(a) Express $3y^2 - 12y - 15$ in the form $3(y + a)^2 + b$, where $a$ and $b$ are constants. (b) Hence find the exact solutions of the equation $3x^4 - 12x^2 - 15 = 0$.

(a) 3(y - 2)^2 - 27 (a = -2, b = -27) · (b) x = ±√5

(a) Prove the identity $$ \frac{\sin^2 x - \cos x - 1}{1 + \cos x} \equiv -\cos x $$ (b) Hence solve the equation $$ \frac{\sin^2 x - \cos x - 1}{2 + 2\cos x} = \frac{1}{4} $$ for $0^\circ \le x \le 360^\circ$.

(a) -cos x · (b) x = 120° or x = 240°

9709/11 May/June 2024 — Mathematics Solved Past Paper

Q1MediumQuadratics

(a)

Express $3y^2 - 12y - 15$ in the form $3(y + a)^2 + b$ , where $a$ and $b$ are constants.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Factor out the leading coefficient $3$ , then complete the square for the quadratic part in $y$ .

Working

Factor out $3$ :

3y^2 - 12y - 15 = 3(y^2 - 4y - 5)

Complete the square for $y^2 - 4y$ :

y^2 - 4y = (y - 2)^2 - 4

Therefore

3(y^2 - 4y - 5) = 3\left((y - 2)^2 - 4 - 5\right) = 3(y - 2)^2 - 27

So $a = -2$ and $b = -27$ .

Answer

3y^2 - 12y - 15 = 3(y - 2)^2 - 27

with $a = -2$ and $b = -27$ .

Final answer

3(y - 2)^2 - 27 (a = -2, b = -27)

Detailed explanation

Walkthrough

We need to rewrite $3y^2 - 12y - 15$ in the form $3(y + a)^2 + b$ . First factor out the $3$ from all terms so that the quadratic inside the bracket has leading coefficient $1$ : $3(y^2 - 4y - 5)$ . Then complete the square on $y^2 - 4y$ : this is $(y - 2)^2 - 4$ . Substituting back gives $3((y - 2)^2 - 4 - 5) = 3(y - 2)^2 - 27$ . Therefore $a = -2$ and $b = -27$ .

Key Takeaways

Completing the square expresses a quadratic in the form $a(x - h)^2 + k$ . When the coefficient of $y^2$ is not $1$ , factor it out first, complete the square inside the bracket, and then multiply the constant term by the factored coefficient.

Common Mistakes

Forgetting to multiply the constant term by $3$ when completing the square.
Sign error: $y^2 - 4y = (y - 2)^2 - 4$ , not $(y - 2)^2 + 4$ .
Writing $a = 2$ instead of $a = -2$ because the requested form is $3(y + a)^2$ .

Things to Be Careful About

Always verify your answer by expanding $3(y - 2)^2 - 27$ to confirm it equals $3y^2 - 12y - 15$ . Note that $a$ can be negative even though the form uses $y + a$ .

Techniques used

factor out the leading coefficientcomplete the squareidentify the constants a and b

(b)

Hence find the exact solutions of the equation $3x^4 - 12x^2 - 15 = 0$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the completed-square form from part a with $y = x^2$ , set the expression equal to zero, solve for $x^2$ , then take square roots and discard any non-real solution.

Working

From part a, with $y = x^2$ :

3x^4 - 12x^2 - 15 = 3(x^2 - 2)^2 - 27

Set the expression equal to $0$ :

3(x^2 - 2)^2 - 27 = 0

(x^2 - 2)^2 = 9

Taking square roots:

x^2 - 2 = \pm 3

So

x^2 = 5 \quad \text{or} \quad x^2 = -1

Since $x^2 \ge 0$ for real $x$ , the equation $x^2 = -1$ has no real solution. Therefore

x^2 = 5

x = \pm \sqrt{5}

Answer

x = \pm \sqrt{5}

Final answer

x = ±√5

Detailed explanation

Walkthrough

We use the result of part a, replacing $y$ by $x^2$ . This gives $3x^4 - 12x^2 - 15 = 3(x^2 - 2)^2 - 27$ . Setting this equal to $0$ gives $(x^2 - 2)^2 = 9$ . Taking square roots gives $x^2 - 2 = \pm 3$ , so $x^2 = 5$ or $x^2 = -1$ . Since $x^2$ cannot be negative for real $x$ , we discard $x^2 = -1$ . Hence $x^2 = 5$ , so $x = \pm \sqrt{5}$ .

Key Takeaways

A quartic that is quadratic in $x^2$ can be solved using the completed-square form.
When a squared variable equals a negative number, there are no real solutions.
Always take both positive and negative square roots when solving $x^2 = k$ .

Common Mistakes

Forgetting to take the negative root, giving only $x = \sqrt{5}$ .
Including $x = \pm i$ from $x^2 = -1$ ; the mark scheme says to ignore $i$ , $-i$ , and $\sqrt{-1}$ .
Using a calculator without showing any working, which scores no marks.
Not using the completed-square form from part a, although factorising is also an accepted method.

Things to Be Careful About

When $x^2 = -1$ , there is no real $x$ , so the answer is only $x = \pm \sqrt{5}$ .
The answer must be exact, so write $\pm \sqrt{5}$ , not a decimal approximation.
Ensure both positive and negative square roots are included.

Techniques used

use the completed-square form with y = x^2solve the quadratic equation in x^2discard the non-real solution

Q2MediumFunctions

Fig. 1

The diagram shows two curves. One curve has equation $y = \sin x$ and the other curve has equation $y = f(x)$ .

(a)

In order to transform the curve $y = \sin x$ to the curve $y = f(x)$ , the curve $y = \sin x$ is first reflected in the $x$ -axis.

Describe fully a sequence of two further transformations which are required.

4M Similar

DifficultyMedium

Worked solution

Approach

Read the amplitude and midline of $f(x)$ from the graph. After the given reflection $y = \sin x \to y = -\sin x$ , determine a stretch and a translation that produce the correct range and position for $f(x)$ .

Working

From the graph, $f(x)$ has:

Maximum value: $y = 1$ at $x = \frac{3}{2}\pi$
Minimum value: $y = -5$ at $x = \frac{1}{2}\pi$
Midline: $y = \frac{1 + (-5)}{2} = -2$
Amplitude: $\frac{1 - (-5)}{2} = 3$

Starting from $y = \sin x$ , after reflection in the $x$ -axis:

y = -\sin x

This has range $[-1, 1]$ .

Method 1: Stretch by factor 3 in the $y$ -direction:

y = -3\sin x

Range becomes $[-3, 3]$ . Then translate by $\begin{pmatrix} 0 \\ -2 \end{pmatrix}$ :

y = -3\sin x - 2

Range becomes $[-5, 1]$ , which matches $f(x)$ . ✓

Method 2: Translate by $\begin{pmatrix} 0 \\ -\frac{2}{3} \end{pmatrix}$ :

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

Range becomes $\left[-\frac{5}{3}, -\frac{1}{3}\right]$ . Then stretch by factor 3 in the $y$ -direction:

y = 3\left(-\sin x - \frac{2}{3}\right) = -3\sin x - 2

Range becomes $[-5, 1]$ , which matches $f(x)$ . ✓

Answer

Stretch by factor 3 in the $y$ -direction and translate by $\begin{pmatrix} 0 \\ -2 \end{pmatrix}$ (in either order).

Final answer

Stretch by factor 3 in the y-direction and translate by (0, -2)

Detailed explanation

Walkthrough

The graph shows $y = \sin x$ (amplitude 1, range $[-1, 1]$ ) and $y = f(x)$ (amplitude 3, range $[-5, 1]$ ). We are told the first transformation is a reflection in the $x$ -axis, giving $y = -\sin x$ .

Step 1: Read features of $f(x)$ . The maximum of $f(x)$ is $1$ and the minimum is $-5$ . The midline (average of max and min) is $\frac{1 + (-5)}{2} = -2$ . The amplitude is half the total height: $\frac{1 - (-5)}{2} = 3$ .

Step 2: Determine the stretch. After reflection, $y = -\sin x$ has amplitude 1. Since $f(x)$ has amplitude 3, we need a stretch by factor 3 in the $y$ -direction, giving $y = -3\sin x$ with range $[-3, 3]$ .

Step 3: Determine the translation. The range of $-3\sin x$ is $[-3, 3]$ , but $f(x)$ has range $[-5, 1]$ . The midline of $-3\sin x$ is $y = 0$ , while the midline of $f(x)$ is $y = -2$ . So we translate down by 2 units: $\begin{pmatrix} 0 \\ -2 \end{pmatrix}$ .

Alternative order: We could translate first by $\begin{pmatrix} 0 \\ -\frac{2}{3} \end{pmatrix}$ to get $y = -\sin x - \frac{2}{3}$ , then stretch by 3 in the $y$ -direction to get $y = -3\sin x - 2$ .

Key Takeaways

The amplitude of a transformed sine curve is the vertical stretch factor.
The midline of the curve gives the vertical translation after accounting for the stretch.
A stretch in the $y$ -direction followed by a translation is not the same as a translation followed by a stretch; the translation amount depends on the order.
When a translation is applied before a stretch, the translation vector must be scaled by the stretch factor in the final expression.

Common Mistakes

Forgetting that the translation vector must be adjusted if the stretch is applied after it (the mark scheme accepts either order, but the translation amounts differ: $-2$ after stretch, or $-\frac{2}{3}$ before stretch).
Describing only one transformation when two are required.
Stating "shift down by 2" without specifying it as a translation vector or being precise about the direction.
Confusing the amplitude of $f(x)$ with the translation amount.

Things to Be Careful About

The question says the curve is first reflected in the $x$ -axis, so you must start from $y = -\sin x$ , not $y = \sin x$ .
The order of stretch and translation matters: if you stretch first then translate, the translation is $\begin{pmatrix} 0 \\ -2 \end{pmatrix}$ ; if you translate first then stretch, the translation is $\begin{pmatrix} 0 \\ -\frac{2}{3} \end{pmatrix}$ .
Both sequences are acceptable, but you must describe two further transformations (the reflection is already given).
Be precise with terminology: use "stretch" with a scale factor and direction, or "translation" with a vector.

Techniques used

read amplitude and midline from graphapply vertical stretchapply vertical translationverify range after transformations

(b)

Find $f(x)$ in terms of $\sin x$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Apply the three transformations in sequence to $y = \sin x$ to obtain $f(x)$ .

Working

Transformation 1: Reflect in the $x$ -axis:

y = \sin x \to y = -\sin x

Transformation 2: Stretch by factor 3 in the $y$ -direction:

y = -\sin x \to y = -3\sin x

Transformation 3: Translate by $\begin{pmatrix} 0 \\ -2 \end{pmatrix}$ (subtract 2 from $y$ ):

y = -3\sin x \to y = -3\sin x - 2

Therefore:

f(x) = -3\sin x - 2

Answer

f(x) = -3\sin x - 2

Final answer

f(x) = -3 sin x - 2

Detailed explanation

Walkthrough

We build $f(x)$ by applying the transformations in order to the original curve $y = \sin x$ .

Step 1: Reflection in the $x$ -axis. This negates the $y$ -values: $y = \sin x$ becomes $y = -\sin x$ .

Step 2: Stretch by factor 3 in the $y$ -direction. Multiply all $y$ -values by 3: $y = -\sin x$ becomes $y = -3\sin x$ .

Step 3: Translation by $\begin{pmatrix} 0 \\ -2 \end{pmatrix}$ . This shifts the curve down by 2 units, which means subtracting 2 from the expression: $y = -3\sin x$ becomes $y = -3\sin x - 2$ .

The final expression is $f(x) = -3\sin x - 2$ .

Key Takeaways

Each transformation can be written as an algebraic operation on the function expression.
Reflection in the $x$ -axis: negate the function.
Stretch by factor $k$ in the $y$ -direction: multiply the function by $k$ .
Translation by $\begin{pmatrix} 0 \\ c \end{pmatrix}$ : add $c$ to the function (so $\begin{pmatrix} 0 \\ -2 \end{pmatrix}$ means subtract 2).
Applying transformations in order and tracking the effect on the expression gives the final function directly.

Common Mistakes

Forgetting the reflection and writing $f(x) = 3\sin x - 2$ instead of $-3\sin x - 2$ .
Applying the stretch to the constant term: writing $f(x) = -3(\sin x - 2) = -3\sin x + 6$ instead of $-3\sin x - 2$ . The stretch only multiplies the function values, not the translation.
Adding extra terms not present in the transformation sequence.

Things to Be Careful About

The mark scheme awards no marks if extra terms are seen in the expression.
The order of transformations matters for the expression: reflection first, then stretch, then translation gives $-3\sin x - 2$ . If you use the alternative order (translate then stretch), you must use the adjusted translation $-\frac{2}{3}$ to get the same result.
Ensure the final expression matches the graph: at $x = \frac{1}{2}\pi$ , $f\left(\frac{1}{2}\pi\right) = -3(1) - 2 = -5$ ✓; at $x = \frac{3}{2}\pi$ , $f\left(\frac{3}{2}\pi\right) = -3(-1) - 2 = 1$ ✓.

Techniques used

apply reflection to sine functionapply vertical stretchapply vertical translationcompose transformations into single expression

Q3MediumSeries

The coefficient of $x^3$ in the expansion of $(3 + ax)^6$ is 160.

(a)

Find the value of the constant $a$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the binomial theorem to write the general term of $(3 + ax)^6$ , extract the term containing $x^3$ , set its coefficient equal to 160, and solve for $a$ .

Working

The general term of $(3 + ax)^6$ is:

\binom{6}{r} 3^{6-r} (ax)^r

For the $x^3$ term, set $r = 3$ :

\binom{6}{3} 3^{3} a^3 x^3 = 20 \times 27 \times a^3 x^3 = 540a^3 x^3

The coefficient of $x^3$ is $540a^3$ . Given that this equals 160:

540a^3 = 160

a^3 = \frac{160}{540} = \frac{8}{27}

Therefore:

a = \frac{2}{3}

Answer

a = \frac{2}{3}

Final answer

a = 2/3

Detailed explanation

Walkthrough

The binomial theorem states that $(3 + ax)^6 = \sum_{r=0}^{6} \binom{6}{r} 3^{6-r} (ax)^r$ . To find the coefficient of $x^3$ , we need the term where $r = 3$ . This gives $\binom{6}{3} 3^{3} a^3 = 20 \times 27 \times a^3 = 540a^3$ . Since the problem states this coefficient is 160, we set up the equation $540a^3 = 160$ . Dividing both sides by 540 gives $a^3 = \frac{8}{27}$ . Taking the cube root of both sides gives $a = \frac{2}{3}$ .

Key Takeaways

The key skill here is extracting a specific term from a binomial expansion. The general term $\binom{6}{r} 3^{6-r} (ax)^r$ allows us to find any term by choosing the appropriate value of $r$ . The coefficient includes the binomial coefficient, the power of the constant term (3), and the power of $a$ .

Common Mistakes

Forgetting to include the $3^{6-r}$ factor — the coefficient is not just $\binom{6}{3} a^3$ .
Making arithmetic errors when simplifying $\frac{160}{540}$ .
Forgetting to take the cube root when solving for $a$ .

Things to Be Careful About

The coefficient of $x^3$ is the full product $\binom{6}{3} 3^3 a^3$ , not just the part involving $a$ .
When solving $a^3 = \frac{8}{27}$ , remember that $a$ could theoretically be negative, but since $a^3$ is positive, $a$ must be positive (the real cube root). Here $\frac{8}{27}$ is positive, so $a = \frac{2}{3}$ .

Techniques used

extract the x^3 term from a binomial expansionsolve the resulting equation for a

(b)

Hence find the coefficient of $x^3$ in the expansion of $(3 + ax)^6(1 - 2x)$ .

3M Similar

DifficultyMedium

Worked solution

Approach

When expanding $(3 + ax)^6(1 - 2x)$ , the $x^3$ coefficient receives contributions from the $x^3$ term of $(3 + ax)^6$ multiplied by the constant 1, and from the $x^2$ term of $(3 + ax)^6$ multiplied by $-2x$ . Find the $x^2$ coefficient using $a = \frac{2}{3}$ , then combine the two contributions.

Working

First, find the coefficient of $x^2$ in $(3 + ax)^6$ . Using the general term with $r = 2$ :

\binom{6}{2} 3^{4} a^2 x^2 = 15 \times 81 \times a^2 x^2 = 1215a^2 x^2

With $a = \frac{2}{3}$ :

1215 \times \left(\frac{2}{3}\right)^2 = 1215 \times \frac{4}{9} = 540

Now combine the two contributions to the $x^3$ coefficient in $(3 + ax)^6(1 - 2x)$ :

160 \times 1 + 540 \times (-2) = 160 - 1080 = -920

Answer

-920

Final answer

-920

Detailed explanation

Walkthrough

When expanding the product $(3 + ax)^6(1 - 2x)$ , each term in the result comes from multiplying one term of $(3 + ax)^6$ by one term of $(1 - 2x)$ . The $x^3$ term can arise in two ways: the $x^3$ term of $(3 + ax)^6$ multiplied by the constant 1, or the $x^2$ term of $(3 + ax)^6$ multiplied by $-2x$ .

We already know the $x^3$ coefficient of $(3 + ax)^6$ is 160 (given in the problem). Now we need the $x^2$ coefficient. Using the general term with $r = 2$ : $\binom{6}{2} 3^4 a^2 = 15 \times 81 \times \left(\frac{2}{3}\right)^2 = 1215 \times \frac{4}{9} = 540$ .

The $x^3$ coefficient of the product is therefore $160 \times 1 + 540 \times (-2) = 160 - 1080 = -920$ .

Key Takeaways

When multiplying two polynomials, the coefficient of $x^n$ in the product is the sum of all products of coefficients $c_i \times d_j$ where $i + j = n$ . This is a fundamental concept that applies to polynomial multiplication in general.

Common Mistakes

Only considering one contribution to the $x^3$ term (e.g., only the $x^3 \times 1$ part).
Forgetting the negative sign in $-2x$ when computing the second contribution.
Using the wrong coefficient for the $x^2$ term of $(3 + ax)^6$ .

Things to Be Careful About

The $x^2$ coefficient of $(3 + ax)^6$ must be computed using the value of $a$ found in part (a).
The sign of the second contribution is negative because of the $-2x$ factor.
The answer is $-920$ , not $920$ .

Techniques used

extract the x^2 term from a binomial expansioncombine contributions from both factors in a productsubstitute a previously found value

Q4Medium-EasyDifferentiation

The equation of a curve is $y = f(x)$ , where $f(x) = (2x - 1)\sqrt{3x - 2} - 2$ . The following points lie on the curve. Non-exact values have been given correct to 5 decimal places.

$A(2, 4)$ , $B(2.0001, k)$ , $C(2.001, 4.00625)$ , $D(2.01, 4.06261)$ , $E(2.1, 4.63566)$ , $F(3, 11.22876)$

(a)

Find the value of $k$ . Give your answer correct to 5 decimal places.

1M Similar

DifficultyEasy

Worked solution

Approach

Evaluate $x = 2.0001$ in $f(x) = (2x - 1)\sqrt{3x - 2} - 2$ . Since $B$ lies on the curve, its $y$ -coordinate $k$ equals $f(2.0001)$ . Round the result to 5 decimal places.

Working

For $x = 2.0001$ :

2x - 1 = 2(2.0001) - 1 = 4.0002 - 1 = 3.0002

3x - 2 = 3(2.0001) - 2 = 6.0003 - 2 = 4.0003

So

k = f(2.0001) = 3.0002\sqrt{4.0003} - 2

Evaluating:

k = 3.0002 \times 2.00007499859\ldots - 2 = 4.00062501078\ldots

Rounded to 5 decimal places:

k = 4.00063

Answer

k = 4.00063

Final answer

k = 4.00063

Detailed explanation

Walkthrough

Because $B(2.0001, k)$ lies on the curve $y = f(x)$ , the value of $k$ is found by substituting $x = 2.0001$ into $f(x)$ . First compute $2x - 1$ : $2(2.0001) - 1 = 3.0002$ . Then compute $3x - 2$ : $3(2.0001) - 2 = 4.0003$ . Therefore $k = 3.0002\sqrt{4.0003} - 2$ . A calculator gives $4.00062501078\ldots$ , which rounds to $4.00063$ at 5 decimal places.

Key Takeaways

This part checks substitution into a function and rounding to a given number of decimal places. It also creates a nearby point $B$ , which will be used to form a chord and approximate the derivative at $A$ .

Common Mistakes

Forgetting to subtract the final $2$ from the whole expression.
Rounding to the wrong number of decimal places; the question asks for 5.
Substituting the wrong value of $x$ , such as using another point's coordinate.
Misreading the expression as $(2x - 1)\sqrt{3x} - 2$ instead of $(2x - 1)\sqrt{3x - 2} - 2$ .

Things to Be Careful About

The $-2$ is outside the product, so it is subtracted after the product is evaluated.
Keep the full calculator value until the final step, then round to 5 decimal places.
The mark scheme accepts $4.00063$ as the correct answer.

Techniques used

substitute an x-value into a functionevaluate a square-root expressionround to a given number of decimal places

(b)

The table shows the gradients of the chords $AB$ , $AC$ , $AD$ and $AF$ .

Chord	$AB$	$AC$	$AD$	$AE$	$AF$
Gradient of chord	6.2501	6.2511	6.2608		7.2288

Find the gradient of the chord $AE$ . Give your answer correct to 4 decimal places.

1M Similar

(c)

Deduce the value of $f'(2)$ using the values in the table.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

The gradient of the chord from $A$ to a point with $x$ -coordinate $2 + h$ is

\frac{f(2 + h) - f(2)}{h}

As $h$ tends to $0$ , this quotient tends to $f'(2)$ . Read the trend in the table to deduce the limit.

Working

The table gives:

$h$	Chord	Gradient
$1$	$AF$	$7.2288$
$0.1$	$AE$	$6.3566$
$0.01$	$AD$	$6.2608$
$0.001$	$AC$	$6.2511$
$0.0001$	$AB$	$6.2501$

As $h$ decreases, the chord gradients approach $6.25$ . Hence the gradient of the tangent at $A$ is $6.25$ .

Answer

f'(2) = 6.25

Final answer

f'(2) = 6.25

Detailed explanation

Walkthrough

The gradient of a chord joining $A$ to a nearby point $(2 + h, f(2 + h))$ is

\frac{f(2 + h) - f(2)}{h}

As $h$ approaches $0$ , this quotient tends to the derivative $f'(2)$ . The table lists chord gradients for decreasing values of $h$ : $7.2288$ for $h = 1$ , $6.3566$ for $h = 0.1$ , $6.2608$ for $h = 0.01$ , $6.2511$ for $h = 0.001$ , and $6.2501$ for $h = 0.0001$ . These values are getting closer and closer to $6.25$ , so the limit is $6.25$ .

Key Takeaways

This is the definition of the derivative as the limit of chord gradients. It shows that $f'(2)$ can be estimated numerically without applying differentiation rules, by letting the chord shrink towards the tangent point.

Common Mistakes

Choosing one of the table values, such as $6.2501$ , instead of the limit $6.25$ .
Treating the closest chord gradient as exactly equal to the derivative.
Trying to differentiate $f(x)$ symbolically when the question asks to deduce the value from the table.
Reading the table in the wrong order.

Things to Be Careful About

The table gives approximations, not the exact derivative, so the answer is the limit they approach.
The correct value is $6.25$ , not $6.2501$ or $6.2511$ .
The mark scheme accepts $6.25$ as the correct answer.

Techniques used

interpret chord gradients as approximations to the derivativeobserve the limiting value as the chord shrinksdeduce the gradient of the tangent at a point

Q5Medium-EasyTrigonometry

(a)

Prove the identity

\frac{\sin^2 x - \cos x - 1}{1 + \cos x} \equiv -\cos x

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the identity $\sin^2 x + \cos^2 x = 1$ to rewrite the numerator in terms of $\cos x$ , factorise, and then cancel the common factor.

Working

Start with the left-hand side:

\frac{\sin^2 x - \cos x - 1}{1 + \cos x}

Replace $\sin^2 x$ by $1 - \cos^2 x$ :

\frac{1 - \cos^2 x - \cos x - 1}{1 + \cos x} = \frac{-\cos^2 x - \cos x}{1 + \cos x}

Factorise the numerator:

\frac{-\cos x(1 + \cos x)}{1 + \cos x}

Cancel the common factor $1 + \cos x$ :

-\cos x

Answer

\frac{\sin^2 x - \cos x - 1}{1 + \cos x} \equiv -\cos x

Final answer

-cos x

Detailed explanation

Walkthrough

We want to show that the left-hand side equals $-\cos x$ . The numerator contains both $\sin^2 x$ and $\cos x$ , so we use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to rewrite $\sin^2 x$ as $1 - \cos^2 x$ . Substituting this into the numerator gives $1 - \cos^2 x - \cos x - 1$ , which simplifies to $-\cos^2 x - \cos x$ . Next, factor out $-\cos x$ from the numerator to get $-\cos x(1 + \cos x)$ . Since the denominator is $1 + \cos x$ , the common factor cancels, leaving $-\cos x$ . This proves the identity.

Key Takeaways

The identity $\sin^2 x + \cos^2 x = 1$ is the key tool for converting between $\sin^2 x$ and $\cos^2 x$ .
You must factorise before cancelling; cancellation is only valid for common factors, not individual terms.
This type of proof is a standard manipulation of trigonometric expressions.

Common Mistakes

Forgetting to replace $\sin^2 x$ with $1 - \cos^2 x$ , leaving the numerator in a form that cannot be simplified.
Making a sign error when factorising $-\cos^2 x - \cos x$ as $-\cos x(1 + \cos x)$ .
Trying to cancel $\cos x$ terms before factorising.

Things to Be Careful About

Cancelling $1 + \cos x$ is only valid when $1 + \cos x \neq 0$ , i.e. when $x \neq 180^\circ$ in the given range. The identity is understood on the domain where the fraction is defined.
Keep the negative sign throughout; a lost sign is a common error.

Techniques used

apply Pythagorean identityfactorise the trigonometric expressioncancel common factors

(b)

Hence solve the equation

\frac{\sin^2 x - \cos x - 1}{2 + 2\cos x} = \frac{1}{4}

for $0^\circ \le x \le 360^\circ$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the identity from part (a). Since $2 + 2\cos x = 2(1 + \cos x)$ , the left-hand side becomes $-\frac{1}{2}\cos x$ . Then solve $-\frac{1}{2}\cos x = \frac{1}{4}$ for $0^\circ \le x \le 360^\circ$ .

Working

From part (a),

\frac{\sin^2 x - \cos x - 1}{1 + \cos x} = -\cos x

Therefore,

\frac{\sin^2 x - \cos x - 1}{2 + 2\cos x} = \frac{1}{2}\cdot\frac{\sin^2 x - \cos x - 1}{1 + \cos x} = \frac{1}{2}(-\cos x) = -\frac{1}{2}\cos x

The equation becomes

-\frac{1}{2}\cos x = \frac{1}{4}

Multiply both sides by $-2$ :

\cos x = -\frac{1}{2}

For $0^\circ \le x \le 360^\circ$ , $\cos x = -\frac{1}{2}$ in the second and third quadrants:

x = 120^\circ, \quad x = 240^\circ

Answer

x = 120^\circ \quad \text{or} \quad x = 240^\circ

Final answer

x = 120° or x = 240°

Detailed explanation

Walkthrough

We start with the equation and use the identity from part (a). The denominator is $2 + 2\cos x = 2(1 + \cos x)$ , so the fraction is exactly half of the left-hand side from part (a). Since that left-hand side simplifies to $-\cos x$ , the whole fraction simplifies to $-\frac{1}{2}\cos x$ . Setting this equal to $\frac{1}{4}$ gives $-\frac{1}{2}\cos x = \frac{1}{4}$ , so $\cos x = -\frac{1}{2}$ . The cosine is negative in the second and third quadrants. Since $\cos 60^\circ = \frac{1}{2}$ , the reference angle is $60^\circ$ . Therefore the solutions in $0^\circ$ to $360^\circ$ are $180^\circ - 60^\circ = 120^\circ$ and $180^\circ + 60^\circ = 240^\circ$ .

Key Takeaways

A result from an earlier part can be substituted directly into a new equation.
To solve $\cos x = k$ , find the principal angle and then use the quadrants to determine all solutions in the required interval.
Exact trigonometric values, such as $\cos 60^\circ = \frac{1}{2}$ , are important for solving without a calculator.

Common Mistakes

Forgetting the factor of $\frac{1}{2}$ when rewriting the denominator $2 + 2\cos x$ .
Giving only one solution, such as $120^\circ$ , instead of both solutions in the interval.
Including extra solutions outside the given interval or values where $\cos x$ is not $-\frac{1}{2}$ .
Giving answers in radians when the question asks for degrees. The mark scheme allows special credit only if both radian answers $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ are given.

Things to Be Careful About

The mark scheme gives follow-through for $360^\circ -$ the candidate's answer, but if extra solutions are included in the range, the final answer mark is lost.
When using a calculator, $\cos^{-1}(-\frac{1}{2})$ returns the principal value $120^\circ$ ; you must also include $240^\circ$ in the interval.
Always check that the final answers satisfy the original equation.

Techniques used

use the identity from part (a)solve a linear equation for cos xfind all angles in a given interval

Q6MediumFunctions

Fig. 2

The function $f$ is defined by $f(x) = \frac{2}{x^2} + 4$ for $x < 0$ . The diagram shows the graph of $y = f(x)$ .

(a)

On this diagram, sketch the graph of $y = f^{-1}(x)$ . Show any relevant mirror line.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$ . Reflect the curve, swap the asymptotes, and mark the mirror line.

Working

The original curve $y = f(x) = \frac{2}{x^2} + 4$ for $x < 0$ has:

Domain $x < 0$ and range $y > 4$
Horizontal asymptote $y = 4$ as $x \to -\infty$
Vertical asymptote $x = 0$ (the $y$ -axis) as $x \to 0^{-}$
Lies in the second quadrant

Reflecting in $y = x$ swaps the roles of $x$ and $y$ , so the inverse curve has:

Domain $x > 4$ and range $y < 0$
Vertical asymptote $x = 4$
Horizontal asymptote $y = 0$ (the $x$ -axis)
Lies in the fourth quadrant

Draw the line $y = x$ through the origin as a dashed line, then sketch the inverse curve in the fourth quadrant: starting just below the $x$ -axis for large $x$ and plunging down to $-\infty$ as $x \to 4^{+}$ .

Answer

The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ , lying in the fourth quadrant with vertical asymptote $x = 4$ and horizontal asymptote $y = 0$ .

Final answer

Sketch of y = f⁻¹(x) in the fourth quadrant: reflection of f in the line y = x, with vertical asymptote x = 4 and horizontal asymptote y = 0.

Detailed explanation

Walkthrough

The graph of the inverse of any one-to-one function is obtained by reflecting the original graph in the line $y = x$ . The point $(a, b)$ on the graph of $f$ corresponds to the point $(b, a)$ on the graph of $f^{-1}$ , and these two points are reflections of each other in $y = x$ .

The original $f(x) = \frac{2}{x^2} + 4$ is defined only for $x < 0$ , so its graph lies entirely in the second quadrant. As $x \to -\infty$ , $\frac{2}{x^2} \to 0$ , so $f(x) \to 4$ from above, giving a horizontal asymptote $y = 4$ . As $x \to 0^{-}$ , $\frac{2}{x^2} \to +\infty$ , so $f(x) \to +\infty$ , giving a vertical asymptote at the $y$ -axis ( $x = 0$ ).

When we reflect in $y = x$ , the two asymptotes swap roles. The horizontal asymptote $y = 4$ of $f$ becomes the vertical asymptote $x = 4$ of $f^{-1}$ . The vertical asymptote $x = 0$ of $f$ becomes the horizontal asymptote $y = 0$ of $f^{-1}$ (i.e., the $x$ -axis). The branch in the second quadrant reflects into the fourth quadrant, and the line $y = x$ should be drawn through the origin to make the mirror visible.

Key Takeaways

The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y = x$ .
Reflection in $y = x$ swaps horizontal and vertical asymptotes.
The domain of $f$ becomes the range of $f^{-1}$ , and vice versa.

Common Mistakes

Placing the inverse in the wrong quadrant (e.g., still in the second quadrant or in the first).
Confusing the asymptotes — giving a horizontal asymptote $y = 4$ for $f^{-1}$ instead of the vertical asymptote $x = 4$ .
Forgetting to draw the mirror line $y = x$ .

Things to Be Careful About

The horizontal asymptote $y = 4$ of $f$ becomes the vertical asymptote $x = 4$ of $f^{-1}$ .
The vertical asymptote $x = 0$ of $f$ becomes the horizontal asymptote $y = 0$ of $f^{-1}$ .
The shape must be a true mirror image of the original across $y = x$ — the perpendicular distance from any point to the line is preserved.

Techniques used

reflect curve in y=xidentify asymptotesidentify correct quadrant

(b)

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

3M Similar

DifficultyMedium

Worked solution

Approach

Write $y = f(x)$ , swap $x$ and $y$ , then solve for $y$ . When the square root appears, use the fact that the range of $f^{-1}$ equals the domain of $f$ (which is $x < 0$ ) to choose the correct sign.

Working

Start with

y = \frac{2}{x^2} + 4

Swap $x$ and $y$ :

x = \frac{2}{y^2} + 4

Subtract $4$ from both sides:

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

Multiply both sides by $y^2$ :

y^2(x - 4) = 2

Divide by $(x - 4)$ :

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

Take the square root:

y = \pm\sqrt{\frac{2}{x - 4}}

The range of $f^{-1}$ is the domain of $f$ , namely $y < 0$ . So we must take the negative root:

f^{-1}(x) = -\sqrt{\frac{2}{x - 4}}

Answer

f^{-1}(x) = -\sqrt{\frac{2}{x - 4}}

Final answer

f⁻¹(x) = -√(2/(x-4))

Detailed explanation

Walkthrough

The standard procedure to find the inverse of a one-to-one function is to swap the variables and solve for the new dependent variable.

Step 1. Write $y = \frac{2}{x^2} + 4$ .

Step 2. Swap $x$ and $y$ . This is the key conceptual step: the equation $x = \frac{2}{y^2} + 4$ now says 'the value of $f$ at $y$ is $x$ ', which is exactly what the inverse means.

Step 3. Isolate $y$ . Subtract $4$ : $x - 4 = \frac{2}{y^2}$ . Cross-multiply: $y^2(x - 4) = 2$ . Divide by $(x - 4)$ : $y^2 = \frac{2}{x - 4}$ .

Step 4. Take the square root: $y = \pm\sqrt{\frac{2}{x - 4}}$ .

Step 5. Choose the correct sign. The original $f$ is defined only for $x < 0$ , so the range of $f^{-1}$ (the possible $y$ -values the inverse can output) must be $y < 0$ . The expression $\sqrt{\frac{2}{x - 4}}$ is non-negative, so to get a negative result we must take the negative root.

The domain of $f^{-1}$ is $x > 4$ (the range of $f$ ), and the range is $y < 0$ (the domain of $f$ ).

Key Takeaways

The procedure for finding the inverse: write $y = f(x)$ , swap $x$ and $y$ , solve for $y$ .
When a square root is involved, use the domain/range of the original function to select the correct sign.
The domain of $f$ becomes the range of $f^{-1}$ , and vice versa.

Common Mistakes

Omitting the sign selection step and giving $\pm$ or the positive root.
Forgetting that the formula implicitly requires $x > 4$ .
Algebra errors when isolating $y^2$ (e.g., writing $y^2 = \frac{x - 4}{2}$ instead of $y^2 = \frac{2}{x - 4}$ ).

Things to Be Careful About

The range of $f$ is $y > 4$ (since $2/x^2 > 0$ ), so the domain of $f^{-1}$ is $x > 4$ .
The range of $f^{-1}$ is $y < 0$ (the domain of $f$ ), so the negative root is required.
The constant $4$ in the original formula is the horizontal asymptote of $f$ , and it becomes the vertical asymptote $x = 4$ of $f^{-1}$ .

Techniques used

swap x and yrearrange to isolate y squaredtake square rootselect sign using range restriction

(c)

Solve the equation $f(x) = 4.5$ .

1M Similar

(d)

Explain why the equation $f^{-1}(x) = f(x)$ has no solution.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

Compare the values $f$ can produce with the values $f^{-1}$ can produce. If those value sets do not overlap, the equation $f^{-1}(x) = f(x)$ cannot hold.

Working

For $f(x) = \frac{2}{x^2} + 4$ , the term $\frac{2}{x^2} > 0$ for every $x$ in the domain, so

f(x) = \frac{2}{x^2} + 4 > 4

i.e. $f(x)$ is always strictly greater than $4$ (and so always positive).

For $f^{-1}(x) = -\sqrt{\frac{2}{x - 4}}$ , the square root is non-negative and the leading minus sign makes $f^{-1}(x)$ strictly negative (for every $x > 4$ ):

f^{-1}(x) < 0

Since $f(x) > 4 > 0$ and $f^{-1}(x) < 0$ , the two functions can never take the same value. Hence the equation $f^{-1}(x) = f(x)$ has no solution.

Answer

$f(x)$ is always $> 4$ (positive) while $f^{-1}(x)$ is always $< 0$ (negative), so the equation has no solution.

Final answer

f(x) is always > 4 (positive) while f⁻¹(x) is always < 0 (negative), so f⁻¹(x) = f(x) has no solution.

Detailed explanation

Walkthrough

The equation $f^{-1}(x) = f(x)$ asks for an $x$ where the inverse and the function itself produce the same value. To see why no such $x$ exists, look at the value sets (ranges) of the two functions.

Range of $f$ . For every $x < 0$ :

f(x) = \frac{2}{x^2} + 4

The term $\frac{2}{x^2}$ is strictly positive, so $f(x) > 4$ . The entire graph of $f$ sits strictly above the line $y = 4$ — it never reaches the $x$ -axis or any negative $y$ -value.

Range of $f^{-1}$ . From part (b):

f^{-1}(x) = -\sqrt{\frac{2}{x - 4}}

The square root is positive for $x > 4$ , and the leading minus sign makes $f^{-1}(x)$ strictly negative. The whole graph of $f^{-1}$ lies below the $x$ -axis.

Since the two graphs lie in entirely different vertical regions (one strictly above $y = 4$ , the other strictly below $y = 0$ ), the curves never meet. So $f^{-1}(x) = f(x)$ has no solution.

Key Takeaways

The range of $f$ and the range of $f^{-1}$ are linked: the range of $f^{-1}$ is the domain of $f$ , and vice versa.
If two functions have disjoint ranges, the equation $g(x) = h(x)$ has no solution.
Sign-of-function arguments are a quick way to rule out solutions.

Common Mistakes

Stating only that 'the curves don't intersect' without giving a reason based on signs or ranges.
Confusing domain and range — saying $f$ is defined only for $x < 0$ does not on its own explain why the equation has no solution; the values it outputs, not the values it accepts, are what matter.

Things to Be Careful About

$f(x) > 4$ strictly, not $f(x) \geq 4$ , because $\frac{2}{x^2} > 0$ strictly.
$f^{-1}(x) < 0$ strictly, because $\sqrt{\frac{2}{x - 4}} > 0$ strictly.
The argument is purely about the outputs (ranges) of the two functions, not about where they are defined (domains).

Techniques used

identify range of fidentify range of inverseconclude impossibility from disjoint ranges

Q7MediumTrigonometryCircular MeasureCoordinate Geometry

Fig. 3

In the diagram, $AOD$ and $BC$ are two parallel straight lines. Arc $AB$ is part of a circle with centre $O$ and radius $15\text{ cm}$ . Angle $BOA = \theta$ radians. Arc $CD$ is part of a circle with centre $O$ and radius $10\text{ cm}$ . Angle $COD = \frac{1}{2}\pi$ radians.

(a)

Show that $\theta = 0.7297$ , correct to 4 decimal places.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

From the diagram, $BC$ is parallel to $AD$ and $\angle COD = \frac{\pi}{2}$ , meaning $OC$ is perpendicular to $AD$ . Since $BC$ is horizontal and $C$ is at a height of $10\text{ cm}$ above $O$ (the radius of the inner arc), point $B$ is also at a vertical height of $10\text{ cm}$ above the line $AD$ . We can form a right-angled triangle with hypotenuse $OB = 15\text{ cm}$ and opposite side $10\text{ cm}$ to find $\theta$ .

Working

Let $E$ be the foot of the perpendicular from $B$ to $AD$ . Then $BE = OC = 10\text{ cm}$ and $OB = 15\text{ cm}$ .

In right-angled triangle $OEB$ :

\sin \theta = \frac{BE}{OB} = \frac{10}{15} = \frac{2}{3}

Solving for $\theta$ :

\theta = \sin^{-1}\left(\frac{2}{3}\right)

\theta \approx 0.7297277\text{ rad}

Rounding to 4 decimal places:

\theta = 0.7297

Answer

$\theta = 0.7297$

Final answer

0.7297

Detailed explanation

Walkthrough

First, we observe the geometric relationships in the diagram. The line $AD$ is horizontal, and $BC$ is parallel to it. The angle $\angle COD$ is a right angle, and $OC$ is a radius of the inner circle with length $10\text{ cm}$ . This means $C$ is located $10\text{ cm}$ directly above $O$ . Because $BC$ is parallel to $AD$ , point $B$ must also be at a vertical height of $10\text{ cm}$ above the line $AD$ .

Next, we consider the right-angled triangle formed by dropping a perpendicular from $B$ to $AD$ . Let's call the intersection point $E$ . The hypotenuse of this triangle is $OB$ , which is a radius of the outer circle, so $OB = 15\text{ cm}$ . The side opposite to angle $\theta$ (which is $\angle BOA$ ) is $BE = 10\text{ cm}$ .

Using the sine ratio in right-angled triangles:

\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{10}{15} = \frac{2}{3}

To find $\theta$ , we apply the inverse sine function:

\theta = \sin^{-1}\left(\frac{2}{3}\right) \approx 0.7297277\text{ rad}

Rounding to 4 decimal places gives $0.7297$ .

Key Takeaways

Parallel lines and perpendicular radii can be used to determine vertical heights in composite geometric shapes.
The sine ratio relates the opposite side and hypotenuse in a right-angled triangle, and inverse sine recovers the angle.

Common Mistakes

Forgetting to convert the result from degrees to radians if using a calculator set to degree mode.
Using the wrong trigonometric ratio (e.g., cosine instead of sine) by misidentifying the sides relative to angle $\theta$ .

Things to Be Careful About

The mark scheme accepts working in degrees as long as it is converted to radians at the end.
Ensure the final answer is given to exactly 4 decimal places as requested.

Techniques used

identify right-angled triangle from parallel linesapply sine ratiouse inverse sine to find angle

(b)

Find the perimeter and the area of the shape $ABCD$ . Give your answers correct to 3 significant figures.

7M Similar

DifficultyMedium

Worked solution

Approach

The perimeter of shape $ABCD$ is the sum of the lengths of its four boundary components: arc $AB$ , line segment $BC$ , arc $CD$ , and line segment $DA$ . The area can be found by decomposing the shape into a right-angled triangle $OBC$ , a sector $AOB$ , and a quarter-circle sector $COD$ .

Working

Perimeter:

Find the length of $BC$ :
Since $B$ is at height $10\text{ cm}$ and $OB = 15\text{ cm}$ , the horizontal distance from $O$ to the vertical line through $B$ is:
$\sqrt{15^2 - 10^2} = \sqrt{225 - 100} = \sqrt{125} = 5\sqrt{5} \approx 11.1803\text{ cm}$
Since $C$ is directly above $O$ , $BC = 5\sqrt{5} \approx 11.1803\text{ cm}$ .
Find the length of arc $AB$ :
$\text{Arc } AB = r\theta = 15 \times 0.7297277 \approx 10.9459\text{ cm}$
Find the length of arc $CD$ :
$\text{Arc } CD = r\phi = 10 \times \frac{\pi}{2} = 5\pi \approx 15.7080\text{ cm}$
Find the length of $DA$ :
$DA = AO + OD = 15 + 10 = 25\text{ cm}$
Total perimeter:
$P = BC + \text{arc } AB + \text{arc } CD + DA$ $P = 5\sqrt{5} + 15(0.7297277) + 5\pi + 25$ $P \approx 11.1803 + 10.9459 + 15.7080 + 25 = 62.8342\text{ cm}$
Correct to 3 significant figures: $P = 62.8\text{ cm}$ .

Area:

Area of triangle $OBC$ :
The base is $OC = 10\text{ cm}$ and the height is $BC = 5\sqrt{5} \approx 11.1803\text{ cm}$ .
$\text{Area}_{\triangle OBC} = \frac{1}{2} \times 10 \times 5\sqrt{5} = 25\sqrt{5} \approx 55.9017\text{ cm}^2$
Area of sector $AOB$ :
$\text{Area}_{\text{sector } AOB} = \frac{1}{2}r^2\theta = \frac{1}{2} \times 15^2 \times 0.7297277 \approx 82.0944\text{ cm}^2$
Area of sector $COD$ :
This is a quarter-circle with radius $10\text{ cm}$ .
$\text{Area}_{\text{sector } COD} = \frac{1}{4} \times \pi \times 10^2 = 25\pi \approx 78.5398\text{ cm}^2$
Total area:
$A = \text{Area}_{\triangle OBC} + \text{Area}_{\text{sector } AOB} + \text{Area}_{\text{sector } COD}$ $A \approx 55.9017 + 82.0944 + 78.5398 = 216.5359\text{ cm}^2$
Correct to 3 significant figures: $A = 217\text{ cm}^2$ .

Answer

Perimeter = $62.8\text{ cm}$
Area = $217\text{ cm}^2$

Final answer

Perimeter = 62.8 cm, Area = 217 cm^2

Detailed explanation

Walkthrough

Perimeter Calculation:
The perimeter is the total length of the boundary of shape $ABCD$ . We break it down into four parts:

Line segment $BC$ : Point $C$ is $10\text{ cm}$ directly above $O$ (since $\angle COD = 90^\circ$ and radius $OC = 10$ ). Point $B$ is on the outer circle (radius $15$ ) and at the same height as $C$ ( $10\text{ cm}$ ). Using Pythagoras' theorem on the right triangle formed by $O$ , $B$ , and the projection of $B$ onto $AD$ , the horizontal distance from $O$ to $B$ is $\sqrt{15^2 - 10^2} = \sqrt{125} = 5\sqrt{5}$ . Since $C$ is directly above $O$ , $BC = 5\sqrt{5} \approx 11.1803\text{ cm}$ .
Arc $AB$ : Using the arc length formula $s = r\theta$ with $r = 15$ and $\theta \approx 0.7297277$ , we get $15 \times 0.7297277 \approx 10.9459\text{ cm}$ .
Arc $CD$ : This is a quarter-circle with radius $10$ , so its length is $\frac{1}{4} \times 2\pi(10) = 5\pi \approx 15.7080\text{ cm}$ .
Line segment $DA$ : This is the straight line along the bottom, composed of $AO = 15$ and $OD = 10$ , giving $25\text{ cm}$ .

Summing these gives the perimeter: $11.1803 + 10.9459 + 15.7080 + 25 = 62.8342\text{ cm}$ , which rounds to $62.8\text{ cm}$ .

Area Calculation:
The shape can be decomposed into three non-overlapping regions meeting at $O$ :

Triangle $OBC$ : A right-angled triangle with base $OC = 10$ and height $BC = 5\sqrt{5}$ . Area $= \frac{1}{2} \times 10 \times 5\sqrt{5} = 25\sqrt{5} \approx 55.9017\text{ cm}^2$ .
Sector $AOB$ : A circular sector with radius $15$ and angle $\theta \approx 0.7297277$ . Area $= \frac{1}{2}r^2\theta = \frac{1}{2}(225)(0.7297277) \approx 82.0944\text{ cm}^2$ .
Sector $COD$ : A quarter-circle with radius $10$ . Area $= \frac{1}{4}\pi(100) = 25\pi \approx 78.5398\text{ cm}^2$ .

Summing these gives the total area: $55.9017 + 82.0944 + 78.5398 = 216.5359\text{ cm}^2$ , which rounds to $217\text{ cm}^2$ .

Key Takeaways

Composite shapes can often be decomposed into simpler geometric figures (triangles, sectors, rectangles) for area calculation.
Arc length and sector area formulas are $s = r\theta$ and $A = \frac{1}{2}r^2\theta$ respectively, where $\theta$ must be in radians.
Pythagoras' theorem is useful for finding horizontal or vertical distances when a point lies on a circle and its height is known.

Common Mistakes

Forgetting to include all boundary segments when calculating the perimeter (e.g., omitting $DA$ or using the wrong arc length).
Using degrees instead of radians in the arc length and sector area formulas.
Incorrectly decomposing the area, such as trying to use a trapezoid formula which doesn't apply to curved boundaries.

Things to Be Careful About

Always ensure angles are in radians when using $s = r\theta$ and $A = \frac{1}{2}r^2\theta$ .
Carry out intermediate calculations with full calculator precision and only round the final answer to the required significant figures to avoid rounding errors.
The mark scheme allows for AWRT (Any Correct Rounding Technique), meaning intermediate rounding is acceptable as long as the final answer is correct to 3 significant figures.

Techniques used

apply Pythagoras theorem to find straight edge lengthcalculate arc length using r*thetasum boundary components for perimeterdecompose area into triangle, sector, and quarter-circleevaluate and round to 3 significant figures

Q8MediumSeries

(a)

The first three terms of an arithmetic progression are $25$ , $4p - 1$ and $13 - p$ , where $p$ is a constant.

Find the value of the tenth term of the progression.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

For an arithmetic progression, consecutive terms differ by the same amount, so the middle term is the average of the first and third terms. This gives an equation for $p$ . Once $p$ is known, the common difference and then the tenth term can be found.

Working

Using the constant-difference property:

2(4p - 1) = 25 + (13 - p)

Expand and simplify:

8p - 2 = 38 - p

9p = 40

p = \frac{40}{9}

The common difference is the second term minus the first:

d = (4p - 1) - 25 = \left(4 \times \frac{40}{9} - 1\right) - 25 = \frac{160}{9} - 1 - 25 = \frac{160}{9} - \frac{234}{9} = -\frac{74}{9}

The tenth term of an arithmetic progression is $a + 9d$ :

T_{10} = 25 + 9\left(-\frac{74}{9}\right) = 25 - 74 = -49

Answer

-49

Final answer

-49

Detailed explanation

Walkthrough

An arithmetic progression has a common difference $d$ , so $T_2 - T_1 = T_3 - T_2$ . Equivalently, the middle term is the average of the first and third terms, giving $2(4p - 1) = 25 + (13 - p)$ . Solve this linear equation: expand to $8p - 2 = 38 - p$ , then $9p = 40$ , so $p = \frac{40}{9}$ . Substitute $p$ into $4p - 1$ and subtract $25$ to get the common difference $d = -\frac{74}{9}$ . Finally use $T_n = a + (n-1)d$ with $n = 10$ : $T_{10} = 25 + 9\left(-\frac{74}{9}\right) = -49$ .

Key Takeaways

An arithmetic progression is characterised by a constant common difference. The nth term is $a + (n-1)d$ . To find an unknown parameter, use the constant-difference property to form an equation, solve it, and then substitute back.

Common Mistakes

Setting up the equation incorrectly, for example forgetting that the middle term must be doubled.
Losing the negative sign when expanding $13 - p$ or $4p - 1$ .
Substituting $p$ directly into the nth term formula instead of first finding the common difference $d$ .
Using $10d$ instead of $9d$ for the tenth term.

Things to Be Careful About

The common difference can be negative; here it is $-\frac{74}{9}$ , so the terms decrease.
Keep all working in exact fractions; do not round $\frac{40}{9}$ or $\frac{74}{9}$ .
The mark scheme requires the method for finding $p$ and $d$ to be shown, so present the equation and substitution clearly.

Techniques used

set up the arithmetic progression propertysolve a linear equation for the unknown constantfind the common differenceapply the nth term formula

(b)

The first three terms of a geometric progression are $25$ , $4q - 1$ and $13 - q$ , where $q$ is a positive constant.

Find the sum to infinity of the progression.

4M Similar

DifficultyMedium

Worked solution

Approach

In a geometric progression, the square of the middle term equals the product of the first and third terms. Use this to form a quadratic in $q$ , solve it, choose the positive value, then find the common ratio. Finally, apply the sum to infinity formula, checking that the common ratio satisfies $|r| < 1$ .

Working

For a geometric progression:

(4q - 1)^2 = 25(13 - q)

Expand and simplify:

16q^2 - 8q + 1 = 325 - 25q

16q^2 + 17q - 324 = 0

Factorise:

(q - 4)(16q + 81) = 0

So $q = 4$ or $q = -\frac{81}{16}$ . Since $q$ is positive, $q = 4$ .

The common ratio is:

r = \frac{4q - 1}{25} = \frac{16 - 1}{25} = \frac{15}{25} = \frac{3}{5}

Since $|r| = \frac{3}{5} < 1$ , the sum to infinity exists:

S_\infty = \frac{a}{1 - r} = \frac{25}{1 - \frac{3}{5}} = \frac{25}{\frac{2}{5}} = \frac{125}{2}

Answer

\frac{125}{2}

Final answer

125/2

Detailed explanation

Walkthrough

In a geometric progression, consecutive terms have the same ratio, so $\frac{4q-1}{25} = \frac{13-q}{4q-1}$ . Cross-multiplying gives $(4q-1)^2 = 25(13-q)$ . Expand to $16q^2 - 8q + 1 = 325 - 25q$ and rearrange to $16q^2 + 17q - 324 = 0$ . Factorising gives $(q-4)(16q+81)=0$ , so $q=4$ or $q=-\frac{81}{16}$ . Since $q$ is positive, only $q=4$ is valid. The common ratio is $r = \frac{4q-1}{25} = \frac{3}{5}$ . Because $|r|<1$ , the sum to infinity is $\frac{25}{1 - \frac{3}{5}} = \frac{125}{2}$ . If you solve directly for $r$ instead, the negative ratio $-\frac{17}{20}$ is rejected because it corresponds to a negative value of $q$ .

Key Takeaways

A geometric progression has a constant ratio, so the middle term squared equals the product of the outer terms. The sum to infinity exists only when $|r|<1$ , and is given by $\frac{a}{1-r}$ . Quadratic equations often arise when finding the parameters of a progression.

Common Mistakes

Forgetting to cross-multiply correctly and writing $(4q-1)^2 = 25(13-q)$ incorrectly.
Expanding $(4q-1)^2$ as $16q^2 - 1$ instead of $16q^2 - 8q + 1$ .
Choosing the negative root $q = -\frac{81}{16}$ without checking the condition that $q$ is positive.
Applying the sum to infinity formula without checking $|r|<1$ .
Giving an unsupported answer for the quadratic; show the factorisation or quadratic formula.

Things to Be Careful About

$q$ is given as positive, so reject $q = -\frac{81}{16}$ .
The common ratio must be between $-1$ and $1$ for a finite sum to infinity.
Keep the final answer as an exact fraction $\frac{125}{2}$ , not a rounded decimal.
The mark scheme ignores the extra negative ratio if solving directly for $r$ , but the positive value must be justified.

Techniques used

use the geometric progression property to form an equationsolve a quadratic equationdetermine the common ratio and check convergenceapply the sum to infinity formula

Q96MMediumIntegration

Fig. 4

The diagram shows part of the curve with equation $y = \frac{1}{(5x - 4)^{\frac{1}{3}}}$ and the lines $x = 2.4$ and $y = 1$ . The curve intersects the line $y = 1$ at the point $(1, 1)$ .

Find the exact volume of the solid generated when the shaded region is rotated through $360^\circ$ about the $x$ -axis.

DifficultyMedium

Worked solution

Approach

The shaded region is bounded above by $y = 1$ , below by $y = \frac{1}{(5x - 4)^{1/3}}$ , and on the right by $x = 2.4$ . The left boundary is the intersection point $(1, 1)$ . When this region is rotated $360^\circ$ about the $x$ -axis, the resulting solid can be viewed as a cylinder of radius $1$ and height $2.4 - 1 = 1.4$ , with the volume generated by rotating the area under the curve removed. We calculate the volume of the cylinder and subtract the volume under the curve.

Working

The volume of the cylinder generated by rotating the rectangle bounded by $y = 0$ , $y = 1$ , $x = 1$ , and $x = 2.4$ about the $x$ -axis is:

V_{\text{cylinder}} = \pi r^2 h = \pi (1)^2 (2.4 - 1) = \pi \times 1.4 = \frac{7}{5}\pi

The volume under the curve $y = \frac{1}{(5x - 4)^{1/3}}$ from $x = 1$ to $x = 2.4$ when rotated about the $x$ -axis is:

V_{\text{curve}} = \int_1^{2.4} \pi y^2 \, dx = \pi \int_1^{2.4} \frac{1}{(5x - 4)^{2/3}} \, dx

To integrate $\frac{1}{(5x - 4)^{2/3}}$ , we use the substitution $u = 5x - 4$ , so $du = 5 \, dx$ or $dx = \frac{du}{5}$ :

\int (5x - 4)^{-2/3} \, dx = \frac{1}{5} \int u^{-2/3} \, du = \frac{1}{5} \left( \frac{u^{1/3}}{1/3} \right) = \frac{3}{5} u^{1/3} = \frac{3}{5} (5x - 4)^{1/3}

Evaluating the definite integral with the limits $x = 1$ to $x = 2.4$ :

V_{\text{curve}} = \pi \left[ \frac{3}{5} (5x - 4)^{1/3} \right]_1^{2.4}

At $x = 2.4$ : $5(2.4) - 4 = 12 - 4 = 8$ , so $(8)^{1/3} = 2$ .
At $x = 1$ : $5(1) - 4 = 1$ , so $(1)^{1/3} = 1$ .

V_{\text{curve}} = \pi \left( \frac{3}{5}(2) - \frac{3}{5}(1) \right) = \pi \left( \frac{6}{5} - \frac{3}{5} \right) = \frac{3}{5}\pi

The volume of the solid generated is the cylinder volume minus the volume under the curve:

V = V_{\text{cylinder}} - V_{\text{curve}} = \frac{7}{5}\pi - \frac{3}{5}\pi = \frac{4}{5}\pi

Answer

\frac{4}{5}\pi

Final answer

4pi/5

Detailed explanation

Walkthrough

The problem asks for the volume of a solid of revolution. The shaded region is bounded by $y = 1$ on top, the curve $y = \frac{1}{(5x - 4)^{1/3}}$ on the bottom, and the vertical line $x = 2.4$ on the right. The left boundary is where the curve meets $y = 1$ , which is at $x = 1$ .

When rotated about the $x$ -axis, the outer boundary is the line $y = 1$ and the inner boundary is the curve. Instead of using the washer method directly, it is simpler to think of the solid as a full cylinder (radius $1$ , height $1.4$ ) with the volume under the curve removed.

First, we calculate the volume of the cylinder:

V_{\text{cylinder}} = \pi r^2 h = \pi (1)^2 (2.4 - 1) = \frac{7}{5}\pi

Next, we find the volume under the curve by integrating $\pi y^2$ from $x = 1$ to $x = 2.4$ :

V_{\text{curve}} = \pi \int_1^{2.4} \left( \frac{1}{(5x - 4)^{1/3}} \right)^2 dx = \pi \int_1^{2.4} (5x - 4)^{-2/3} dx

Using the substitution $u = 5x - 4$ , the integral becomes $\frac{1}{5} \int u^{-2/3} du = \frac{3}{5} u^{1/3} = \frac{3}{5}(5x - 4)^{1/3}$ . Evaluating this from $1$ to $2.4$ gives $\frac{3}{5}(8^{1/3} - 1^{1/3}) = \frac{3}{5}(2 - 1) = \frac{3}{5}$ . Multiplying by $\pi$ gives $\frac{3}{5}\pi$ .

Finally, subtracting the volume under the curve from the cylinder volume gives $\frac{7}{5}\pi - \frac{3}{5}\pi = \frac{4}{5}\pi$ .

Key Takeaways

When a region between a curve and a horizontal line is rotated about the $x$ -axis, the solid can often be viewed as a cylinder minus the volume under the curve.
Integration of $(ax + b)^n$ requires the reverse chain rule (or substitution), giving $\frac{1}{a} \frac{(ax + b)^{n+1}}{n+1}$ .
Always check the limits of integration and evaluate the antiderivative at both bounds carefully.

Common Mistakes

Forgetting to square the $y$ -expression when setting up the volume integral ( $\int \pi y^2 dx$ instead of $\int \pi y dx$ ).
Incorrectly applying the chain rule in reverse for the integration, forgetting to divide by the coefficient of $x$ (which is $5$ ).
Subtracting the wrong way around or miscalculating the cylinder height ( $2.4 - 1 = 1.4 = \frac{7}{5}$ , not $\frac{6}{5}$ ).

Things to Be Careful About

The limits of integration are $x = 1$ and $x = 2.4$ . Ensure $5(2.4) - 4 = 8$ and $5(1) - 4 = 1$ are computed correctly.
The cube root of $8$ is $2$ , and the cube root of $1$ is $1$ . Do not confuse these with square roots.
The mark scheme awards a method mark for using $\int y$ instead of $\int y^2$ only if it is used to find the volume of the cylinder directly; using $\int y$ for the curve volume will not earn further marks.

Techniques used

identify region as difference of two solids of revolutioncalculate volume of cylinderset up integral for volume under curveuse substitution to integrate fractional powerevaluate definite integral and subtract volumes

Approach

The line passes through the point $(0,-9)$ , so its equation is $y = mx - 9$ . Substitute this into the circle equation to obtain a quadratic in $x$ . For the line to be tangent, this quadratic must have exactly one solution, so its discriminant must be zero. Solve for $m$ , then substitute each value back to find the $x$ -coordinate of the tangent point and use the line equation to find the corresponding $y$ -coordinate.

Working

The line through $(0,-9)$ with gradient $m$ is

y = mx - 9

Substitute into the circle equation:

(x - 3)^2 + (mx - 9)^2 = 18

Expand and collect all terms on one side:

x^2 - 6x + 9 + m^2x^2 - 18mx + 81 = 18

(m^2 + 1)x^2 - (6 + 18m)x + 72 = 0

For tangency, set the discriminant equal to zero:

(6 + 18m)^2 - 4(m^2 + 1)(72) = 0

Simplify:

36 + 216m + 324m^2 - 288m^2 - 288 = 0

36m^2 + 216m - 252 = 0

Divide by $36$ :

m^2 + 6m - 7 = 0

Factorise:

(m - 1)(m + 7) = 0

So

m = 1 \quad \text{or} \quad m = -7

For $m = 1$ , the quadratic in $x$ becomes:

2x^2 - 24x + 72 = 0

x^2 - 12x + 36 = 0

(x - 6)^2 = 0

Hence $x = 6$ . Then

y = 1(6) - 9 = -3

So one tangent point is $(6, -3)$ .

For $m = -7$ , the quadratic in $x$ becomes:

50x^2 + 120x + 72 = 0

Divide by $2$ :

25x^2 + 60x + 36 = 0

(5x + 6)^2 = 0

Hence $x = -\frac{6}{5}$ . Then

y = -7\left(-\frac{6}{5}\right) - 9 = \frac{42}{5} - \frac{45}{5} = -\frac{3}{5}

So the other tangent point is $\left(-\frac{6}{5}, -\frac{3}{5}\right)$ .

Answer

The two possible values of $m$ are $1$ and $-7$ . The corresponding tangent points are $(6, -3)$ and $\left(-\frac{6}{5}, -\frac{3}{5}\right)$ .

Final answer

m = 1 or m = -7; tangent points (6, -3) and (-6/5, -3/5)

Detailed explanation

Walkthrough

The line passes through $(0,-9)$ , so in $y = mx + c$ we must have $c = -9$ , giving $y = mx - 9$ .

A tangent line touches the circle at exactly one point. Therefore, when we substitute the line into the circle equation, the resulting quadratic in $x$ must have exactly one repeated root. This happens precisely when its discriminant is zero.

Substituting $y = mx - 9$ into $(x-3)^2 + y^2 = 18$ gives

(x - 3)^2 + (mx - 9)^2 = 18

Expanding and collecting terms:

(m^2 + 1)x^2 - (6 + 18m)x + 72 = 0

Apply the discriminant condition with $a = m^2+1$ , $b = -(6+18m)$ and $c = 72$ :

(6 + 18m)^2 - 4(m^2 + 1)(72) = 0

This simplifies to $m^2 + 6m - 7 = 0$ , which factorises as $(m-1)(m+7)=0$ , so $m = 1$ or $m = -7$ .

For each value of $m$ , substitute back into the quadratic in $x$ . The quadratic should be a perfect square; its repeated root is the $x$ -coordinate of the tangent point. Then use $y = mx - 9$ to find the corresponding $y$ -coordinate.

For $m = 1$ , $x = 6$ and $y = -3$ . For $m = -7$ , $x = -\frac{6}{5}$ and $y = -\frac{3}{5}$ .

Key Takeaways

A line is tangent to a circle when it intersects it at exactly one point, so the substitution gives a quadratic with a double root.
The discriminant condition $b^2 - 4ac = 0$ is the algebraic way to enforce tangency.
After finding the gradient $m$ , the tangent point is found by solving the quadratic's repeated root and then using the line equation.
This question connects coordinate geometry of circles with algebraic manipulation of quadratics.

Common Mistakes

Forgetting that the line passes through $(0,-9)$ , so $c = -9$ .
Not collecting all terms on one side of the equation before using the discriminant.
Sign errors when expanding $(6+18m)^2$ or when subtracting $4(m^2+1)(72)$ .
Stopping after finding $m$ without finding the tangent point coordinates.
Using the quadratic formula instead of the discriminant to enforce tangency; the mark scheme specifically requires the discriminant.

Things to Be Careful About

The discriminant must be set equal to zero for tangency.
Keep $m$ symbolic until the discriminant step; do not substitute a numeric value too early.
The coefficient of $x^2$ is $m^2+1$ , not $m^2$ .
For $m = -7$ , the quadratic $50x^2 + 120x + 72 = 0$ simplifies to $(5x+6)^2 = 0$ , giving $x = -\frac{6}{5}$ .
Check that the tangent points satisfy the original circle equation.

Techniques used

substitute the line equation into the circle equationexpand and collect terms into a quadratic in xapply the discriminant condition for tangencysolve the resulting quadratic for the gradientfind the repeated root and tangent point coordinates

Approach

To find where $f(x)$ is decreasing, we need $\frac{dy}{dx} < 0$ . We differentiate $f(x)$ , set the derivative equal to zero to find stationary points, then determine the intervals where the derivative is negative, remembering that $x \neq 0$ .

Working

Differentiate $f(x) = 4x^{-3} - 3x^{-1} + 2$ :

\frac{dy}{dx} = -12x^{-4} + 3x^{-2} = -\frac{12}{x^4} + \frac{3}{x^2}

Set $\frac{dy}{dx} = 0$ to find stationary points:

-\frac{12}{x^4} + \frac{3}{x^2} = 0

Multiply through by $x^4$ (valid since $x \neq 0$ ):

-12 + 3x^2 = 0

3x^2 = 12

x^2 = 4

x = \pm 2

For $f(x)$ to be decreasing, we require $\frac{dy}{dx} < 0$ :

-\frac{12}{x^4} + \frac{3}{x^2} < 0

Combine over a common denominator:

\frac{-12 + 3x^2}{x^4} < 0

Since $x^4 > 0$ for all $x \neq 0$ , the sign of the fraction depends only on the numerator:

-12 + 3x^2 < 0

3x^2 < 12

x^2 < 4

-2 < x < 2

Since $x \neq 0$ , the function is decreasing for:

-2 < x < 0 \quad \text{or} \quad 0 < x < 2

Answer

-2 < x < 0 \quad \text{or} \quad 0 < x < 2

Final answer

-2 < x < 0 or 0 < x < 2

Detailed explanation

Walkthrough

Step 1: Differentiate the function.
We rewrite $f(x) = \frac{4}{x^3} - \frac{3}{x} + 2$ as $4x^{-3} - 3x^{-1} + 2$ and apply the power rule term by term. The derivative of $4x^{-3}$ is $-12x^{-4}$ , and the derivative of $-3x^{-1}$ is $3x^{-2}$ . The constant $2$ differentiates to $0$ .

Step 2: Find stationary points.
We set $\frac{dy}{dx} = 0$ . Multiplying through by $x^4$ (which is positive and non-zero) clears the denominators, giving $-12 + 3x^2 = 0$ . This is a simple quadratic in $x^2$ , yielding $x^2 = 4$ , so $x = \pm 2$ .

Step 3: Determine where the function is decreasing.
We need $\frac{dy}{dx} < 0$ , which means $\frac{-12 + 3x^2}{x^4} < 0$ . Since $x^4 > 0$ always (for $x \neq 0$ ), the inequality reduces to $-12 + 3x^2 < 0$ , giving $x^2 < 4$ , i.e., $-2 < x < 2$ . We must exclude $x = 0$ because the function is undefined there.

Step 4: Write the final answer.
The function is decreasing on $(-2, 0)$ and $(0, 2)$ , or equivalently $-2 < x < 0$ and $0 < x < 2$ .

Key Takeaways

To find where a function is decreasing, set $\frac{dy}{dx} < 0$ .
When solving $\frac{dy}{dx} < 0$ with rational expressions, consider the sign of each factor separately.
Always check the domain of the original function and exclude points where it is undefined.

Common Mistakes

Forgetting to exclude $x = 0$ from the final answer, giving $-2 < x < 2$ without the domain restriction.
Making sign errors when multiplying through by $x^4$ or when simplifying the fraction.
Only finding stationary points and not checking the sign of $\frac{dy}{dx}$ to determine decreasing intervals.
Writing $-2 \leq x \leq 2$ including the endpoints where $\frac{dy}{dx} = 0$ ; the question asks where $f(x)$ is decreasing, which is where $\frac{dy}{dx} < 0$ .

Things to Be Careful About

The function is undefined at $x = 0$ , so this point must be excluded from any interval.
$x^4$ is always positive for $x \neq 0$ , so multiplying or dividing by it does not change the inequality direction.
The mark scheme awards B1FT marks for the final interval, but only if the correct $\frac{dy}{dx}$ expression was used earlier.

Techniques used

differentiate using power ruleset derivative equal to zerosolve quadratic equation in x squareddetermine sign of derivative to find decreasing intervals

(b)

A triangle is bounded by the $y$ -axis, the normal to the curve at the point where $x = 1$ and the tangent to the curve at the point where $x = -1$ .

Find the area of the triangle. Give your answer correct to 3 significant figures.

8M Similar

DifficultyMedium-Hard

Worked solution

Approach

The triangle is bounded by three lines: the $y$ -axis ( $x = 0$ ), the normal to the curve at $x = 1$ , and the tangent to the curve at $x = -1$ . We find the equations of these two lines, then determine the three vertices of the triangle by finding where they intersect. Finally, we compute the area using the base along the $y$ -axis and the horizontal distance to the third vertex.

Working

Step 1: Find the point and gradient on the curve at $x = 1$ .

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

Point: $(1, 3)$

\left.\frac{dy}{dx}\right|_{x=1} = -\frac{12}{1^4} + \frac{3}{1^2} = -12 + 3 = -9

Step 2: Equation of the normal at $x = 1$ .

The normal gradient is the negative reciprocal of the tangent gradient:

m_{\text{norm}} = -\frac{1}{-9} = \frac{1}{9}

Equation of the normal through $(1, 3)$ :

y - 3 = \frac{1}{9}(x - 1)

y = \frac{1}{9}x - \frac{1}{9} + 3 = \frac{1}{9}x + \frac{26}{9}

Step 3: Find the point and gradient on the curve at $x = -1$ .

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

Point: $(-1, 1)$

\left.\frac{dy}{dx}\right|_{x=-1} = -\frac{12}{(-1)^4} + \frac{3}{(-1)^2} = -12 + 3 = -9

Step 4: Equation of the tangent at $x = -1$ .

Equation of the tangent through $(-1, 1)$ :

y - 1 = -9(x + 1)

y = -9x - 9 + 1 = -9x - 8

Step 5: Find the vertices of the triangle.

The triangle is bounded by:

The $y$ -axis: $x = 0$
The normal: $y = \frac{1}{9}x + \frac{26}{9}$
The tangent: $y = -9x - 8$

Vertex A: Intersection of the normal and the $y$ -axis ( $x = 0$ ):

y = \frac{1}{9}(0) + \frac{26}{9} = \frac{26}{9}

Point A: $\left(0, \frac{26}{9}\right)$

Vertex B: Intersection of the tangent and the $y$ -axis ( $x = 0$ ):

y = -9(0) - 8 = -8

Point B: $(0, -8)$

Vertex C: Intersection of the normal and the tangent:

\frac{1}{9}x + \frac{26}{9} = -9x - 8

Multiply through by $9$ :

x + 26 = -81x - 72

82x = -98

x = -\frac{98}{82} = -\frac{49}{41}

Substitute back to find $y$ :

y = -9\left(-\frac{49}{41}\right) - 8 = \frac{441}{41} - \frac{328}{41} = \frac{113}{41}

Point C: $\left(-\frac{49}{41}, \frac{113}{41}\right)$

Step 6: Calculate the area of the triangle.

The base lies along the $y$ -axis from $y = -8$ to $y = \frac{26}{9}$ :

\text{Base} = \frac{26}{9} - (-8) = \frac{26}{9} + \frac{72}{9} = \frac{98}{9}

The height is the horizontal distance from the $y$ -axis to vertex C:

\text{Height} = \left| -\frac{49}{41} \right| = \frac{49}{41}

\text{Area} = \frac{1}{2} \times \frac{98}{9} \times \frac{49}{41} = \frac{1}{2} \times \frac{4802}{369} = \frac{2401}{369}

\text{Area} \approx 6.50677\ldots

Correct to 3 significant figures:

\text{Area} = 6.51

Answer

6.51

Final answer

6.51

Detailed explanation

Walkthrough

Step 1: Evaluate the function and its derivative at $x = 1$ .
Substitute $x = 1$ into $f(x)$ to get $y = 3$ , giving the point $(1, 3)$ . Substitute into $\frac{dy}{dx}$ to get the tangent gradient $m = -9$ .

Step 2: Find the equation of the normal at $(1, 3)$ .
The normal is perpendicular to the tangent, so its gradient is $-\frac{1}{m} = \frac{1}{9}$ . Using point-slope form $y - y_1 = m(x - x_1)$ , we get $y = \frac{1}{9}x + \frac{26}{9}$ .

Step 3: Evaluate the function and its derivative at $x = -1$ .
Substitute $x = -1$ into $f(x)$ to get $y = 1$ , giving the point $(-1, 1)$ . The derivative at $x = -1$ is also $-9$ (since $\frac{dy}{dx}$ depends on $x^2$ and $x^4$ , which are the same for $x = 1$ and $x = -1$ ).

Step 4: Find the equation of the tangent at $(-1, 1)$ .
Using point-slope form with gradient $-9$ : $y - 1 = -9(x + 1)$ , giving $y = -9x - 8$ .

Step 5: Find the three vertices of the triangle.
The triangle is bounded by the $y$ -axis ( $x = 0$ ), the normal, and the tangent. The three vertices are:

Where the normal meets the $y$ -axis: $(0, \frac{26}{9})$
Where the tangent meets the $y$ -axis: $(0, -8)$
Where the normal meets the tangent: solve $\frac{1}{9}x + \frac{26}{9} = -9x - 8$ to get $x = -\frac{49}{41}$ , $y = \frac{113}{41}$

Step 6: Compute the area.
Use the $y$ -axis segment as the base (length $\frac{98}{9}$ ) and the horizontal distance from the $y$ -axis to the third vertex as the height ( $\frac{49}{41}$ ). Area $= \frac{1}{2} \times \frac{98}{9} \times \frac{49}{41} = \frac{2401}{369} \approx 6.51$ .

Key Takeaways

The gradient of the normal is the negative reciprocal of the tangent gradient: $m_{\text{norm}} = -\frac{1}{m_{\text{tan}}}$ .
When a triangle is bounded by a coordinate axis and two lines, use the axis as the base and the perpendicular distance to the third vertex as the height.
Always verify that the intersection point lies on both lines by substituting back.

Common Mistakes

Using the wrong sign for the normal gradient (forgetting the negative in $-\frac{1}{m}$ ).
Making algebra errors when solving $\frac{1}{9}x + \frac{26}{9} = -9x - 8$ ; multiplying through by $9$ first avoids fraction errors.
Computing the area using integration instead of the simple triangle formula, and then making substitution errors.
Forgetting that the base is along the $y$ -axis, so the height is measured horizontally (the $|x|$ -coordinate of the third vertex), not vertically.
Rounding too early; keep exact fractions until the final step.

Things to Be Careful About

The mark scheme allows integration if set up correctly, but the triangle area formula is simpler and less error-prone.
Give the final answer to 3 significant figures as stated: $6.51$ (not $6.507$ or $6.5$ ).
The base length is $\frac{26}{9} - (-8) = \frac{98}{9}$ , not $\frac{26}{9} + 8$ written without combining fractions — both are correct but combining avoids errors.
The $y$ -intercepts of the normal and tangent are $\frac{26}{9}$ and $-8$ respectively; these are the $y$ -coordinates of two vertices, not the base length itself.

Techniques used

evaluate function and derivative at given pointsfind equation of normal using negative reciprocal gradientfind equation of tangentsolve simultaneous equations to find intersectioncalculate triangle area from vertices

Mathematics 9709/11 — May/June 2024

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