3}{2}$ .

Key Takeaways

Completing the square for a quadratic $ax^2 + bx + c$ with $a < 0$ gives a maximum value.
The range of a downward-opening parabola $a - b(x-c)^2$ (with $b > 0$ ) is $y \leq a$ .

Common Mistakes

Forgetting to include the constant term $+3$ when completing the square.
Sign errors when distributing the negative coefficient $-2$ .
Writing the range with $<$ instead of $\leq$ — the maximum is actually attained at $x = \frac{3}{2}$ .

Things to Be Careful About

The marking scheme accepts $\leq$ given or clearly implied; using $<$ would lose the mark.
The values of $a$ , $b$ , $c$ must match the form $a - b(x-c)^2$ exactly — note the minus sign before $b(x-c)^2$ .

Techniques used

factor out leading coefficientcomplete the squareuse non-negativity of squared term to find range

(b)

The graph of $y = f(x)$ is transformed to the graph of $y = h(x)$ by a reflection in one of the axes followed by a translation. It is given that the graph of $y = h(x)$ has a minimum point at the origin.

Give details of the reflection and translation involved.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

From part (a), $f(x) = \frac{15}{2} - 2\left(x - \frac{3}{2}\right)^2$ has a maximum at $\left(\frac{3}{2}, \frac{15}{2}\right)$ . To obtain a minimum at the origin, first reflect in the $x$ -axis to convert the maximum to a minimum, then translate to move the minimum to $(0, 0)$ .

Working

Reflection: Reflect $y = f(x)$ in the $x$ -axis to get $y = -f(x)$ :

y = -f(x) = -\frac{15}{2} + 2\left(x - \frac{3}{2}\right)^2 = 2\left(x - \frac{3}{2}\right)^2 - \frac{15}{2}

This is an upward-opening parabola with a minimum at $\left(\frac{3}{2}, -\frac{15}{2}\right)$ .

Translation: To move the minimum from $\left(\frac{3}{2}, -\frac{15}{2}\right)$ to the origin $(0, 0)$ , translate by:

\begin{pmatrix} 0 - \frac{3}{2} \\ 0 - \left(-\frac{15}{2}\right) \end{pmatrix} = \begin{pmatrix} -\frac{3}{2} \\ \frac{15}{2} \end{pmatrix}

Answer

Reflection in the $x$ -axis, followed by a translation by $\begin{pmatrix} -\frac{3}{2} \\ \frac{15}{2} \end{pmatrix}$ .

(Note: the order of reflection and translation can be reversed.)

Final answer

Reflection in the x-axis, then translation by (-3/2, 15/2)

Detailed explanation

Walkthrough

The original function $f(x) = \frac{15}{2} - 2\left(x - \frac{3}{2}\right)^2$ is a downward-opening parabola with vertex (maximum) at $\left(\frac{3}{2}, \frac{15}{2}\right)$ .

Step 1 — Reflection: Reflecting in the $x$ -axis replaces $y$ with $-y$ , giving $y = -f(x) = 2\left(x - \frac{3}{2}\right)^2 - \frac{15}{2}$ . This is now an upward-opening parabola with a minimum at $\left(\frac{3}{2}, -\frac{15}{2}\right)$ .

Step 2 — Translation: We need the minimum at the origin $(0, 0)$ . The current minimum is at $\left(\frac{3}{2}, -\frac{15}{2}\right)$ . To move it to $(0, 0)$ , we shift left by $\frac{3}{2}$ and up by $\frac{15}{2}$ , i.e., by the vector $\begin{pmatrix} -3/2 \\ 15/2 \end{pmatrix}$ .

The order of reflection and translation can be swapped — both give the same result.

Key Takeaways

Reflecting in the $x$ -axis changes a maximum to a minimum (and vice versa).
A translation by vector $\begin{pmatrix} p \\ q \end{pmatrix}$ moves the vertex from $(h, k)$ to $(h+p, k+q)$ .

Common Mistakes

Reflecting in the $y$ -axis instead of the $x$ -axis — this would not convert the maximum to a minimum.
Getting the translation vector signs wrong (e.g., $\begin{pmatrix} 3/2 \\ -15/2 \end{pmatrix}$ instead of $\begin{pmatrix} -3/2 \\ 15/2 \end{pmatrix}$ ).

Things to Be Careful About

The marking scheme accepts transformations in any order, so stating translation first then reflection is also valid.
The translation vector must follow from the student's own values in part (a). If part (a) is wrong, part (b) can still earn marks via follow-through.

Techniques used

reflect in x-axis to convert maximum to minimumtranslate to move vertex to origin

(c)

The function $g$ is defined by $g(x) = 3 + 6x - 2x^2$ for $x \le 0$ .

Sketch the graph of $y = g(x)$ and explain why $g$ is a one-one function. You are not required to find the coordinates of any intersections with the axes.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The function $g(x) = 3 + 6x - 2x^2$ is defined for $x \leq 0$ . From part (a), the full parabola has its vertex (maximum) at $x = \frac{3}{2}$ . Since $x \leq 0$ is entirely to the left of the vertex, $g(x)$ is strictly increasing on this domain. Sketch the left branch of the parabola and use the horizontal line test to show $g$ is one-one.

Working

Sketch: The function $g(x) = \frac{15}{2} - 2\left(x - \frac{3}{2}\right)^2$ for $x \leq 0$ :

At $x = 0$ : $g(0) = 3$ , so the graph passes through $(0, 3)$ .
As $x \to -\infty$ : $g(x) \to -\infty$ .
The graph is the left branch of the downward-opening parabola, lying in the second quadrant (for $-\sqrt{3/2} < x \leq 0$ , where $g(x) > 0$ ) and the third quadrant (for $x < -\sqrt{3/2}$ , where $g(x) < 0$ ).

The curve is strictly increasing on $x \leq 0$ because it lies entirely to the left of the vertex at $x = \frac{3}{2}$ , and the parabola opens downward.

Why $g$ is one-one:

Since $g$ is strictly increasing on its domain $x \leq 0$ , each $y$ -value is associated with exactly one $x$ -value. Equivalently, any horizontal line intersects the graph of $y = g(x)$ at most once (the horizontal line test is satisfied).

Answer

The graph of $y = g(x)$ is the left branch of the parabola $y = \frac{15}{2} - 2\left(x - \frac{3}{2}\right)^2$ for $x \leq 0$ , passing through $(0, 3)$ and extending downward into the third quadrant. $g$ is one-one because it is strictly increasing on $x \leq 0$ , so each $y$ -value corresponds to a unique $x$ -value (horizontal line test).

Final answer

g is one-one because it is strictly increasing on x ≤ 0; each y-value corresponds to a single x-value.

Detailed explanation

Walkthrough

We are given $g(x) = 3 + 6x - 2x^2$ for $x \leq 0$ .

Step 1 — Understand the shape: From part (a), the full parabola $f(x) = \frac{15}{2} - 2\left(x - \frac{3}{2}\right)^2$ has its vertex at $\left(\frac{3}{2}, \frac{15}{2}\right)$ and opens downward. For $x \leq 0$ , we are looking at the portion of the parabola to the left of the $y$ -axis.

Step 2 — Key points for the sketch:

At $x = 0$ : $g(0) = 3$ , so the curve meets the $y$ -axis at $(0, 3)$ .
The $x$ -intercept (where $g(x) = 0$ ): $\frac{15}{2} - 2\left(x - \frac{3}{2}\right)^2 = 0 \Rightarrow \left(x - \frac{3}{2}\right)^2 = \frac{15}{4} \Rightarrow x = \frac{3}{2} - \frac{\sqrt{15}}{2} \approx -0.936$ . This is in the second/third quadrant boundary region.
As $x \to -\infty$ , $g(x) \to -\infty$ .

Step 3 — Why one-one: On the domain $x \leq 0$ , the function lies entirely to the left of the vertex. Since the parabola opens downward, moving left from the vertex means the function is increasing (as $x$ increases toward $\frac{3}{2}$ , $g(x)$ increases). Therefore $g$ is strictly increasing on $x \leq 0$ , which means each $y$ -value is produced by exactly one $x$ -value. This satisfies the horizontal line test.

Key Takeaways

Restricting the domain of a quadratic to one side of its vertex makes it one-one.
A strictly monotonic function on its domain is always one-one.
The horizontal line test is the geometric way to verify one-oneness.

Common Mistakes

Sketching the full parabola instead of only the branch for $x \leq 0$ .
Saying the function passes the vertical line test — this is always true for any function and does not prove one-oneness.
Not mentioning that the function is increasing or that horizontal lines meet the graph at most once.

Things to Be Careful About

The question says you are not required to find axis intersection coordinates, so do not spend time calculating them.
The sketch should clearly show the curve only in the second and third quadrants (for $x \leq 0$ ).

Techniques used

restrict domain to left of vertexevaluate at boundary x = 0apply horizontal line test

(d)

Sketch the graph of $y = g^{-1}(x)$ on your diagram in (c), and find an expression for $g^{-1}(x)$ . You should label the two graphs in your diagram appropriately and show any relevant mirror line.

4M Similar

DifficultyMedium

Worked solution

Approach

Sketch $y = g^{-1}(x)$ by reflecting the graph of $y = g(x)$ in the line $y = x$ . To find the expression for $g^{-1}(x)$ , set $y = g(x)$ , swap $x$ and $y$ , and solve for $y$ using the quadratic formula, selecting the branch consistent with the domain $x \leq 0$ .

Working

Sketch: Reflect the curve $y = g(x)$ (from part (c)) in the line $y = x$ . The resulting curve $y = g^{-1}(x)$ lies in the third and fourth quadrants only, passing through $(3, 0)$ (the image of $(0, 3)$ ) and extending downward to the right.

Finding $g^{-1}(x)$ :

Set $y = 3 + 6x - 2x^2$ and swap $x$ and $y$ :

x = 3 + 6y - 2y^2

Rearrange into standard quadratic form in $y$ :

2y^2 - 6y + (x - 3) = 0

Apply the quadratic formula:

y = \frac{6 \pm \sqrt{36 - 8(x - 3)}}{4} = \frac{6 \pm \sqrt{60 - 8x}}{4} = \frac{3 \pm \sqrt{15 - 2x}}{2}

Since the domain of $g$ is $x \leq 0$ , the range of $g^{-1}$ must satisfy $y \leq 0$ . We need:

\frac{3 \pm \sqrt{15 - 2x}}{2} \leq 0

Since $\sqrt{15 - 2x} \geq 0$ , the $+$ branch gives $y \geq \frac{3}{2} > 0$ , which is invalid. The $-$ branch gives:

y = \frac{3 - \sqrt{15 - 2x}}{2} = \frac{3}{2} - \sqrt{\frac{15}{4} - \frac{x}{2}}

Check: at $x = 3$ (the image of $y = 0$ ), $y = \frac{3}{2} - \sqrt{\frac{15}{4} - \frac{3}{2}} = \frac{3}{2} - \sqrt{\frac{9}{4}} = \frac{3}{2} - \frac{3}{2} = 0$ . ✓

Answer

g^{-1}(x) = \frac{3}{2} - \sqrt{\frac{15}{4} - \frac{x}{2}}

Final answer

g^{-1}(x) = 3/2 - sqrt(15/4 - x/2)

Detailed explanation

Walkthrough

We need to sketch $y = g^{-1}(x)$ and find its expression.

Step 1 — Sketch the inverse: The graph of $y = g^{-1}(x)$ is the reflection of $y = g(x)$ in the line $y = x$ . Since $g(x)$ is defined for $x \leq 0$ and has range $y \leq 3$ , the inverse $g^{-1}(x)$ is defined for $x \leq 3$ and has range $y \leq 0$ . The curve passes through $(3, 0)$ (reflection of $(0, 3)$ ) and extends into the third and fourth quadrants.

Step 2 — Find the expression algebraically:

Start with $y = 3 + 6x - 2x^2$ .
Swap $x$ and $y$ : $x = 3 + 6y - 2y^2$ .
Rearrange: $2y^2 - 6y + (x - 3) = 0$ .
Quadratic formula: $y = \frac{6 \pm \sqrt{36 - 8(x-3)}}{4} = \frac{3 \pm \sqrt{15 - 2x}}{2}$ .

Step 3 — Choose the correct branch: Since the range of $g^{-1}$ is $y \leq 0$ (because the domain of $g$ is $x \leq 0$ ), we need the branch that gives $y \leq 0$ . The $+$ branch gives $y \geq \frac{3}{2}$ , which is wrong. The $-$ branch gives $y = \frac{3 - \sqrt{15 - 2x}}{2} \leq \frac{3}{2}$ , and for $x \leq 3$ , $\sqrt{15 - 2x} \geq 3$ , so $y \leq 0$ . ✓

Step 4 — Simplify:
$y = \frac{3}{2} - \frac{\sqrt{15 - 2x}}{2} = \frac{3}{2} - \sqrt{\frac{15 - 2x}{4}} = \frac{3}{2} - \sqrt{\frac{15}{4} - \frac{x}{2}}$

Key Takeaways

The inverse function is found by swapping $x$ and $y$ and solving.
When solving a quadratic to find the inverse, always check which branch is consistent with the domain/range restrictions.
The graph of $y = g^{-1}(x)$ is the reflection of $y = g(x)$ in $y = x$ .

Common Mistakes

Choosing the $+$ branch of the square root, which gives $y \geq \frac{3}{2}$ , violating the range restriction $y \leq 0$ .
Forgetting to swap $x$ and $y$ before solving (solving for $x$ in terms of $y$ without swapping gives the inverse relation but not the function $g^{-1}(x)$ ).
Not drawing the line $y = x$ on the sketch.

Things to Be Careful About

The final expression must involve $x$ , not $y$ .
The domain of $g^{-1}$ is $x \leq 3$ (from the range of $g$ ), and the range is $y \leq 0$ (from the domain of $g$ ). The sketch should reflect this — the curve appears in the third and fourth quadrants only.
The mark scheme accepts equivalent forms such as $\frac{3 - \sqrt{15 - 2x}}{2}$ .

Techniques used

reflect curve in y = x lineswap x and ysolve quadratic for xselect correct branch using domain restriction

Mathematics 9709/11 — October/November 2024

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