(a) Prove the identity $$ \frac{\tan \theta + 7}{\tan^2 \theta - 3} \equiv \frac{\sin \theta \cos \theta + 7\cos^2 \theta}{1 - 4\cos^2 \theta}. $$ (b) Hence solve the equation $$ \frac{\sin \theta \cos \theta + 7\cos^2 \theta}{1 - 4\cos^2 \theta} = \frac{5}{\tan \theta} $$ for $0^\circ \le \theta \le 180^\circ$.

(a) The identity is proven. · (b) θ = 71.6° and θ = 128.7°

The diagram shows the circle with equation $x^2 + y^2 - 14x + 8y + 36 = 0$ and the line $y = -2$. The line intersects the circle at the points $A$ and $B$. The centre of the circle is $C$. (a) Find the coordinates of $A$, $B$ and $C$. (b) Find the angle $ACB$ in radians. Give your answer correct to 3 significant figures. (c) The chord $AB$ divides the circle into two segments. Find the area of the larger segment.

(a) A = (2, -2), B = (12, -2), C = (7, -4) · (b) 2.38 · (c) 66.6

(a) The first, second and third terms of an arithmetic progression are $4k$, $k^2$ and $8k$ respectively, where $k$ is a non-zero constant. (i) Find the value of $k$. (ii) Find the sum of the first 20 terms of the progression. (b) The fourth and sixth terms of a geometric progression are 36 and 6 respectively. The common ratio of the progression is positive. Find the sum to infinity of the progression. Give your answer in the form $\frac{a}{\sqrt{b}-c}$, where $a$, $b$ and…

(a)(i) k = 6 · (a)(ii) S20 = 2760 · (b) S∞ = 1296/(√6 - 1)

9709/12 May/June 2025 — Mathematics Solved Past Paper

Q14MMedium-EasyFunctions

Fig. 1

The diagram shows the graphs with equations $y = f(x)$ and $y = g(x)$ .

Describe fully a sequence of two transformations which transforms the graph of $y = f(x)$ to the graph of $y = g(x)$ . Make clear the order in which the transformations should be applied.

DifficultyMedium-Easy

Worked solution

Approach

Compare key points on the graphs of $y = f(x)$ and $y = g(x)$ to determine the transformations. Since the x-coordinates of corresponding points are unchanged, the transformations must be purely vertical (parallel to the y-axis).

Working

Method 1: Stretch then translation

Consider a vertical stretch by a factor of $k$ parallel to the y-axis. The y-coordinates are multiplied by $k$ .
Taking the point $(0, 6)$ on $f(x)$ , stretching by a factor of 2 gives $(0, 12)$ .
Now apply a vertical translation by $\begin{pmatrix} 0 \\ c \end{pmatrix}$ .
The point $(0, 12)$ must map to $(0, -2)$ on $g(x)$ .

12 + c = -2 \implies c = -14

Check with another point: $(3, 3)$ on $f(x)$ .
Stretch by 2: $(3, 6)$ .
Translate by $\begin{pmatrix} 0 \\ -14 \end{pmatrix}$ : $(3, 6 - 14) = (3, -8)$ . This matches $g(x)$ .

Method 2: Translation then stretch

Consider a vertical translation by $\begin{pmatrix} 0 \\ c \end{pmatrix}$ .
Taking the point $(0, 6)$ on $f(x)$ , translating by $-7$ gives $(0, -1)$ .
Now apply a vertical stretch by a factor of 2.
The point $(0, -1)$ maps to $(0, -2)$ , which matches $g(x)$ .
Check with $(3, 3)$ : translate by $-7$ gives $(3, -4)$ , stretch by 2 gives $(3, -8)$ . This matches.

Answer

A stretch with scale factor 2 parallel to the y-axis, followed by a translation by $\begin{pmatrix} 0 \\ -14 \end{pmatrix}$ .\n(Alternatively: a translation by $\begin{pmatrix} 0 \\ -7 \end{pmatrix}$ followed by a stretch with scale factor 2 parallel to the y-axis.)

Final answer

Stretch with scale factor 2 parallel to the y-axis, then translation by (0, -14)

Detailed explanation

Walkthrough

First, observe the graphs of $y = f(x)$ and $y = g(x)$ . The x-coordinates of all corresponding points (such as the peaks and corners) are identical, meaning there is no horizontal transformation. The transformations must therefore be purely vertical.

We can determine the vertical transformations by comparing the y-coordinates of key points. For example, $f(0) = 6$ and $g(0) = -2$ ; $f(3) = 3$ and $g(3) = -8$ .

Method 1 (Stretch then Translation):
If we stretch $f(x)$ vertically by a factor of 2, the point $(0, 6)$ becomes $(0, 12)$ and $(3, 3)$ becomes $(3, 6)$ . To map $(0, 12)$ to $(0, -2)$ , we must translate downwards by 14 units. This translation also maps $(3, 6)$ to $(3, -8)$ , confirming the scale factor and translation are correct.

Method 2 (Translation then Stretch):
Alternatively, if we first translate $f(x)$ downwards by 7 units, the point $(0, 6)$ becomes $(0, -1)$ and $(3, 3)$ becomes $(3, -4)$ . Stretching these vertically by a factor of 2 gives $(0, -2)$ and $(3, -8)$ , which exactly match the corresponding points on $g(x)$ .

Key Takeaways

When x-coordinates are unchanged, transformations are purely vertical (parallel to the y-axis).
A vertical stretch by factor $k$ multiplies all y-coordinates by $k$ .
A vertical translation by $c$ adds $c$ to all y-coordinates.
The order of stretch and translation matters; if a stretch is applied first, the translation must compensate for the stretched value, whereas if a translation is applied first, the stretch will also scale the translation amount.

Common Mistakes

Forgetting to specify the direction of the stretch or translation (e.g., saying "stretch by factor 2" without stating "parallel to the y-axis").
Using the wrong translation value because the stretch was not accounted for (e.g., stating a translation of $-2$ instead of $-14$ when stretching first).
Confusing horizontal and vertical transformations when the x-coordinates happen to be the same.

Things to Be Careful About

Always state the direction of transformations clearly: "parallel to the y-axis", "in the y-direction", or "vertically".
Ensure the order of transformations is explicitly stated, as different orders require different numerical values for the translation.
Check your answer against at least two points on the graph to ensure consistency.

Techniques used

identify vertical stretchidentify vertical translationcombine transformations

Similar questions

Q24MMedium-EasyQuadratics

Find the coordinates of the points of intersection of the curve and the line with equations

2xy + 5y^2 = 24 \quad \text{and} \quad 2x + y + 4 = 0.

DifficultyMedium-Easy

Worked solution

Approach

Eliminate one variable by substituting the linear equation into the curve equation. This gives a quadratic in one variable. Solve the quadratic, then substitute each solution back into the linear equation to find the corresponding coordinates.

Working

From the line equation, rearrange to make $x$ the subject:

2x + y + 4 = 0 \quad \Rightarrow \quad x = \frac{-y - 4}{2}.

Substitute $x = \frac{-y - 4}{2}$ into $2xy + 5y^2 = 24$ :

2\left(\frac{-y - 4}{2}\right)y + 5y^2 = 24

(-y - 4)y + 5y^2 = 24

-y^2 - 4y + 5y^2 = 24

4y^2 - 4y = 24

Divide by 4 and rearrange:

y^2 - y - 6 = 0

Factorise:

(y - 3)(y + 2) = 0

So

y = 3 \quad \text{or} \quad y = -2.

Now use $x = \frac{-y - 4}{2}$ to find the corresponding $x$ -coordinates.

If $y = -2$ :

x = \frac{-(-2) - 4}{2} = \frac{2 - 4}{2} = -1.

If $y = 3$ :

x = \frac{-3 - 4}{2} = -\frac{7}{2}.

Answer

The points of intersection are

(-1, -2) \quad \text{and} \quad \left(-\frac{7}{2}, 3\right).

Final answer

(-1, -2) and (-7/2, 3)

Detailed explanation

Walkthrough

We are given two equations: a curve equation and a line equation. Since the line equation is linear, we can rearrange it to express one variable in terms of the other. Here it is easiest to write $x$ in terms of $y$ :

$x = \frac{-y - 4}{2}.$

Substituting this into the curve equation eliminates $x$ , leaving a single equation in $y$ . This is the standard method for solving simultaneous equations where one equation is linear and the other is quadratic.

After substitution, the expression becomes

$(-y - 4)y + 5y^2 = 24.$

Expanding gives

$-y^2 - 4y + 5y^2 = 24,$

which simplifies to

$4y^2 - 4y = 24.$

Dividing by 4 and rearranging gives

$y^2 - y - 6 = 0.$

This quadratic factorises as

$(y - 3)(y + 2) = 0,$

so $y = 3$ or $y = -2$ . There are at most two solutions because the curve is quadratic.

Finally, substitute each $y$ -value back into the linear equation to find the matching $x$ -value. For $y = -2$ , we get $x = -1$ . For $y = 3$ , we get $x = -\frac{7}{2}$ . Therefore the two intersection points are $(-1, -2)$ and $\left(-\frac{7}{2}, 3\right)$ .

Key Takeaways

When solving simultaneous equations with one linear and one quadratic equation, solve the linear equation for one variable and substitute into the quadratic.
The resulting quadratic can be solved by factorisation, completing the square, or the quadratic formula.
After finding one coordinate, substitute back into the linear equation to find the other coordinate.
The final answers must be written as ordered pairs $(x, y)$ .

Common Mistakes

Substituting incorrectly because of a sign error when rearranging the line equation. For example, writing $x = \frac{y + 4}{2}$ instead of $x = \frac{-y - 4}{2}$ .
Simplifying the substituted expression incorrectly, especially when distributing the $2$ and the $y$ .
Finding the $y$ -values but forgetting to substitute back to find the corresponding $x$ -values.
Mixing up which $x$ belongs to which $y$ when writing the final ordered pairs.

Things to Be Careful About

The mark scheme allows sign errors in the rearrangement or expansion, but the final values must be correct.
Check that each $(x, y)$ pair satisfies both original equations.
When factorising $y^2 - y - 6 = 0$ , be careful with the signs: the factors are $(y - 3)(y + 2)$ , not $(y + 3)(y - 2)$ .
The final answer should list both points clearly, with the $x$ -coordinate first in each ordered pair.

Techniques used

Substitute the linear expression into the quadratic equationSimplify and collect terms into a single quadratic equationSolve the quadratic by factorisationSubstitute each solution back to find the corresponding coordinate

Similar questions

Q34MMedium-EasySeries

The coefficient of $x^7$ in the expansion of $\left(px^2 + \frac{4}{p}x\right)^5$ is 1280.

Find the value of the constant $p$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the general term of the binomial expansion of $\left(px^2 + \frac{4}{p}x\right)^5$ , find the term containing $x^7$ , extract its coefficient, set it equal to 1280, and solve for $p$ .

Working

The general term is

\binom{5}{r} (px^2)^{5-r} \left(\frac{4}{p}x\right)^r

The power of $x$ in this term is

2(5-r) + r = 10 - r

We require the power of $x^7$ , so

10 - r = 7 \quad \Rightarrow \quad r = 3

Therefore the required term is

\binom{5}{3}(px^2)^2 \left(\frac{4}{p}x\right)^3

Simplify this term:

\begin{aligned} \binom{5}{3}(px^2)^2 \left(\frac{4}{p}x\right)^3 &= 10 \cdot p^2 x^4 \cdot \frac{64}{p^3} x^3 \\ &= \frac{640}{p} x^7 \end{aligned}

So the coefficient of $x^7$ is $\frac{640}{p}$ . Since this coefficient is 1280,

\frac{640}{p} = 1280

p = \frac{640}{1280} = \frac{1}{2}

Answer

p = \frac{1}{2}

Final answer

p = 1/2

Detailed explanation

Walkthrough

We need the coefficient of $x^7$ in the expansion of $\left(px^2 + \frac{4}{p}x\right)^5$ . Start by writing the general term of the binomial expansion:

\binom{5}{r} (px^2)^{5-r} \left(\frac{4}{p}x\right)^r

Here $r$ is the number of times the second term $\frac{4}{p}x$ is chosen. The first term contributes $x^2$ each time it is chosen, and the second term contributes $x$ each time it is chosen. So the total power of $x$ in a term is $2(5-r) + r = 10 - r$ .

We want the power to be 7, so we solve $10-r=7$ , giving $r=3$ . This tells us which term to use: the term in which the second factor is chosen three times.

Now substitute $r=3$ into the general term and simplify:

\binom{5}{3}(px^2)^2 \left(\frac{4}{p}x\right)^3 = 10 \cdot p^2 x^4 \cdot \frac{64}{p^3} x^3 = \frac{640}{p} x^7

The coefficient of $x^7$ is therefore $\frac{640}{p}$ . Setting this equal to 1280 gives

\frac{640}{p} = 1280

so $p = \frac{1}{2}$ .

Key Takeaways

The binomial expansion can be written term by term using the general term $\binom{n}{r} a^{n-r}b^r$ . The index $r$ is determined by matching the power of the variable. Once the correct term is found, its coefficient can be written in terms of the unknown constant and used to solve an equation.

Common Mistakes

Choosing the wrong value of $r$ without checking that the power of $x$ is 7.
Forgetting to include the powers of $x$ when simplifying the term, so the coefficient is incorrect.
Dropping the $x^7$ factor and comparing the whole expression to 1280 instead of comparing only the coefficient.
Losing the factor of $p$ when simplifying $\frac{p^2}{p^3}$ .

Things to Be Careful About

The required term is $\binom{5}{3}(px^2)^2 \left(\frac{4}{p}x\right)^3$ , not $\binom{5}{2}(px^2)^3 \left(\frac{4}{p}x\right)^2$ , because $r=2$ gives $x^{10}$ .
Write brackets carefully, especially $\left(\frac{4}{p}x\right)^3$ , because the cube applies to both $\frac{4}{p}$ and $x$ .
The mark scheme accepts the binomial coefficient as $\binom{5}{3}$ or $\binom{5}{2}$ , but the simplified coefficient must be clearly identified.
When forming the equation, use the coefficient $\frac{640}{p}$ and set it equal to 1280; do not include the $x^7$ factor.

Techniques used

identify the general term in a binomial expansionmatch powers of x to find the required termsimplify the coefficient and solve the resulting equation

Similar questions

Q4MediumDifferentiation

A point $P$ is moving along the curve with equation $y = ax^{\frac{3}{2}} - 12x$ in such a way that the $x$ -coordinate of $P$ is increasing at a constant rate of 5 units per second.

(a)

Find the rate at which the $y$ -coordinate of $P$ is changing when $x = 9$ . Give your answer in terms of the constant $a$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $y$ with respect to $x$ to obtain $\frac{dy}{dx}$ . Then use the chain rule

\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}

with $\frac{dx}{dt} = 5$ and $x = 9$ to find the rate of change of the $y$ -coordinate.

Working

Differentiate the curve:

\frac{dy}{dx} = \frac{3}{2}ax^{\frac{1}{2}} - 12

When $x = 9$ :

\frac{dy}{dx} = \frac{3}{2}a \cdot 9^{\frac{1}{2}} - 12 = \frac{3}{2}a \cdot 3 - 12 = \frac{9}{2}a - 12

Using the chain rule with $\frac{dx}{dt} = 5$ :

\frac{dy}{dt} = \left(\frac{9}{2}a - 12\right) \cdot 5

Simplify:

\frac{dy}{dt} = 5\left(\frac{9}{2}a - 12\right) = \frac{45}{2}a - 60

Answer

\frac{dy}{dt} = \frac{45}{2}a - 60

Final answer

dy/dt = 45a/2 - 60 units per second

Detailed explanation

Walkthrough

The curve is $y = ax^{\frac{3}{2}} - 12x$ , and the $x$ -coordinate is increasing at a constant rate $\frac{dx}{dt} = 5$ units per second. We need the corresponding rate of change of $y$ , i.e. $\frac{dy}{dt}$ .

The chain rule connects the rates:

\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}

First find $\frac{dy}{dx}$ by differentiating term by term. For $ax^{\frac{3}{2}}$ , multiply by the power $\frac{3}{2}$ and reduce the power by 1, giving $\frac{3}{2}ax^{\frac{1}{2}}$ . For $-12x$ , the derivative is $-12$ . So:

\frac{dy}{dx} = \frac{3}{2}ax^{\frac{1}{2}} - 12

At $x = 9$ , $9^{\frac{1}{2}} = 3$ , so:

\frac{dy}{dx} = \frac{3}{2}a \cdot 3 - 12 = \frac{9}{2}a - 12

Now multiply by $\frac{dx}{dt} = 5$ :

\frac{dy}{dt} = 5\left(\frac{9}{2}a - 12\right) = \frac{45}{2}a - 60

This is the rate at which the $y$ -coordinate is changing.

Key Takeaways

This problem combines differentiation with the chain rule for related rates. The key idea is that $y$ depends on $x$ , and $x$ depends on $t$ , so the rate of change of $y$ with respect to $t$ is the product of $\frac{dy}{dx}$ and $\frac{dx}{dt}$ . Fractional powers are differentiated using the same power rule as integer powers.

Common Mistakes

Forgetting to multiply by $\frac{dx}{dt} = 5$ after finding $\frac{dy}{dx}$ .
Differentiating $ax^{\frac{3}{2}}$ incorrectly, for example writing $\frac{3}{2}ax^{-\frac{1}{2}}$ instead of $\frac{3}{2}ax^{\frac{1}{2}}$ .
Losing the constant $-12$ when differentiating $-12x$ .
Substituting $x = 9$ into $y$ instead of into $\frac{dy}{dx}$ .

Things to Be Careful About

Apply the chain rule in the correct order: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ .
Keep the constant $a$ throughout; the answer is required in terms of $a$ .
Show the differentiation and the chain-rule substitution clearly to earn the method marks.

Techniques used

differentiate the curve with respect to xapply the chain rule to relate dy/dt and dx/dtsubstitute the given x-coordinate and rate of change

(b)

Given that the curve has a minimum point when $x = \frac{1}{4}$ , find the value of $a$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

At a minimum point, the gradient is zero, so $\frac{dy}{dx} = 0$ . Substitute $x = \frac{1}{4}$ into the derivative and solve for $a$ .

Working

The derivative is:

\frac{dy}{dx} = \frac{3}{2}ax^{\frac{1}{2}} - 12

At the minimum point, $x = \frac{1}{4}$ and $\frac{dy}{dx} = 0$ :

\frac{3}{2}a\left(\frac{1}{4}\right)^{\frac{1}{2}} - 12 = 0

Since $\left(\frac{1}{4}\right)^{\frac{1}{2}} = \frac{1}{2}$ :

\frac{3}{2}a \cdot \frac{1}{2} - 12 = 0

\frac{3}{4}a - 12 = 0

Solve for $a$ :

\frac{3}{4}a = 12

a = 12 \times \frac{4}{3} = 16

Answer

a = 16

Final answer

a = 16

Detailed explanation

Walkthrough

A minimum point is a stationary point, so its derivative is zero. The derivative of the curve is:

\frac{dy}{dx} = \frac{3}{2}ax^{\frac{1}{2}} - 12

At the minimum point, $x = \frac{1}{4}$ . Substitute this value and set the derivative equal to zero:

\frac{3}{2}a\left(\frac{1}{4}\right)^{\frac{1}{2}} - 12 = 0

Since $\left(\frac{1}{4}\right)^{\frac{1}{2}} = \frac{1}{2}$ , this becomes:

\frac{3}{2}a \cdot \frac{1}{2} - 12 = 0

\frac{3}{4}a - 12 = 0

Then:

\frac{3}{4}a = 12, \quad a = 12 \times \frac{4}{3} = 16

Key Takeaways

At any stationary point, $\frac{dy}{dx} = 0$ . When the $x$ -coordinate of a stationary point is given, substitute it into the derivative and solve for any unknown parameter.

Common Mistakes

Forgetting to set the derivative equal to zero.
Substituting $x = \frac{1}{4}$ into the original equation for $y$ instead of into $\frac{dy}{dx}$ .
Making fraction arithmetic errors: $\left(\frac{1}{4}\right)^{\frac{1}{2}} = \frac{1}{2}$ , so $\frac{3}{2}a \cdot \frac{1}{2} = \frac{3}{4}a$ .

Things to Be Careful About

Although the problem says the point is a minimum, the only condition needed is that it is stationary, so $\frac{dy}{dx} = 0$ .
Keep the factor $a$ with the derivative; it is easy to drop it when solving.
The mark scheme allows a restart for $\frac{dy}{dx}$ if two terms are seen and at least one is correct, so write the derivative clearly.

Techniques used

set the derivative equal to zero at the stationary pointsubstitute x = 1/4 and solve for a

Q5MediumTrigonometry

The equation of a curve is $y = 4\cos 2x + 3$ for $0 \le x \le 2\pi$ .

(a)

State the greatest and least possible values of $y$ .

2M Similar

DifficultyEasy

Worked solution

Approach

The function is $y = 4\cos 2x + 3$ . Since $\cos 2x$ oscillates between $-1$ and $1$ , we can find the greatest and least values of $y$ by substituting these extreme values.

Working

The range of $\cos 2x$ is:

-1 \leq \cos 2x \leq 1

Multiply by the amplitude $4$ :

-4 \leq 4\cos 2x \leq 4

Add the vertical shift $3$ :

-4 + 3 \leq 4\cos 2x + 3 \leq 4 + 3

-1 \leq y \leq 7

Answer

Greatest value: $7$

Least value: $-1$

Final answer

Greatest value: 7, Least value: -1

Detailed explanation

Walkthrough

The function $y = 4\cos 2x + 3$ is a cosine wave with amplitude $4$ and vertical shift $+3$ . The cosine function $\cos 2x$ always lies in the interval $[-1, 1]$ regardless of the argument. Multiplying by $4$ scales this to $[-4, 4]$ , and adding $3$ shifts the entire range up by $3$ , giving $[-1, 7]$ . The greatest value $7$ occurs when $\cos 2x = 1$ (e.g., at $x = 0, \pi, 2\pi$ ), and the least value $-1$ occurs when $\cos 2x = -1$ (e.g., at $x = \pi/2, 3\pi/2$ ).

Key Takeaways

The range of $a\cos(bx) + c$ is $[c - |a|, c + |a|]$ .
The vertical shift $c$ moves the midline of the wave, and the amplitude $|a|$ determines how far the curve extends above and below that midline.

Common Mistakes

Forgetting to add the vertical shift when computing the extremes.
Confusing the amplitude with the maximum value directly.

Things to Be Careful About

The question asks for the greatest and least values of $y$ , not the values of $x$ at which they occur. Only provide the $y$ -values.

Techniques used

identify range of cosine functionapply amplitude and vertical shift

(b)

Sketch the curve.

Fig. 2

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Sketch $y = 4\cos 2x + 3$ for $0 \leq x \leq 2\pi$ . Determine the period, amplitude, and vertical shift, then plot key points at maxima, minima, and midline crossings over two complete cycles.

Working

Period: The basic cosine function $\cos x$ has period $2\pi$ . For $\cos 2x$ , the period is $\frac{2\pi}{2} = \pi$ . Over $[0, 2\pi]$ , there are $\frac{2\pi}{\pi} = 2$ complete cycles.

Amplitude and vertical shift: Amplitude $= 4$ , vertical shift $= +3$ . The curve oscillates between $y = -1$ (minimum) and $y = 7$ (maximum), with midline $y = 3$ .

Key points:

\begin{aligned} x = 0 &: y = 4\cos 0 + 3 = 7 \quad \text{(maximum)} \\ x = \frac{\pi}{4} &: y = 4\cos\frac{\pi}{2} + 3 = 3 \quad \text{(midline, decreasing)} \\ x = \frac{\pi}{2} &: y = 4\cos\pi + 3 = -1 \quad \text{(minimum)} \\ x = \frac{3\pi}{4} &: y = 4\cos\frac{3\pi}{2} + 3 = 3 \quad \text{(midline, increasing)} \\ x = \pi &: y = 4\cos 2\pi + 3 = 7 \quad \text{(maximum)} \\ x = \frac{5\pi}{4} &: y = 4\cos\frac{5\pi}{2} + 3 = 3 \quad \text{(midline, decreasing)} \\ x = \frac{3\pi}{2} &: y = 4\cos 3\pi + 3 = -1 \quad \text{(minimum)} \\ x = \frac{7\pi}{4} &: y = 4\cos\frac{7\pi}{2} + 3 = 3 \quad \text{(midline, increasing)} \\ x = 2\pi &: y = 4\cos 4\pi + 3 = 7 \quad \text{(maximum)} \end{aligned}

The curve starts at $(0, 7)$ , decreases to a minimum of $-1$ at $x = \frac{\pi}{2}$ , rises to a maximum of $7$ at $x = \pi$ , decreases to a minimum of $-1$ at $x = \frac{3\pi}{2}$ , and rises back to $7$ at $x = 2\pi$ . The curve should be smooth and rounded at the extrema, starting and finishing with a horizontal tangent (level off).

Answer

A smooth cosine curve completing two full cycles from $(0, 7)$ to $(2\pi, 7)$ , with minima at $\left(\frac{\pi}{2}, -1\right)$ and $\left(\frac{3\pi}{2}, -1\right)$ , and midline crossings at $y = 3$ .

Final answer

Sketch showing two complete cycles of y = 4cos 2x + 3 from x = 0 to x = 2π, with maxima at y = 7 and minima at y = -1.

Detailed explanation

Walkthrough

To sketch $y = 4\cos 2x + 3$ , we first identify the transformation parameters:

Amplitude: $4$ , so the curve extends $4$ units above and below the midline.
Vertical shift: $+3$ , so the midline is $y = 3$ .
Period: $\frac{2\pi}{2} = \pi$ , meaning one complete cycle takes $\pi$ units of $x$ .
Domain: $[0, 2\pi]$ , which contains exactly $2$ complete cycles.

Since this is a cosine function (not sine), it starts at its maximum value at $x = 0$ . The key points within each cycle of length $\pi$ are:

Maximum at the start: $x = 0, \pi, 2\pi$ with $y = 7$ .
Midline crossing (decreasing): $x = \frac{\pi}{4}, \frac{5\pi}{4}$ with $y = 3$ .
Minimum: $x = \frac{\pi}{2}, \frac{3\pi}{2}$ with $y = -1$ .
Midline crossing (increasing): $x = \frac{3\pi}{4}, \frac{7\pi}{4}$ with $y = 3$ .

Plot these points and join them with a smooth, rounded curve. The curve should start and finish with a horizontal tangent (level off) at $x = 0$ and $x = 2\pi$ .

Key Takeaways

For $y = a\cos(bx) + c$ , the period is $\frac{2\pi}{|b|}$ , amplitude is $|a|$ , and midline is $y = c$ .
A cosine graph starts at a maximum (not a midline crossing like sine).
Always check the domain to determine how many complete cycles are shown.

Common Mistakes

Drawing only one cycle instead of two over $[0, 2\pi]$ .
Starting the curve at a midline crossing instead of at the maximum.
Using straight lines between key points instead of a smooth curve.
Not letting the curve level off (become horizontal) at the endpoints $x = 0$ and $x = 2\pi$ .

Things to Be Careful About

The sketch should only show the curve for $0 \leq x \leq 2\pi$ ; do not extend beyond this domain.
The curve must be smooth and rounded at the maxima and minima, not angular.
Incorrect $x$ -axis intercepts are condoned, but the overall shape and key $y$ -values must be correct.

Techniques used

determine period of transformed cosineidentify key points (maxima, minima, midline crossings)sketch curve over given domain

(c)

Hence determine the number of solutions of the equation $4\cos 2x + 3 = 2x - 1$ for $0 \le x \le 2\pi$ .

1M Similar

DifficultyMedium

Worked solution

Approach

The equation $4\cos 2x + 3 = 2x - 1$ asks where the curve $y = 4\cos 2x + 3$ meets the straight line $y = 2x - 1$ . Use the sketch from part (b) and evaluate the line at key $x$ -values to count intersections.

Working

The line $y = 2x - 1$ has gradient $2$ and $y$ -intercept $-1$ .

Evaluate the line at the key $x$ -values from the sketch:

\begin{aligned} x = 0 &: y = 2(0) - 1 = -1 \\ x = \frac{\pi}{2} \approx 1.57 &: y = 2\left(\frac{\pi}{2}\right) - 1 = \pi - 1 \approx 2.14 \\ x = \pi \approx 3.14 &: y = 2\pi - 1 \approx 5.28 \\ x = \frac{3\pi}{2} \approx 4.71 &: y = 3\pi - 1 \approx 8.42 \\ x = 2\pi \approx 6.28 &: y = 4\pi - 1 \approx 11.57 \end{aligned}

Compare with the curve $y = 4\cos 2x + 3$ at these points:

\begin{aligned} x = 0 &: \text{curve} = 7, \; \text{line} = -1 \quad \Rightarrow \; \text{curve} > \text{line} \\ x = \frac{\pi}{2} &: \text{curve} = -1, \; \text{line} \approx 2.14 \quad \Rightarrow \; \text{line} > \text{curve} \\ x = \pi &: \text{curve} = 7, \; \text{line} \approx 5.28 \quad \Rightarrow \; \text{curve} > \text{line} \\ x = \frac{3\pi}{2} &: \text{curve} = -1, \; \text{line} \approx 8.42 \quad \Rightarrow \; \text{line} > \text{curve} \\ x = 2\pi &: \text{curve} = 7, \; \text{line} \approx 11.57 \quad \Rightarrow \; \text{line} > \text{curve} \end{aligned}

The curve and line change relative position between consecutive key points:

Between $x = 0$ and $x = \frac{\pi}{2}$ : curve goes from above to below the line → 1 intersection.
Between $x = \frac{\pi}{2}$ and $x = \pi$ : line goes from above to below the curve → 1 intersection.
Between $x = \pi$ and $x = \frac{3\pi}{2}$ : curve goes from above to below the line → 1 intersection.
Between $x = \frac{3\pi}{2}$ and $x = 2\pi$ : curve ( $-1$ to $7$ ) is always below the line ( $8.42$ to $11.57$ ) since the curve's maximum $7 < 8.42$ → no intersection.

Answer

The number of solutions is $3$ .

Final answer

3

Detailed explanation

Walkthrough

The equation $4\cos 2x + 3 = 2x - 1$ can be interpreted as finding the $x$ -values where the curve $y = 4\cos 2x + 3$ (sketched in part b) intersects the straight line $y = 2x - 1$ .

The line $y = 2x - 1$ has gradient $2$ and passes through $(0, -1)$ . At $x = 2\pi \approx 6.28$ , the line reaches $y = 4\pi - 1 \approx 11.57$ , which is well above the curve's maximum of $7$ .

By evaluating both functions at the key points $x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ :

At $x = 0$ : curve is at $7$ , line is at $-1$ . Curve is above.
At $x = \frac{\pi}{2}$ : curve is at $-1$ , line is at $\pi - 1 \approx 2.14$ . Line is above. → They must cross once in $(0, \frac{\pi}{2})$ .
At $x = \pi$ : curve is at $7$ , line is at $2\pi - 1 \approx 5.28$ . Curve is above. → They must cross once in $(\frac{\pi}{2}, \pi)$ .
At $x = \frac{3\pi}{2}$ : curve is at $-1$ , line is at $3\pi - 1 \approx 8.42$ . Line is above. → They must cross once in $(\pi, \frac{3\pi}{2})$ .
At $x = 2\pi$ : curve is at $7$ , line is at $4\pi - 1 \approx 11.57$ . Line is above. In the interval $(\frac{3\pi}{2}, 2\pi)$ , the curve rises from $-1$ to $7$ while the line rises from $8.42$ to $11.57$ . Since the curve's maximum value $7$ is less than the line's minimum value in this interval ( $8.42$ ), they never cross. → No intersection.

Total intersections: $3$ .

Key Takeaways

Equations of the form $f(x) = g(x)$ can be solved graphically by finding intersections of $y = f(x)$ and $y = g(x)$ .
When counting intersections, evaluate both functions at key points and use the intermediate value theorem to detect sign changes in $f(x) - g(x)$ .
Always check whether the functions can actually cross in an interval by comparing their ranges.

Common Mistakes

Assuming there is an intersection in every interval where the relative positions change, without verifying that the functions actually cross (e.g., in $(\frac{3\pi}{2}, 2\pi)$ where the curve never reaches the line).
Miscounting by not checking all intervals carefully.
Forgetting that the line continues to rise and may exceed the curve's maximum value.

Things to Be Careful About

The word 'hence' means you should use the sketch from part (b); do not solve the equation algebraically.
Be careful in the last interval $(\frac{3\pi}{2}, 2\pi)$ : even though the curve is rising, its maximum ( $7$ ) is below the line's minimum value in that interval ( $\approx 8.42$ ), so there is no intersection.
Only count intersections within the domain $0 \leq x \leq 2\pi$ .

Techniques used

recognise equation as intersection of two graphsanalyse line y = 2x - 1 against curve rangecount intersections using sketch

Q66MMediumIntegration

Fig. 3

The diagram shows the curve with equation $y = \frac{9}{(5x+4)^{\frac{1}{2}}}$ and the line $y = 6 - 3x$ . The line and the curve intersect at the point $P$ which has $y$ -coordinate 3.

Find the area of the shaded region.

DifficultyMedium

Worked solution

Approach

The shaded region can be split into two parts:

The area under the curve $y = \dfrac{9}{(5x+4)^{1/2}}$ from $x = 0$ to $x = 1$ (the $x$ -coordinate of $P$ ).
The triangular region between the line $y = 6 - 3x$ and the $x$ -axis from $x = 1$ to $x = 2$ (where the line meets the $x$ -axis).

We integrate the curve between $x = 0$ and $x = 1$ , then add the triangle area.

Working

Step 1: Find the $x$ -coordinate of $P$ .

$P$ lies on the line with $y = 3$ :

6 - 3x = 3 \Rightarrow 3x = 3 \Rightarrow x = 1

So $P = (1, 3)$ .

Step 2: Integrate the curve.

\int \frac{9}{(5x+4)^{1/2}}\, dx = 9 \cdot \frac{(5x+4)^{1/2}}{\frac{1}{2} \cdot 5} = \frac{18}{5}(5x+4)^{1/2}

Step 3: Evaluate the definite integral from $0$ to $1$ .

\int_0^1 \frac{9}{(5x+4)^{1/2}}\, dx = \left[\frac{18}{5}(5x+4)^{1/2}\right]_0^1

At $x = 1$ : $\frac{18}{5}(5+4)^{1/2} = \frac{18}{5} \cdot 3 = \frac{54}{5}$ .

At $x = 0$ : $\frac{18}{5}(0+4)^{1/2} = \frac{18}{5} \cdot 2 = \frac{36}{5}$ .

So the area under the curve is:

\frac{54}{5} - \frac{36}{5} = \frac{18}{5}

Step 4: Area of the triangle.

The line $y = 6 - 3x$ meets the $x$ -axis when $6 - 3x = 0$ , i.e. at $x = 2$ .

The triangle has vertices at $P(1, 3)$ , $(1, 0)$ and $(2, 0)$ , so its base is $1$ (along the $x$ -axis) and its height is $3$ :

\text{Triangle area} = \frac{1}{2} \times 1 \times 3 = \frac{3}{2}

Step 5: Total shaded area.

\text{Total area} = \frac{18}{5} + \frac{3}{2} = \frac{36}{10} + \frac{15}{10} = \frac{51}{10}

Answer

\frac{51}{10}

Final answer

51/10

Detailed explanation

Walkthrough

The shaded region in the figure is bounded on the left by the $y$ -axis, above partly by the curve and partly by the line, and below by the $x$ -axis. A clean way to compute its area is to split it at the vertical line $x = 1$ (which passes through $P$ ):

From $x = 0$ to $x = 1$ : the upper boundary is the curve. We compute $\int_0^1 \frac{9}{(5x+4)^{1/2}}\, dx$ .
From $x = 1$ to $x = 2$ : the upper boundary is the straight line $y = 6 - 3x$ (it touches the $x$ -axis at $x = 2$ ). The region between the line and the $x$ -axis here is a right-angled triangle with vertices $P(1,3)$ , $(1,0)$ and $(2,0)$ .

Finding $P$ . Since $P$ has $y$ -coordinate $3$ and lies on the line, $6 - 3x = 3$ gives $x = 1$ , so $P = (1, 3)$ . This tells us the upper limit of the curve's definite integral.

Integrating the curve. Rewrite $9(5x+4)^{-1/2}$ as a function of the form $(5x+4)^n$ with $n = -\tfrac{1}{2}$ . Using $\int (ax+b)^n\, dx = \frac{(ax+b)^{n+1}}{a(n+1)}$ for $n \neq -1$ :

\int 9(5x+4)^{-1/2}\, dx = 9 \cdot \frac{(5x+4)^{1/2}}{5 \cdot \frac{1}{2}} = \frac{18}{5}(5x+4)^{1/2}

At the upper limit $x = 1$ : $\frac{18}{5}\sqrt{9} = \frac{18}{5} \cdot 3 = \frac{54}{5}$ .

At the lower limit $x = 0$ : $\frac{18}{5}\sqrt{4} = \frac{18}{5} \cdot 2 = \frac{36}{5}$ .

Subtracting gives the area under the curve as $\frac{54-36}{5} = \frac{18}{5}$ .

Triangle area. The line $y = 6 - 3x$ hits the $x$ -axis when $6 - 3x = 0$ , i.e. $x = 2$ . The triangle has a horizontal base from $(1, 0)$ to $(2, 0)$ of length $1$ , and a vertical height equal to the $y$ -value of $P$ , which is $3$ . So its area is $\tfrac{1}{2} \times 1 \times 3 = \tfrac{3}{2}$ .

Combining. Total area $= \frac{18}{5} + \frac{3}{2}$ . Using a common denominator of $10$ : $\frac{36}{10} + \frac{15}{10} = \frac{51}{10}$ .

Key Takeaways

When a bounded region is partly under a curve and partly triangular, split it at a convenient vertical line and add the pieces.
The standard formula $\int (ax+b)^n\, dx = \frac{(ax+b)^{n+1}}{a(n+1)}$ (for $n \neq -1$ ) handles fractional powers like $(5x+4)^{-1/2}$ cleanly.
Always check where any straight-line boundary meets the $x$ -axis so you can identify triangles correctly.
For areas described by a curve from $x = 0$ to a vertical line, $\int$ gives the area between the curve and the $x$ -axis directly.

Common Mistakes

Forgetting to multiply the chain-rule constant: integrating $(5x+4)^{1/2}$ gives $\frac{2}{5}(5x+4)^{3/2}$ , not $\frac{1}{5}(5x+4)^{3/2}$ . The factor of $\frac{1}{5}$ from the derivative of $5x+4$ must be divided into the result.
Using the wrong $y$ -coordinate to find the $x$ -coordinate of $P$ : the question states the $y$ -coordinate is $3$ , so substitute $3$ , not $9$ , into the line equation.
Treating the entire shaded region as a single integral from $0$ to $2$ : the upper boundary changes at $x = 1$ (curve to line), so a single integral would not give the correct area.
Forgetting the $\frac{1}{2}$ in the triangle area formula.

Things to Be Careful About

The integration is between $x = 0$ and $x = 1$ , not between $x = 0$ and the $x$ -intercept of the curve (the curve is only used to $x = 1$ here because $P$ is at $x = 1$ ).
The triangle is bounded by $x = 1$ and $x = 2$ , so its base is $1$ , not $2$ . Using $2$ would give a triangle twice as large.
The answer is $\frac{51}{10}$ as an exact fraction — leave it in this form unless the question requests a decimal.
Make sure the limits are substituted in the correct order: upper limit minus lower limit, not the other way around.

Techniques used

solve linear equation for intersection x-coordinateintegrate a function of the form (ax+b)^(-1/2)evaluate a definite integral between limitsidentify a triangle and compute its area using base and heightcombine separate area contributions

Similar questions

Q7MediumTrigonometry

(a)

Prove the identity

\frac{\tan \theta + 7}{\tan^2 \theta - 3} \equiv \frac{\sin \theta \cos \theta + 7\cos^2 \theta}{1 - 4\cos^2 \theta}.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Start with the left side and rewrite $\tan\theta$ as $\frac{\sin\theta}{\cos\theta}$ . Combine the fractions, then use $\sin^2\theta + \cos^2\theta = 1$ to simplify the denominator.

Working

\frac{\tan\theta + 7}{\tan^2\theta - 3} = \frac{\frac{\sin\theta}{\cos\theta} + 7}{\frac{\sin^2\theta}{\cos^2\theta} - 3}

Combine the numerator and denominator:

= \frac{\frac{\sin\theta + 7\cos\theta}{\cos\theta}}{\frac{\sin^2\theta - 3\cos^2\theta}{\cos^2\theta}} = \frac{\cos\theta(\sin\theta + 7\cos\theta)}{\sin^2\theta - 3\cos^2\theta}

Simplify the numerator:

= \frac{\sin\theta\cos\theta + 7\cos^2\theta}{\sin^2\theta - 3\cos^2\theta}

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

\sin^2\theta - 3\cos^2\theta = 1 - \cos^2\theta - 3\cos^2\theta = 1 - 4\cos^2\theta

Therefore:

\frac{\tan\theta + 7}{\tan^2\theta - 3} \equiv \frac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta}

Answer

The identity is proven.

Final answer

The identity is proven.

Detailed explanation

Walkthrough

Start with the left-hand side and replace $\tan\theta$ with $\frac{\sin\theta}{\cos\theta}$ in both the numerator and denominator. This creates a compound fraction. Write the numerator as a single fraction and the denominator as a single fraction. Dividing by the denominator fraction is the same as multiplying by its reciprocal, so one factor of $\cos\theta$ cancels. The numerator becomes $\sin\theta\cos\theta + 7\cos^2\theta$ , and the denominator becomes $\sin^2\theta - 3\cos^2\theta$ . Finally, use $\sin^2\theta = 1 - \cos^2\theta$ to rewrite the denominator as $1 - 4\cos^2\theta$ , which matches the right-hand side.

Key Takeaways

This question tests the two fundamental trigonometric identities $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta = 1$ . It also tests the algebraic skill of simplifying compound fractions. Recognising when to substitute $\tan\theta$ and when to apply the Pythagorean identity is essential for proving trigonometric identities.

Common Mistakes

Forgetting to combine the numerator and denominator before simplifying, which can prevent the cancellation of $\cos\theta$ .
Incorrectly simplifying $\frac{\frac{a}{\cos\theta}}{\frac{b}{\cos^2\theta}}$ ; the result is $\frac{a\cos\theta}{b}$ , not $\frac{a}{b\cos\theta}$ .
Using the Pythagorean identity incorrectly, for example writing $\sin^2\theta = 1 + \cos^2\theta$ .
Working on both sides simultaneously without reaching a common correct expression; the mark scheme only allows full marks if one side is transformed into the other.

Things to Be Careful About

The proof can be done from left to right or right to left, but it is usually easier to start with the more complicated side.
Keep brackets when substituting, to avoid sign errors.
The mark scheme condones missing brackets if they are recovered, but it is safer to include them throughout.

Techniques used

express tanθ as sinθ/cosθcombine compound fractionsapply the Pythagorean identity

(b)

Hence solve the equation

\frac{\sin \theta \cos \theta + 7\cos^2 \theta}{1 - 4\cos^2 \theta} = \frac{5}{\tan \theta}

for $0^\circ \le \theta \le 180^\circ$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the identity from part (a) to replace the left-hand fraction. Then cross-multiply and solve the resulting quadratic in $\tan\theta$ . Finally, find the angles in $0^\circ \le \theta \le 180^\circ$ .

Working

From part (a),

\frac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta} = \frac{\tan\theta + 7}{\tan^2\theta - 3}

So the equation becomes:

\frac{\tan\theta + 7}{\tan^2\theta - 3} = \frac{5}{\tan\theta}

Cross-multiply, noting that $\tan\theta \neq 0$ and $\tan^2\theta \neq 3$ :

\tan\theta(\tan\theta + 7) = 5(\tan^2\theta - 3)

Expand and rearrange:

\tan^2\theta + 7\tan\theta = 5\tan^2\theta - 15

4\tan^2\theta - 7\tan\theta - 15 = 0

Factorise:

(4\tan\theta + 5)(\tan\theta - 3) = 0

Thus

\tan\theta = -\frac{5}{4} \quad \text{or} \quad \tan\theta = 3

For $0^\circ \le \theta \le 180^\circ$ , $\tan\theta = 3$ gives an angle in the first quadrant:

\theta = \tan^{-1}(3) \approx 71.6^\circ

For $\tan\theta = -\frac{5}{4}$ , the angle is in the second quadrant:

\theta = 180^\circ - \tan^{-1}\left(\frac{5}{4}\right) \approx 128.7^\circ

Answer

\theta = 71.6^\circ \quad \text{and} \quad 128.7^\circ

Final answer

θ = 71.6° and θ = 128.7°

Detailed explanation

Walkthrough

The key is to use the identity from part (a) to replace the left-hand side by $\frac{\tan\theta + 7}{\tan^2\theta - 3}$ . The equation then becomes a rational equation in $\tan\theta$ . Cross-multiply by $\tan\theta$ and by $\tan^2\theta - 3$ ; this is valid for the solutions because the denominators are not zero there. Expanding gives a quadratic in $\tan\theta$ . Factorise it to get $\tan\theta = -\frac{5}{4}$ or $\tan\theta = 3$ . Then solve each equation over $0^\circ \le \theta \le 180^\circ$ . Since tangent is positive in the first quadrant, $\tan\theta = 3$ gives $71.6^\circ$ . Since tangent is negative in the second quadrant, $\tan\theta = -\frac{5}{4}$ gives $180^\circ - 51.3^\circ = 128.7^\circ$ .

Key Takeaways

This question combines a trigonometric identity with solving a quadratic. A trigonometric equation can be reduced to an algebraic equation by letting $x = \tan\theta$ . It also reinforces the sign convention: tangent is positive in the first quadrant and negative in the second, so a negative tangent value in $0^\circ$ to $180^\circ$ gives a second-quadrant angle.

Common Mistakes

Forgetting to use part (a) and trying to solve the original complicated fraction directly.
Cross-multiplying incorrectly, for example writing $\tan\theta(\tan\theta + 7) = 5(\tan^2\theta - 3)$ incorrectly.
Making sign errors when expanding the equation to form the quadratic.
Factorising the quadratic incorrectly. The mark scheme condones errors made in forming the three-term quadratic, but not errors in solving it.
Giving only the first-quadrant solution and missing the second-quadrant solution for the negative $\tan\theta$ .
Including solutions outside the range $0^\circ \le \theta \le 180^\circ$ .

Things to Be Careful About

$\tan\theta = 0$ would make $\frac{5}{\tan\theta}$ undefined, so $\theta = 0^\circ$ and $\theta = 180^\circ$ are not solutions.
The interval is $0^\circ \le \theta \le 180^\circ$ , so only the first and second quadrants are used.
The mark scheme accepts answers within $1^\circ$ of the correct values, so give $71.6^\circ$ and $128.7^\circ$ .
Extra answers outside the range are ignored, but they should not be included in the final answer.

Techniques used

replace a fraction using a proven identitycross-multiply to form a quadratic in tanθfactorise and solve a quadratic in tanθsolve tanθ equations over a fixed interval

Q8MediumCoordinate GeometryTrigonometry

Fig. 4

The diagram shows the circle with equation $x^2 + y^2 - 14x + 8y + 36 = 0$ and the line $y = -2$ . The line intersects the circle at the points $A$ and $B$ . The centre of the circle is $C$ .

(a)

Find the coordinates of $A$ , $B$ and $C$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Complete the square for the circle equation to find the centre $C$ and radius. Then substitute $y = -2$ into the circle equation to find the $x$ -coordinates of the intersection points $A$ and $B$ .

Working

Finding the centre $C$ and radius:

\begin{aligned} x^2 + y^2 - 14x + 8y + 36 &= 0 \\ (x^2 - 14x) + (y^2 + 8y) + 36 &= 0 \\ (x - 7)^2 - 49 + (y + 4)^2 - 16 + 36 &= 0 \\ (x - 7)^2 + (y + 4)^2 &= 29 \end{aligned}

The centre is $C = (7, -4)$ and the radius is $r = \sqrt{29}$ .

Finding points $A$ and $B$ :

Substitute $y = -2$ into the original circle equation:

x^2 + (-2)^2 - 14x + 8(-2) + 36 = 0

x^2 + 4 - 14x - 16 + 36 = 0

x^2 - 14x + 24 = 0

Factorise the quadratic:

(x - 2)(x - 12) = 0

So $x = 2$ or $x = 12$ . The points are $A = (2, -2)$ and $B = (12, -2)$ .

Answer

A = (2, -2), \quad B = (12, -2), \quad C = (7, -4)

Final answer

A = (2, -2), B = (12, -2), C = (7, -4)

Detailed explanation

Walkthrough

First, we rewrite the circle equation in standard form by completing the square for both $x$ and $y$ . This gives us the centre $C = (7, -4)$ and radius $r = \sqrt{29}$ . Next, to find where the line $y = -2$ intersects the circle, we substitute $y = -2$ into the original equation. This produces a quadratic in $x$ , which we solve by factorising to get $x = 2$ and $x = 12$ . These are the $x$ -coordinates of $A$ and $B$ .

Key Takeaways

Completing the square is essential for extracting the centre and radius from a general circle equation. Substituting a line equation into a circle equation is the standard algebraic method for finding intersection points.

Common Mistakes

Forgetting to add the constant from completing the square to the right-hand side (e.g., writing $(x-7)^2 + (y+4)^2 = 0$ instead of $29$ ).
Sign errors when completing the square, such as writing $(x+7)^2$ instead of $(x-7)^2$ .
Forgetting that both points have $y = -2$ .

Things to Be Careful About

The mark scheme concedes $x = 2$ and $y = -2$ if seen together without parentheses, but full coordinate notation is preferred. Ensure the quadratic is correctly simplified before factorising.

Techniques used

complete the square to find centre and radiussubstitute line equation into circle equationsolve resulting quadratic

(b)

Find the angle $ACB$ in radians. Give your answer correct to 3 significant figures.

2M Similar

DifficultyMedium

Worked solution

Approach

Use the coordinates from part (a) to find the lengths of the sides of $\triangle ACB$ . Then use either the cosine rule or a half-angle right triangle to find $\angle ACB$ .

Working

Side lengths:

\begin{aligned} AC &= \sqrt{(7 - 2)^2 + (-4 - (-2))^2} = \sqrt{25 + 4} = \sqrt{29} \\ BC &= \sqrt{(12 - 7)^2 + (-2 - (-4))^2} = \sqrt{25 + 4} = \sqrt{29} \\ AB &= 12 - 2 = 10 \end{aligned}

Finding $\angle ACB$ :

Let $\theta = \angle ACB$ . Using the cosine rule in $\triangle ACB$ :

AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos\theta

10^2 = 29 + 29 - 2(\sqrt{29})(\sqrt{29})\cos\theta

100 = 58 - 58\cos\theta

\cos\theta = \frac{58 - 100}{58} = -\frac{42}{58} = -\frac{21}{29}

\theta = \cos^{-1}\left(-\frac{21}{29}\right) \approx 2.38 \text{ radians}

Alternative method using half-angle:

The perpendicular from $C(7, -4)$ to $AB$ (which lies on $y = -2$ ) meets $AB$ at the midpoint $(7, -2)$ . This splits $\triangle ACB$ into two right-angled triangles with opposite side $5$ and adjacent side $2$ .

\tan\left(\frac{\theta}{2}\right) = \frac{5}{2} \implies \theta = 2\tan^{-1}\left(\frac{5}{2}\right) \approx 2.38 \text{ radians}

Answer

\angle ACB = 2.38 \text{ radians}

Final answer

2.38

Detailed explanation

Walkthrough

We first calculate the lengths of $AC$ , $BC$ , and $AB$ using the distance formula. Since $AC = BC = \sqrt{29}$ , the triangle is isosceles. We can apply the cosine rule with $AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos\theta$ to solve for $\cos\theta$ , giving $\cos\theta = -21/29$ . Taking the inverse cosine yields $\theta \approx 2.38$ radians. Alternatively, dropping a perpendicular from $C$ to $AB$ creates a right triangle with legs $5$ and $2$ , allowing us to use $\tan(\theta/2) = 5/2$ .

Key Takeaways

The cosine rule is a powerful tool for finding angles in triangles when all three side lengths are known. Recognising an isosceles triangle and using a half-angle right triangle can simplify the calculation.

Common Mistakes

Using $r = 29$ instead of $r = \sqrt{29}$ in the cosine rule (the mark scheme awards 0 marks for this).
Forgetting to double the half-angle if using the right-triangle method.
Giving the answer in degrees ( $136.4^\circ$ ) when radians are requested; this scores a maximum of 1/2.

Things to Be Careful About

The question asks for the answer in radians to 3 significant figures. Ensure your calculator is in radian mode. The mark scheme explicitly states to ignore the degree symbol if present, but the final numerical value must be correct for radians.

Techniques used

calculate lengths of triangle sides using distance formulaapply cosine rule or half-angle tangentevaluate inverse trigonometric function

(c)

The chord $AB$ divides the circle into two segments.

Find the area of the larger segment.

4M Similar

DifficultyMedium

Worked solution

Approach

The chord $AB$ divides the circle into two segments. The larger segment corresponds to the reflex angle at the centre. Its area is the area of the larger sector plus the area of $\triangle ACB$ . Alternatively, find the area of the smaller segment and subtract from the total circle area.

Working

Area of $\triangle ACB$ :

Base $AB = 10$ . The perpendicular height from $C(7, -4)$ to the line $y = -2$ is $|-4 - (-2)| = 2$ .

\text{Area of } \triangle ACB = \frac{1}{2} \times 10 \times 2 = 10

Area of the larger sector:

The reflex angle is $2\pi - \theta = 2\pi - 2.38 \approx 3.903$ radians. The radius is $r = \sqrt{29}$ .

\text{Area of larger sector} = \frac{1}{2} r^2 (2\pi - \theta) = \frac{1}{2}(29)(2\pi - 2.38) \approx 56.587

Area of the larger segment:

\text{Area} = \text{Area of larger sector} + \text{Area of } \triangle ACB = 56.587 + 10 = 66.587

Rounding to 3 significant figures:

\text{Area} = 66.6

Alternative method:

\text{Area of circle} = \pi r^2 = 29\pi \approx 91.106

\text{Area of smaller sector} = \frac{1}{2}(29)(2.38) \approx 34.518

\text{Area of smaller segment} = 34.518 - 10 = 24.518

\text{Area of larger segment} = 91.106 - 24.518 = 66.588 \approx 66.6

Answer

66.6

Final answer

66.6

Detailed explanation

Walkthrough

The chord $AB$ splits the circle into a smaller segment (containing the minor arc) and a larger segment (containing the major arc). The area of the larger segment is the area of the major sector (the sector with the reflex angle $2\pi - \theta$ ) plus the area of $\triangle ACB$ . We already know $\theta \approx 2.38$ rad and $r = \sqrt{29}$ . The triangle area is easily found using base $AB = 10$ and height $2$ . Adding the sector area $\frac{1}{2}r^2(2\pi - \theta)$ and the triangle area $10$ gives the final answer.

Key Takeaways

The area of a circular segment is found by combining sector and triangle areas. For the larger segment, you add the major sector area to the triangle area, or subtract the smaller segment area from the total circle area.

Common Mistakes

Using the angle $\theta$ for the larger sector instead of the reflex angle $2\pi - \theta$ .
Forgetting to add the triangle area when calculating the larger segment (this gives the sector area, not the segment area).
Awarding marks for $\sqrt{29}$ sighted in earlier parts; it must be explicitly seen in part (c) to earn the mark.

Things to Be Careful About

The mark scheme requires the final answer to be given to an appropriate number of significant figures (3 s.f. gives $66.6$ ). Ensure you use the exact value of $\theta$ or at least 4 significant figures in intermediate steps to avoid rounding errors.

Techniques used

calculate sector area using angle and radiuscalculate triangle area using base and heightsubtract or add areas to find segment area

Q9MediumIntegrationDifferentiationCoordinate Geometry

The equation of a curve is such that $\frac{d^2y}{dx^2} = -\frac{24}{x^3}$ . It is given that the curve has a stationary point at $(-2, 19)$ .

(a)

Find an expression for $\frac{dy}{dx}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

We integrate $\frac{d^2y}{dx^2}$ with respect to $x$ to obtain $\frac{dy}{dx}$ , including a constant of integration. Since the curve has a stationary point at $(-2,19)$ , the gradient $\frac{dy}{dx}$ is $0$ when $x=-2$ ; this determines the constant.

Working

Integrate:

\frac{dy}{dx} = \int -\frac{24}{x^3}\,dx = \int -24x^{-3}\,dx = \frac{-24x^{-2}}{-2} + c = \frac{12}{x^2} + c

At the stationary point $x=-2$ , $\frac{dy}{dx}=0$ :

0 = \frac{12}{(-2)^2} + c = \frac{12}{4} + c = 3 + c

So $c=-3$ . Therefore:

\frac{dy}{dx} = \frac{12}{x^2} - 3

Answer

\frac{dy}{dx} = \frac{12}{x^2} - 3

Final answer

dy/dx = 12/x^2 - 3

Detailed explanation

Walkthrough

We are given the second derivative $\frac{d^2y}{dx^2}=-\frac{24}{x^3}$ . To go from a second derivative to a first derivative, integrate once. Rewriting $-\frac{24}{x^3}$ as $-24x^{-3}$ makes the power rule for integration clear: increase the power by 1 and divide by the new power, giving $12x^{-2}+c$ . The constant $c$ is unknown, so we use the fact that the curve has a stationary point at $(-2,19)$ . At a stationary point the gradient is zero, so substituting $x=-2$ into $\frac{dy}{dx}$ and setting it equal to 0 gives $c=-3$ .

Key Takeaways

This question tests integration as reverse differentiation and the use of a boundary/initial condition to find a constant of integration. It also reinforces that stationary points have zero gradient.

Common Mistakes

Forgetting the constant of integration $c$ .
Making a sign error when integrating a negative power: $-24x^{-3}$ integrates to $12x^{-2}$ , not $-12x^{-2}$ .
Substituting $y=19$ instead of using the gradient condition $\frac{dy}{dx}=0$ at $x=-2$ .

Things to Be Careful About

The stationary point gives $\frac{dy}{dx}=0$ at $x=-2$ , not $y=0$ . Simplify $\frac{-24}{-2}$ to $12$ before writing the final answer.

Techniques used

integrate a power of xapply stationary point conditiondetermine constant of integration

(b)

Find the $x$ -coordinate of the other stationary point of the curve, and determine the nature of this stationary point.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Stationary points occur where $\frac{dy}{dx}=0$ . We already know that $x=-2$ gives one stationary point; solve $\frac{dy}{dx}=0$ to find the other. Then use $\frac{d^2y}{dx^2}$ at that point to decide whether it is a maximum or minimum.

Working

Set $\frac{dy}{dx}=0$ :

\frac{12}{x^2} - 3 = 0

\frac{12}{x^2} = 3

x^2 = 4

x = \pm 2

One stationary point is at $x=-2$ , so the other is $x=2$ .

At $x=2$ :

\frac{d^2y}{dx^2} = -\frac{24}{2^3} = -\frac{24}{8} = -3

Since $\frac{d^2y}{dx^2}<0$ , the stationary point at $x=2$ is a maximum.

Answer

$x=2$ ; maximum

Final answer

x = 2; maximum

Detailed explanation

Walkthrough

Stationary points are points where the gradient is zero. We set $\frac{dy}{dx}=0$ using the expression from part (a). Solving $\frac{12}{x^2}-3=0$ gives $x^2=4$ , so $x=\pm 2$ . Since the question says there is already a stationary point at $x=-2$ , the other stationary point is $x=2$ . To determine its nature, substitute $x=2$ into the given second derivative $\frac{d^2y}{dx^2}=-\frac{24}{x^3}$ . The value is $-3$ , which is negative, so the stationary point is a maximum.

Key Takeaways

A stationary point satisfies $\frac{dy}{dx}=0$ . The sign of $\frac{d^2y}{dx^2}$ determines whether it is a maximum (negative) or minimum (positive).

Common Mistakes

Forgetting that $x=-2$ is already known and giving only $x=\pm 2$ as the answer.
Using the second derivative at $x=-2$ instead of at $x=2$ .
Misreading the sign of $\frac{d^2y}{dx^2}$ and calling the point a minimum.

Things to Be Careful About

The question asks for the other stationary point, so the answer must be $x=2$ , not $x=\pm 2$ . The second derivative is negative at $x=2$ , so it is a maximum. The mark scheme allows other valid methods for determining nature, but the second-derivative test is the most direct.

Techniques used

set derivative equal to zerosolve for xuse second derivative to classify stationary point

(c)

Find the equation of the curve.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Integrate $\frac{dy}{dx}$ to obtain $y$ , introducing a new constant of integration $d$ . Substitute the known point $(-2,19)$ to find $d$ .

Working

Integrate:

y = \int \left(\frac{12}{x^2} - 3\right)\,dx = \int \left(12x^{-2} - 3\right)\,dx = -12x^{-1} - 3x + d = -\frac{12}{x} - 3x + d

Substitute $x=-2$ and $y=19$ :

19 = -\frac{12}{-2} - 3(-2) + d = 6 + 6 + d = 12 + d

d = 7

Therefore:

y = -\frac{12}{x} - 3x + 7

Answer

y = -\frac{12}{x} - 3x + 7

Final answer

y = -12/x - 3x + 7

Detailed explanation

Walkthrough

We now integrate the first derivative to recover $y$ . The expression $\frac{12}{x^2}-3$ is rewritten as $12x^{-2}-3$ , which integrates to $-12x^{-1}-3x+d$ , or $-\frac{12}{x}-3x+d$ . The new constant $d$ is found by substituting the given point $(-2,19)$ into this expression. This gives $19=12+d$ , so $d=7$ . The final curve equation is $y=-\frac{12}{x}-3x+7$ .

Key Takeaways

Each integration introduces a new constant. A point on the curve provides the condition needed to determine that constant. The final equation of a curve is obtained by integrating the gradient and using given coordinates.

Common Mistakes

Forgetting the second constant of integration $d$ .
Using the stationary point condition $\frac{dy}{dx}=0$ again instead of substituting the point into $y$ .
Arithmetic errors when evaluating $-\frac{12}{-2}-3(-2)$ .

Things to Be Careful About

Use the point $(-2,19)$ in the expression for $y$ , not in the derivative. Evaluate carefully: $-\frac{12}{-2}=6$ and $-3(-2)=6$ , so the sum is $12$ . The final answer must include the constant $7$ .

Techniques used

integrate first derivativesubstitute given pointdetermine constant of integration

(d)

Find the equation of the normal to the curve at the point where $\frac{dy}{dx} = -\frac{9}{4}$ and $x$ is positive. Express your answer in the form $px + qy + r = 0$ , where $p$ , $q$ and $r$ are integers.

4M Similar

DifficultyMedium

Worked solution

Approach

Set the derivative equal to $-\frac{9}{4}$ and solve for the positive value of $x$ . Substitute this into the curve equation to find $y$ . The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of $-\frac{9}{4}$ . Use the point-gradient form and rearrange into $px+qy+r=0$ .

Working

Set $\frac{dy}{dx}=-\frac{9}{4}$ :

\frac{12}{x^2} - 3 = -\frac{9}{4}

\frac{12}{x^2} = \frac{3}{4}

x^2 = 16

Since $x$ is positive, $x=4$ .

Find the corresponding $y$ -coordinate:

y = -\frac{12}{4} - 3(4) + 7 = -3 - 12 + 7 = -8

The tangent gradient is $-\frac{9}{4}$ , so the normal gradient is:

m = \frac{4}{9}

Equation of the normal through $(4,-8)$ :

\frac{y+8}{x-4} = \frac{4}{9}

9(y+8) = 4(x-4)

9y + 72 = 4x - 16

4x - 9y - 88 = 0

Answer

4x - 9y - 88 = 0

Final answer

4x - 9y - 88 = 0

Detailed explanation

Walkthrough

To find the required point, set the derivative equal to $-\frac{9}{4}$ . Solving $\frac{12}{x^2}-3=-\frac{9}{4}$ gives $\frac{12}{x^2}=\frac{3}{4}$ , so $x^2=16$ . Since the question specifies $x$ is positive, $x=4$ . Substitute into the curve equation to get $y=-8$ . The gradient of the tangent at this point is $-\frac{9}{4}$ , so the gradient of the normal is the negative reciprocal, $\frac{4}{9}$ . Using the point-gradient form with $(4,-8)$ gives $\frac{y+8}{x-4}=\frac{4}{9}$ . Cross-multiply and rearrange: $9y+72=4x-16$ , so $4x-9y-88=0$ .

Key Takeaways

To find a normal, use the perpendicular gradient relationship: if the tangent gradient is $m$ , the normal gradient is $-\frac{1}{m}$ . The equation of a line can be written from a point and gradient, then rearranged into the required form $px+qy+r=0$ .

Common Mistakes

Using $x=-4$ as well as $x=4$ ; the question specifies $x$ is positive.
Taking the normal gradient as $-\frac{9}{4}$ instead of $\frac{4}{9}$ .
Making a sign error when rearranging to $4x-9y-88=0$ .

Things to Be Careful About

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal. When rearranging, keep all terms on one side to match $px+qy+r=0$ . The mark scheme accepts other equivalent forms, but the requested integer form is $4x-9y-88=0$ .

Techniques used

solve derivative equation for xfind corresponding y-coordinateuse perpendicular gradient for normalwrite equation of line in required form

Q10Medium-HardSeriesQuadratics

(a)

The first, second and third terms of an arithmetic progression are $4k$ , $k^2$ and $8k$ respectively, where $k$ is a non-zero constant.

5M

(i)

Find the value of $k$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the fact that consecutive terms of an arithmetic progression have a common difference. Equate the difference between the second and first terms to the difference between the third and second terms, then solve the resulting quadratic equation.

Working

For an arithmetic progression, the common difference is constant, so

k^2 - 4k = 8k - k^2

Rearrange:

2k^2 - 12k = 0

Factorise:

2k(k - 6) = 0

Hence $k = 0$ or $k = 6$ . Since $k$ is given as non-zero,

Answer

k = 6

Final answer

k = 6

Detailed explanation

Walkthrough

In an arithmetic progression, consecutive terms differ by the same amount. The first three terms are $4k$ , $k^2$ and $8k$ , so the difference between the second and first terms must equal the difference between the third and second terms:

k^2 - 4k = 8k - k^2

Add $k^2$ to both sides and subtract $8k$ from both sides:

2k^2 - 12k = 0

Factorise:

2k(k - 6) = 0

This gives $k = 0$ or $k = 6$ . The question states that $k$ is non-zero, so $k = 0$ is not allowed. Therefore $k = 6$ .

Key Takeaways

An arithmetic progression is identified by its constant common difference. When a quadratic arises from a context, check every solution against the conditions given in the question.

Common Mistakes

Equating the terms themselves instead of the differences between consecutive terms.
Making a sign error when rearranging $k^2 - 4k = 8k - k^2$ .
Forgetting that $k$ is stated to be non-zero and leaving $k = 0$ as the answer.

The mark scheme allows the extra solution $k = 0$ to be written down, but the final answer must be $k = 6$ .

Things to Be Careful About

The condition $k \neq 0$ is essential. Also, the method mark is awarded for forming an equation in $k$ only, so write the equality of the two common differences clearly.

Techniques used

equate common differences in an arithmetic progressionsolve a quadratic equation by factorisationdiscard a root using a given condition

(ii)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the value $k = 6$ to write the first term and common difference. Then substitute into the arithmetic series sum formula $S_n = \frac{n}{2}(2a + (n-1)d)$ with $n = 20$ .

Working

With $k = 6$ , the first term is

a = 4k = 24

The common difference is

d = k^2 - 4k = 36 - 24 = 12

Using the sum formula for $n = 20$ :

\begin{aligned} S_{20} &= \frac{20}{2}(2(24) + 19(12)) \\ &= 10(48 + 228) \\ &= 10(276) \\ &= 2760 \end{aligned}

Answer

S_{20} = 2760

Final answer

S20 = 2760

Detailed explanation

Walkthrough

From part (a)(i), $k = 6$ . The first term of the arithmetic progression is $a = 4k = 24$ . The common difference is the difference between consecutive terms, for example $d = k^2 - 4k = 36 - 24 = 12$ . The sum of the first $n$ terms of an arithmetic progression is

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

Put $n = 20$ , $a = 24$ and $d = 12$ :

S_{20} = \frac{20}{2}(2(24) + 19(12)) = 10(48 + 228) = 2760

Key Takeaways

The sum formula for an arithmetic progression needs the first term, the common difference and the number of terms. Here the first term is $4k$ , not $k$ , and the common difference can be found from the terms once $k$ is known.

Common Mistakes

Using $k = 0$ instead of $k = 6$ .
Using $n = 19$ in the factor outside the bracket instead of $n = 20$ .
Confusing the first term with $k$ rather than $4k$ .
Forgetting to multiply by $\frac{n}{2}$ after computing the bracket.

Things to Be Careful About

The formula uses $(n-1)d$ inside the bracket, so for $n = 20$ the multiplier of $d$ is $19$ , not $20$ . The method mark requires using their values of $a$ and $d$ in the correct sum formula.

Techniques used

determine the first term and common differenceapply the arithmetic series sum formula

(b)

The fourth and sixth terms of a geometric progression are 36 and 6 respectively. The common ratio of the progression is positive.

Find the sum to infinity of the progression. Give your answer in the form $\frac{a}{\sqrt{b}-c}$ , where $a$ , $b$ and $c$ are integers.

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Use the general term of a geometric progression, $u_n = ar^{n-1}$ , to write equations for the fourth and sixth terms. Divide them to eliminate $a$ and find $r$ , then find $a$ and apply the sum to infinity formula.

Working

Let the first term be $a$ and the common ratio be $r$ .

ar^3 = 36

ar^5 = 6

Dividing the second equation by the first:

r^2 = \frac{6}{36} = \frac{1}{6}

Since $r$ is positive,

r = \frac{1}{\sqrt{6}}

Now find $a$ :

a = \frac{36}{r^3} = \frac{36}{(\frac{1}{\sqrt{6}})^3} = 36(\sqrt{6})^3 = 216\sqrt{6}

Since $|r| < 1$ , the sum to infinity is

S_\infty = \frac{a}{1-r} = \frac{216\sqrt{6}}{1 - \frac{1}{\sqrt{6}}}

Rationalise the denominator:

S_\infty = \frac{216\sqrt{6}}{\frac{\sqrt{6}-1}{\sqrt{6}}} = \frac{216\sqrt{6} \cdot \sqrt{6}}{\sqrt{6}-1} = \frac{1296}{\sqrt{6}-1}

Answer

S_\infty = \frac{1296}{\sqrt{6}-1}

Final answer

S∞ = 1296/(√6 - 1)

Detailed explanation

Walkthrough

For a geometric progression with first term $a$ and common ratio $r$ , the $n$ th term is $u_n = ar^{n-1}$ . The fourth term is $ar^3 = 36$ and the sixth term is $ar^5 = 6$ .

Dividing the sixth-term equation by the fourth-term equation eliminates $a$ :

\frac{ar^5}{ar^3} = \frac{6}{36} \Rightarrow r^2 = \frac{1}{6}

Since the common ratio is positive, $r = \frac{1}{\sqrt{6}}$ . Now use the fourth term to find $a$ :

a = \frac{36}{r^3} = \frac{36}{(\frac{1}{\sqrt{6}})^3} = 36(\sqrt{6})^3 = 216\sqrt{6}

Because $|r| = \frac{1}{\sqrt{6}} < 1$ , the sum to infinity exists and is

S_\infty = \frac{a}{1-r} = \frac{216\sqrt{6}}{1 - \frac{1}{\sqrt{6}}}

To write this in the required form, multiply numerator and denominator by $\sqrt{6}$ :

S_\infty = \frac{216\sqrt{6} \cdot \sqrt{6}}{\sqrt{6}-1} = \frac{1296}{\sqrt{6}-1}

So $a = 1296$ , $b = 6$ and $c = 1$ .

Key Takeaways

A geometric progression is described by its first term and common ratio. Dividing two term equations is a quick way to eliminate the first term. The sum to infinity formula is only valid when $|r| < 1$ , and surd manipulation is often needed to present the answer in a required form.

Common Mistakes

Using the wrong index for the fourth term, such as $ar^4$ instead of $ar^3$ .
Taking $r = -\frac{1}{\sqrt{6}}$ even though the question says the common ratio is positive.
Stopping at $\frac{216\sqrt{6}}{1 - \frac{1}{\sqrt{6}}}$ without rationalising into the required form.
Forgetting to check that $|r| < 1$ before using the sum to infinity formula.

Things to Be Careful About

The answer must be in the form $\frac{a}{\sqrt{b}-c}$ with integers $a$ , $b$ and $c$ . Here $b$ is the number inside the square root, so $b = 6$ , not $36$ . The mark scheme condones using $\pm$ when first finding $r$ , but the positive value must be used because the common ratio is stated to be positive.

Techniques used

write equations for given GP termsdivide equations to find the common ratiorecover the first termapply the sum to infinity formularationalise the denominator into the required form

Q11Medium-HardQuadraticsFunctions

(a)

Express $x^2 + 4x + 2$ in the form $(x + a)^2 + b$ , where $a$ and $b$ are integers.

2M Similar

(b)

The functions $f$ and $g$ are defined as follows.

\begin{aligned} f(x) &= x^2 + 4x + 2 && \text{for } x \le -2 \\ g(x) &= -x - 4 && \text{for } x \ge -2 \end{aligned}

7M

(i)

Find an expression for $f^{-1}(x)$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Write $f(x)$ in its completed-square form, set $y = f(x)$ , solve for $x$ in terms of $y$ , then use the domain $x \le -2$ to choose the correct sign of the square root.

From part (a),

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

Working

Set $y = (x + 2)^2 - 2$ . Then

y + 2 = (x + 2)^2

Taking square roots,

x + 2 = \pm\sqrt{y + 2}

Because the domain of $f$ is $x \le -2$ , we have $x + 2 \le 0$ , so the negative root applies:

x + 2 = -\sqrt{y + 2}

x = -\sqrt{y + 2} - 2

Hence, replacing $y$ with $x$ ,

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

Answer

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

Final answer

f^-1(x) = -sqrt(x+2) - 2

Detailed explanation

Walkthrough

From part (a) we know $f(x) = (x+2)^2 - 2$ for $x \le -2$ . To invert, write $y = (x+2)^2 - 2$ , so $y+2 = (x+2)^2$ . Taking square roots of both sides gives two possibilities, $x+2 = +\sqrt{y+2}$ or $x+2 = -\sqrt{y+2}$ . The domain tells us $x \le -2$ , i.e. $x+2 \le 0$ , which forces the negative branch. Hence $x = -\sqrt{y+2} - 2$ . Finally we swap $y$ back to $x$ : $f^{-1}(x) = -\sqrt{x+2} - 2$ .

The mark scheme allows interchanging $x$ and $y$ at any stage, so you could equally start with $x = (y+2)^2 - 2$ and solve for $y$ .

Key Takeaways

An inverse function swaps the roles of the input and output. The domain of $f$ decides which square-root branch is valid; with $x \le -2$ we need the negative root.

Common Mistakes

Dropping the minus sign and writing $f^{-1}(x) = \sqrt{x+2} - 2$ — this ignores the domain restriction.
Giving the answer as $x = -\sqrt{y+2} - 2$ without replacing $y$ by $x$ ; the mark scheme does not condone $x = \dots$ as the final answer.

Things to Be Careful About

The final answer must be written as $f^{-1}(x) = \dots$ . Note also that the domain of $f^{-1}$ is $x \ge -2$ (so that $x+2 \ge 0$ ), which matches the range of $f$ .

Techniques used

write the function in completed-square formrearrange to isolate xapply the domain restriction to select the negative root

(ii)

Find an expression for $(gf)^{-1}(x)$ .

4M Similar

DifficultyMedium-Hard

Worked solution

Approach

Form the composite $gf(x) = g(f(x))$ , write it in completed-square form, then invert by setting $y = gf(x)$ , solving for $x$ and choosing the negative root from the domain $x \le -2$ .

Working

First form $gf$ :

\begin{aligned} gf(x) &= g(f(x)) \\ &= -(x^2 + 4x + 2) - 4 \\ &= -\left((x+2)^2 - 2\right) - 4 \\ &= -(x+2)^2 - 2 \end{aligned}

Now set $y = -(x+2)^2 - 2$ . Then

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

x + 2 = \pm\sqrt{-y - 2}

Since the relevant domain is $x \le -2$ , $x + 2 \le 0$ , so

x + 2 = -\sqrt{-y - 2}

x = -\sqrt{-y - 2} - 2

Replacing $y$ with $x$ ,

(gf)^{-1}(x) = -\sqrt{-x - 2} - 2

Answer

(gf)^{-1}(x) = -\sqrt{-x - 2} - 2

Final answer

(gf)^-1(x) = -sqrt(-x-2) - 2

Detailed explanation

Walkthrough

First compose: $gf(x) = g(f(x))$ . Since $g(x) = -x - 4$ , we feed the whole of $f(x)$ in: $gf(x) = -(x^2 + 4x + 2) - 4$ . Using the completed-square form, $x^2 + 4x + 2 = (x+2)^2 - 2$ , so

gf(x) = -\left((x+2)^2 - 2\right) - 4 = -(x+2)^2 - 2

Quick check by expanding: $-(x+2)^2 - 2 = -x^2 - 4x - 6$ , and $-(x^2 + 4x + 2) - 4 = -x^2 - 4x - 6$ . The two match.

Now invert. Set $y = -(x+2)^2 - 2$ . Then

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

so

x + 2 = \pm\sqrt{-y - 2}

The relevant domain is still $x \le -2$ (inherited from $f$ ), so $x+2 \le 0$ and we take the negative root: $x = -\sqrt{-y-2} - 2$ . Swapping $y$ back to $x$ :

(gf)^{-1}(x) = -\sqrt{-x-2} - 2

An alternative is to use the rule $(gf)^{-1} = f^{-1}g^{-1}$ . Here $g^{-1}(x) = -x - 4$ , so $f^{-1}g^{-1}(x) = f^{-1}(-x-4) = -\sqrt{(-x-4)+2} - 2 = -\sqrt{-x-2} - 2$ , the same result.

Key Takeaways

To invert a composite you can either form the composite first and then invert, or invert in reverse order. Both routes need the completed-square form and a correct choice of root sign.

Common Mistakes

Using $fg(x)$ instead of $gf(x)$ — the mark scheme awards zero for $fg(x)$ , so the order of composition is crucial.
Sign errors when simplifying: $gf(x) = -(x^2+4x+2) - 4 = -(x+2)^2 - 2$ , with the minus distributed correctly through the brackets.
Picking the positive root, which ignores the domain $x \le -2$ .

Things to Be Careful About

The final answer must be written as $(gf)^{-1}(x) = \dots$ , not $x = \dots$ . The DM1 step in the mark scheme requires a completed-square form containing a $-(x+k)^2$ term; an invalid intermediate such as $x = -\sqrt{y+2} - 2$ earns no further marks. Note also that the domain of $(gf)^{-1}$ is $x \le -2$ .

Techniques used

compose the two functionswrite the composite in completed-square formrearrange to isolate xapply the domain restriction to select the negative root

Mathematics 9709/12 — May/June 2025

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