(a) Show that $\tan^4 \theta - 1 \equiv \frac{1 - 2\cos^2 \theta}{\cos^4 \theta}$. (b) Hence solve the equation $\cos^2 \theta (\tan^4 \theta - 1) = 7$ for $0^\circ < \theta < 180^\circ$.

(a) tan^4 θ - 1 ≡ (1 - 2cos^2 θ) / cos^4 θ · (b) θ = 70.5° and θ = 109.5°

A curve passes through the point $P(4, 3)$ and is such that $$ \frac{dy}{dx} = \frac{8}{x^2} - \frac{10}{(2x-3)^2}. $$ (a) Find the equation of the normal to the curve at $P$. Give your answer in the form $y = mx + c$. (b) Find the rate of change of the gradient of the curve when $x = 4$. (c) Given that the curve also passes through the point $(-1, q)$, find the value of $q$.

(a) y = -10x + 43 · (b) 7/100 · (c) q = 11

9709/11 Oct/Nov 2025 — Mathematics Solved Past Paper

Q14MMediumQuadratics

Find the set of values of the constant $k$ for which the quadratic equation

3kx^2 + (k+8)x + 3 = 0

has two distinct real roots.

DifficultyMedium

Worked solution

Approach

For a quadratic equation $ax^2 + bx + c = 0$ , two distinct real roots occur when the discriminant is positive:

b^2 - 4ac > 0

Here $a = 3k$ , $b = k+8$ , $c = 3$ . Substitute these into the discriminant and solve the resulting quadratic inequality.

Working

The given quadratic is

3kx^2 + (k+8)x + 3 = 0

So $a = 3k$ , $b = k+8$ , $c = 3$ . For two distinct real roots:

(k+8)^2 - 4(3k)(3) > 0

Simplify:

(k+8)^2 - 36k > 0

Expand $(k+8)^2$ :

k^2 + 16k + 64 - 36k > 0

k^2 - 20k + 64 > 0

Factor the quadratic:

(k-4)(k-16) > 0

The critical values are $k = 4$ and $k = 16$ . Since the quadratic opens upwards, the inequality is satisfied outside the interval between the roots:

k < 4 \quad \text{or} \quad k > 16

Answer

The set of values of $k$ is

k < 4 \quad \text{or} \quad k > 16

Final answer

k < 4 or k > 16

Detailed explanation

Walkthrough

We need the quadratic equation $3kx^2 + (k+8)x + 3 = 0$ to have two distinct real roots. For a quadratic $ax^2 + bx + c = 0$ , the discriminant $b^2 - 4ac$ determines the number of roots:

b^2 - 4ac > 0

gives two distinct real roots.

Here $a = 3k$ , $b = k+8$ , and $c = 3$ . Substituting into the discriminant gives

(k+8)^2 - 4(3k)(3) > 0

Simplify the expression. Expand $(k+8)^2$ to get $k^2 + 16k + 64$ , and compute $4(3k)(3) = 36k$ . Then:

k^2 + 16k + 64 - 36k > 0

k^2 - 20k + 64 > 0

Factor the quadratic:

(k-4)(k-16) > 0

The critical points are $k = 4$ and $k = 16$ . The product is positive when both factors have the same sign. Test the intervals:

For $k < 4$ , both $k-4$ and $k-16$ are negative, so the product is positive.
For $4 < k < 16$ , one factor is negative and one is positive, so the product is negative.
For $k > 16$ , both factors are positive, so the product is positive.

Therefore the solution is $k < 4$ or $k > 16$ . The endpoints $k=4$ and $k=16$ are excluded because the discriminant is zero there, giving one repeated root rather than two distinct real roots.

Key Takeaways

The discriminant $b^2 - 4ac$ is the key tool for determining the nature of roots of a quadratic equation.
Two distinct real roots require a strictly positive discriminant.
Solving a quadratic inequality involves finding the roots and testing the sign in each interval.
The leading coefficient of the quadratic in $k$ is positive, so the inequality $>0$ is satisfied outside the interval between the roots.

Common Mistakes

Using $\geqslant$ or $\leqslant$ instead of strict $>$ or $<$ . The mark scheme explicitly gives B0 if the endpoints are included.
Incorrectly computing $4(3k)(3)$ as $12k$ instead of $36k$ .
Expanding $(k+8)^2$ incorrectly as $k^2 + 8k + 64$ instead of $k^2 + 16k + 64$ .
Failing to factor the quadratic correctly or misidentifying the sign intervals.

Things to Be Careful About

The discriminant must be strictly greater than zero; equal to zero gives a repeated root.
The critical values $k=4$ and $k=16$ are not included in the answer.
If $k=0$ , the equation is not quadratic, but $k=0$ is not in the solution set, so no separate exception is needed.
The answer must be written as two separate strict inequalities, not as a single combined interval.

Techniques used

apply the discriminant condition for two distinct real rootssimplify the discriminant expressionsolve a quadratic inequality by factorisationdetermine the sign intervals of the quadratic

Approach

Write the second and fifth terms of the geometric progression as $ar$ and $ar^4$ , where $r=\cos\theta$ . Dividing the terms eliminates $a$ and gives an equation for $r$ . Solve for $\cos\theta$ using the inverse cosine in the given interval.

Working

Let $r=\cos\theta$ . The second term is

ar = 8

and the fifth term is

ar^4 = \frac{1}{8}

Dividing the fifth term by the second term:

\frac{ar^4}{ar} = \frac{\frac{1}{8}}{8}

so

r^3 = \frac{1}{64}

Taking cube roots:

r = \sqrt[3]{\frac{1}{64}} = \frac{1}{4}

Since $r=\cos\theta$ ,

\cos\theta = \frac{1}{4}

Therefore

\theta = \cos^{-1}\left(\frac{1}{4}\right) = 1.318116\ldots

Correct to 3 significant figures, $\theta = 1.32$ radians.

Answer

\theta = 1.32

Final answer

θ = 1.32 radians (3 s.f.)

Detailed explanation

Walkthrough

In a geometric progression, the $n$ th term is $ar^{n-1}$ . Here $r=\cos\theta$ , so the second term is $a\cos\theta$ and the fifth term is $a\cos^4\theta$ . The question tells us these are $8$ and $\frac{1}{8}$ .

Dividing the fifth term by the second term removes the unknown first term $a$ :

$\frac{ar^4}{ar}=r^3=\frac{\frac{1}{8}}{8}=\frac{1}{64}$ .

Taking the cube root gives $r=\frac{1}{4}$ , so $\cos\theta=\frac{1}{4}$ . Since $0<\theta<\frac{1}{2}\pi$ , the angle is in the first quadrant and there is only one solution. Therefore $\theta=\cos^{-1}\left(\frac{1}{4}\right)=1.318116\ldots$ radians, which is $1.32$ to 3 significant figures.

Key Takeaways

The $n$ th term of a GP is $ar^{n-1}$ .
Dividing two terms of a GP can eliminate the first term and leave an equation for the common ratio.
To solve $\cos^3\theta=k$ , take the cube root before applying the inverse cosine.
The given interval determines which trigonometric solution is valid.

Common Mistakes

Giving the answer in degrees, e.g. $75.5^\circ$ , is not accepted because the interval is in radians.
Dividing the wrong way round: the fifth term divided by the second term gives $r^3=\frac{1}{64}$ , not $64$ .
Forgetting to take the cube root and treating $\cos^3\theta=\frac{1}{64}$ as if it were $\cos\theta=\frac{1}{64}$ .
Rounding $r$ too early, instead of keeping $r=\frac{1}{4}$ exactly for part (b).

Things to Be Careful About

The interval $0<\theta<\frac{1}{2}\pi$ is in radians, so the final answer must be in radians.
The angle is acute, so the inverse cosine is unique in this interval.
The mark scheme allows greater accuracy such as $1.318116\ldots$ , but the 3 significant figure answer is $1.32$ .

Techniques used

divide two terms of a geometric progression to eliminate the first termsolve an equation of the form cos^3 theta = kuse inverse cosine to find an angle in a given interval

(b)

Find the exact value of the sum to infinity.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the exact value $r=\frac{1}{4}$ from part (a) and the given second term to find $a$ . Then apply the sum-to-infinity formula for a geometric progression, valid because $|r|<1$ .

Working

From part (a), $r=\cos\theta=\frac{1}{4}$ . The second term is $ar=8$ , so

a\cdot\frac{1}{4}=8

and hence

a=32

Since $|r|=\frac{1}{4}<1$ , the sum to infinity exists and is

S_\infty=\frac{a}{1-r}=\frac{32}{1-\frac{1}{4}}=\frac{32}{\frac{3}{4}}=\frac{128}{3}

Answer

S_\infty=\frac{128}{3}

Final answer

S∞ = 128/3

Detailed explanation

Walkthrough

From part (a), the common ratio is exactly $r=\frac{1}{4}$ . The second term of the GP is $ar=8$ , so with $r=\frac{1}{4}$ we have $a\cdot\frac{1}{4}=8$ , giving $a=32$ .

A geometric progression has a sum to infinity only when $|r|<1$ . Here $|r|=\frac{1}{4}<1$ , so $S_\infty=\frac{a}{1-r}$ . Substituting $a=32$ and $r=\frac{1}{4}$ gives $S_\infty=\frac{32}{1-\frac{1}{4}}=\frac{32}{\frac{3}{4}}=\frac{128}{3}$ .

Key Takeaways

The sum to infinity formula is $S_\infty=\frac{a}{1-r}$ and requires $|r|<1$ .
Once the common ratio is known, the first term can be found from any given term.
Using exact values throughout gives an exact final answer.

Common Mistakes

Using an approximate value of $r$ instead of the exact $r=\frac{1}{4}$ , which may lose the exact fraction in the final answer.
Substituting into $S_\infty=\frac{a}{1-r}$ without checking that $|r|<1$ .
Arithmetic errors when dividing by $\frac{3}{4}$ : $32\div\frac{3}{4}=32\times\frac{4}{3}=\frac{128}{3}$ .

Things to Be Careful About

The mark scheme allows an attempt using an approximate value of $r$ , but the exact final answer requires using $r=\frac{1}{4}$ .
The fraction should be simplified: $\frac{128}{3}$ is already in simplest form.
The final answer is $S_\infty=\frac{128}{3}$ , not an unsimplified expression such as $\frac{32}{\frac{3}{4}}$ .

Techniques used

find the first term from a known term of a geometric progressioncheck the convergence condition for an infinite geometric seriesapply the sum-to-infinity formula

Q35MMediumSeriesQuadratics

In the expansion of

(px + 3)^5 - \left(x^3 + \frac{p}{x}\right)^4,

the coefficient of $x^4$ is 216.

Find the value of the positive constant $p$ .

DifficultyMedium

Worked solution

Approach

Expand each binomial using the binomial theorem, extract the terms that give $x^4$ , combine their coefficients, and solve the resulting quadratic in $p^2$ .

Working

For $(px + 3)^5$ , the general term is

\binom{5}{r}(px)^{5-r}3^r

The power of $x$ is $5 - r$ . To obtain $x^4$ , take $r = 1$ :

\binom{5}{1}(px)^4 \cdot 3 = 5p^4x^4 \cdot 3 = 15p^4x^4

For $\left(x^3 + \frac{p}{x}\right)^4$ , the general term is

\binom{4}{r}(x^3)^{4-r}\left(\frac{p}{x}\right)^r

The power of $x$ is

3(4-r) - r = 12 - 4r

To obtain $x^4$ , solve $12 - 4r = 4$ , so $r = 2$ :

\binom{4}{2}(x^3)^2\left(\frac{p}{x}\right)^2 = 6p^2x^4

The second bracket is subtracted, so the contribution to the coefficient of $x^4$ is $-6p^2$ .

Therefore the total coefficient of $x^4$ is

15p^4 - 6p^2 = 216

15p^4 - 6p^2 - 216 = 0

Divide by 3:

5p^4 - 2p^2 - 72 = 0

Let $y = p^2$ . Then

5y^2 - 2y - 72 = 0

Factorise:

(5y + 18)(y - 4) = 0

So $y = 4$ or $y = -\frac{18}{5}$ . Since $y = p^2 \geq 0$ , only $y = 4$ is valid:

p^2 = 4

Because $p$ is positive,

p = 2

Answer

p = 2

Final answer

p = 2

Detailed explanation

Walkthrough

We need the coefficient of $x^4$ in the whole expression.

For $(px + 3)^5$ , the binomial theorem gives terms of the form $\binom{5}{r}(px)^{5-r}3^r$ . The power of $x$ is $5-r$ , so to get $x^4$ we need $r=1$ . This gives $5(px)^4 \cdot 3 = 15p^4x^4$ , so the coefficient of the first bracket is $15p^4$ .

For $\left(x^3 + \frac{p}{x}\right)^4$ , the general term is $\binom{4}{r}(x^3)^{4-r}\left(\frac{p}{x}\right)^r$ . Its power of $x$ is $3(4-r)-r = 12-4r$ . Setting this equal to 4 gives $r=2$ , and the term is $6p^2x^4$ . Since the second bracket is subtracted, the coefficient contributed is $-6p^2$ .

Adding the two coefficients and setting the total equal to 216 gives $15p^4 - 6p^2 = 216$ . Rearranging and dividing by 3 gives $5p^4 - 2p^2 - 72 = 0$ . Letting $y = p^2$ turns this into a quadratic in $y$ : $5y^2 - 2y - 72 = 0$ , which factorises as $(5y+18)(y-4)=0$ . Thus $y=4$ or $y=-\frac{18}{5}$ . Since $p^2$ cannot be negative, only $y=4$ is possible, so $p=2$ because $p$ is positive.

Key Takeaways

The binomial theorem lets us choose any individual term by selecting the appropriate value of $r$ .
When an expansion is subtracted, the sign of every coefficient in that expansion is reversed.
An equation such as $15p^4 - 6p^2 - 216 = 0$ is quadratic in $p^2$ ; substitute $y = p^2$ to solve it.
Reject roots that are impossible for the given conditions, such as a negative value of $p^2$ .

Common Mistakes

Writing the coefficient of $x^4$ from the first expansion without the factor of 3, or using an unsimplified binomial coefficient such as $\binom{5}{1}$ .
Forgetting the minus sign before the second bracket, which would incorrectly give $15p^4 + 6p^2 = 216$ .
Treating the equation as quadratic in $p$ rather than in $p^2$ .
Forgetting to reject $p^2 = -\frac{18}{5}$ and therefore reporting a non-real or negative value.

Things to Be Careful About

Check the exponent of $x$ in each general term before selecting $r$ : it is $5-r$ in the first expansion and $12-4r$ in the second.
The second bracket contributes $-6p^2x^4$ , not $+6p^2x^4$ .
Since $p$ is specified to be positive, only $p=2$ is accepted, not $p=-2$ .
Show the full factorisation or quadratic formula step, because the mark scheme requires a valid method for solving the quadratic in $p^2$ .

Techniques used

expand a binomial using the binomial theoremselect the term with the required power of xcombine coefficients with the correct signsolve a quadratic in p^2select the positive root

Approach

Complete the square on $-x^2 - 6x + 1$ by first taking out a factor of $-1$ from the $x$ -terms.

Working

1 - 6x - x^2 = -(x^2 + 6x) + 1

Complete the square inside the bracket:

x^2 + 6x = (x+3)^2 - 9

Therefore:

1 - 6x - x^2 = -[(x+3)^2 - 9] + 1 = -(x+3)^2 + 9 + 1 = 10 - (x+3)^2

So $a = 10$ and $b = 3$ .

Answer

1 - 6x - x^2 = 10 - (x+3)^2, \quad a = 10,\; b = 3

Final answer

a = 10, b = 3

Detailed explanation

Walkthrough

We need to write $1 - 6x - x^2$ in the form $a - (x+b)^2$ . The expression has a negative $x^2$ term, so first take out $-1$ from the terms involving $x$ : $1 - 6x - x^2 = -(x^2 + 6x) + 1$ . Inside the bracket, complete the square: $x^2 + 6x = (x+3)^2 - 9$ . Substituting back gives $-(x+3)^2 + 9 + 1 = 10 - (x+3)^2$ . Comparing with $a - (x+b)^2$ gives $a = 10$ and $b = 3$ .

Key Takeaways

Completing the square works even when the coefficient of $x^2$ is negative: factor out the negative sign first and complete the square inside the bracket. The form $a - (x+b)^2$ is useful for identifying transformations.

Common Mistakes

Forgetting to multiply the constant $-9$ by the factor $-1$ when removing the bracket, leading to $1 - (x+3)^2 - 9$ instead of $1 - (x+3)^2 + 9$ . Also, a sign error in $b$ : since the bracket is $(x+3)^2$ , $b = 3$ , not $-3$ .

Things to Be Careful About

The form is $a - (x+b)^2$ , so the sign inside the bracket is already included. Check by expanding $10 - (x+3)^2 = 10 - (x^2 + 6x + 9) = 1 - 6x - x^2$ .

Techniques used

factor out the negative leading coefficientcomplete the square inside the bracketcompare with the target form a - (x+b)^2

(b)

The graph of $y = x^2$ is transformed to the graph of $y = 1 - 6x - x^2$ by a reflection followed by a translation of $\begin{pmatrix} m \\ n \end{pmatrix}$ . Give details of the reflection and determine the values of $m$ and $n$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Reflect the graph of $y = x^2$ in the $x$ -axis, then apply the translation vector. Compare the resulting equation with the completed-square form from part (a).

Working

Reflecting $y = x^2$ in the $x$ -axis gives:

y = -x^2

Translating by $\begin{pmatrix} m \\ n \end{pmatrix}$ gives:

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

From part (a), the target is:

y = 1 - 6x - x^2 = 10 - (x+3)^2

Comparing:

-(x-m)^2 + n = -(x+3)^2 + 10

So $x-m = x+3$ , giving $m = -3$ , and $n = 10$ .

Answer

Reflection in the $x$ -axis, followed by translation by $\begin{pmatrix} -3 \\ 10 \end{pmatrix}$ .

So $m = -3$ and $n = 10$ .

Final answer

Reflection in x-axis; m = -3, n = 10

Detailed explanation

Walkthrough

Start with $y = x^2$ . A reflection in the $x$ -axis changes $y$ to $-y$ , so the graph becomes $y = -x^2$ . Then a translation by $\begin{pmatrix} m \\ n \end{pmatrix}$ moves the graph $m$ units horizontally and $n$ units vertically. For $y = f(x)$ , this translation gives $y = f(x-m) + n$ , so here $y = -(x-m)^2 + n$ . We want $y = 1 - 6x - x^2$ , which from part (a) is $10 - (x+3)^2$ . Compare $-(x-m)^2 + n$ with $-(x+3)^2 + 10$ . The brackets match when $x-m = x+3$ , so $m = -3$ , and the vertical shift is $n = 10$ . Therefore the required transformation is a reflection in the $x$ -axis followed by translation by $\begin{pmatrix} -3 \\ 10 \end{pmatrix}$ .

Key Takeaways

Transformations of graphs are applied in the order stated. A translation vector $\begin{pmatrix} m \\ n \end{pmatrix}$ replaces $x$ by $x-m$ and adds $n$ to the function. The completed-square form makes the translation visible.

Common Mistakes

Getting the sign of $m$ wrong: a translation by $m$ moves the graph right if $m > 0$ and left if $m < 0$ , but the equation uses $x-m$ . Since the target has $x+3$ , we need $m = -3$ . Also, applying the translation before the reflection would give a different result; the order matters.

Things to Be Careful About

The reflection must be stated as "in the $x$ -axis". The translation vector is $\begin{pmatrix} -3 \\ 10 \end{pmatrix}$ ; note the horizontal component is negative. If part (a) had a different value of $b$ or $a$ , the mark scheme allows follow-through for $m$ and $n$ .

Techniques used

reflect the graph in the x-axisapply a translation vector to the transformed graphcompare the transformed equation with the completed-square form

Q5MediumTrigonometry

(a)

Show that $\tan^4 \theta - 1 \equiv \frac{1 - 2\cos^2 \theta}{\cos^4 \theta}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Express $\tan^4\theta$ as $\frac{\sin^4\theta}{\cos^4\theta}$ , combine the two terms over the common denominator $\cos^4\theta$ , then use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to simplify the numerator to $1 - 2\cos^2\theta$ .

Working

Start from the left-hand side:

\tan^4\theta - 1 = \frac{\sin^4\theta}{\cos^4\theta} - 1

Combine over the common denominator $\cos^4\theta$ :

\tan^4\theta - 1 = \frac{\sin^4\theta - \cos^4\theta}{\cos^4\theta}

Factorise the numerator as a difference of two squares:

\sin^4\theta - \cos^4\theta = \left(\sin^2\theta - \cos^2\theta\right)\left(\sin^2\theta + \cos^2\theta\right)

Since $\sin^2\theta + \cos^2\theta = 1$ :

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

Now substitute $\sin^2\theta = 1 - \cos^2\theta$ :

\sin^4\theta - \cos^4\theta = \left(1 - \cos^2\theta\right) - \cos^2\theta = 1 - 2\cos^2\theta

Therefore:

\tan^4\theta - 1 = \frac{1 - 2\cos^2\theta}{\cos^4\theta}

Answer

\tan^4\theta - 1 \equiv \frac{1 - 2\cos^2\theta}{\cos^4\theta}

Final answer

tan^4 θ - 1 ≡ (1 - 2cos^2 θ) / cos^4 θ

Detailed explanation

Walkthrough

The goal is to prove the identity $\tan^4\theta - 1 \equiv \frac{1 - 2\cos^2\theta}{\cos^4\theta}$ . The key strategy is to express everything in terms of $\cos\theta$ only.

First, write $\tan^4\theta$ as $\frac{\sin^4\theta}{\cos^4\theta}$ using the definition $\tan\theta = \frac{\sin\theta}{\cos\theta}$ . Then combine the two terms over the common denominator $\cos^4\theta$ , giving $\frac{\sin^4\theta - \cos^4\theta}{\cos^4\theta}$ .

The numerator is a difference of two squares: $a^2 - b^2 = (a-b)(a+b)$ with $a = \sin^2\theta$ and $b = \cos^2\theta$ . This factorises to $(\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta)$ . The second factor equals 1 by the Pythagorean identity, leaving $\sin^2\theta - \cos^2\theta$ .

Finally, replace $\sin^2\theta$ with $1 - \cos^2\theta$ (again using the Pythagorean identity) to get $1 - 2\cos^2\theta$ . This gives the required result.

Key Takeaways

The Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ is the fundamental tool for converting between sine and cosine. The difference-of-squares factorisation $a^2 - b^2 = (a-b)(a+b)$ is extremely useful when working with even powers of trigonometric functions. Identity proofs require showing every step clearly, since the answer is given in the question.

Common Mistakes

Omitting intermediate steps. Since this is an "AG" (answer given) question, the marking scheme requires "necessary detail" — a B1 for the initial substitution, M1 for combining into a single fraction, and A1 for the final confirmation.
Working on both sides of the identity simultaneously without reaching a common correct expression — the mark scheme limits credit in this case.
Forgetting that $\tan\theta = \frac{\sin\theta}{\cos\theta}$ raised to the fourth power gives $\frac{\sin^4\theta}{\cos^4\theta}$ .

Things to Be Careful About

The identity must be proven in full; simply stating the result earns no marks. The mark scheme accepts working from either side (LHS to RHS or RHS to LHS) for full marks. An alternative valid approach is to start from the RHS: $\frac{1 - 2\cos^2\theta}{\cos^4\theta} = \frac{(1 - \cos^2\theta)^2}{\cos^4\theta} - 1 = \tan^4\theta - 1$ .

Techniques used

express tan in terms of sin and coscombine terms over a common denominatorfactorise a difference of squaresapply the Pythagorean identity

(b)

Hence solve the equation $\cos^2 \theta (\tan^4 \theta - 1) = 7$ for $0^\circ < \theta < 180^\circ$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Substitute the identity from part (a) into the equation, simplify to obtain a value for $\cos^2\theta$ , then take square roots and find all solutions in the interval $0^\circ < \theta < 180^\circ$ .

Working

Using the identity from part (a):

\cos^2\theta\left(\tan^4\theta - 1\right) = 7

\cos^2\theta \cdot \frac{1 - 2\cos^2\theta}{\cos^4\theta} = 7

Cancel $\cos^2\theta$ from numerator and denominator:

\frac{1 - 2\cos^2\theta}{\cos^2\theta} = 7

Multiply through by $\cos^2\theta$ :

1 - 2\cos^2\theta = 7\cos^2\theta

Collect terms:

1 = 9\cos^2\theta

\cos^2\theta = \frac{1}{9}

Take square roots:

\cos\theta = \pm\frac{1}{3}

For $\cos\theta = \frac{1}{3}$ , in $0^\circ < \theta < 180^\circ$ :

\theta = \cos^{-1}\left(\frac{1}{3}\right) = 70.5^\circ \quad (1 \text{ d.p.})

For $\cos\theta = -\frac{1}{3}$ , in $0^\circ < \theta < 180^\circ$ :

\theta = 180^\circ - \cos^{-1}\left(\frac{1}{3}\right) = 180^\circ - 70.5^\circ = 109.5^\circ \quad (1 \text{ d.p.})

Answer

\theta = 70.5^\circ \quad \text{and} \quad \theta = 109.5^\circ

Final answer

θ = 70.5° and θ = 109.5°

Detailed explanation

Walkthrough

The equation uses the expression $\tan^4\theta - 1$ , which we proved in part (a) equals $\frac{1 - 2\cos^2\theta}{\cos^4\theta}$ . Substituting this in:

\cos^2\theta \cdot \frac{1 - 2\cos^2\theta}{\cos^4\theta} = 7

The $\cos^2\theta$ in the numerator cancels with part of the $\cos^4\theta$ in the denominator, leaving $\frac{1 - 2\cos^2\theta}{\cos^2\theta} = 7$ .

Multiplying through by $\cos^2\theta$ (valid since $\cos^2\theta \neq 0$ in the given interval — at $\theta = 90^\circ$ the original expression would be undefined) gives $1 - 2\cos^2\theta = 7\cos^2\theta$ , which rearranges to $9\cos^2\theta = 1$ , so $\cos^2\theta = \frac{1}{9}$ .

Taking square roots, $\cos\theta = \pm\frac{1}{3}$ . In the interval $0^\circ < \theta < 180^\circ$ , cosine is positive in the first quadrant and negative in the second. So $\cos\theta = \frac{1}{3}$ gives the acute angle $\theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.5^\circ$ , and $\cos\theta = -\frac{1}{3}$ gives the second-quadrant angle $\theta = 180^\circ - 70.5^\circ = 109.5^\circ$ .

Key Takeaways

When solving $\cos^2\theta = k$ , always consider both $\cos\theta = \pm\sqrt{k}$ . In the interval $0^\circ < \theta < 180^\circ$ , a positive cosine value gives one solution in the first quadrant, and the corresponding negative value gives one solution in the second quadrant ( $180^\circ$ minus the acute angle). The identity from part (a) was essential to convert the equation into a form solvable for $\cos\theta$ .

Common Mistakes

Forgetting the negative square root: $\cos\theta = -\frac{1}{3}$ also gives a valid solution.
Including $\theta = 90^\circ$ as a solution. If the equation is written as $\cos^2\theta(1 - 9\cos^2\theta) = 0$ , one might incorrectly conclude $\cos^2\theta = 0$ gives a solution — but $\cos^2\theta = 0$ makes the original expression undefined (division by zero), and the mark scheme awards A0 in this case.
Giving answers in radians when the question specifies degrees. The mark scheme gives only B1 (instead of A1) for $1.23^\text{c}$ and $1.91^\text{c}$ only.

Things to Be Careful About

The interval is strict: $0^\circ < \theta < 180^\circ$ , so $\theta = 0^\circ$ and $\theta = 180^\circ$ are excluded. Both solutions $70.5^\circ$ and $109.5^\circ$ lie strictly inside the interval. The mark scheme accepts greater accuracy ( $70.528\ldots^\circ$ and $109.471\ldots^\circ$ ). When simplifying $\frac{1 - 2\cos^2\theta}{\cos^2\theta} = 7$ to $1 - 2\cos^2\theta = 7\cos^2\theta$ , the multiplication by $\cos^2\theta$ requires $\cos^2\theta \neq 0$ , which holds throughout the interval except at $\theta = 90^\circ$ — and at that point the original equation is not satisfied anyway.

Techniques used

substitute the proven identitysimplify to isolate cos squared thetatake square roots with both signsfind all solutions in the given interval

Q6MediumFunctionsQuadratics

Functions $f$ and $g$ are defined by

\begin{aligned} f(x) &= (x+3)^2 - 12 && \text{for } x \geq 0, \\ g(x) &= 2x - 5 && \text{for } x \in \mathbb{R}. \end{aligned}

(a)

State the range of $f$ .

1M Similar

DifficultyEasy

Worked solution

Approach

For $x \geq 0$ , the squared term $(x+3)^2$ is always non-negative and is smallest when $x = 0$ . Substituting this gives the minimum value of $f$ , so the range is all values from that minimum upwards.

Working

Since $x \geq 0$ , we have $x+3 \geq 3$ , so

(x+3)^2 \geq 9

Therefore

f(x) = (x+3)^2 - 12 \geq 9 - 12 = -3

As $x$ increases without bound, $f(x)$ increases without bound, so the range is all values at least $-3$ .

Answer

f(x) \geq -3

Final answer

f(x) >= -3

Detailed explanation

Walkthrough

The function is $f(x) = (x+3)^2 - 12$ , but its domain is restricted to $x \geq 0$ . Although the vertex of the parabola is at $x = -3$ , that value is not in the domain. On $x \geq 0$ , the smallest possible value of $x+3$ is $3$ , so the smallest value of $(x+3)^2$ is $9$ . Therefore the smallest value of $f(x)$ is $9 - 12 = -3$ . As $x$ gets larger, $f(x)$ gets larger without bound, so the range is $f(x) \geq -3$ .

Key Takeaways

The range of a quadratic function depends on its domain. When the domain excludes the vertex, the minimum or maximum may occur at an endpoint of the domain. Here the squared term is smallest at $x = 0$ , not at $x = -3$ .

Common Mistakes

Stating $x \geq -3$ instead of $f(x) \geq -3$ ; the range is about output values, not input values.
Using $>$ instead of $\geq$ ; because $x = 0$ is included, $f(x) = -3$ is actually achieved.
Assuming the vertex gives the minimum without checking whether it lies in the domain.

Things to Be Careful About

The mark scheme requires the inequality to be $\geq$ , not $>$ , and the variable in the range statement must be $f(x)$ or $y$ , not $x$ .

Techniques used

identify the minimum value of a squared expression over a restricted domainstate the range using an inequality

(b)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To find the inverse, write $y = f(x)$ and rearrange to make $x$ the subject. Because the domain is $x \geq 0$ , the expression $x+3$ is positive, so we take the positive square root. Finally, interchange $x$ and $y$ to express the inverse in terms of $x$ .

Working

Let

y = (x+3)^2 - 12

Add $12$ to both sides:

y + 12 = (x+3)^2

Since $x \geq 0$ , we have $x+3 \geq 3 > 0$ , so

x + 3 = \sqrt{y+12}

Hence

x = -3 + \sqrt{y+12}

Interchanging $x$ and $y$ gives the inverse function:

f^{-1}(x) = -3 + \sqrt{x+12}

Answer

f^{-1}(x) = -3 + \sqrt{x+12}

Final answer

f^{-1}(x) = -3 + sqrt(x+12)

Detailed explanation

Walkthrough

Start by setting $y = f(x)$ . To reverse the operations, add $12$ to isolate the squared bracket, then take square roots. The domain $x \geq 0$ tells us $x+3$ is positive, so only the positive square root is valid. After solving for $x$ in terms of $y$ , swap the variable names to obtain $f^{-1}(x)$ .

Key Takeaways

An inverse function reverses the original function. For a quadratic, the domain must be restricted to make the function one-one, and that same restriction tells you which square-root sign to choose when finding the inverse.

Common Mistakes

Writing $\pm$ in the final answer; the domain $x \geq 0$ forces the positive square root.
Stopping at $x = -3 + \sqrt{y+12}$ without swapping variables; the final expression must be in terms of $x$ .
Making a sign error when rearranging, such as $x = 3 + \sqrt{y+12}$ .

Things to Be Careful About

The mark scheme allows the final answer written as $y = -3 + \sqrt{x+12}$ , but not with a $\pm$ sign. The rearrangement must be a genuine attempt to make $x$ the subject; sign errors are the only errors condoned at the method mark.

Techniques used

write y = f(x) and rearrange to make x the subjecttake the positive square root using the domaininterchange x and y to state the inverse

(c)

Solve the equation $gf(x) = 69$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Form the composite function $gf(x) = g(f(x))$ by substituting $f(x)$ into $g$ . Set the resulting expression equal to $69$ and solve the quadratic equation. Finally, check that any solution lies in the domain $x \geq 0$ of $f$ .

Working

First form the composite:

\begin{aligned} gf(x) &= g\left((x+3)^2 - 12\right) \\ &= 2\left((x+3)^2 - 12\right) - 5 \\ &= 2(x+3)^2 - 24 - 5 \\ &= 2(x+3)^2 - 29 \end{aligned}

Set this equal to $69$ :

2(x+3)^2 - 29 = 69

Solve:

2(x+3)^2 = 98

(x+3)^2 = 49

x+3 = \pm 7

So

x = 4 \quad \text{or} \quad x = -10

Since the domain of $f$ is $x \geq 0$ , discard $x = -10$ .

Answer

x = 4

Final answer

x = 4

Detailed explanation

Walkthrough

To find $gf(x)$ , apply $g$ to the output of $f$ . Since $g(t) = 2t - 5$ , replace $t$ with $(x+3)^2 - 12$ . Simplify to $2(x+3)^2 - 29$ . Setting this equal to $69$ gives a quadratic equation in $x$ . Solve by isolating the squared bracket and taking square roots, or by expanding and factorising. The two algebraic solutions are $x=4$ and $x=-10$ , but $f$ is only defined for $x \geq 0$ , so only $x=4$ is valid.

Key Takeaways

Composite functions are evaluated by substituting the inner function into the outer function. After solving an equation involving a composite function, always check the domain of the inner function, because extraneous solutions may be introduced by the algebra.

Common Mistakes

Forming $gf(x)$ in the wrong order, e.g. substituting $g(x)$ into $f$ instead of $f(x)$ into $g$ .
Forgetting to check the domain and giving both $x=4$ and $x=-10$ ; the mark scheme awards the final mark only for $x=4$ .
Making a sign error when simplifying $2((x+3)^2 - 12) - 5$ .

Things to Be Careful About

The composite must be $2((x+3)^2 - 12) - 5$ , not $2(x+3)^2 - 12 - 5$ (the $-12$ is inside the bracket and must also be multiplied by $2$ ). The final answer must be only $x=4$ ; $x=-10$ is not in the domain of $f$ .

Techniques used

form the composite function gf(x)simplify the composite expressionsolve the resulting quadratic equationdiscard any solution outside the domain of f

Q7MediumCircular MeasureDifferentiation

Fig. 7.1

The diagram shows a sector of a circle with centre $O$ and radius $r$ cm. The shaded region is bounded by the chord $AB$ and the arc $AB$ . The size of angle $AOB$ is $\frac{2}{3}\pi$ radians.

(a)

Show that the area of the shaded region is approximately $0.614r^2$ cm $^2$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The shaded region is a circular segment. Its area equals the area of the sector OAB minus the area of the triangle OAB.

Working

Area of sector OAB:

\frac{1}{2}r^2\theta = \frac{1}{2}r^2 \times \frac{2}{3}\pi = \frac{\pi r^2}{3}

Area of triangle OAB:

\frac{1}{2}r^2\sin\theta = \frac{1}{2}r^2\sin\frac{2\pi}{3} = \frac{1}{2}r^2 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}r^2}{4}

Area of shaded region:

A = \frac{\pi r^2}{3} - \frac{\sqrt{3}r^2}{4} = r^2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)

Calculating the numerical coefficient:

\frac{\pi}{3} - \frac{\sqrt{3}}{4} \approx 1.04720 - 0.43301 = 0.61419

Therefore:

A \approx 0.614r^2 \text{ cm}^2

Answer

A \approx 0.614r^2 \text{ cm}^2

Final answer

0.614r^2

Detailed explanation

Walkthrough

The shaded region in Fig. 7.1 is a circular segment — the region between the chord AB and the arc AB. To find its area, we subtract the area of triangle OAB from the area of sector OAB.

Step 1: Sector area. The formula for the area of a sector with radius $r$ and angle $\theta$ (in radians) is $\frac{1}{2}r^2\theta$ . Substituting $\theta = \frac{2\pi}{3}$ gives $\frac{\pi r^2}{3}$ .

Step 2: Triangle area. The formula for the area of a triangle with two sides $r$ and included angle $\theta$ is $\frac{1}{2}r^2\sin\theta$ . Since $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ , the triangle area is $\frac{\sqrt{3}r^2}{4}$ .

Step 3: Segment area. Subtracting the triangle area from the sector area gives $r^2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)$ . Evaluating the constant factor numerically yields approximately $0.614$ .

Key Takeaways

The area of a circular segment equals the sector area minus the triangle area: $A = \frac{1}{2}r^2(\theta - \sin\theta)$ .
Always ensure angles are in radians when using sector formulae.
The sine formula $\frac{1}{2}ab\sin C$ is essential for finding the triangle area when two sides and the included angle are known.

Common Mistakes

Using degrees instead of radians in the sector area formula.
Forgetting that $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ , not $\frac{1}{2}$ .
Adding the sector and triangle areas instead of subtracting.

Things to Be Careful About

The mark scheme accepts greater accuracy such as $0.61418\dots r^2$ , but $0.614r^2$ is the required answer to 3 significant figures.
Units are cm², which should be stated in the final answer.

Techniques used

calculate sector area using formulacalculate triangle area using sine formulasubtract to find segment area

(b)

It is given that the radius of the circle is increasing at a rate of 0.4 cm s $^{-1}$ .

6M

(i)

Find the rate of increase of the area of the shaded region at the instant when $r = 20$ . Give your answer correct to 2 significant figures.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the result from part (a), $A \approx 0.614r^2$ , to find $\frac{dA}{dr}$ . Then apply the chain rule $\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$ with $r = 20$ and $\frac{dr}{dt} = 0.4$ .

Working

From part (a):

A \approx 0.614r^2

Differentiate with respect to $r$ :

\frac{dA}{dr} = 2 \times 0.614r = 1.228r

At $r = 20$ :

\frac{dA}{dr} = 1.228 \times 20 = 24.56

Apply the chain rule:

\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = 24.56 \times 0.4 = 9.824

Rounding to 2 significant figures:

\frac{dA}{dt} \approx 9.8 \text{ cm}^2\text{s}^{-1}

Answer

9.8 \text{ cm}^2\text{s}^{-1}

Final answer

9.8 cm^2 s^{-1}

Detailed explanation

Walkthrough

We are given that the radius is increasing at $\frac{dr}{dt} = 0.4$ cm s⁻¹ and need to find how fast the shaded area is increasing when $r = 20$ .

Step 1: Differentiate the area formula. From part (a), $A \approx 0.614r^2$ . Differentiating with respect to $r$ gives $\frac{dA}{dr} = 1.228r$ .

Step 2: Evaluate at $r = 20$ . Substituting gives $\frac{dA}{dr} = 1.228 \times 20 = 24.56$ cm² per cm of radius increase.

Step 3: Apply the chain rule. Since $r$ changes with time, we use $\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$ . Multiplying $24.56 \times 0.4$ gives $9.824$ cm² s⁻¹.

Step 4: Round to 2 significant figures. The answer is $9.8$ cm² s⁻¹.

Key Takeaways

The chain rule connects rates of change: $\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$ .
When the area is given as a function of $r$ , differentiate with respect to $r$ first, then multiply by $\frac{dr}{dt}$ .
Always check units: here the result is in cm² s⁻¹.

Common Mistakes

Forgetting the chain rule and using $\frac{dA}{dr}$ directly as the answer.
Using $r = 0.4$ instead of $\frac{dr}{dt} = 0.4$ .
Rounding too early, which can affect the final answer.

Things to Be Careful About

The mark scheme awards a mark for stating or implying $\frac{dA}{dr} = 1.228r$ (or greater accuracy), then for applying the chain rule correctly.
The answer must be given to 2 significant figures: $9.8$ , not $9.82$ or $10$ .
Using the exact value $\frac{\pi}{3} - \frac{\sqrt{3}}{4}$ instead of $0.614$ gives a slightly more accurate intermediate result but the final answer to 2 s.f. is the same.

Techniques used

differentiate area with respect to radiusapply chain rule for related rates

(ii)

Find the rate of increase of the length of the arc $AB$ . Give your answer correct to 2 significant figures.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the arc length formula $l = r\theta$ to express the length of arc AB in terms of $r$ , differentiate with respect to $r$ , then apply the chain rule with $\frac{dr}{dt} = 0.4$ .

Working

Length of arc AB:

l = r\theta = r \times \frac{2}{3}\pi = \frac{2\pi r}{3}

Differentiate with respect to $r$ :

\frac{dl}{dr} = \frac{2\pi}{3}

Apply the chain rule:

\frac{dl}{dt} = \frac{dl}{dr} \times \frac{dr}{dt} = \frac{2\pi}{3} \times 0.4 = \frac{0.8\pi}{3}

Calculating:

\frac{0.8\pi}{3} \approx 0.83776

Rounding to 2 significant figures:

\frac{dl}{dt} \approx 0.84 \text{ cm s}^{-1}

Answer

0.84 \text{ cm s}^{-1}

Final answer

0.84 cm s^{-1}

Detailed explanation

Walkthrough

We need to find how fast the arc length AB is increasing. The arc length depends on the radius $r$ and the angle $\theta = \frac{2\pi}{3}$ , which is constant.

Step 1: Arc length formula. For a circle of radius $r$ and angle $\theta$ in radians, the arc length is $l = r\theta$ . Here $l = \frac{2\pi r}{3}$ .

Step 2: Differentiate. Since $\theta$ is constant, $\frac{dl}{dr} = \frac{2\pi}{3}$ .

Step 3: Chain rule. $\frac{dl}{dt} = \frac{dl}{dr} \times \frac{dr}{dt} = \frac{2\pi}{3} \times 0.4 = \frac{0.8\pi}{3} \approx 0.8378$ cm s⁻¹.

Step 4: Round to 2 significant figures. The answer is $0.84$ cm s⁻¹.

Key Takeaways

Arc length is $l = r\theta$ where $\theta$ is in radians.
When only one variable ( $r$ ) changes and $\theta$ is constant, $\frac{dl}{dr}$ is simply $\theta$ .
The chain rule applies identically: $\frac{dl}{dt} = \frac{dl}{dr} \cdot \frac{dr}{dt}$ .

Common Mistakes

Forgetting that the angle is constant, so $\frac{dl}{dr} = \theta = \frac{2\pi}{3}$ , not a function of $r$ .
Using the wrong angle (e.g., degrees instead of radians).
Not applying the chain rule and giving $\frac{dl}{dr}$ as the final answer.

Things to Be Careful About

The mark scheme accepts any method that differentiates correctly and applies the chain rule.
The answer must be to 2 significant figures: $0.84$ , not $0.838$ or $0.8$ .
Units are cm s⁻¹, not cm² s⁻¹.

Techniques used

use arc length formuladifferentiate arc length with respect to radiusapply chain rule for related rates

Q8MediumIntegration

Fig. 8.1

The diagram shows the curve with equation $y = \frac{1}{2}\sqrt{x}$ and the point $P$ with coordinates $(9, \frac{3}{2})$ . The shaded region is bounded by the curve and the lines $x = 0$ and $y = \frac{3}{2}$ .

(a)

Find the area of the shaded region.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The shaded region is bounded by the $y$ -axis ( $x=0$ ), the horizontal line $y = \frac{3}{2}$ , and the curve $y = \frac{1}{2}\sqrt{x}$ . We can find the area by calculating the area of the rectangle bounded by $x=0$ , $x=9$ , $y=0$ , $y=\frac{3}{2}$ and subtracting the area under the curve from $x=0$ to $x=9$ . Alternatively, we can express $x$ in terms of $y$ and integrate with respect to $y$ .

Working

Method 1: Integrate with respect to $x$

The area of the rectangle from $x=0$ to $x=9$ and $y=0$ to $y=\frac{3}{2}$ is:

9 \times \frac{3}{2} = \frac{27}{2}

The area under the curve $y = \frac{1}{2}x^{1/2}$ from $x=0$ to $x=9$ is:

\int_0^9 \frac{1}{2}x^{1/2} \, dx = \left[ \frac{1}{2} \cdot \frac{x^{3/2}}{3/2} \right]_0^9 = \left[ \frac{1}{3}x^{3/2} \right]_0^9

Evaluating at the limits:

\frac{1}{3}(9)^{3/2} - 0 = \frac{1}{3}(27) = 9

The shaded area is the rectangle area minus the area under the curve:

\frac{27}{2} - 9 = \frac{27}{2} - \frac{18}{2} = \frac{9}{2}

Method 2: Integrate with respect to $y$

From $y = \frac{1}{2}\sqrt{x}$ , we have $\sqrt{x} = 2y$ , so $x = 4y^2$ .

The shaded region is bounded by $x=0$ on the left and $x=4y^2$ on the right, for $y$ from $0$ to $\frac{3}{2}$ .

\text{Area} = \int_0^{3/2} 4y^2 \, dy = \left[ \frac{4}{3}y^3 \right]_0^{3/2}

Evaluating at the limits:

\frac{4}{3} \left( \frac{3}{2} \right)^3 = \frac{4}{3} \cdot \frac{27}{8} = \frac{9}{2}

Answer

\frac{9}{2}

Final answer

9/2

Detailed explanation

Walkthrough

The shaded region is bounded on the left by the $y$ -axis, on the top by the horizontal line $y = \frac{3}{2}$ , and on the right/bottom by the curve $y = \frac{1}{2}\sqrt{x}$ .

Method 1 (integrating with respect to $x$ ): We can find the area by taking the total area of the bounding rectangle (from $x=0$ to $x=9$ and $y=0$ to $y=\frac{3}{2}$ ) and subtracting the area under the curve. The rectangle has area $9 \times \frac{3}{2} = \frac{27}{2}$ . The area under the curve is found by integrating $y = \frac{1}{2}x^{1/2}$ from $0$ to $9$ , which gives $9$ . Subtracting this from the rectangle area gives $\frac{27}{2} - 9 = \frac{9}{2}$ .

Method 2 (integrating with respect to $y$ ): We can also integrate the horizontal width of the shaded region. Solving $y = \frac{1}{2}\sqrt{x}$ for $x$ gives $x = 4y^2$ . The width at height $y$ is simply $x = 4y^2$ . Integrating this from $y=0$ to $y=\frac{3}{2}$ gives $\int_0^{3/2} 4y^2 \, dy = \left[ \frac{4}{3}y^3 \right]_0^{3/2} = \frac{4}{3} \cdot \frac{27}{8} = \frac{9}{2}$ .

Key Takeaways

Area between a curve and the $y$ -axis can be found by integrating $x$ with respect to $y$ .
Alternatively, area can be found by subtracting the area under the curve (integrated with respect to $x$ ) from the bounding rectangle.
When integrating powers, remember to add $1$ to the exponent and divide by the new exponent.

Common Mistakes

Forgetting to subtract the area under the curve from the rectangle area when using Method 1.
Incorrectly integrating $x^{1/2}$ ; the integral is $\frac{2}{3}x^{3/2}$ , so $\frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$ .
Using the wrong limits of integration.

Things to Be Careful About

Ensure the limits of integration match the variable you are integrating with respect to. For $x$ , limits are $0$ to $9$ ; for $y$ , limits are $0$ to $\frac{3}{2}$ .
The shaded region is above the curve, not below it, when integrating with respect to $x$ .

Techniques used

set up definite integral for areaintegrate power functionsubtract from rectangle area

(b)

The shaded region is rotated through $360^\circ$ about the $y$ -axis.

Find the exact volume of the solid produced.

4M Similar

DifficultyMedium

Worked solution

Approach

When a region is rotated about the $y$ -axis, the volume of the solid produced is given by $V = \pi \int x^2 \, dy$ . We must express $x^2$ in terms of $y$ and determine the limits of integration for $y$ .

Working

From the equation of the curve $y = \frac{1}{2}\sqrt{x}$ , we solve for $x$ :

\sqrt{x} = 2y \implies x = 4y^2

Squaring this expression gives:

x^2 = (4y^2)^2 = 16y^4

The shaded region extends from $y = 0$ to $y = \frac{3}{2}$ . The volume of revolution about the $y$ -axis is:

V = \pi \int_0^{3/2} x^2 \, dy = \pi \int_0^{3/2} 16y^4 \, dy

Integrating:

V = \pi \left[ \frac{16}{5}y^5 \right]_0^{3/2}

Evaluating at the limits:

V = \pi \left( \frac{16}{5} \left( \frac{3}{2} \right)^5 - 0 \right)

Calculate $\left( \frac{3}{2} \right)^5$ :

\left( \frac{3}{2} \right)^5 = \frac{243}{32}

Substitute back:

V = \pi \left( \frac{16}{5} \cdot \frac{243}{32} \right) = \pi \left( \frac{1}{5} \cdot \frac{243}{2} \right) = \frac{243}{10}\pi

Answer

\frac{243}{10}\pi

Final answer

243pi/10

Detailed explanation

Walkthrough

The shaded region is rotated $360^\circ$ about the $y$ -axis. The formula for the volume of revolution about the $y$ -axis is $V = \pi \int x^2 \, dy$ .

First, we need $x^2$ in terms of $y$ . From $y = \frac{1}{2}\sqrt{x}$ , we get $x = 4y^2$ , so $x^2 = 16y^4$ .

The limits of integration are the $y$ -values that bound the shaded region, which are from $y = 0$ (at the origin) to $y = \frac{3}{2}$ (at point $P$ ).

We set up the integral $V = \pi \int_0^{3/2} 16y^4 \, dy$ . Integrating $16y^4$ gives $\frac{16}{5}y^5$ . Evaluating this from $0$ to $\frac{3}{2}$ gives $\frac{16}{5} \left( \frac{3}{2} \right)^5 = \frac{16}{5} \cdot \frac{243}{32} = \frac{243}{10}$ . Multiplying by $\pi$ gives the final volume $\frac{243}{10}\pi$ .

Key Takeaways

Volume of revolution about the $y$ -axis requires integrating $\pi x^2$ with respect to $y$ .
Always express the function in the form $x = f(y)$ when rotating about the $y$ -axis.
Remember to square the expression for $x$ before integrating.

Common Mistakes

Attempting to integrate $\pi y^2 \, dx$ (which is for rotation about the $x$ -axis) instead of $\pi x^2 \, dy$ .
Forgetting to square $x$ when setting up the integral (i.e., using $4y^2$ instead of $16y^4$ ).
Omitting the $\pi$ factor in the final answer.

Things to Be Careful About

Ensure you are using the correct volume formula for the axis of rotation. About the $y$ -axis, it is $\int \pi x^2 \, dy$ .
The limits must be in terms of $y$ , not $x$ .

Techniques used

express x in terms of yset up volume integral about y-axisintegrate power functionevaluate definite integral

Q9MediumSeries

An arithmetic progression has first term 2 and common difference $d$ . The sum of the first $n$ terms is denoted by $S_n$ .

(a)

It is given that $(S_2 - 1)$ , $S_4$ , $S_9$ are the first three terms of a second arithmetic progression.

Find the value of $d$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Write $S_2 - 1$ , $S_4$ and $S_9$ in terms of $d$ using the sum formula for the first $n$ terms of an arithmetic progression with first term $a = 2$ :

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

Since these three values form the first three terms of a second arithmetic progression, the middle value is the average of the outer two, so $2S_4 = (S_2 - 1) + S_9$ . Solving the resulting linear equation gives $d$ .

Working

The first progression has $a = 2$ and common difference $d$ , so

S_n = \frac{n}{2}\left[2(2) + (n-1)d\right] = \frac{n}{2}\left[4 + (n-1)d\right]

For $n = 2$ :

S_2 = \frac{2}{2}\left[4 + d\right] = 4 + d

Therefore

S_2 - 1 = 3 + d

For $n = 4$ :

S_4 = \frac{4}{2}\left[4 + 3d\right] = 2(4 + 3d) = 8 + 6d

For $n = 9$ :

S_9 = \frac{9}{2}\left[4 + 8d\right] = 18 + 36d

Since $(S_2 - 1)$ , $S_4$ , $S_9$ are the first three terms of an arithmetic progression, the middle term is the mean of the outer two:

(S_2 - 1) + S_9 = 2S_4

Substitute the expressions in terms of $d$ :

(3 + d) + (18 + 36d) = 2(8 + 6d)

Simplify:

21 + 37d = 16 + 12d

25d = -5

d = -\frac{1}{5}

Answer

d = -\frac{1}{5}

Final answer

d = -1/5

Detailed explanation

Walkthrough

This part gives an arithmetic progression with first term 2 and common difference $d$ , and asks for $d$ given that three particular sums form the first three terms of another arithmetic progression.

Step 1: Recall the sum formula. For an arithmetic progression the sum of the first $n$ terms is

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

With $a = 2$ this becomes

S_n = \frac{n}{2}\left[4 + (n-1)d\right]

Step 2: Compute each needed sum. For $n = 2$ : $S_2 = 4 + d$ , so $S_2 - 1 = 3 + d$ . For $n = 4$ : $S_4 = 8 + 6d$ . For $n = 9$ : $S_9 = 18 + 36d$ . Every quantity is now written in terms of the unknown $d$ .

Step 3: Use the AP condition. Three numbers $p$ , $q$ , $r$ are in arithmetic progression precisely when the middle is the average of the outer two, i.e. $2q = p + r$ . Applying this:

(S_2 - 1) + S_9 = 2S_4

Step 4: Solve. Substitute and simplify:

(3 + d) + (18 + 36d) = 2(8 + 6d)

21 + 37d = 16 + 12d \;\Rightarrow\; 25d = -5 \;\Rightarrow\; d = -\frac{1}{5}

The key step is translating "first three terms of an arithmetic progression" into the double-the-middle condition, which turns the problem into a single linear equation in $d$ .

Key Takeaways

The sum formula for an arithmetic progression must be applied with the correct first term and common difference.
Three consecutive terms of an AP are equally spaced: twice the middle term equals the sum of the two outer terms; this is the condition that links the given sums.
Expressing every unknown piece in terms of one parameter ( $d$ ) reduces the problem to a linear equation.

Common Mistakes

Forgetting to subtract 1 from $S_2$ , producing the wrong first term $4 + d$ instead of $3 + d$ .
Misapplying the sum formula, e.g. omitting the factor of 2 in front of the first term or using the wrong multiplier for $n$ .
Setting up the wrong linkage equation, such as arranging the middle and outer terms in the wrong order.
Arithmetic slips when evaluating $8 + 6d$ and $18 + 36d$ .

Things to Be Careful About

The mark scheme awards A1 for the simplified forms of $S_2 - 1$ , $S_4$ and $S_9$ (namely $3 + d$ , $8 + 6d$ and $18 + 36d$ ); unsimplified forms can be implied by later work but may lose the mark.
The condition that the three values form an AP must be written as $2 \times S_4 = (S_2 - 1) + S_9$ ; stating it correctly earns DM1.
Watch the sign of $d$ : losing the negative sign in $25d = -5$ is a common way to reach the wrong $d = \frac{1}{5}$ .

Techniques used

apply the sum formula for an arithmetic progressionexpress the given sums in terms of the common differenceuse the arithmetic-progression condition on three consecutive termsform and solve a linear equation

(b)

Hence find the difference between the values of the 15th terms of the two arithmetic progressions.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the nth term formula for an arithmetic progression:

T_n = a + (n-1)d

Apply it to both progressions. For the first progression use $a = 2$ and the value of $d$ found in part (a). For the second progression, identify its first term and its common difference from the three given terms, then compute its 15th term. Finally take the difference between the two 15th terms.

Working

First progression: $a = 2$ , $d = -\frac{1}{5}$ .

T_{15} = 2 + 14\left(-\frac{1}{5}\right) = 2 - \frac{14}{5} = \frac{10}{5} - \frac{14}{5} = -\frac{4}{5} = -0.8

Second progression: the first term is $S_2 - 1$ and the common difference is the gap between consecutive terms, $S_4 - (S_2 - 1)$ .

\text{first term} = S_2 - 1 = 3 + d = 3 - \frac{1}{5} = \frac{14}{5}

\text{common difference} = S_4 - (S_2 - 1) = (8 + 6d) - (3 + d) = 5 + 5d = 5 + 5\left(-\frac{1}{5}\right) = 4

15th term of the second progression:

T_{15} = \frac{14}{5} + 14(4) = \frac{14}{5} + 56 = \frac{14}{5} + \frac{280}{5} = \frac{294}{5} = 58.8

Difference between the two 15th terms:

58.8 - (-0.8) = 59.6

Answer

59.6

Final answer

59.6

Detailed explanation

Walkthrough

This part reuses $d = -\frac{1}{5}$ from part (a) to compare the 15th terms of the two progressions.

Step 1: 15th term of the first progression. Apply the nth term formula

T_n = a + (n-1)d

with $a = 2$ and $d = -\frac{1}{5}$ :

T_{15} = 2 + 14\left(-\frac{1}{5}\right) = -\frac{4}{5} = -0.8

Step 2: Identify the second progression. Its first term is the first of the three given values:

S_2 - 1 = 3 + d = \frac{14}{5}

Its common difference is the constant gap between consecutive given values:

S_4 - (S_2 - 1) = 5 + 5d = 4

Step 3: 15th term of the second progression. Apply the same formula:

T_{15} = \frac{14}{5} + 14(4) = \frac{294}{5} = 58.8

Step 4: Difference.

58.8 - (-0.8) = 59.6

The word "hence" signals that the answer from part (a) must be carried through, so no new value of $d$ should be introduced.

Key Takeaways

The nth term of an AP is obtained from its first term and common difference via the formula $T_n = a + (n-1)d$ , which applies identically to any AP.
A second progression is fully determined from just three contained values: the first term is the first value and the common difference is the constant gap between consecutive values.
"Difference between values" means the positive (absolute) difference of the two 15th terms.

Common Mistakes

Introducing a new value of $d$ instead of using $d = -\frac{1}{5}$ from part (a).
Taking the second progression's first term as $S_2 = 4 + d$ instead of $S_2 - 1 = 3 + d$ .
Using the wrong common difference for the second progression, e.g. $S_9 - S_4$ instead of $S_4 - (S_2 - 1)$ .
Using the wrong index when forming the 15th term, e.g. adding 15 differences instead of 14.

Things to Be Careful About

The B1 for the second progression's first term ( $\frac{14}{5}$ ) or common difference (4) is only credited if it follows from the correct $d = -\frac{1}{5}$ .
Use exactly $a + 14d$ for the 15th term — index errors here cost the mark.
The final answer is the absolute difference $59.6$ ; giving $-59.6$ or stopping at $58.8$ loses the last mark.

Techniques used

apply the nth term formula for an arithmetic progressiondetermine the first term and common difference of a second progressioncompute the difference between two identified terms

Q10MediumQuadraticsCoordinate Geometry

A circle has equation $x^2 + y^2 + 4y - 21 = 0$ and a straight line has equation $2x + y - 8 = 0$ . The line intersects the circle at two points.

(a)

Find the coordinates of these two points of intersection.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute the linear equation into the circle equation to eliminate one variable, forming a quadratic equation. Solve the quadratic, then substitute back to find the corresponding coordinates.

Working

From the line equation $2x + y - 8 = 0$ , express $y$ in terms of $x$ :

y = 8 - 2x

Substitute into the circle equation $x^2 + y^2 + 4y - 21 = 0$ :

x^2 + (8 - 2x)^2 + 4(8 - 2x) - 21 = 0

Expand:

x^2 + 64 - 32x + 4x^2 + 32 - 8x - 21 = 0

Collect like terms:

5x^2 - 40x + 75 = 0

Divide through by 5:

x^2 - 8x + 15 = 0

Factorise:

(x - 3)(x - 5) = 0

So $x = 3$ or $x = 5$ .

Substitute back into $y = 8 - 2x$ :

For $x = 3$ : $y = 8 - 6 = 2$ .

For $x = 5$ : $y = 8 - 10 = -2$ .

Answer

The two points of intersection are $(3, 2)$ and $(5, -2)$ .

Final answer

(3, 2) and (5, -2)

Detailed explanation

Walkthrough

We need to find the points where the line and the circle meet. Both equations must be satisfied at the same time. Since the line is linear, we express $y$ in terms of $x$ from the line equation ( $y = 8 - 2x$ ) and substitute into the circle equation. This produces a quadratic in $x$ . Solving it gives the $x$ -coordinates of the two intersection points. Substituting each $x$ -value back into the line equation gives the matching $y$ -coordinates.

Key Takeaways

To solve a linear and a quadratic equation simultaneously, substitute the linear expression into the quadratic.
The intersection points of a line and a circle are the simultaneous solutions of their equations.
Factorising a quadratic is the quickest way to find its roots.

Common Mistakes

Sign errors when expanding $(8 - 2x)^2$ .
Forgetting to substitute the $x$ -values back to find the $y$ -coordinates.
Not simplifying the quadratic before factorising.

Things to be Careful About

The substitution must come from the line equation: $y = 8 - 2x$ .
Both $x$ -values must be used to find both $y$ -values.
Write the answers as ordered pairs $(x, y)$ .

Techniques used

substitute the linear equation into the circle equationsolve the resulting quadratic equation by factorisationsubstitute the solutions back to find the coordinates

(b)

The circle has centre $C$ and the two points of intersection are denoted by $A$ and $B$ .

Find the area of the triangle $ABC$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Complete the square on the circle equation to find the centre $C$ . Since $A$ and $B$ are points on the circle, $CA = CB$ , so the perpendicular from $C$ to the chord $AB$ bisects $AB$ . Using the midpoint of $AB$ to find the perpendicular height, then compute the area as $\frac{1}{2} \times \text{base} \times \text{height}$ .

Working

Complete the square on the circle equation:

x^2 + y^2 + 4y - 21 = 0

x^2 + (y + 2)^2 - 4 - 21 = 0

x^2 + (y + 2)^2 = 25

So the centre is $C(0, -2)$ and the radius is 5.

From part (a), $A = (3, 2)$ and $B = (5, -2)$ .

The midpoint $M$ of $AB$ is:

M = \left( \frac{3+5}{2}, \frac{2 + (-2)}{2} \right) = (4, 0)

The length of $AB$ is:

AB = \sqrt{(5-3)^2 + (-2-2)^2} = \sqrt{4 + 16} = \sqrt{20}

The height of the triangle is the distance from $C$ to $M$ :

CM = \sqrt{(4-0)^2 + (0-(-2))^2} = \sqrt{16 + 4} = \sqrt{20}

Therefore the area is:

\text{Area} = \frac{1}{2} \times AB \times CM = \frac{1}{2} \times \sqrt{20} \times \sqrt{20} = 10

Answer

The area of triangle $ABC$ is $10$ square units.

Final answer

10

Detailed explanation

Walkthrough

First, find the centre of the circle by completing the square. The $y$ -terms $y^2 + 4y$ become $(y + 2)^2 - 4$ . So the circle equation becomes $x^2 + (y + 2)^2 = 25$ , which is the standard form. The centre is $C(0, -2)$ and the radius is 5.

Now consider triangle $ABC$ . Since $A$ and $B$ are on the circle, $CA = CB = 5$ , so the triangle is isosceles. The perpendicular from the centre $C$ to the chord $AB$ bisects the chord. Therefore the foot of the perpendicular is the midpoint of $AB$ , which is $(4, 0)$ .

The base $AB$ has length $\sqrt{20}$ and the height $CM$ is also $\sqrt{20}$ . The area is $\frac{1}{2} \times \sqrt{20} \times \sqrt{20} = 10$ .

Key Takeaways

Completing the square reveals the centre and radius of a circle.
The perpendicular from the centre of a circle to a chord bisects the chord.
The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$ .

Common Mistakes

Sign errors when completing the square (e.g., getting the centre as $(0, 2)$ ).
Using the radius as the height of the triangle.
Forgetting that the foot of the perpendicular is the midpoint of the chord.

Things to be Careful About

The centre is $(0, -2)$ , not $(0, 2)$ .
The height of the triangle is the perpendicular distance from $C$ to $AB$ , which equals $\sqrt{20}$ , not the radius 5.
The area is $10$ square units.

Techniques used

complete the square to find the centre of the circlefind the midpoint of the chord ABcompute the area of a triangle using base and perpendicular height

Q11MediumDifferentiationIntegration

A curve passes through the point $P(4, 3)$ and is such that

\frac{dy}{dx} = \frac{8}{x^2} - \frac{10}{(2x-3)^2}.

(a)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The gradient of the curve at $P(4,3)$ is found by substituting $x=4$ into $\frac{dy}{dx}$ . The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient. Then use the point $(4,3)$ to write the equation of the normal.

Working

The gradient of the curve at $P$ is

m = \frac{8}{4^2} - \frac{10}{(2(4)-3)^2} = \frac{8}{16} - \frac{10}{25} = \frac{1}{2} - \frac{2}{5} = \frac{1}{10}.

The normal has gradient

m_N = -\frac{1}{m} = -\frac{1}{1/10} = -10.

Using $y - y_1 = m_N(x - x_1)$ with $(4,3)$ :

y - 3 = -10(x - 4)

y = -10x + 40 + 3 = -10x + 43.

Answer

y = -10x + 43

Final answer

y = -10x + 43

Detailed explanation

Walkthrough

The derivative $\frac{dy}{dx}$ gives the gradient of the curve at any point. To find the gradient at $P(4,3)$ , substitute $x=4$ . Since the normal is perpendicular to the tangent, its gradient is the negative reciprocal of the tangent gradient. Finally, use the point-slope equation of a straight line and rearrange to the form $y=mx+c$ .

Key Takeaways

The gradient of a curve at a point is found by substituting the $x$ -coordinate into $\frac{dy}{dx}$ .
The normal at a point is perpendicular to the tangent, so its gradient is $-\frac{1}{m}$ .
The equation of a line with gradient $m$ through $(x_1,y_1)$ is $y-y_1=m(x-x_1)$ .

Common Mistakes

Using the tangent gradient instead of the normal gradient.
Forgetting to take the reciprocal of the gradient.
Arithmetic errors when simplifying $\frac{10}{25}$ to $\frac{2}{5}$ .

Things to Be Careful About

The mark scheme requires the gradient of the curve at $P$ to be shown as $\frac{1}{10}$ .
The normal gradient must be $-10$ , not $-\frac{1}{10}$ .
Give the final answer in the requested form $y=mx+c$ .

Techniques used

substitute the x-coordinate into the derivative to find the gradienttake the negative reciprocal of the gradient to find the normal gradientuse the point-slope form to write the equation of the normal

(b)

Find the rate of change of the gradient of the curve when $x = 4$ .

3M Similar

DifficultyMedium

Worked solution

Approach

The rate of change of the gradient of the curve is the second derivative $\frac{d^2y}{dx^2}$ . Differentiate $\frac{dy}{dx}$ term by term, then substitute $x=4$ .

Working

\frac{dy}{dx} = 8x^{-2} - 10(2x-3)^{-2}.

Differentiate with respect to $x$ :

\frac{d^2y}{dx^2} = 8(-2)x^{-3} - 10(-2)(2x-3)^{-3}\cdot 2

\frac{d^2y}{dx^2} = -16x^{-3} + 40(2x-3)^{-3}.

At $x=4$ :

\frac{d^2y}{dx^2} = -\frac{16}{4^3} + \frac{40}{(2(4)-3)^3} = -\frac{16}{64} + \frac{40}{125} = -\frac{1}{4} + \frac{8}{25} = \frac{7}{100}.

Answer

\frac{7}{100}

Final answer

7/100

Detailed explanation

Walkthrough

The rate of change of the gradient is the derivative of the gradient, i.e. $\frac{d^2y}{dx^2}$ . Differentiate $\frac{dy}{dx}=8x^{-2}-10(2x-3)^{-2}$ term by term. For $-10(2x-3)^{-2}$ , use the chain rule: the derivative of $(2x-3)^{-2}$ is $-2(2x-3)^{-3}\cdot 2$ , so the whole term becomes $+40(2x-3)^{-3}$ . Then substitute $x=4$ and simplify the fractions.

Key Takeaways

The rate of change of the gradient is the second derivative.
Differentiating powers of $x$ uses $\frac{d}{dx}(x^n)=nx^{n-1}$ .
The chain rule is needed for terms like $(2x-3)^{-2}$ .

Common Mistakes

Forgetting the chain-rule factor of $2$ when differentiating $(2x-3)^{-2}$ .
Sign errors when differentiating the negative coefficient $-10$ .
Incorrect arithmetic when combining $-\frac{1}{4}$ and $\frac{8}{25}$ .

Things to Be Careful About

The second derivative must be found before substituting $x=4$ .
At $x=4$ , $2x-3=5$ , so the denominator becomes $5^3=125$ .
The final answer is a rate, $\frac{7}{100}$ , not a point or equation.

Techniques used

differentiate the first derivative to obtain the second derivativeapply the chain rule to the term involving 2x - 3substitute x = 4 into the second derivative

(c)

Given that the curve also passes through the point $(-1, q)$ , find the value of $q$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Integrate $\frac{dy}{dx}$ to obtain $y$ as a function of $x$ , including a constant of integration. Use the point $P(4,3)$ to determine that constant, then substitute $x=-1$ to find $q$ .

Working

\frac{dy}{dx} = 8x^{-2} - 10(2x-3)^{-2}.

Integrate term by term:

y = \int \left(8x^{-2} - 10(2x-3)^{-2}\right)\,dx

y = -8x^{-1} + 5(2x-3)^{-1} + c.

Use $P(4,3)$ :

3 = -\frac{8}{4} + \frac{5}{2(4)-3} + c = -2 + \frac{5}{5} + c = -2 + 1 + c = -1 + c.

So $c=4$ . Hence

y = -8x^{-1} + 5(2x-3)^{-1} + 4.

At $x=-1$ :

q = -\frac{8}{-1} + \frac{5}{2(-1)-3} + 4 = 8 + \frac{5}{-5} + 4 = 8 - 1 + 4 = 11.

Answer

q = 11

Final answer

q = 11

Detailed explanation

Walkthrough

To find $y$ from $\frac{dy}{dx}$ , integrate each term. The integral of $8x^{-2}$ is $-8x^{-1}$ , and the integral of $-10(2x-3)^{-2}$ is $5(2x-3)^{-1}$ ; this follows because differentiating $5(2x-3)^{-1}$ gives $-10(2x-3)^{-2}$ . Remember to include the constant $c$ . Use $P(4,3)$ to find $c=4$ , then substitute $x=-1$ to obtain $q=11$ .

Key Takeaways

Integration reverses differentiation and introduces a constant of integration.
For $(ax+b)^n$ , integrate by increasing the power by $1$ and dividing by the new power and by $a$ .
A given point on the curve is used to find the constant of integration.

Common Mistakes

Omitting the constant of integration.
Incorrectly integrating $(2x-3)^{-2}$ : the factor of $\frac{1}{2}$ must appear.
Substituting $x=-1$ into $\frac{dy}{dx}$ instead of into $y$ .

Things to Be Careful About

The final answer must be $q=11$ , not $y=11$ .
When substituting $x=-1$ , $2(-1)-3=-5$ , so the term becomes $\frac{5}{-5}=-1$ .
The mark scheme requires the integrated expression to be shown before substituting the point.

Techniques used

integrate the derivative term by termuse the given point to find the constant of integrationsubstitute x = -1 to find the y-coordinate q

Mathematics 9709/11 — October/November 2025

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