(a) Find the first three terms in the expansion of $\left( 2 - \frac{3}{2}x \right)^5$ in ascending powers of $x$. (b) Use your answer to part (a), with a suitable value of $x$, to find an approximation to $1.985^5$.

(a) 32 - 120x + 180x^2 · (b) 30.818

A curve is such that $\frac{dy}{dx} = 3x^2 + 10x - 8$. (a) Find the set of values of $x$ for which $y$ decreases as $x$ increases. (b) It is given that the maximum point of the curve has $y$-coordinate 27. Find the equation of the curve.

(a) -4 < x < 2/3 · (b) y = x^3 + 5x^2 - 8x - 21

Three points $P$, $Q$ and $R$ have coordinates $P(-13, 5)$, $Q(5, 1)$ and $R(2, k)$, where $k$ is a constant. It is given that the angle $PRQ$ is a right angle. (a) Show that one of the possible values of $k$ is 10, and find the other possible value. (b) It is now given that $k = 10$. A circle passes through the points $P$, $Q$ and $R$. Find the equation of the tangent to the circle at $R$. Give your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers.

(a) k = 10 or k = -4 · (b) 6x + 7y - 82 = 0

9709/13 May/June 2025 — Mathematics Solved Past Paper

Q14MMedium-EasyDifferentiation

A curve has equation $y = 2x + \frac{12}{x^2}$ .

Find the equation of the tangent to the curve at the point $(-2, -1)$ . Give your answer in the form $y = mx + c$ .

DifficultyMedium-Easy

Worked solution

Approach

Differentiate $y = 2x + \frac{12}{x^2}$ term by term. Then substitute $x = -2$ into $\frac{dy}{dx}$ to find the gradient of the tangent. Finally use the point-slope form of a line with the given point $(-2, -1)$ .

Working

y = 2x + 12x^{-2}

Differentiate:

\frac{dy}{dx} = 2 - 24x^{-3}

At $x = -2$ :

\frac{dy}{dx} = 2 - 24(-2)^{-3} = 2 - 24\left(-\frac{1}{8}\right) = 2 + 3 = 5

So the tangent has gradient $m = 5$ . Using $y - y_1 = m(x - x_1)$ with $(-2, -1)$ :

y - (-1) = 5(x - (-2))

y + 1 = 5x + 10

y = 5x + 9

Answer

y = 5x + 9

Final answer

y = 5x + 9

Detailed explanation

Walkthrough

We need the equation of the tangent line at the point $(-2, -1)$ . The gradient of the tangent is the value of the derivative $\frac{dy}{dx}$ at $x = -2$ .

First rewrite the curve equation:

y = 2x + \frac{12}{x^2} = 2x + 12x^{-2}

Differentiate term by term using the power rule. The derivative of $2x$ is $2$ . The derivative of $12x^{-2}$ is $12 \times (-2)x^{-3} = -24x^{-3}$ . So:

\frac{dy}{dx} = 2 - 24x^{-3}

Now substitute $x = -2$ :

\frac{dy}{dx} = 2 - 24(-2)^{-3}

Since $(-2)^{-3} = -\frac{1}{8}$ , we get:

\frac{dy}{dx} = 2 - 24\left(-\frac{1}{8}\right) = 2 + 3 = 5

So the gradient of the tangent is $m = 5$ . The tangent passes through $(-2, -1)$ , so using point-slope form:

y - (-1) = 5(x - (-2))

y + 1 = 5x + 10

y = 5x + 9

Key Takeaways

A term like $\frac{12}{x^2}$ can be rewritten as $12x^{-2}$ and differentiated using the power rule.
The gradient of the tangent to a curve at a point is the value of $\frac{dy}{dx}$ at that point.
The equation of a line can be found using $y - y_1 = m(x - x_1)$ once the gradient and one point are known.

Common Mistakes

Forgetting to differentiate the $\frac{12}{x^2}$ term correctly, especially the negative sign from the power rule.
Substituting the $y$ -coordinate into the derivative instead of the $x$ -coordinate.
Making a sign error when evaluating $(-2)^{-3}$ .
Not simplifying the tangent equation to the required form $y = mx + c$ .

Things to Be Careful About

$(-2)^{-3}$ is negative: $(-2)^{-3} = -\frac{1}{8}$ , not $\frac{1}{8}$ .
The derivative of $x^{-2}$ is $-2x^{-3}$ , so the coefficient becomes $-24$ .
When using point-slope form, be careful with the double negatives: $y - (-1)$ and $x - (-2)$ .
The final answer must be written as $y = 5x + 9$ , with $m = 5$ and $c = 9$ .

Techniques used

rewrite the term with a negative exponentdifferentiate term by term using the power ruleevaluate the derivative at the given x-coordinateform the tangent equation using point-slope form

Similar questions

Q24MMediumSeriesQuadraticsTrigonometry

The first two terms of a geometric progression are

4\sin^2\theta, \ 8\sin^3\theta,

where $\theta$ is an angle such that $0 < \theta < \frac{1}{6}\pi$ .

Given that the sum to infinity of the progression is $\frac{1}{2}$ , find the value of $\theta$ . Give your answer in the form $\sin^{-1} k$ , where $k$ is a rational number.

DifficultyMedium

Worked solution

Approach

Identify the common ratio of the geometric progression, write the sum-to-infinity formula, set it equal to $\frac{1}{2}$ , form a quadratic in $\sin\theta$ , solve it, and select the solution consistent with the given interval.

Working

The common ratio is the ratio of the second term to the first:

r = \frac{8\sin^3\theta}{4\sin^2\theta} = 2\sin\theta

The sum to infinity of a geometric progression is $S_\infty = \frac{a}{1 - r}$ , so:

S_\infty = \frac{4\sin^2\theta}{1 - 2\sin\theta}

Given $S_\infty = \frac{1}{2}$ :

\frac{4\sin^2\theta}{1 - 2\sin\theta} = \frac{1}{2}

Cross-multiplying:

8\sin^2\theta = 1 - 2\sin\theta

8\sin^2\theta + 2\sin\theta - 1 = 0

Let $x = \sin\theta$ . Then:

8x^2 + 2x - 1 = 0

Using the quadratic formula:

x = \frac{-2 \pm \sqrt{4 - 4(8)(-1)}}{16} = \frac{-2 \pm \sqrt{36}}{16} = \frac{-2 \pm 6}{16}

So $x = \frac{4}{16} = \frac{1}{4}$ or $x = \frac{-8}{16} = -\frac{1}{2}$ .

Since $0 < \theta < \frac{\pi}{6}$ , we have $\sin\theta > 0$ , so $\sin\theta = \frac{1}{4}$ .

Answer

\theta = \sin^{-1}\left(\frac{1}{4}\right)

So $k = \frac{1}{4}$ .

Final answer

θ = sin⁻¹(1/4), so k = 1/4

Detailed explanation

Walkthrough

First, we need the common ratio of the geometric progression. The common ratio is found by dividing the second term by the first term: dividing $8\sin^3\theta$ by $4\sin^2\theta$ gives $2\sin\theta$ .

The sum to infinity of a geometric progression exists only when $|r| < 1$ , and equals $\frac{a}{1 - r}$ . Here $a = 4\sin^2\theta$ and $r = 2\sin\theta$ , so the sum is $\frac{4\sin^2\theta}{1 - 2\sin\theta}$ .

We are told this sum equals $\frac{1}{2}$ . Setting the two equal and cross-multiplying gives $8\sin^2\theta = 1 - 2\sin\theta$ , which rearranges to the quadratic $8\sin^2\theta + 2\sin\theta - 1 = 0$ .

Treating $\sin\theta$ as the unknown $x$ , we solve $8x^2 + 2x - 1 = 0$ with the quadratic formula. The discriminant is $4 - 4(8)(-1) = 36$ , so $x = \frac{-2 \pm 6}{16}$ , giving $x = \frac{1}{4}$ or $x = -\frac{1}{2}$ .

Finally, the interval $0 < \theta < \frac{\pi}{6}$ means $\theta$ is a small positive acute angle, so $\sin\theta$ must be positive. Hence $\sin\theta = \frac{1}{4}$ , and $\theta = \sin^{-1}\left(\frac{1}{4}\right)$ .

Key Takeaways

This question tests the sum-to-infinity formula for a geometric progression, solving a quadratic equation, and selecting a trigonometric solution consistent with a restricted interval. It also shows the importance of checking that the common ratio satisfies the convergence condition $|r| < 1$ .

Common Mistakes

Selecting the negative root $\sin\theta = -\frac{1}{2}$ , which is outside the interval $0 < \theta < \frac{\pi}{6}$ .
Forgetting to check the convergence condition $|r| < 1$ for the sum to infinity. Here $r = 2\sin\theta = \frac{1}{2}$ , which is valid.
Not showing the quadratic equation or the solving method — the mark scheme requires the method to be shown.

Things to Be Careful About

The interval $0 < \theta < \frac{\pi}{6}$ restricts $\sin\theta$ to the range $(0, \frac{1}{2})$ , so only the positive root is valid.
The mark scheme awards the final mark for clearly identifying one answer only — give $\theta = \sin^{-1}\left(\frac{1}{4}\right)$ , not both roots.
The common ratio $r = 2\sin\theta = \frac{1}{2}$ satisfies $|r| < 1$ , confirming that the sum to infinity is valid.

Techniques used

find the common ratio of a geometric progressionapply the sum to infinity formulaform and solve a quadratic equation in sin θselect the valid solution in the given interval

Similar questions

Q34MMediumIntegration

Given that $\int_1^3 \left( \frac{a}{(4x-3)^2} + 2 \right) dx = 12$ , find the value of the constant $a$ .

DifficultyMedium

Worked solution

Approach

We integrate the expression term by term, apply the limits $x=3$ and $x=1$ , form an equation in $a$ , and solve.

Working

\int_1^3 \left( \frac{a}{(4x-3)^2} + 2 \right) \, dx = 12

Rewrite the integrand using a negative power:

\int_1^3 \left( a(4x-3)^{-2} + 2 \right) \, dx = 12

Integrate term by term. For $\int (4x-3)^{-2} \, dx$ , increase the power by 1 and divide by the new power and by the derivative of $4x-3$ :

\int a(4x-3)^{-2} \, dx = -\frac{a}{4(4x-3)}

and

\int 2 \, dx = 2x

So the integral is

\left[ -\frac{a}{4(4x-3)} + 2x \right]_1^3

Apply the limits. At $x = 3$ :

-\frac{a}{4(4(3)-3)} + 2(3) = -\frac{a}{36} + 6

At $x = 1$ :

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

Subtract the value at $x=1$ from the value at $x=3$ and set equal to 12:

\left( -\frac{a}{36} + 6 \right) - \left( -\frac{a}{4} + 2 \right) = 12

Simplify:

-\frac{a}{36} + 6 + \frac{a}{4} - 2 = 12

\frac{a}{4} - \frac{a}{36} + 4 = 12

\frac{9a - a}{36} = 8

\frac{8a}{36} = 8

\frac{2a}{9} = 8

a = 36

Answer

a = 36

Final answer

a = 36

Detailed explanation

Walkthrough

We are told that the definite integral of $\frac{a}{(4x-3)^2} + 2$ from $x=1$ to $x=3$ equals 12. The unknown constant $a$ is inside the integrand, so we first integrate normally, treating $a$ as a constant, and then use the given value of the integral to solve for $a$ .

The key integration is $\int (4x-3)^{-2} \, dx$ . This is a composite power. We increase the power from $-2$ to $-1$ , divide by the new power $-1$ , and also divide by the derivative of the inside, which is $4$ . This gives $-\frac{1}{4}(4x-3)^{-1}$ . Multiplying by the constant $a$ gives $-\frac{a}{4(4x-3)}$ . The integral of the constant $2$ is simply $2x$ .

Then we evaluate the antiderivative at the upper limit $x=3$ and the lower limit $x=1$ , and subtract. This creates an expression involving $a$ : $-\frac{a}{36} + 6$ minus $-\frac{a}{4} + 2$ . Simplifying gives $\frac{2a}{9} + 4$ . Setting this equal to 12 and solving gives $a = 36$ .

Key Takeaways

To integrate $(ax+b)^n$ with $n \neq -1$ , increase the power by 1, divide by the new power and by $a$ :

\int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C

Definite integrals use the Fundamental Theorem: evaluate the antiderivative at the upper limit minus the lower limit.
A constant factor $a$ can be pulled through the integral.
After applying limits, solve the resulting equation for the unknown constant.

Common Mistakes

Forgetting to divide by 4 when integrating $(4x-3)^{-2}$ , because the derivative of the inside is 4.
Incorrectly handling the negative power: after integrating, the exponent is $-1$ , not $-2$ .
Sign errors when subtracting the lower-limit value: $\left(-\frac{a}{4}+2\right)$ is subtracted, so the minus signs must be handled carefully.
Arithmetic errors when combining fractions: $\frac{a}{4} - \frac{a}{36} = \frac{8a}{36}$ .
Forgetting to include the integral of $2$ , which is $2x$ .

Things to Be Careful About

The derivative of $4x-3$ is 4, so the integral of $(4x-3)^{-2}$ has a factor $\frac{1}{4}$ .
When applying limits, the lower-limit value must be subtracted from the upper-limit value. In this problem, the lower value contains $-\frac{a}{4}$ , so subtracting it effectively adds $\frac{a}{4}$ .
The final equation is linear in $a$ ; solve by isolating $a$ .
The mark scheme requires the antiderivative to contain $(4x-3)^{-1}$ before limits are applied, so keep the expression in that form.

Techniques used

integrate a composite power of the form (ax+b)^napply limits of a definite integralform and solve a linear equation in a

Similar questions

Q4Medium-EasySeries

(a)

Find the first three terms in the expansion of $\left( 2 - \frac{3}{2}x \right)^5$ in ascending powers of $x$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the binomial theorem for $(a+b)^5$ , taking $a=2$ and $b=-\frac{3}{2}x$ , and keep only the first three terms in ascending powers of $x$ .

Working

The binomial expansion is

(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + \cdots

Substitute $a=2$ and $b=-\frac{3}{2}x$ :

2^5 + 5(2)^4\left(-\frac{3}{2}x\right) + 10(2)^3\left(-\frac{3}{2}x\right)^2

Simplify each term:

32 + 5(16)\left(-\frac{3}{2}x\right) + 10(8)\left(\frac{9}{4}x^2\right)

32 - 120x + 180x^2

Answer

32 - 120x + 180x^2

Final answer

32 - 120x + 180x^2

Detailed explanation

Walkthrough

The binomial theorem for $(a+b)^5$ gives coefficients $1,5,10,10,5,1$ . Since only the first three terms are needed, use the coefficients $1$ , $5$ and $10$ . Taking $a=2$ and $b=-\frac{3}{2}x$ :

The constant term is $2^5=32$ .
The $x$ term is $5(2)^4\left(-\frac{3}{2}x\right)=-120x$ .
The $x^2$ term is $10(2)^3\left(-\frac{3}{2}x\right)^2=180x^2$ .

The negative sign in $b$ makes the $x$ term negative, but the $x^2$ term is positive because the negative sign is squared.

Key Takeaways

This question tests the binomial expansion for a positive integer power. You must identify the correct binomial coefficients, use the powers of $x$ in ascending order, and simplify coefficients fully.

Common Mistakes

Forgetting the negative sign in $-\frac{3}{2}x$ , which would give the wrong sign for the $x$ term.
Not simplifying coefficients, for example leaving $5(16)\left(-\frac{3}{2}x\right)$ instead of writing $-120x$ .
Including terms beyond the first three when only the first three are required.

Things to Be Careful About

The mark scheme requires the coefficients to be simplified. An equivalent factorised form such as $2^5\left(1-\frac{15}{4}x+\frac{45}{8}x^2+\cdots\right)$ is acceptable. Write the terms in ascending powers of $x$ : constant first, then $x$ , then $x^2$ .

Techniques used

apply the binomial theoremsimplify coefficients and powersgenerate the first three ascending terms

(b)

Use your answer to part (a), with a suitable value of $x$ , to find an approximation to $1.985^5$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Set $2 - \frac{3}{2}x$ equal to $1.985$ to find the correct value of $x$ , then substitute that value into the expansion from part (a).

Working

Solve for $x$ :

2 - \frac{3}{2}x = 1.985

\frac{3}{2}x = 0.015

x = 0.01

Substitute $x=0.01$ into $32 - 120x + 180x^2$ :

32 - 120(0.01) + 180(0.01)^2

= 32 - 1.2 + 0.018

= 30.818

Answer

1.985^5 \approx 30.818

Final answer

30.818

Detailed explanation

Walkthrough

To use the expansion in part (a), choose $x$ such that $2-\frac{3}{2}x=1.985$ . Solving gives $x=0.01$ . Substituting this into the first three terms of the expansion gives $32-120(0.01)+180(0.01)^2=32-1.2+0.018=30.818$ . This is an approximation because the expansion is truncated after the $x^2$ term.

Key Takeaways

This is a common use of binomial expansions: choose a value of $x$ that turns the base of the expansion into the number you want, then use the first few terms as an approximation.

Common Mistakes

Substituting $x=1.985$ directly into the expansion instead of solving $2-\frac{3}{2}x=1.985$ .
Choosing $x=0.015$ ; this gives $2-\frac{3}{2}(0.015)=1.9775$ , not $1.985$ .
Writing the final answer without showing the substitution and simplification; the mark scheme requires correct working for the final mark.

Things to Be Careful About

The mark scheme condones substitution into further terms, but the answer should come from substituting into at least three terms. Give the approximation to at least 3 significant figures; $30.818$ is correct.

Techniques used

solve for the substitution valuesubstitute into the binomial expansionevaluate the decimal approximation

Q56MMediumTrigonometryQuadratics

Solve the equation

4\sin\theta\tan\theta = 1 + 5\cos\theta

for $-180^\circ < \theta < 180^\circ$ .

DifficultyMedium

Worked solution

Approach

Rewrite $\tan\theta$ in terms of $\sin\theta$ and $\cos\theta$ , use the Pythagorean identity to obtain a quadratic in $\cos\theta$ , solve the quadratic, then find all $\theta$ values in the given interval.

Working

Start with:

4\sin\theta\tan\theta = 1 + 5\cos\theta

Replace $\tan\theta = \frac{\sin\theta}{\cos\theta}$ :

4\sin\theta \cdot \frac{\sin\theta}{\cos\theta} = 1 + 5\cos\theta

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

Multiply through by $\cos\theta$ (valid since $\cos\theta \neq 0$ ; otherwise $\tan\theta$ is undefined):

4\sin^2\theta = \cos\theta + 5\cos^2\theta

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

4(1 - \cos^2\theta) = \cos\theta + 5\cos^2\theta

4 - 4\cos^2\theta = \cos\theta + 5\cos^2\theta

Rearrange:

9\cos^2\theta + \cos\theta - 4 = 0

Let $c = \cos\theta$ . Solve:

9c^2 + c - 4 = 0

c = \frac{-1 \pm \sqrt{1 + 144}}{18} = \frac{-1 \pm \sqrt{145}}{18}

So:

c = \frac{-1 + \sqrt{145}}{18} \approx 0.6134 \quad \text{or} \quad c = \frac{-1 - \sqrt{145}}{18} \approx -0.7245

Now find $\theta$ in $(-180^\circ, 180^\circ)$ .

For $\cos\theta = 0.6134$ : $\theta = \pm 52.2^\circ$ .

For $\cos\theta = -0.7245$ : $\theta = \pm 136.4^\circ$ .

Answer

\theta = \pm 52.2^\circ, \pm 136.4^\circ

Final answer

θ = ±52.2°, ±136.4°

Detailed explanation

Walkthrough

We start with the equation $4\sin\theta\tan\theta = 1 + 5\cos\theta$ . The first step is to replace $\tan\theta$ with $\frac{\sin\theta}{\cos\theta}$ , since $\tan\theta$ is defined as $\frac{\sin\theta}{\cos\theta}$ . This gives $\frac{4\sin^2\theta}{\cos\theta} = 1 + 5\cos\theta$ .

Next, we multiply both sides by $\cos\theta$ to clear the fraction. This is valid because if $\cos\theta = 0$ , then $\tan\theta$ would be undefined, so such values cannot be solutions anyway. This gives $4\sin^2\theta = \cos\theta + 5\cos^2\theta$ .

Now we have both $\sin^2\theta$ and $\cos^2\theta$ . To reduce to a single variable, we use the Pythagorean identity $\sin^2\theta = 1 - \cos^2\theta$ . Substituting gives $4(1 - \cos^2\theta) = \cos\theta + 5\cos^2\theta$ , which simplifies to $9\cos^2\theta + \cos\theta - 4 = 0$ .

This is a quadratic in $\cos\theta$ . Letting $c = \cos\theta$ , we solve $9c^2 + c - 4 = 0$ using the quadratic formula:

c = \frac{-1 \pm \sqrt{1 + 144}}{18} = \frac{-1 \pm \sqrt{145}}{18}

giving $c \approx 0.6134$ or $c \approx -0.7245$ .

Finally, we find all angles $\theta$ in $(-180^\circ, 180^\circ)$ with these cosine values. For $\cos\theta = 0.6134$ , the angles are $\pm 52.2^\circ$ . For $\cos\theta = -0.7245$ , the angles are $\pm 136.4^\circ$ . These are all within the required interval.

Key Takeaways

When an equation involves $\tan\theta$ , rewriting it as $\frac{\sin\theta}{\cos\theta}$ often helps.
The identity $\sin^2\theta + \cos^2\theta = 1$ lets you express everything in terms of one trigonometric function.
Solving $\cos\theta = k$ for a given interval requires finding all angles with that cosine value, not just the principal value.
The quadratic formula is used when the quadratic does not factorise neatly.

Common Mistakes

Forgetting to multiply the right-hand side by $\cos\theta$ when clearing the fraction.
Sign errors when rearranging to $9\cos^2\theta + \cos\theta - 4 = 0$ .
Only giving one angle for each cosine value (e.g., only $52.2^\circ$ instead of $\pm 52.2^\circ$ ).
Not checking that all answers lie within $(-180^\circ, 180^\circ)$ .

Things to Be Careful About

$\cos\theta = 0$ would make $\tan\theta$ undefined, so it cannot be a solution; the solutions found do not include it.
The interval is open, so the endpoints $-180^\circ$ and $180^\circ$ are not included.
The mark scheme requires all four values; missing one costs marks.

Techniques used

rewrite tan in terms of sin and cosapply the Pythagorean identity sin^2 + cos^2 = 1form and solve a quadratic in cos θfind all solutions in the given interval

Similar questions

Q66MMediumSeries

An arithmetic progression has first term $a$ and common difference 2. The $N$ th term is 55 and the sum of the first $3N$ terms is 5760.

Find the values of $N$ and $a$ .

DifficultyMedium

Worked solution

Approach

Use the formula for the $N$ th term of an arithmetic progression and the formula for the sum of $3N$ terms. This gives two equations in $a$ and $N$ . Eliminate one variable and solve the resulting quadratic.

Working

The $N$ th term is

a + (N-1)(2) = 55

so

a + 2N - 2 = 55 \quad \Rightarrow \quad a = 57 - 2N.

The sum of the first $3N$ terms is

\frac{3N}{2}\left[2a + (3N-1)(2)\right] = 5760.

Simplify:

3N(a + 3N - 1) = 5760.

Substitute $a = 57 - 2N$ :

3N(57 - 2N + 3N - 1) = 5760

3N(N + 56) = 5760

N(N + 56) = 1920

N^2 + 56N - 1920 = 0.

Factorising:

(N - 24)(N + 80) = 0

so $N = 24$ or $N = -80$ . Since $N$ is a positive integer, $N = 24$ .

Then

a = 57 - 2(24) = 9.

Answer

N = 24, \quad a = 9.

Final answer

N = 24, a = 9

Detailed explanation

Walkthrough

We are told the progression is arithmetic with common difference $d = 2$ . For an arithmetic progression:

The $N$ th term is $a + (N-1)d$ . With $d = 2$ , this gives the first equation.
The sum of the first $m$ terms is $\frac{m}{2}(2a + (m-1)d)$ . Here $m = 3N$ , so we can write the second equation.

The first equation lets us express $a$ in terms of $N$ : $a = 57 - 2N$ . We substitute this into the sum equation. Before substituting, simplify the sum equation: $\frac{3N}{2}(2a + (3N-1)\cdot 2) = 3N(a + 3N - 1)$ . Then substitute to get a quadratic in $N$ . Solve by factorising or using the quadratic formula. The positive root gives $N = 24$ , and back-substitution gives $a = 9$ .

Key Takeaways

Know the two key arithmetic progression formulae: $u_n = a + (n-1)d$ and $S_n = \frac{n}{2}(2a + (n-1)d)$ .
Translate worded progression problems into equations before solving.
When two unknowns are involved, use substitution to reduce to one equation.
Always check that a mathematical solution makes sense in the context (here, $N$ must be positive).

Common Mistakes

Using $a + Nd$ instead of $a + (N-1)d$ for the $N$ th term.
Using $a + 3Nd$ instead of $a + (3N-1)d$ in the sum formula, or using the wrong number of terms.
Making sign errors when substituting $a = 57 - 2N$ into the sum equation.
Not simplifying $\frac{3N}{2}(2a + (3N-1)\cdot 2)$ before substituting, which makes the algebra harder.
Accepting $N = -80$ without noticing that $N$ must be a positive integer. The mark scheme says extra solutions may be ignored, but in context the negative root is invalid.
If eliminating $N$ instead, not checking the extra root $a = 217$ .

Things to Be Careful About

The first term is $a$ , not the zeroth term, so the $N$ th term is $a + (N-1)d$ .
The sum is over the first $3N$ terms, so the number of terms is $3N$ and the last term is $a + (3N-1)d$ .
The sum formula simplifies to $3N(a + 3N - 1)$ ; using this simplified form reduces algebra errors.
After solving the quadratic, check the root in the context: $N$ must be positive, and then $a$ follows from $a = 57 - 2N$ .
The mark scheme awards marks for showing the two equations, the substitution, the quadratic, and the final values, so do not skip steps. The quadratic may be written with terms not all on one side and still be accepted.

Techniques used

write the Nth term formula for an arithmetic progressionwrite the sum formula for the first 3N termssubstitute to eliminate a variablesolve the resulting quadratic equation

Similar questions

Q7MediumDifferentiationQuadraticsIntegration

A curve is such that $\frac{dy}{dx} = 3x^2 + 10x - 8$ .

(a)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $y$ decreases as $x$ increases, the gradient $\frac{dy}{dx}$ must be negative. So we solve the quadratic inequality $3x^2+10x-8<0$ .

Working

3x^2+10x-8<0

Factorising:

(3x-2)(x+4)<0

The critical values are $x=-4$ and $x=\frac{2}{3}$ . The quadratic has a positive coefficient of $x^2$ , so its graph lies below the $x$ -axis between the two roots:

-4<x<\frac{2}{3}

Answer

-4<x<\frac{2}{3}

Final answer

-4 < x < 2/3

Detailed explanation

Walkthrough

A function $y$ decreases as $x$ increases exactly when its gradient $\frac{dy}{dx}$ is negative. Here the gradient is given as $3x^2+10x-8$ , so we need to solve $3x^2+10x-8<0$ .

First factorise the quadratic: $3x^2+10x-8 = (3x-2)(x+4)$ . Setting each factor to zero gives the boundary points $x=-4$ and $x=\frac{2}{3}$ . Because the coefficient of $x^2$ is positive, the quadratic is negative between these two roots and positive outside them. Hence the required set is $-4<x<\frac{2}{3}$ .

Key Takeaways

This question tests the link between the sign of the derivative and whether a function is increasing or decreasing. It also tests solving a quadratic inequality: factorise, find the critical values, then choose the correct interval using the shape of a positive quadratic.

Common Mistakes

Solving the quadratic equality instead of an inequality.
Choosing the wrong interval, e.g. $x<-4$ or $x>\frac{2}{3}$ , which would be the region where the quadratic is positive.
Writing the final answer with $\leqslant$ when the derivative must be strictly negative for the function to be decreasing.

Things to Be Careful About

The answer must use strict inequalities because the function is neither increasing nor decreasing at the stationary points themselves. The mark scheme allows the inequality sign to be mistyped as $\leqslant$ in the working, but the final interval must use strict signs.

Techniques used

set the derivative less than zero to find where y is decreasingfactorise a quadratic expressionstate the interval between the roots for a negative quadratic inequality

(b)

It is given that the maximum point of the curve has $y$ -coordinate 27.

Find the equation of the curve.

4M Similar

DifficultyMedium

Worked solution

Approach

To find the curve, integrate $\frac{dy}{dx}$ . The maximum point is a stationary point where $\frac{dy}{dx}=0$ ; we identify which stationary point is the maximum and use the given point $(-4,27)$ to find the constant of integration.

Working

First find the stationary points by setting $\frac{dy}{dx}=0$ :

3x^2+10x-8=0

(3x-2)(x+4)=0

So $x=-4$ or $x=\frac{2}{3}$ .

The second derivative is

\frac{d^2y}{dx^2}=6x+10

At $x=-4$ :

6(-4)+10=-14<0

so $x=-4$ gives a maximum point. Therefore the maximum point has $x=-4$ .

Integrate the derivative:

y=\int (3x^2+10x-8)\,dx=x^3+5x^2-8x+c

Use the given point $(-4,27)$ :

27=(-4)^3+5(-4)^2-8(-4)+c

27=-64+80+32+c

27=48+c

c=-21

Hence the equation of the curve is

y=x^3+5x^2-8x-21

Answer

y=x^3+5x^2-8x-21

Final answer

y = x^3 + 5x^2 - 8x - 21

Detailed explanation

Walkthrough

We need the equation of the curve, so we integrate the given derivative. But the curve is not unique until we know the constant of integration. The question tells us the maximum point has $y$ -coordinate 27, which lets us determine that constant.

First locate the stationary points. Set $3x^2+10x-8=0$ . This factorises as $(3x-2)(x+4)=0$ , giving $x=-4$ and $x=\frac{2}{3}$ . To decide which one is the maximum, compute the second derivative $\frac{d^2y}{dx^2}=6x+10$ . At $x=-4$ it is $-14$ , which is negative, so this is a maximum. At $x=\frac{2}{3}$ it is positive, so that point is a minimum.

Now integrate term by term:

$\int (3x^2+10x-8)\,dx = x^3+5x^2-8x+c$

Since the maximum is at $x=-4$ and its $y$ -coordinate is 27, substitute $x=-4$ and $y=27$ into the integrated expression. This gives $27=48+c$ , so $c=-21$ . Therefore the curve is $y=x^3+5x^2-8x-21$ .

Key Takeaways

A given point on a curve determines the constant of integration.
A maximum or minimum occurs where $\frac{dy}{dx}=0$ ; the second derivative tells us which type it is.
When a maximum point is described, its $x$ -coordinate must be found from the derivative before using the $y$ -coordinate.

Common Mistakes

Using the wrong stationary point, such as $x=\frac{2}{3}$ , and so finding the wrong constant.
Forgetting the constant of integration entirely.
Substituting $x=-4$ into the original derivative instead of into the integrated expression.
Making an arithmetic error when evaluating $(-4)^3+5(-4)^2-8(-4)$ .

Things to Be Careful About

The second derivative test is essential to identify the maximum. The mark scheme allows the maximum point to be identified from the lower critical value from part (a) if the factorisation is consistent; however, the correct mathematical reason is that $\frac{d^2y}{dx^2}<0$ at $x=-4$ . The final equation must be stated as a cubic; continuing to find a straight line would lose marks.

Techniques used

identify the maximum stationary point using the second derivativeintegrate the derivative term by termsubstitute a given point to find the constant of integration

Q8MediumTrigonometryCircular MeasureIntegration

Fig. 1

The diagram shows a square $ABCD$ where each side has length $12\text{ cm}$ . Points $E$ and $F$ lie on the sides $BC$ and $CD$ respectively and are such that $BE = \frac{1}{3}BC$ and $DF = \frac{1}{3}DC$ . The arc $EF$ is part of a circle with centre $A$ . The shaded region is bounded by the arc $EF$ and the line segments $EC$ and $FC$ .

(a)

Show that the size of angle $EAF$ is 0.9273 radians, correct to 4 significant figures.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The angle $\angle BAD = \frac{\pi}{2}$ since $ABCD$ is a square. By symmetry, $\angle EAB = \angle FAD$ . We find $\angle EAB$ using right-angled triangle $ABE$ , then compute $\angle EAF = \frac{\pi}{2} - 2\angle EAB$ .

Working

Since $BE = \frac{1}{3}BC = \frac{1}{3}(12) = 4\text{ cm}$ and $AB = 12\text{ cm}$ , in right-angled triangle $ABE$ :

\tan(\angle EAB) = \frac{BE}{AB} = \frac{4}{12} = \frac{1}{3}

\angle EAB = \tan^{-1}\left(\frac{1}{3}\right) = 0.32175\ldots\text{ rad}

By symmetry, $\angle FAD = \angle EAB = 0.32175\ldots\text{ rad}$ .

Therefore:

\angle EAF = \frac{\pi}{2} - 2\tan^{-1}\left(\frac{1}{3}\right)

\angle EAF = 1.57080\ldots - 2(0.32175\ldots)

\angle EAF = 1.57080\ldots - 0.64350\ldots = 0.92729\ldots\text{ rad}

Rounding to 4 significant figures:

\angle EAF = 0.9273\text{ rad}

Answer

\angle EAF = 0.9273\text{ rad}

Final answer

0.9273 radians

Detailed explanation

Walkthrough

We are given a square $ABCD$ with side length $12\text{ cm}$ . Points $E$ and $F$ lie on $BC$ and $CD$ respectively, with $BE = \frac{1}{3}BC = 4\text{ cm}$ and $DF = \frac{1}{3}DC = 4\text{ cm}$ . The arc $EF$ has centre $A$ , so $AE = AF$ is the radius of that circle.

To find $\angle EAF$ , we use the fact that $\angle BAD = \frac{\pi}{2}$ (a corner of a square). The angle $\angle EAF$ sits inside this right angle, with $\angle EAB$ on one side and $\angle FAD$ on the other. By the symmetry of the configuration ( $BE = DF = 4\text{ cm}$ and $AB = AD = 12\text{ cm}$ ), we have $\angle EAB = \angle FAD$ .

In right-angled triangle $ABE$ , the side opposite $\angle EAB$ is $BE = 4$ and the adjacent side is $AB = 12$ . So $\tan(\angle EAB) = \frac{4}{12} = \frac{1}{3}$ , giving $\angle EAB = \tan^{-1}\left(\frac{1}{3}\right) \approx 0.32175\text{ rad}$ .

Subtracting both $\angle EAB$ and $\angle FAD$ from $\frac{\pi}{2}$ gives $\angle EAF = \frac{\pi}{2} - 2(0.32175\ldots) = 0.92729\ldots \approx 0.9273\text{ rad}$ to 4 significant figures.

Key Takeaways

Angles in a square's corner sum to $\frac{\pi}{2}$ .
Inverse tangent can be used to find angles in right-angled triangles when two sides are known.
Symmetry often simplifies angle calculations in geometric figures.

Common Mistakes

Forgetting that the angles must be in radians when the final answer is required in radians.
Using $r = 12$ instead of the correct radius $AE = \sqrt{160}$ in later parts.
Not rounding to 4 significant figures at the end (e.g., $53.14°$ does not convert correctly to $0.9273$ rad).

Things to Be Careful About

Ensure at least one intermediate value (like $0.32175\ldots$ or $0.6435$ ) is shown to justify the final answer.
The angle $53.14°$ converts to approximately $0.9278$ rad, which is not $0.9273$ , so working in degrees with early rounding will lose the accuracy mark.

Techniques used

use right-angled triangle to find angleapply inverse tangentsubtract angles from pi/2

(b)

Find the perimeter of the shaded region.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The perimeter of the shaded region consists of the arc $EF$ , the segment $EC$ , and the segment $FC$ . We find the radius $AE = AF$ using Pythagoras' theorem, then use the arc length formula $r\theta$ with the angle from part (a).

Working

In right-angled triangle $ABE$ :

AE = \sqrt{AB^2 + BE^2} = \sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160} \approx 12.649\text{ cm}

The arc length $EF$ is:

r\theta = \sqrt{160} \times 0.9273 = 12.649\ldots \times 0.9273 = 11.729\ldots\text{ cm}

The straight segments are:

EC = BC - BE = 12 - 4 = 8\text{ cm}

FC = DC - DF = 12 - 4 = 8\text{ cm}

The perimeter of the shaded region is:

\text{Perimeter} = \text{arc } EF + EC + FC = 11.729\ldots + 8 + 8 = 27.729\ldots\text{ cm}

Rounding to 3 significant figures:

\text{Perimeter} = 27.7\text{ cm}

Answer

\text{Perimeter} = 27.7\text{ cm}

Final answer

27.7 cm

Detailed explanation

Walkthrough

The shaded region is bounded by three parts: the arc $EF$ , the line segment $EC$ , and the line segment $FC$ . To find the perimeter, we add the lengths of these three boundaries.

First, we need the radius of the circle, which is $AE$ (or $AF$ , since both are radii from centre $A$ to points on the arc). Using Pythagoras' theorem in right-angled triangle $ABE$ with $AB = 12$ and $BE = 4$ :

$AE = \sqrt{12^2 + 4^2} = \sqrt{160} \approx 12.649\text{ cm}$ .

The arc length is given by $r\theta$ where $r = \sqrt{160}$ and $\theta = 0.9273\text{ rad}$ (from part a):

Arc $EF = \sqrt{160} \times 0.9273 \approx 11.729\text{ cm}$ .

The straight segments $EC$ and $FC$ are each $12 - 4 = 8\text{ cm}$ .

Total perimeter $= 11.729 + 8 + 8 = 27.729 \approx 27.7\text{ cm}$ .

Key Takeaways

The arc length formula $s = r\theta$ requires $\theta$ in radians and $r$ as the actual radius, not a side of the square.
The perimeter of a region bounded by curves and lines is the sum of all boundary lengths.

Common Mistakes

Using $r = 12$ (the side of the square) instead of the correct radius $\sqrt{160} \approx 12.65$ .
Forgetting to include all three boundary components (arc + two straight segments).

Things to Be Careful About

The mark scheme explicitly notes that using $r = 12$ scores $M0$ for the arc length step.
Use the unrounded radius $\sqrt{160}$ in intermediate calculations to avoid accumulation of rounding errors.

Techniques used

calculate radius using Pythagorasapply arc length formula r thetaadd arc length and straight segments

(c)

Find the area of the shaded region.

3M Similar

DifficultyMedium

Worked solution

Approach

The shaded region can be found by subtracting the unshaded areas from the total area of the square. The unshaded areas are: triangle $ABE$ , triangle $ADF$ , and the circular sector $AEF$ . Alternatively, the shaded region equals triangle $ECF$ minus the circular segment between chord $EF$ and arc $EF$ .

Working

Method 1: Square minus parts

Area of square $ABCD$ :

12^2 = 144\text{ cm}^2

Area of triangle $ABE$ :

\frac{1}{2} \times AB \times BE = \frac{1}{2} \times 12 \times 4 = 24\text{ cm}^2

Area of triangle $ADF$ :

\frac{1}{2} \times AD \times DF = \frac{1}{2} \times 12 \times 4 = 24\text{ cm}^2

Area of sector $AEF$ (using $r = \sqrt{160}$ and $\theta = 0.9273$ ):

\frac{1}{2}r^2\theta = \frac{1}{2} \times 160 \times 0.9273 = 80 \times 0.9273 = 74.184\text{ cm}^2

Area of shaded region:

144 - 24 - 24 - 74.184 = 21.816\text{ cm}^2

Rounding to 3 significant figures:

\text{Area} = 21.8\text{ cm}^2

Answer

\text{Area} = 21.8\text{ cm}^2

Final answer

21.8 cm^2

Detailed explanation

Walkthrough

The shaded region is bounded by arc $EF$ , segment $EC$ , and segment $FC$ . To find its area, we use a decomposition approach: the total area of the square minus the three unshaded regions (triangle $ABE$ , triangle $ADF$ , and sector $AEF$ ).

The square has area $12^2 = 144\text{ cm}^2$ .

Triangle $ABE$ has base $AB = 12$ and height $BE = 4$ , so its area is $\frac{1}{2} \times 12 \times 4 = 24\text{ cm}^2$ . By symmetry, triangle $ADF$ has the same area: $24\text{ cm}^2$ .

The sector $AEF$ has radius $r = AE = \sqrt{160}$ and angle $\theta = 0.9273\text{ rad}$ . Its area is $\frac{1}{2}r^2\theta = \frac{1}{2} \times 160 \times 0.9273 = 74.184\text{ cm}^2$ .

Subtracting: $144 - 24 - 24 - 74.184 = 21.816 \approx 21.8\text{ cm}^2$ .

Alternative Method 2: Triangle minus segment

Triangle $ECF$ has $EC = FC = 8$ , so its area is $\frac{1}{2} \times 8 \times 8 = 32\text{ cm}^2$ .

The circular segment (between chord $EF$ and arc $EF$ ) has area = sector $AEF$ - triangle $AEF$ .

Triangle $AEF$ has area $\frac{1}{2}r^2\sin\theta = \frac{1}{2} \times 160 \times \sin(0.9273) = 80 \times 0.8 = 64\text{ cm}^2$ .

Segment area $= 74.184 - 64 = 10.184\text{ cm}^2$ .

Shaded area $= 32 - 10.184 = 21.816 \approx 21.8\text{ cm}^2$ .

Key Takeaways

Area of a circular sector is $\frac{1}{2}r^2\theta$ with $\theta$ in radians.
Decomposing a complex region into simpler known shapes (square, triangles, sectors) is a powerful technique.
The segment area (between a chord and an arc) equals sector area minus triangle area.

Common Mistakes

Using $r = 12$ instead of $r = \sqrt{160}$ for the sector area (this scores $M0$ in the mark scheme).
Forgetting to subtract all three unshaded regions from the square.
Rounding intermediate values too early, leading to loss of accuracy.

Things to Be Careful About

The mark scheme condones using $r = 12$ in part (c) but explicitly notes it scores $M0$ for the sector area step in part (b).
Ensure $\theta$ is in radians when using $\frac{1}{2}r^2\theta$ .
Both methods should give the same answer: $21.8\text{ cm}^2$ to 3 significant figures.

Techniques used

calculate sector area using 1/2 r^2 thetasubtract triangles and sector from square areadecompose region into known shapes

Q9MediumCoordinate GeometryQuadratics

Three points $P$ , $Q$ and $R$ have coordinates $P(-13, 5)$ , $Q(5, 1)$ and $R(2, k)$ , where $k$ is a constant. It is given that the angle $PRQ$ is a right angle.

(a)

Show that one of the possible values of $k$ is 10, and find the other possible value.

4M Similar

DifficultyMedium

Worked solution

Approach

Since angle $PRQ$ is a right angle, the lines $PR$ and $RQ$ are perpendicular. Use the condition that the product of their gradients is $-1$ , form a quadratic equation in $k$ , and solve it.

Working

Gradient of $PR$ :

m_{PR} = \frac{k-5}{2-(-13)} = \frac{k-5}{15}

Gradient of $RQ$ :

m_{RQ} = \frac{k-1}{2-5} = \frac{k-1}{-3} = -\frac{k-1}{3}

For a right angle at $R$ , $PR \perp RQ$ , so:

m_{PR} \times m_{RQ} = -1

\frac{k-5}{15} \times \left(-\frac{k-1}{3}\right) = -1

Simplify:

-\frac{(k-5)(k-1)}{45} = -1

(k-5)(k-1) = 45

Expand:

k^2 - 6k + 5 = 45

k^2 - 6k - 40 = 0

Factorise:

(k-10)(k+4) = 0

So:

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

Thus one possible value is $10$ , and the other possible value is $-4$ .

Answer

k = 10 \text{ or } k = -4

Final answer

k = 10 or k = -4

Detailed explanation

Walkthrough

The angle $PRQ$ is the angle at $R$ between the lines $RP$ and $RQ$ . If this angle is a right angle, the two lines are perpendicular. For perpendicular lines, the product of their gradients is $-1$ . So the key is to write the gradients of $PR$ and $RQ$ in terms of $k$ .

The gradient of $PR$ uses the change in $y$ divided by the change in $x$ from $P$ to $R$ :

m_{PR} = \frac{k-5}{2-(-13)} = \frac{k-5}{15}

The gradient of $RQ$ uses the change from $R$ to $Q$ :

m_{RQ} = \frac{k-1}{2-5} = -\frac{k-1}{3}

Multiplying these and setting the product equal to $-1$ gives an equation. After clearing fractions and expanding, we get a quadratic. Factorising it gives $k=10$ and $k=-4$ , so the statement is verified and the other value is found.

Key Takeaways

This question combines coordinate geometry with algebra. The main idea is the perpendicular gradient condition $m_1 m_2 = -1$ . It also shows how a geometric condition can be converted into an algebraic equation.

Common Mistakes

Using the gradient of $RQ$ as $\frac{k-1}{2-5}$ and forgetting the negative sign from the denominator.
Setting the product of gradients equal to $1$ instead of $-1$ .
Expanding $(k-5)(k-1)$ incorrectly.
Stopping after finding only $k=10$ instead of giving both solutions.

Things to Be Careful About

The gradient formula must use the same order of coordinates in the numerator and denominator.
The mark scheme allows sign errors during simplification but requires the final quadratic and both values.
An unsupported answer is not enough; show the perpendicular condition and the equation.

Techniques used

calculate gradients of two linesapply the perpendicular gradient conditionform and solve a quadratic equation

(b)

It is now given that $k = 10$ . A circle passes through the points $P$ , $Q$ and $R$ .

Find the equation of the tangent to the circle at $R$ . Give your answer in the form $ax + by + c = 0$ , where $a$ , $b$ and $c$ are integers.

5M Similar

DifficultyMedium

Worked solution

Approach

With $k=10$ , the angle $PRQ$ is a right angle, so $PQ$ is a diameter of the circle. Find the centre as the midpoint of $PQ$ , then use the fact that the tangent at $R$ is perpendicular to the radius $CR$ .

Working

When $k=10$ , $R=(2,10)$ .

Since angle $PRQ$ is a right angle, the chord $PQ$ is a diameter of the circle. Therefore the centre $C$ is the midpoint of $P(-13,5)$ and $Q(5,1)$ :

C = \left( \frac{-13+5}{2}, \frac{5+1}{2} \right) = (-4,3)

Gradient of the radius $CR$ :

m_{CR} = \frac{10-3}{2-(-4)} = \frac{7}{6}

The tangent at $R$ is perpendicular to the radius, so its gradient is:

m_T = -\frac{1}{m_{CR}} = -\frac{6}{7}

Using the point-slope form through $R(2,10)$ :

y - 10 = -\frac{6}{7}(x-2)

Multiply through by $7$ :

7y - 70 = -6x + 12

Rearrange into the form $ax+by+c=0$ :

6x + 7y - 82 = 0

Answer

6x + 7y - 82 = 0

Final answer

6x + 7y - 82 = 0

Detailed explanation

Walkthrough

With $k=10$ , $R=(2,10)$ . Since $P$ , $Q$ and $R$ lie on a circle and angle $PRQ = 90^\circ$ , the chord $PQ$ must be a diameter. This is because an angle subtended by a diameter at the circumference is a right angle, and conversely a right angle at the circumference subtends a diameter. Therefore the centre of the circle is the midpoint of $P$ and $Q$ :

C = \left(\frac{-13+5}{2}, \frac{5+1}{2}\right) = (-4,3)

The radius to $R$ has gradient:

m_{CR} = \frac{10-3}{2-(-4)} = \frac{7}{6}

A tangent is perpendicular to the radius at the point of contact, so its gradient is the negative reciprocal:

m_T = -\frac{6}{7}

Using the point-slope form through $R(2,10)$ :

y-10 = -\frac{6}{7}(x-2)

Multiplying through by $7$ and rearranging gives the required integer form.

Key Takeaways

In a circle, a right angle subtended by a chord means that chord is a diameter.
The centre is the midpoint of a diameter.
The tangent and radius are perpendicular, so use negative reciprocal gradients.
A linear equation can be rearranged into the form $ax+by+c=0$ .

Common Mistakes

Using the midpoint of $PR$ or $QR$ as the centre instead of the midpoint of $PQ$ .
Taking the gradient of the tangent as the same as the radius gradient, instead of the negative reciprocal.
Forgetting to convert the fractional equation into integer coefficients.
Using the wrong point in the point-slope equation.

Things to Be Careful About

The centre must be found from the diameter $PQ$ , not from any other pair of points.
The radius gradient uses the centre and $R$ , not $P$ or $Q$ .
The final answer must be in the form $ax+by+c=0$ with integer $a$ , $b$ and $c$ ; here $6x+7y-82=0$ .

Techniques used

locate the circle centre as the midpoint of a diametercalculate the gradient of a radiususe the negative reciprocal gradient for the tangentwrite the tangent equation in integer linear form

Q10MediumDifferentiationFunctions

A curve $C$ has equation $y = \frac{9}{2x-5} + 2x - 5$ .

(a)

Find the coordinates of the two stationary points.

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate the function using the power rule and chain rule, then set the derivative equal to zero and solve for $x$ . Substitute each $x$ -value back into $y$ to find the coordinates.

Working

y = \frac{9}{2x-5} + 2x - 5 = 9(2x-5)^{-1} + 2x - 5

\frac{dy}{dx} = -9 \times 2(2x-5)^{-2} + 2 = -\frac{18}{(2x-5)^2} + 2

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

-\frac{18}{(2x-5)^2} + 2 = 0

\frac{18}{(2x-5)^2} = 2

(2x-5)^2 = 9

2x-5 = \pm 3

So $x = 4$ or $x = 1$ .

For $x = 1$ :

y = \frac{9}{2(1)-5} + 2(1) - 5 = \frac{9}{-3} - 3 = -3 - 3 = -6

For $x = 4$ :

y = \frac{9}{2(4)-5} + 2(4) - 5 = \frac{9}{3} + 3 = 3 + 3 = 6

Answer

The stationary points are $(1, -6)$ and $(4, 6)$ .

Final answer

(1, -6) and (4, 6)

Detailed explanation

Walkthrough

We are given $y = \frac{9}{2x-5} + 2x - 5$ . Rewrite the fraction as $9(2x-5)^{-1}$ so we can differentiate it with the power rule. Because the inside is $2x-5$ , the chain rule contributes a factor of $2$ , giving $\frac{dy}{dx} = -\frac{18}{(2x-5)^2} + 2$ .

Stationary points occur where $\frac{dy}{dx} = 0$ . Setting the derivative equal to zero gives $\frac{18}{(2x-5)^2} = 2$ , so $(2x-5)^2 = 9$ . Taking square roots gives $2x-5 = \pm 3$ , so $x = 4$ or $x = 1$ . Substituting these back into the original equation gives the $y$ -coordinates.

Key Takeaways

Rewriting a fraction as a negative power makes differentiation easier. Stationary points are found by setting the first derivative to zero. Always substitute the $x$ -values back into the original equation, not the derivative.

Common Mistakes

Forgetting the factor of $2$ from differentiating $2x-5$ .
Taking only the positive square root of $(2x-5)^2 = 9$ and losing $x = 1$ .
Substituting into $\frac{dy}{dx}$ instead of into $y$ when finding the coordinates.

Things to Be Careful About

The curve is not differentiable at $x = \frac{5}{2}$ , but neither stationary point is there. Both $x = 1$ and $x = 4$ must be obtained from the square root.

Techniques used

rewrite the fraction as a negative powerdifferentiate using the chain ruleset the derivative to zerosolve the resulting equation for xsubstitute back into the original equation

(b)

Find $\frac{d^2y}{dx^2}$ and hence determine the nature of each stationary point.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate $\frac{dy}{dx}$ using the chain rule to obtain $\frac{d^2y}{dx^2}$ . Substitute the $x$ -coordinates of the stationary points and use the sign of the second derivative to classify each point.

Working

From part (a),

\frac{dy}{dx} = -18(2x-5)^{-2} + 2

Differentiate again:

\frac{d^2y}{dx^2} = -18 \times (-2) \times 2(2x-5)^{-3} = \frac{72}{(2x-5)^3}

At $x = 1$ :

\frac{d^2y}{dx^2} = \frac{72}{(2-5)^3} = \frac{72}{-27} = -\frac{8}{3} < 0

So $(1, -6)$ is a maximum point.

At $x = 4$ :

\frac{d^2y}{dx^2} = \frac{72}{(8-5)^3} = \frac{72}{27} = \frac{8}{3} > 0

So $(4, 6)$ is a minimum point.

Answer

$(1, -6)$ is a maximum and $(4, 6)$ is a minimum.

Final answer

(1, -6) is a maximum; (4, 6) is a minimum

Detailed explanation

Walkthrough

We differentiate $\frac{dy}{dx}$ with respect to $x$ again. Since $\frac{dy}{dx} = -18(2x-5)^{-2} + 2$ , applying the chain rule again gives $\frac{d^2y}{dx^2} = 72(2x-5)^{-3}$ . We then evaluate this at each stationary point.

At $x = 1$ , the denominator is negative, so the second derivative is negative, meaning the curve is concave down and the point is a maximum. At $x = 4$ , the denominator is positive, so the second derivative is positive, meaning the curve is concave up and the point is a minimum.

Key Takeaways

The second derivative tells us the concavity of the curve: negative means a maximum, positive means a minimum. The chain rule is needed again because the denominator is $2x-5$ .

Common Mistakes

Forgetting the extra factor of $2$ from the chain rule when differentiating $(2x-5)^{-2}$ .
Using the sign of the second derivative incorrectly, for example calling a negative value a minimum.
Not evaluating the second derivative at both stationary points.

Things to Be Careful About

The denominator $(2x-5)^3$ changes sign between $x = 1$ and $x = 4$ , so the two values of $\frac{d^2y}{dx^2}$ have opposite signs. The nature of each point must follow from the signs of the second derivative, and unsupported answers may lose marks.

Techniques used

differentiate the first derivative againevaluate the second derivative at each stationary pointuse the sign of the second derivative to classify the points

(c)

The curve $C$ is transformed to the curve $C_1$ using a translation of $\begin{pmatrix} -3 \\ 7 \end{pmatrix}$ followed by reflection in the $x$ -axis.

4M

(i)

State the coordinates of the maximum point of $C_1$ .

1M Similar

DifficultyMedium

Worked solution

Approach

The original curve has stationary points from parts (a) and (b). Apply the translation and then the reflection to these key points. A translation by $(-3,7)$ moves $x$ to $x-3$ and $y$ to $y+7$ ; reflection in the $x$ -axis changes $y$ to $-y$ .

Working

Original maximum: $(1,-6)$ .

After translation:

(1,-6) \rightarrow (-2,1)

After reflection:

(-2,1) \rightarrow (-2,-1)

So this is a minimum on $C_1$ .

Original minimum: $(4,6)$ .

After translation:

(4,6) \rightarrow (1,13)

After reflection:

(1,13) \rightarrow (1,-13)

So this is the maximum point of $C_1$ .

Answer

The maximum point of $C_1$ is $(1,-13)$ .

Final answer

(1, -13)

Detailed explanation

Walkthrough

From part (b), the original curve has a maximum at $(1,-6)$ and a minimum at $(4,6)$ . A translation by the vector $(-3,7)$ moves every point left by $3$ and up by $7$ , so the original maximum becomes $(-2,1)$ . Reflection in the $x$ -axis then changes the sign of the $y$ -coordinate, giving $(-2,-1)$ , which is a minimum.

Similarly, the original minimum $(4,6)$ moves to $(1,13)$ under the translation, and reflection in the $x$ -axis gives $(1,-13)$ . Since the maximum of $C_1$ comes from the original minimum, the maximum point is $(1,-13)$ .

Key Takeaways

Transformations can be applied directly to key points. A translation shifts both coordinates, and a reflection in the $x$ -axis changes the sign of the $y$ -coordinate. A reflection can swap maxima and minima.

Common Mistakes

Applying the transformation to the original maximum instead of recognising that the maximum of $C_1$ comes from the original minimum.
Forgetting to change the sign of the $y$ -coordinate after reflection.
Mixing up the direction of the translation.

Things to Be Careful About

The translation vector $(-3,7)$ means $x$ decreases by $3$ and $y$ increases by $7$ . After reflection in the $x$ -axis, the roles of maximum and minimum are reversed.

Techniques used

apply the translation to key pointsreflect points in the x-axisidentify which original stationary point becomes the maximum

(ii)

Find the equation of $C_1$ in the form $y = \frac{a}{bx+c} + dx + e$ , where $a$ , $b$ , $c$ , $d$ and $e$ are integers.

3M Similar

DifficultyMedium

Worked solution

Approach

Apply the translation to the original equation by replacing $x$ with $x+3$ and adding $7$ , then apply the reflection in the $x$ -axis by multiplying the whole expression by $-1$ . Finally simplify into the required form.

Working

Original:

f(x) = \frac{9}{2x-5} + 2x - 5

After translation by $(-3,7)$ :

y = f(x+3) + 7 = \frac{9}{2(x+3)-5} + 2(x+3) - 5 + 7

Simplify:

y = \frac{9}{2x+1} + 2x + 6 - 5 + 7 = \frac{9}{2x+1} + 2x + 8

Reflection in the $x$ -axis:

y = -\left(\frac{9}{2x+1} + 2x + 8\right)

y = -\frac{9}{2x+1} - 2x - 8

Answer

y = -\frac{9}{2x+1} - 2x - 8

Final answer

y = -9/(2x+1) - 2x - 8

Detailed explanation

Walkthrough

Start with the original equation. To apply the translation by $(-3,7)$ , replace $x$ by $x+3$ and add $7$ to the whole expression. This gives $\frac{9}{2(x+3)-5}+2(x+3)-5+7$ .

Simplify the denominator and the linear terms: $2(x+3)-5 = 2x+1$ and $2(x+3)-5+7 = 2x+8$ . Then apply the reflection in the $x$ -axis by multiplying the whole right-hand side by $-1$ , giving $y = -\frac{9}{2x+1}-2x-8$ .

Key Takeaways

For a translation by $(-3,7)$ , replace $x$ by $x+3$ and add $7$ to the whole expression. Reflection in the $x$ -axis multiplies the whole expression by $-1$ .

Common Mistakes

Using $x-3$ instead of $x+3$ for the horizontal translation.
Adding the $7$ after the reflection instead of before it.
Reflecting only the fraction and not the whole expression.
Not simplifying $2(x+3)-5$ correctly.

Things to Be Careful About

The final answer must be in the exact form $y = \frac{a}{bx+c}+dx+e$ with integer coefficients. Here $a = -9$ , $b = 2$ , $c = 1$ , $d = -2$ and $e = -8$ .

Techniques used

replace x in the equation to apply the horizontal translationadd the vertical translationmultiply by -1 for the reflectionsimplify the expression into the required form

Q11MediumQuadraticsFunctions

The function $f$ is defined by $f(x) = x^2 + 4ax + a$ for $x \in \mathbb{R}$ , where $a$ is a constant.

The function $g$ is such that $g^{-1}(x) = \sqrt[3]{2x-4}$ for $x \in \mathbb{R}$ .

(a)

Given that the range of $f$ is $f(x) \ge -33$ , find the possible values of $a$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Complete the square on $f(x)$ to find its minimum value, then equate it to the range bound $-33$ and solve the resulting quadratic equation in $a$ .

Working

Complete the square:

\begin{aligned} f(x) &= x^2 + 4ax + a \\ &= (x + 2a)^2 - 4a^2 + a \end{aligned}

The minimum value of $f(x)$ is $-4a^2 + a$ , occurring at $x = -2a$ .

Since the range is $f(x) \geq -33$ , equate the minimum value to $-33$ :

-4a^2 + a = -33

Rearranging:

4a^2 - a - 33 = 0

Factorising:

(4a + 11)(a - 3) = 0

Hence:

a = -\frac{11}{4} \quad \text{or} \quad a = 3

Answer

a = -\frac{11}{4} \quad \text{or} \quad a = 3

Final answer

a = -11/4 or a = 3

Detailed explanation

Walkthrough

Since $f(x) = x^2 + 4ax + a$ has a positive leading coefficient, its graph is a parabola opening upwards, so it has a minimum value. The condition that the range is $f(x) \geq -33$ means the minimum value of $f$ is exactly $-33$ .

To locate the minimum, complete the square. The coefficient of $x$ is $4a$ , so we write $(x + 2a)^2 = x^2 + 4ax + 4a^2$ . Subtracting $4a^2$ and adding $a$ gives $f(x) = (x+2a)^2 - 4a^2 + a$ . Because $(x+2a)^2 \geq 0$ for all real $x$ , the minimum value is $-4a^2 + a$ .

Equating the minimum to $-33$ gives $-4a^2 + a = -33$ , which rearranges to $4a^2 - a - 33 = 0$ . Factorising gives $(4a + 11)(a - 3) = 0$ , so $a = -\frac{11}{4}$ or $a = 3$ . Both values are valid and give the required range.

Key Takeaways

Completing the square reveals the vertex (minimum or maximum) of a quadratic.
The range of an upward-opening quadratic is $[\text{minimum}, \infty)$ .
Solving a quadratic equation in a parameter is exactly the same as solving in a variable.

Common Mistakes

Forgetting to subtract $4a^2$ when completing the square, writing $(x+2a)^2 + a$ instead of $(x+2a)^2 - 4a^2 + a$ .
Sign errors when rearranging $-4a^2 + a = -33$ into $4a^2 - a - 33 = 0$ .
Incorrectly factorising $4a^2 - a - 33$ .
Rejecting one of the two valid values of $a$ .

Things to Be Careful About

The condition $f(x) \geq -33$ means the minimum equals $-33$ , not that the minimum is greater than $-33$ .
The mark scheme warns against giving the final answer as a range or rejecting one of the two values ("Do not ISW if their final answer is given as a range or if one of the answers is rejected").

Techniques used

complete the square to find the vertexequate the minimum value to the range boundsolve the resulting quadratic equation in a

(b)

Given instead that $fgg(0) = 96$ , find the value of $a$ .

6M Similar

DifficultyMedium

Worked solution

Approach

First find $g(x)$ from the given $g^{-1}(x)$ , then evaluate $g(0)$ , $gg(0)$ , and $f(gg(0))$ step by step, set the result equal to 96, and solve for $a$ .

Working

Given $g^{-1}(x) = \sqrt[3]{2x - 4}$ , find $g(x)$ .

Let $y = g^{-1}(x) = \sqrt[3]{2x - 4}$ . Then $x = g(y)$ :

y^3 = 2x - 4

x = \frac{y^3 + 4}{2}

So

g(x) = \frac{x^3 + 4}{2}

Now evaluate $g(0)$ :

g(0) = \frac{0^3 + 4}{2} = \frac{4}{2} = 2

Then $gg(0) = g(g(0)) = g(2)$ :

g(2) = \frac{2^3 + 4}{2} = \frac{8 + 4}{2} = \frac{12}{2} = 6

Now $f(gg(0)) = f(6)$ :

f(6) = 6^2 + 4a(6) + a = 36 + 24a + a = 36 + 25a

Given $fgg(0) = 96$ :

36 + 25a = 96

25a = 60

a = \frac{60}{25} = \frac{12}{5}

Answer

a = \frac{12}{5}

Final answer

a = 12/5

Detailed explanation

Walkthrough

We are given the inverse function $g^{-1}(x) = \sqrt[3]{2x-4}$ and need the original function $g$ . To reverse the inverse, set $y = g^{-1}(x)$ and solve for $x$ : $y = \sqrt[3]{2x-4}$ , so $y^3 = 2x - 4$ and $x = (y^3 + 4)/2$ . Since $x = g(y)$ , we have $g(y) = (y^3 + 4)/2$ , i.e. $g(x) = (x^3 + 4)/2$ .

Now evaluate the composite $fgg$ at $x = 0$ step by step. First $g(0) = (0^3 + 4)/2 = 2$ . Then $gg(0) = g(g(0)) = g(2) = (2^3 + 4)/2 = (8+4)/2 = 6$ . Finally $f(gg(0)) = f(6) = 6^2 + 4a(6) + a = 36 + 24a + a = 36 + 25a$ .

Setting $fgg(0) = 96$ gives $36 + 25a = 96$ , so $25a = 60$ and $a = 60/25 = 12/5$ .

Key Takeaways

To find a function from its inverse, swap the roles of $x$ and $y$ and solve for $x$ .
Composite functions are evaluated from the inside out: $fgg(x) = f(g(g(x)))$ .
Substituting a value into a composite function can be done step by step, which is often simpler than forming the full composite expression.

Common Mistakes

Confusing $g^{-1}(x)$ with $1/g(x)$ .
Forgetting to cube both sides when solving $y = \sqrt[3]{2x-4}$ .
Evaluating $gg(0)$ as $g(0) \cdot g(0)$ instead of $g(g(0))$ .
Sign errors in the final linear equation $36 + 25a = 96$ .

Things to be Careful About

The order of composition matters: $fgg(0)$ means $f(g(g(0)))$ , not $f(0) \cdot g(0) \cdot g(0)$ .
The mark scheme requires the expression for $g(x)$ to be shown or implied (SOI).
The final answer is a fraction; express it exactly as $a = \frac{12}{5}$ .

Techniques used

find the original function from its inverseevaluate the composite function step by stepsubstitute and solve the resulting linear equation

Mathematics 9709/13 — May/June 2025

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