The first three terms of a geometric progression are $a$, $b$ and $c$ respectively, where $a$, $b$ and $c$ are positive constants. The first three terms of an arithmetic progression are $a$, $b$ and $-3c$ respectively. (a) Show that $a^2 - 10ac + 9c^2 = 0$. (b) It is now given that $a = 9$ and $c$ takes the smaller of its two possible values. (i) Find the sum to infinity of the geometric progression. (ii) Find the sum of the first 20 terms of the arithmetic progression.

(a) a^2 - 10ac + 9c^2 = 0 · (b)(i) 27/2 · (b)(ii) -960

The function $f$ is defined by $f(x) = \frac{4}{(3x - 6)^2} + \frac{1}{(3x - 6)^3}$ for $x > 2$. (a) Find an expression for $f'(x)$ and hence determine whether $f$ is an increasing function, a decreasing function or neither. (b) State whether $f^{-1}$ exists. Give a reason for your answer. (c) The function $g$ is defined by $g(x) = 4x - 3$ for $x > a$. Find the range of $g$ in terms of the constant $a$. (d) Find the set of values of $a$ for which the composite function $fg$…

(a) f'(x) = -24/(3x-6)^3 - 9/(3x-6)^4; f is decreasing. · (b) Yes, f^-1 exists because f is decreasing (one-to-one). · (c) g(x) > 4a - 3, i.e. (4a - 3, ∞) · (d) a ≥ 5/4, i.e. [5/4, ∞)

9709/12 Oct/Nov 2025 — Mathematics Solved Past Paper

Q1Medium-EasyQuadratics

(a)

Express $9x^2 - 36x + 8$ in the form $p(x + q)^2 + r$ , where $p$ , $q$ and $r$ are constants.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Factor the leading coefficient $9$ from the first two terms, then complete the square inside the bracket, remembering to adjust the constant term.

Working

9x^2 - 36x + 8 = 9(x^2 - 4x) + 8

Complete the square inside the bracket:

9[(x-2)^2 - 4] + 8 = 9(x-2)^2 - 36 + 8 = 9(x-2)^2 - 28

So the expression is $9(x-2)^2 - 28$ , giving $p = 9$ , $q = -2$ and $r = -28$ .

Answer

9x^2 - 36x + 8 = 9(x-2)^2 - 28

with $p = 9$ , $q = -2$ , $r = -28$ .

Final answer

p = 9, q = -2, r = -28

Detailed explanation

Walkthrough

The aim is to rewrite the quadratic in vertex form $p(x+q)^2 + r$ . Start by factoring $9$ out of the $x^2$ and $x$ terms only: $9x^2 - 36x + 8 = 9(x^2 - 4x) + 8$ . Inside the bracket, complete the square for $x^2 - 4x$ : half of $-4$ is $-2$ , so $x^2 - 4x = (x-2)^2 - 4$ . Multiplying back by $9$ gives $9(x-2)^2 - 36$ , and adding the original $+8$ gives $9(x-2)^2 - 28$ . Thus $p=9$ , $q=-2$ , $r=-28$ .

Key Takeaways

This question tests completing the square when the coefficient of $x^2$ is not $1$ . Factor out the coefficient before completing the square, and carefully combine the constant terms. The completed-square form is useful for finding minimum or maximum values and for solving equations.

Common Mistakes

Forgetting to multiply the subtracted constant inside the bracket by the factored coefficient $9$ ; this changes the constant term.
Writing $q=2$ instead of $q=-2$ because the form is $p(x+q)^2$ .
Losing the $+8$ when completing the square.

Things to Be Careful About

The expression is $p(x+q)^2 + r$ , so with $(x-2)^2$ the value of $q$ is $-2$ , not $2$ . Also ensure the final constant is $-28$ , not $-4$ or $-36$ .

Techniques used

factor out the leading coefficientcomplete the square on the quadratic expressionidentify the constants p, q and r

(b)

Hence find the set of values of the constant $k$ for which the equation $9x^2 - 36x + 8 = k$ has no real roots.

1M Similar

DifficultyEasy

Worked solution

Approach

Use the completed-square form to identify the minimum value of $9x^2 - 36x + 8$ . The equation has no real roots exactly when the constant $k$ is below this minimum.

Working

From part (a),

9x^2 - 36x + 8 = 9(x-2)^2 - 28

Since $9(x-2)^2 \geq 0$ for all real $x$ , the minimum value of the left-hand side is $-28$ . The equation $9x^2 - 36x + 8 = k$ has no real roots when the horizontal line $y = k$ lies below this minimum:

k < -28

Answer

k < -28

Final answer

k < -28

Detailed explanation

Walkthrough

From part (a), the quadratic can be written as $9(x-2)^2 - 28$ . The squared term $9(x-2)^2$ is always non-negative, so the smallest possible value of the whole expression is $-28$ , occurring at $x=2$ . The equation $9x^2 - 36x + 8 = k$ asks for the $x$ -values where the quadratic equals $k$ . If $k$ is less than the minimum value, the horizontal line $y=k$ never meets the upward-opening parabola, so there are no real roots. Hence the set of values is $k < -28$ .

Key Takeaways

The completed-square form immediately gives the vertex and therefore the minimum or maximum value of a quadratic. Comparing a constant $k$ with this extremum determines whether the equation has real roots.

Common Mistakes

Writing $k \leq -28$ instead of $k < -28$ . At $k = -28$ there is one repeated root, so it is excluded.
Using the original constant $8$ instead of $-28$ .

Things to Be Careful About

The condition is strict: $k < -28$ . When $k = -28$ , the equation becomes $9(x-2)^2 = 0$ , which has the real root $x=2$ . Also note that the mark scheme allows follow-through from the value of $r$ found in part (a).

Techniques used

use the completed-square minimumapply the condition for no real roots

(c)

Find the exact roots of the equation $9x^2 - 36x + 8 = -15$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute $k = -15$ into the completed-square form, isolate $(x-2)^2$ , and take square roots to obtain the exact roots.

Working

Set the expression equal to $-15$ :

9(x-2)^2 - 28 = -15

Add $28$ to both sides:

9(x-2)^2 = 13

Divide by $9$ :

(x-2)^2 = \frac{13}{9}

Take square roots:

x - 2 = \pm \frac{\sqrt{13}}{3}

Therefore

x = 2 \pm \frac{\sqrt{13}}{3}

Answer

x = 2 \pm \frac{\sqrt{13}}{3}

or equivalently $x = \frac{6 \pm \sqrt{13}}{3}$ .

Final answer

x = 2 + sqrt(13)/3 or x = 2 - sqrt(13)/3

Detailed explanation

Walkthrough

Use the completed-square form from part (a): $9(x-2)^2 - 28 = -15$ . Add $28$ to both sides to get $9(x-2)^2 = 13$ . Dividing by $9$ gives $(x-2)^2 = \frac{13}{9}$ . Taking square roots introduces the $\pm$ sign, giving $x-2 = \pm \frac{\sqrt{13}}{3}$ . Finally add $2$ to both sides to obtain $x = 2 \pm \frac{\sqrt{13}}{3}$ , which is the same as $\frac{6 \pm \sqrt{13}}{3}$ .

Key Takeaways

Completing the square turns a quadratic equation into a simple squared-bracket equation. Taking square roots of both sides is a quick way to solve it exactly, provided the surd is simplified correctly.

Common Mistakes

Forgetting the $\pm$ when taking the square root.
Making a sign error when moving $-28$ to the other side.
Not simplifying $\sqrt{13/9}$ to $\sqrt{13}/3$ if exact roots are required.
Using decimals instead of exact surd forms; the mark scheme requires exact roots.

Things to Be Careful About

The roots can be written as $2 \pm \frac{\sqrt{13}}{3}$ or $\frac{6 \pm \sqrt{13}}{3}$ ; these are equivalent. The mark scheme accepts any equivalent exact form. If decimals are given after exact answers, ignore subsequent working (ISW).

Techniques used

substitute the given value into the completed-square formisolate the squared brackettake square roots and simplify the surd

Q23MMedium-EasySeries

Find the term independent of $x$ in the expansion of

\left( 2x^2 - \frac{3}{x} \right)^6

DifficultyMedium-Easy

Worked solution

Approach

Use the binomial theorem to write the general term of the expansion, then identify the term where the power of $x$ is zero, and evaluate it.

Working

The general term in the expansion of $\left(2x^2 - \frac{3}{x}\right)^6$ is:

\binom{6}{r} (2x^2)^{6-r} \left(-\frac{3}{x}\right)^r

The power of $x$ in this term is:

2(6-r) - r = 12 - 3r

For the term to be independent of $x$ , set this power to zero:

12 - 3r = 0 \implies r = 4

So the relevant term is:

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

Since $\binom{6}{4} = \binom{6}{2} = 15$ , this becomes:

15 \cdot (2x^2)^2 \cdot \left(-\frac{3}{x}\right)^4

Now evaluate each factor:

(2x^2)^2 = 4x^4, \qquad \left(-\frac{3}{x}\right)^4 = \frac{81}{x^4}

So:

15 \cdot 4x^4 \cdot \frac{81}{x^4} = 15 \cdot 4 \cdot 81 = 4860

Answer

4860

Final answer

4860

Detailed explanation

Walkthrough

We need to expand the binomial $\left(2x^2 - \frac{3}{x}\right)^6$ and find the term that has no $x$ in it. The binomial theorem states that each term in the expansion of $(a+b)^n$ is of the form $\binom{n}{r} a^{n-r} b^r$ . Here, $a = 2x^2$ and $b = -\frac{3}{x}$ , with $n = 6$ .

The general term is $\binom{6}{r} (2x^2)^{6-r} \left(-\frac{3}{x}\right)^r$ . We need the value of $r$ that makes the power of $x$ equal to zero. The factor $(2x^2)^{6-r}$ contributes $x^{2(6-r)}$ , and the factor $\left(-\frac{3}{x}\right)^r$ contributes $x^{-r}$ . So the total power of $x$ is $2(6-r) - r = 12 - 3r$ . Setting $12 - 3r = 0$ gives $r = 4$ .

Now substitute $r = 4$ into the general term: the binomial coefficient is $\binom{6}{4} = 15$ . Then $(2x^2)^2 = 4x^4$ and $\left(-\frac{3}{x}\right)^4 = \frac{81}{x^4}$ . The $x^4$ and $\frac{1}{x^4}$ cancel, leaving $15 \times 4 \times 81 = 4860$ .

Key Takeaways

The key idea is to find the term independent of $x$ by setting the total power of $x$ in the general term to zero and solving for $r$ .
This requires careful handling of the exponents from both parts of the binomial.
The binomial coefficient $\binom{n}{r}$ can be evaluated using symmetry, e.g., $\binom{6}{4} = \binom{6}{2} = 15$ .

Common Mistakes

Forgetting the $x^{-1}$ from the $\frac{3}{x}$ factor when computing the total power of $x$ .
Miscomputing the exponent: the power is $2(6-r) - r$ , not $2(6-r) + r$ .
Forgetting that $\left(-\frac{3}{x}\right)^4$ is positive because the exponent is even.

Things to Be Careful About

The exponent of $x$ must be set to zero, not the coefficient. The final answer is a positive number with no $x$ factor. The mark scheme also accepts $4860x^0$ .

Techniques used

apply the binomial theorem to write the general termset the power of x to zero to locate the constant termevaluate the binomial coefficient and simplify the coefficient

Approach

Compare the given equation $y = f(3x) + 2$ with the original $y = f(x)$ and identify each change separately. The argument $3x$ affects the horizontal direction, and the constant $+2$ affects the vertical direction.

Working

Starting from $y = f(x)$ :

The change from $f(x)$ to $f(3x)$ replaces $x$ by $3x$ . For the same output value, the new $x$ -coordinate satisfies $3x' = x$ , so $x' = \frac{x}{3}$ . Hence the graph is stretched by scale factor $\frac{1}{3}$ parallel to the $x$ -axis (horizontally).

The $+2$ adds 2 to every $y$ -value, so the graph is translated by 2 units in the positive $y$ -direction. This is a translation by the vector:

\begin{pmatrix} 0 \\ 2 \end{pmatrix}

Answer

The two transformations, combined in the order they appear, are:

A stretch with scale factor $\frac{1}{3}$ parallel to the $x$ -axis (horizontal stretch).
A translation by 2 units in the $y$ -direction, i.e. by the vector $(0, 2)$ .

Final answer

Stretch with scale factor 1/3 parallel to the x-axis, and translation by 2 units in the y-direction (vector (0, 2)).

Detailed explanation

Walkthrough

We are told that the graph $y = f(x)$ is transformed to $y = f(3x) + 2$ , and we must describe the two transformations.

First, focus on the argument of the function. In $f(3x)$ , the input $x$ is multiplied by 3 before the function is applied. To obtain the same output as before, the new $x$ -coordinate must be one third of the original. So the graph is compressed horizontally — a stretch by scale factor $\frac{1}{3}$ parallel to the $x$ -axis. A common mistake is to think $f(3x)$ stretches by factor 3, but it actually compresses by factor $\frac{1}{3}$ .

Second, the $+2$ outside the function adds 2 to every output value. This shifts the whole graph upward by 2 units, a translation by the vector $(0, 2)$ , i.e. in the $y$ -direction.

The order of the two transformations follows the structure of the equation: the stretch acts on the argument, and the translation acts on the output.

Key Takeaways

The transformation $f(x) \to f(ax)$ is a horizontal stretch by scale factor $\frac{1}{a}$ .
The transformation $f(x) \to f(x) + c$ is a vertical translation by $c$ units.
Always state both the scale factor and the direction of a stretch, and the direction and amount of a translation.

Common Mistakes

Thinking $f(3x)$ is a stretch by scale factor 3 instead of $\frac{1}{3}$ .
Describing the translation as 'up' without specifying the direction — the mark scheme requires 'parallel to the $y$ -axis', 'in the $y$ -direction', or 'vertically'.
Omitting the direction of the stretch (parallel to the $x$ -axis).

Things to Be Careful About

The translation must be described as being in the $y$ -direction (or parallel to the $y$ -axis), not just 'up'.
The stretch must be given with its scale factor ( $\frac{1}{3}$ ) and its direction (parallel to the $x$ -axis).
The mark scheme awards B1 for the translation and B2 for the stretch; partial credit is available if two of the three components of the stretch are correct.

Techniques used

identify horizontal stretch from argument changeidentify vertical translation from added constantdescribe the combined transformation

(b)

A different graph has equation $y = g(x)$ . This graph is stretched by scale factor 3 in the $y$ -direction and then reflected in the $y$ -axis.

Write down the equation of the transformed graph in terms of the function $g$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Apply the transformations to $y = g(x)$ in the order given: first the stretch by scale factor 3 in the $y$ -direction, then the reflection in the $y$ -axis.

Working

Start with $y = g(x)$ .

A stretch by scale factor 3 in the $y$ -direction multiplies the output by 3:

y = 3g(x)

A reflection in the $y$ -axis replaces $x$ by $-x$ :

y = 3g(-x)

Answer

The equation of the transformed graph is $y = 3g(-x)$ .

Final answer

y = 3g(-x)

Detailed explanation

Walkthrough

The graph $y = g(x)$ is transformed in two steps.

First, a stretch by scale factor 3 in the $y$ -direction multiplies every $y$ -value by 3, so the function becomes $y = 3g(x)$ .

Second, a reflection in the $y$ -axis maps each point $(x, y)$ to $(-x, y)$ , so we replace $x$ by $-x$ in the function: $y = 3g(-x)$ .

The order matters: we stretch first, then reflect. In this case the algebra works out the same either way because the stretch factor is outside and the reflection is inside, but we must apply the transformations in the order stated.

Key Takeaways

A vertical stretch by scale factor $k$ multiplies the function by $k$ : $y = kg(x)$ .
A reflection in the $y$ -axis replaces $x$ by $-x$ : $y = g(-x)$ .
Apply transformations in the order given and combine them correctly.

Common Mistakes

Reflecting in the $x$ -axis instead of the $y$ -axis (replacing $y$ by $-y$ instead of $x$ by $-x$ ).
Putting the stretch factor inside the function (e.g. $g(3x)$ ) instead of outside.
Adding an extra transformation not asked for, such as a translation.

Things to Be Careful About

The scale factor 3 must appear outside the function: $3g(-x)$ .
The reflection must appear as $-x$ inside the function.
The mark scheme awards B1 for a single correct component (either the 3 outside or the $-x$ inside) but B2 only for both.
The final answer must be written in terms of $g$ , not $f$ .

Techniques used

apply vertical stretch by multiplying the functionapply reflection in the y-axis by replacing x with -xcombine transformations in the given order

Q4MediumDifferentiationIntegration

The equation of a curve is such that

\frac{dy}{dx} = kx^3 + \frac{2}{x^2}

where $k$ is a constant. The curve passes through the point $S(2, 20)$ and the gradient of the curve at $S$ is $\frac{65}{2}$ .

(a)

Find the value of $k$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

At the point $S(2,20)$ , the gradient is the value of $\frac{dy}{dx}$ when $x=2$ . Substitute $x=2$ into the given derivative, set it equal to the stated gradient, and solve for $k$ .

Working

The gradient at $S$ is

\frac{dy}{dx}\Big|_{x=2} = k(2)^3 + \frac{2}{2^2}

Since $\frac{2}{2^2} = \frac{2}{4} = \frac{1}{2}$ and the gradient is $\frac{65}{2}$ :

8k + \frac{1}{2} = \frac{65}{2}

Subtract $\frac{1}{2}$ from both sides:

8k = \frac{64}{2} = 32

Hence

k = 4

Answer

k = 4

Final answer

k = 4

Detailed explanation

Walkthrough

The derivative $\frac{dy}{dx}$ is the gradient function: substituting the $x$ -coordinate of a point gives the gradient of the curve there. Here the point is $S(2,20)$ , so substitute $x=2$ into $kx^3 + \frac{2}{x^2}$ . This gives $8k + \frac{1}{2}$ . The question states that the gradient at $S$ is $\frac{65}{2}$ , so set these equal and solve for $k$ : first subtract $\frac{1}{2}$ to get $8k = 32$ , then divide by $8$ to obtain $k=4$ .

Key Takeaways

The derivative of a curve is a gradient function. To find the gradient at a particular point, evaluate the derivative at the point's $x$ -coordinate, not at its $y$ -coordinate. You should also be comfortable solving linear equations containing fractions.

Common Mistakes

A common error is to use the $y$ -coordinate $20$ in the derivative expression; it is not needed here. Another possible mistake is an arithmetic slip when subtracting $\frac{1}{2}$ from $\frac{65}{2}$ , so always rewrite both fractions with a common denominator.

Things to Be Careful About

The point $S(2,20)$ has two coordinates, but the gradient condition only uses $x=2$ . The $y$ -coordinate $20$ is used later in part (b) when finding the constant of integration, not in part (a).

Techniques used

substitute x-coordinate into derivativesolve a linear equation for the constant

(b)

The coordinates of a point $T$ on the curve are $(1, t)$ .

Find the value of $t$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use $k=4$ from part (a). Integrate the derivative term by term and add a constant of integration $c$ . Then use the point $S(2,20)$ to find $c$ . Finally, substitute $x=1$ into the equation of the curve to find $t$ .

Working

With $k=4$ , the gradient is

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

Integrate using the power rule for $n \neq -1$ :

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

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

Use $S(2,20)$ , so $x = 2$ and $y = 20$ :

20 = (2)^4 - \frac{2}{2} + c = 16 - 1 + c

Thus

20 = 15 + c

c = 5

Therefore the curve is

y = x^4 - \frac{2}{x} + 5

At $T(1,t)$ , substitute $x = 1$ :

t = (1)^4 - \frac{2}{1} + 5 = 1 - 2 + 5 = 4

Answer

t = 4

Final answer

t = 4

Detailed explanation

Walkthrough

First rewrite $\frac{2}{x^2}$ as $2x^{-2}$ so both terms are powers of $x$ . Integrate $4x^3$ to get $x^4$ , and integrate $2x^{-2}$ using the power rule: increase the exponent by $1$ (from $-2$ to $-1$ ) and divide by the new exponent, giving $-2x^{-1}$ , which is $-\frac{2}{x}$ . Since this is an indefinite integral, add a constant $c$ . The point $S(2,20)$ lies on the curve, so substitute $x=2$ and $y=20$ into $y = x^4 - \frac{2}{x} + c$ . This gives $20 = 16 - 1 + c$ , so $c=5$ . The full equation is $y = x^4 - \frac{2}{x} + 5$ . Finally, put $x=1$ to get $t = 1 - 2 + 5 = 4$ .

Key Takeaways

To recover a curve from its gradient, integrate the derivative and include an arbitrary constant $+c$ . A point on the curve supplies enough information to determine $c$ . Once the equation of the curve is known, its $y$ -coordinate at any $x$ -value can be found by substitution. This question also reinforces integrating negative powers by the same power rule.

Common Mistakes

Forgetting the constant of integration; without $+c$ it is impossible to use $S(2,20)$ and the final answer would be wrong.
Writing $\int \frac{2}{x^2}\,dx = 2 \ln x$ ; this is incorrect because the power is $-2$ , not $-1$ . The power rule applies for every exponent except $-1$ .
A sign error when integrating $2x^{-2}$ to $-2x^{-1}$ : because the new exponent is negative, the coefficient is divided by $-1$ , giving a negative term.
Substituting $y=20$ into the derivative instead of the integrated expression; the point must be used after integration.

Things to Be Careful About

Use the correct value of $k$ from part (a), $k=4$ . Be careful at $x=1$ : the term $-\frac{2}{x}$ becomes $-2$ , so $t = 1 - 2 + 5 = 4$ . The original derivative is undefined at $x=0$ , but the points given have $x=2$ and $x=1$ , so there is no issue here. When finding $c$ , show both the substitution and the resulting equation to ensure the mark for using the point is awarded.

Techniques used

integrate a power function term by termadd a constant of integrationuse a given point to determine the constantevaluate the curve equation at a new x-value

Q5MediumDifferentiationIntegration

Fig. 1

The equation of a curve is $y = 4x^{\frac{1}{2}} - x$ . The curve has a maximum point when $x = a$ and crosses the $x$ -axis at the point with coordinates $(b, 0)$ , where $b > 0$ . The shaded region is bounded by the curve, the line $x = a$ and the $x$ -axis (see diagram).

(a)

Find the value of $a$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

To find the x-coordinate of the maximum point, differentiate the curve equation and set the derivative equal to zero, then solve for $x$ .

Working

Differentiate $y = 4x^{\frac{1}{2}} - x$ with respect to $x$ :

\frac{dy}{dx} = 4 \cdot \frac{1}{2} x^{-\frac{1}{2}} - 1

\frac{dy}{dx} = 2x^{-\frac{1}{2}} - 1

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

2x^{-\frac{1}{2}} - 1 = 0

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

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

x^{\frac{1}{2}} = 2

x = 4

Since the curve has a maximum at this point, $a = 4$ .

Answer

a = 4

Final answer

a = 4

Detailed explanation

Walkthrough

The problem asks for the x-coordinate of the maximum point on the curve $y = 4x^{\frac{1}{2}} - x$ . At a maximum (or minimum) point, the gradient of the curve is zero, so we differentiate $y$ with respect to $x$ and set the result equal to zero.

First, differentiate using the power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$ . For $4x^{\frac{1}{2}}$ , the derivative is $4 \cdot \frac{1}{2} x^{-\frac{1}{2}} = 2x^{-\frac{1}{2}}$ . For $-x$ , the derivative is $-1$ . So $\frac{dy}{dx} = 2x^{-\frac{1}{2}} - 1$ .

Setting this equal to zero gives $2x^{-\frac{1}{2}} = 1$ , which rearranges to $x^{-\frac{1}{2}} = \frac{1}{2}$ . Taking the reciprocal of both sides gives $x^{\frac{1}{2}} = 2$ , and squaring both sides yields $x = 4$ . This is the x-coordinate of the maximum point, so $a = 4$ .

Key Takeaways

At a stationary point (maximum or minimum), the first derivative equals zero.
The power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ applies to fractional powers as well.
Solving equations involving fractional powers may require taking reciprocals and squaring.

Common Mistakes

Forgetting the negative sign when differentiating $-x$ .
Incorrectly applying the power rule to $x^{\frac{1}{2}}$ , e.g., writing $\frac{1}{2}x^{-\frac{1}{2}}$ instead of $2x^{-\frac{1}{2}}$ .
Making algebraic errors when solving $2x^{-\frac{1}{2}} = 1$ , such as forgetting to invert both sides.
Not verifying that the stationary point is indeed a maximum (though in this context, the question states it is).

Things to Be Careful About

The mark scheme condones sign errors in the step from $2x^{-\frac{1}{2}} - 1 = 0$ to $x^{\frac{1}{2}} = d$ , so small sign slips here may not cost marks if the method is clear.
Ensure the final answer is clearly stated as $a = 4$ , not just $x = 4$ .

Techniques used

differentiate the curve equationset dy/dx equal to zerosolve for x to find the x-coordinate of the maximum

(b)

Find the exact area of the shaded region.

5M Similar

DifficultyMedium

Worked solution

Approach

The shaded region is bounded by the curve $y = 4x^{\frac{1}{2}} - x$ , the vertical line $x = a = 4$ , and the x-axis. To find its area, first determine the x-intercept $b$ by setting $y = 0$ , then evaluate the definite integral $\int_a^b (4x^{\frac{1}{2}} - x)\,dx$ .

Working

Find $b$ by setting $y = 0$ :

4x^{\frac{1}{2}} - x = 0

Factor out $x^{\frac{1}{2}}$ :

x^{\frac{1}{2}}(4 - x^{\frac{1}{2}}) = 0

This gives $x^{\frac{1}{2}} = 0$ or $x^{\frac{1}{2}} = 4$ , so $x = 0$ or $x = 16$ . Since $b > 0$ , we have $b = 16$ .

Integrate the curve:

\int (4x^{\frac{1}{2}} - x)\,dx = 4 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^2}{2}

= \frac{8}{3}x^{\frac{3}{2}} - \frac{1}{2}x^2

Evaluate the definite integral from $x = 4$ to $x = 16$ :

\text{Area} = \left[\frac{8}{3}x^{\frac{3}{2}} - \frac{1}{2}x^2\right]_4^{16}

Substitute the upper limit $x = 16$ :

\frac{8}{3} \times 16^{\frac{3}{2}} - \frac{1}{2} \times 16^2 = \frac{8}{3} \times 64 - \frac{1}{2} \times 256 = \frac{512}{3} - 128 = \frac{128}{3}

Substitute the lower limit $x = 4$ :

\frac{8}{3} \times 4^{\frac{3}{2}} - \frac{1}{2} \times 4^2 = \frac{8}{3} \times 8 - \frac{1}{2} \times 16 = \frac{64}{3} - 8 = \frac{40}{3}

Subtract:

\text{Area} = \frac{128}{3} - \frac{40}{3} = \frac{88}{3}

Answer

\text{Area} = \frac{88}{3}

Final answer

88/3

Detailed explanation

Walkthrough

The shaded region lies between $x = a = 4$ and $x = b$ , bounded above by the curve and below by the x-axis. We need to find $b$ first, then integrate.

Finding $b$ : Set $y = 0$ : $4x^{\frac{1}{2}} - x = 0$ . Factor out $x^{\frac{1}{2}}$ to get $x^{\frac{1}{2}}(4 - x^{\frac{1}{2}}) = 0$ . This gives $x = 0$ (the origin) or $x = 16$ . Since $b > 0$ , we take $b = 16$ .

Integrating: Use the power rule for integration: $\int x^n\,dx = \frac{x^{n+1}}{n+1}$ . For $4x^{\frac{1}{2}}$ , the integral is $4 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{8}{3}x^{\frac{3}{2}}$ . For $-x$ , the integral is $-\frac{x^2}{2}$ . So the antiderivative is $\frac{8}{3}x^{\frac{3}{2}} - \frac{1}{2}x^2$ .

Evaluating: Substitute $x = 16$ and $x = 4$ into the antiderivative and subtract. Note that $16^{\frac{3}{2}} = (\sqrt{16})^3 = 4^3 = 64$ and $4^{\frac{3}{2}} = (\sqrt{4})^3 = 2^3 = 8$ . This gives $\frac{512}{3} - 128 = \frac{128}{3}$ for the upper limit and $\frac{64}{3} - 8 = \frac{40}{3}$ for the lower limit. The difference is $\frac{88}{3}$ .

Key Takeaways

The x-intercepts of a curve are found by setting $y = 0$ and solving.
Integration with fractional powers follows the same power rule as integer powers.
When evaluating definite integrals, always compute $F(b) - F(a)$ where $F$ is the antiderivative.
Fractional powers like $x^{\frac{3}{2}}$ can be evaluated as $(\sqrt{x})^3$ for simplicity.

Common Mistakes

Forgetting to find $b$ before setting up the integral.
Incorrect integration of $4x^{\frac{1}{2}}$ , e.g., writing $\frac{4}{\frac{3}{2}}x^{\frac{3}{2}}$ without simplifying to $\frac{8}{3}x^{\frac{3}{2}}$ .
Arithmetic errors when evaluating $16^{\frac{3}{2}}$ or $4^{\frac{3}{2}}$ .
Forgetting to subtract the lower limit evaluation from the upper limit evaluation.
Giving a decimal answer instead of the exact fraction $\frac{88}{3}$ .

Things to Be Careful About

The answer must be exact: $\frac{88}{3}$ or $29\frac{1}{3}$ . Do not give a decimal approximation.
The mark scheme requires $0 < a < b$ for full marks on the substitution step; if the limits are in the wrong order, the method mark may be lost.
Allow missing brackets if they can be recovered in subsequent working.
The value $b = 16$ must be explicitly shown in part (b) even if found in part (a).

Techniques used

find x-intercept by solving y = 0integrate the curve equation using power ruleevaluate definite integral using upper and lower limitssubstitute limits and simplify fractional arithmetic

Q6Medium-HardTrigonometry

(a)

Sketch the graph of $y = 3\sin x + 2$ for $0 \le x \le 2\pi$ .

Fig. 2

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The graph of $y = 3\sin x + 2$ is a transformation of $y = \sin x$ : vertically stretched by a factor of $3$ (amplitude becomes $3$ ) and translated upwards by $2$ units. The period remains $2\pi$ . We identify the key points at the quarter-period intervals and sketch one full cycle on $0 \leq x \leq 2\pi$ .

Working

The midline is $y = 2$ and the amplitude is $3$ , so the maximum value is $5$ and the minimum value is $-1$ .

Key points on the curve:

At $x = 0$ : $y = 3\sin 0 + 2 = 2$ (start at the midline going up).
At $x = \frac{1}{2}\pi$ : $y = 3\sin\left(\frac{\pi}{2}\right) + 2 = 5$ (maximum).
At $x = \pi$ : $y = 3\sin\pi + 2 = 2$ (back to midline going down).
At $x = \frac{3}{2}\pi$ : $y = 3\sin\left(\frac{3\pi}{2}\right) + 2 = -1$ (minimum).
At $x = 2\pi$ : $y = 3\sin(2\pi) + 2 = 2$ (back to midline going up).

The $y$ -axis is labelled with $-1$ , $0$ , $2$ and $5$ , and the $x$ -axis is labelled with $0$ , $\frac{1}{2}\pi$ , $\pi$ , $\frac{3}{2}\pi$ and $2\pi$ . The curve crosses the $x$ -axis between $0$ and $\frac{1}{2}\pi$ and again between $\pi$ and $\frac{3}{2}\pi$ .

Answer

The sketch is a sine curve with maximum $(\frac{1}{2}\pi, 5)$ , minimum $(\frac{3}{2}\pi, -1)$ , passing through $(0, 2)$ and $(2\pi, 2)$ .

Final answer

Sine curve: max (π/2, 5), min (3π/2, −1), through (0, 2) and (2π, 2).

Detailed explanation

Walkthrough

The function $y = 3\sin x + 2$ is built from $y = \sin x$ in two stages. First the amplitude is multiplied by $3$ , so the curve oscillates between $-3$ and $3$ instead of between $-1$ and $1$ . Then the whole graph is shifted up by $2$ units, which moves the midline from $y = 0$ to $y = 2$ , the maximum from $1$ to $5$ , and the minimum from $-1$ to $-1$ . The period is unchanged at $2\pi$ since the coefficient of $x$ inside the sine is still $1$ .

For the sketch we only need the values at the quarter-period marks: $x = 0, \frac{1}{2}\pi, \pi, \frac{3}{2}\pi, 2\pi$ . Substituting gives $y = 2, 5, 2, -1, 2$ . The curve must start at $(0, 2)$ going up, reach the maximum $(\frac{1}{2}\pi, 5)$ , return to the midline at $(\pi, 2)$ going down, reach the minimum $(\frac{3}{2}\pi, -1)$ , and finish at $(2\pi, 2)$ going up. The mark scheme requires these key values (or clear indications of them) on the y-axis and a curve that is horizontal only at the maximum and minimum points.

Key Takeaways

The graph of $y = a\sin x + k$ has amplitude $|a|$ , midline $y = k$ , and the same period $2\pi$ as $y = \sin x$ .
A sketch over one full period only requires the five key points at multiples of $\frac{1}{2}\pi$ .

Common Mistakes

Drawing a curve that starts at the origin or ends at $(2\pi, 0)$ — these score zero, since the curve must start at $(0, 2)$ .
Forgetting the vertical shift and drawing $y = 3\sin x$ instead of $y = 3\sin x + 2$ (max would be $3$ , min would be $-3$ ).
Labelling the maximum and minimum at the wrong $x$ -values (e.g. max at $\pi$ and min at $2\pi$ ).

Things to Be Careful About

The maximum and minimum must be horizontal tangents at $x = \frac{1}{2}\pi$ and $x = \frac{3}{2}\pi$ respectively; the mark scheme requires the curve to be flat only at these two points.
The $y$ -axis does not need to be to scale, but the values $-1$ , $2$ and $5$ must be clearly indicated.

Techniques used

identify amplitude and vertical shift of a sine functionlocate key points (max, min, intercepts)sketch one full cycle

(b)

Determine the number of solutions in the interval $0 \le x \le 2\pi$ of each of the following equations.

2M

(i)

$3\sin x + 2 = x$

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

The number of solutions of $3\sin x + 2 = x$ in $0 \le x \le 2\pi$ equals the number of intersections of $y = 3\sin x + 2$ and $y = x$ in this interval. We track the sign of $\left(3\sin x + 2 - x\right)$ at a few key points.

Working

At $x = 0$ : $\;3\sin 0 + 2 - 0 = 2 > 0$ , so the curve lies above the line.
At $x = 2\pi$ : $\;3\sin(2\pi) + 2 - 2\pi = 2 - 2\pi \approx -4.28 < 0$ , so the curve lies below the line.

Since the difference changes sign only once and the oscillating curve stays between $-1$ and $5$ while $y = x$ grows steadily from $0$ to $2\pi \approx 6.28$ , the two graphs cross exactly once in $[0, 2\pi]$ .

Answer

$1$ solution.

Final answer

1

Detailed explanation

Walkthrough

We need to count how many times the graph $y = 3\sin x + 2$ meets the straight line $y = x$ on the interval $0 \le x \le 2\pi$ . The line $y = x$ starts at the origin and ends at $(2\pi, 2\pi) \approx (6.28, 6.28)$ . The sine curve oscillates between $-1$ and $5$ and finishes at $(2\pi, 2)$ . At $x = 0$ the curve sits at $y = 2$ above the line, and at $x = 2\pi$ the curve sits at $y = 2$ well below the line. The line is steep enough that once it overtakes the curve it stays above it, so exactly one crossing occurs.

Key Takeaways

Equations of the form $f(x) = g(x)$ can be solved graphically by counting intersections of the two graphs.
A useful trick is to evaluate $f(x) - g(x)$ at a few key points; every sign change indicates a crossing.

Common Mistakes

Forgetting that $y = x$ reaches $2\pi \approx 6.28$ at the right end of the interval, well above the maximum of the sine curve, and concluding there might be multiple crossings.
Counting the starting point $x = 0$ as a solution: the equation is $3\sin x + 2 = x$ , which at $x = 0$ gives $2 = 0$ , so $x = 0$ is not a solution.

Things to Be Careful About

The mark scheme says to "ignore any working out seen or graphs drawn" — the answer is awarded purely for the correct number $1$ .

Techniques used

compare an oscillating curve with a straight linecount intersections by tracking sign changes

(ii)

$3\sin x + 2 = 5 - x$

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

The number of solutions of $3\sin x + 2 = 5 - x$ in $0 \le x \le 2\pi$ equals the number of intersections of $y = 3\sin x + 2$ and $y = 5 - x$ in this interval. We track the sign of $\left(3\sin x + 2 - (5 - x)\right) = \left(3\sin x + x - 3\right)$ at the key points $0, \frac{1}{2}\pi, \pi, \frac{3}{2}\pi, 2\pi$ .

Working

At $x = 0$ : $\;3\sin 0 + 0 - 3 = -3 < 0$ (line above curve).
At $x = \frac{1}{2}\pi$ : $\;3(1) + \frac{\pi}{2} - 3 = \frac{\pi}{2} \approx 1.57 > 0$ (curve above line). Sign change — first intersection.
At $x = \pi$ : $\;3(0) + \pi - 3 \approx 0.14 > 0$ (curve above line). No new intersection yet.
At $x = \frac{3}{2}\pi$ : $\;3(-1) + \frac{3\pi}{2} - 3 = \frac{3\pi}{2} - 6 \approx -1.29 < 0$ (line above curve). Sign change — second intersection.
At $x = 2\pi$ : $\;3(0) + 2\pi - 3 \approx 3.28 > 0$ (curve above line). Sign change — third intersection.

Three sign changes in total, giving three intersections.

Answer

$3$ solutions.

Final answer

3

Detailed explanation

Walkthrough

We count intersections of $y = 3\sin x + 2$ (oscillating between $-1$ and $5$ ) with the line $y = 5 - x$ (passing through $(0, 5)$ and $(2\pi, 5 - 2\pi) \approx (6.28, -1.28)$ ). The line is decreasing with slope $-1$ , so it cuts through the oscillating region of the sine curve multiple times.

We use the equivalent test on $h(x) = 3\sin x + x - 3$ , which has the same zeros as $3\sin x + 2 = 5 - x$ . A sign change of $h(x)$ between two consecutive test points means the graphs cross somewhere in between.

$h(0) = -3$ : line above curve.
$h(\frac{1}{2}\pi) \approx 1.57 > 0$ : curve above line. One crossing in $(0, \frac{1}{2}\pi)$ .
$h(\pi) \approx 0.14 > 0$ : curve still above line; no crossing in $(\frac{1}{2}\pi, \pi)$ .
$h(\frac{3}{2}\pi) \approx -1.29 < 0$ : line above curve. Second crossing in $(\pi, \frac{3}{2}\pi)$ .
$h(2\pi) \approx 3.28 > 0$ : curve above line again. Third crossing in $(\frac{3}{2}\pi, 2\pi)$ .

Three crossings in total.

Key Takeaways

When counting solutions graphically, sample the difference at every quarter-period; between any two consecutive samples the curve cannot have more than one crossing if the two graphs are a slowly varying line and an oscillation.
A line with slope $-1$ that starts above the maximum of the oscillation and ends well below the curve must cross three times.

Common Mistakes

Stopping after a single sign change and reporting one intersection.
Forgetting that the line $y = 5 - x$ at $x = 2\pi$ is at $y \approx -1.28$ , which is below the minimum of the curve, so the line must cross the curve a third time on the way back up.

Things to Be Careful About

The mark scheme simply awards one mark for the number $3$ ; no working is required to earn it.

Techniques used

compare an oscillating curve with a straight linecount intersections by tracking sign changes at key points

(c)

Solve the equation $3\sin x + 2 = 5\cos^2 x - 1$ for $0 \le x \le 2\pi$ .

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Use the Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ to convert the equation into a quadratic in $\sin x$ , solve the quadratic, and then find the corresponding $x$ -values in $0 \le x \le 2\pi$ .

Working

Substitute $\cos^2 x = 1 - \sin^2 x$ :

3\sin x + 2 = 5(1 - \sin^2 x) - 1

Expand and simplify:

3\sin x + 2 = 5 - 5\sin^2 x - 1

3\sin x + 2 = 4 - 5\sin^2 x

Rearrange into a three-term quadratic in $\sin x$ :

5\sin^2 x + 3\sin x - 2 = 0

Factorise the quadratic:

(5\sin x - 2)(\sin x + 1) = 0

So either

5\sin x - 2 = 0 \quad\Rightarrow\quad \sin x = 0.4

or

\sin x + 1 = 0 \quad\Rightarrow\quad \sin x = -1

Case 1: $\sin x = 0.4$ .
The principal value is $x = \sin^{-1}(0.4) \approx 0.4115$ . In $0 \le x \le 2\pi$ the solutions are

x \approx 0.4115, \quad \pi - 0.4115 \approx 2.7301

Case 2: $\sin x = -1$ .
In $0 \le x \le 2\pi$ this gives $x = \frac{3}{2}\pi$ .

All three values lie in the required interval.

Answer

x \approx 0.4115, \; 2.7301, \; \frac{3}{2}\pi

(AWRT $0.41, 2.73, \frac{3}{2}\pi$ .)

Final answer

x = 0.4115, 2.7301, 3π/2 (AWRT 0.41, 2.73, 3π/2)

Detailed explanation

Walkthrough

The equation $3\sin x + 2 = 5\cos^2 x - 1$ mixes $\sin x$ and $\cos^2 x$ , so we use the Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ to express everything in terms of $\sin x$ . This turns the equation into a quadratic in $\sin x$ , which we can solve with the usual methods.

After substitution, the equation becomes $3\sin x + 2 = 4 - 5\sin^2 x$ , i.e. $5\sin^2 x + 3\sin x - 2 = 0$ . This factors cleanly as $(5\sin x - 2)(\sin x + 1) = 0$ , giving $\sin x = 0.4$ and $\sin x = -1$ .

For $\sin x = 0.4$ , the principal value is $x \approx 0.4115$ . Since sine is positive in both the first and second quadrants, there is a second solution at $\pi - 0.4115 \approx 2.7301$ . For $\sin x = -1$ , the only solution in $[0, 2\pi]$ is $x = \frac{3}{2}\pi$ . All three values lie in the required interval.

Key Takeaways

The identity $\cos^2 x = 1 - \sin^2 x$ is the standard tool for converting an equation in both $\sin x$ and $\cos x$ into a single-variable equation in $\sin x$ .
Each value of $\sin x = k$ (with $|k| < 1$ ) generally yields two solutions in $[0, 2\pi]$ — one in the first or second quadrant and one in the supplementary position.
$\sin x = -1$ has the unique solution $x = \frac{3}{2}\pi$ in $[0, 2\pi]$ .

Common Mistakes

Forgetting to substitute the $-1$ on the right-hand side, giving an equation of degree one in $\sin x$ . The mark scheme notes that candidates who omit the $-1$ can still earn M1 DM1 but no accuracy marks.
Finding only the principal value of $\sin^{-1}(0.4)$ and missing the second solution in the second quadrant.
Using $\cos x = \sqrt{1 - \sin^2 x}$ with the wrong sign, which is a sign-error trap when not using the squared version of the identity.

Things to Be Careful About

Final answers must be AWRT (within $0.005$ or so) of $0.41$ , $2.73$ and $\frac{3}{2}\pi$ . The mark scheme accepts any equivalent forms such as $0.13\pi$ , $0.87\pi$ , $\frac{3}{2}\pi$ or fractions of $\pi$ with denominator $450$ (e.g. $\frac{59}{450}\pi$ ).
A degree sign written alongside a radian answer can be ignored.
Any value outside $[0, 2\pi]$ should be ignored (and is silently discarded by the mark scheme).

Techniques used

use the identity cos^2 x = 1 - sin^2 xform and solve a quadratic in sin xfactor the quadraticapply arcsin twice within the given interval

Q7MediumCoordinate Geometry

The coordinates of the points $P$ and $Q$ are $(1, 1)$ and $(7, 11)$ respectively. The line segment $PQ$ forms a diameter of a circle.

(a)

Find the equation of the circle.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $PQ$ is a diameter, the centre of the circle is the midpoint of $PQ$ , and the radius is half the length of $PQ$ . Substitute the centre and radius into the general equation of a circle.

Working

The centre $C$ is the midpoint of $P(1,1)$ and $Q(7,11)$ :

C = \left( \frac{1+7}{2}, \frac{1+11}{2} \right) = (4,6)

The squared length of the diameter is

PQ^2 = (7-1)^2 + (11-1)^2 = 6^2 + 10^2 = 136

The radius is half the diameter, so the radius squared is

r^2 = \frac{PQ^2}{4} = \frac{136}{4} = 34

Therefore the equation of the circle is

(x-4)^2 + (y-6)^2 = 34

Expanding gives the equivalent form

x^2 + y^2 - 8x - 12y + 18 = 0

Answer

The equation of the circle is

(x-4)^2 + (y-6)^2 = 34

or $x^2 + y^2 - 8x - 12y + 18 = 0$ .

Final answer

(x-4)^2 + (y-6)^2 = 34

Detailed explanation

Walkthrough

The line segment $PQ$ is given as a diameter, so the centre of the circle is exactly halfway between $P$ and $Q$ . Find the midpoint by averaging the $x$ -coordinates and the $y$ -coordinates, giving $(4,6)$ .

The radius is half the length of the diameter. Rather than finding the length itself, work with squared lengths to avoid square roots. The squared length of the diameter is

PQ^2 = (7-1)^2 + (11-1)^2 = 136

so the squared radius is $136/4 = 34$ .

The general circle with centre $(a,b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$ . Substitute $a=4$ , $b=6$ , $r^2=34$ to obtain the required equation.

Key Takeaways

The centre of a circle through two endpoints of a diameter is the midpoint of those endpoints. The radius is half the diameter. Always identify the centre and radius before writing the equation of a circle.

Common Mistakes

Using $P(1,1)$ or $Q(7,11)$ as the centre of the circle.
Using the full length of $PQ$ as the radius instead of half of it.
Writing $r^2 = 136$ instead of $r^2 = 34$ .
Giving the final answer in the form $(\sqrt{34})^2$ or $5.83^2$ , which is not accepted by the mark scheme.

Things to Be Careful About

The radius may be left unsimplified during working, but the final equation should be exact. Because $PQ^2 = 136$ , the correct squared radius is $34$ . If you prefer the expanded form, expand carefully:

(x-4)^2 + (y-6)^2 = 34

gives

x^2 - 8x + 16 + y^2 - 12y + 36 = 34

so $x^2+y^2-8x-12y+18=0$ .

Techniques used

find the midpoint of a line segmentcalculate the squared length of a segmentwrite the equation of a circle from its centre and radius

(b)

Find the equation of the tangent to the circle at the point $Q$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The radius to $Q$ has the same gradient as $PQ$ . The tangent at $Q$ is perpendicular to this radius, so use $m_1m_2=-1$ to find the gradient of the tangent, then write the line through $Q(7,11)$ .

Working

The gradient of $PQ$ is

m_{PQ} = \frac{11-1}{7-1} = \frac{10}{6} = \frac{5}{3}

For perpendicular lines, $m_1m_2=-1$ , so the gradient of the tangent is

m = -\frac{1}{m_{PQ}} = -\frac{3}{5}

Using the point $Q(7,11)$ , the tangent has equation

y-11 = -\frac{3}{5}(x-7)

Rearranging,

y = -\frac{3}{5}x + \frac{21}{5} + 11

so

y = -\frac{3}{5}x + \frac{76}{5}

Equivalently,

\frac{y-11}{x-7} = -\frac{3}{5}

Answer

The equation of the tangent at $Q$ is

y = -\frac{3}{5}x + \frac{76}{5}

or $\frac{y-11}{x-7} = -\frac{3}{5}$ .

Final answer

y = -3/5 x + 76/5

Detailed explanation

Walkthrough

The centre of the circle is $(4,6)$ , so the radius to $Q(7,11)$ is the line segment from the centre to $Q$ . This radius lies along the same straight line as $PQ$ , so its gradient is the gradient of $PQ$ :

\frac{11-1}{7-1} = \frac{5}{3}

A tangent to a circle at a point is perpendicular to the radius at that point. Therefore the tangent's gradient is the negative reciprocal of $5/3$ , namely $-3/5$ . Use the point-slope form with $(x_1,y_1)=(7,11)$ :

y-11 = -\frac{3}{5}(x-7)

Then rearrange to the simplified form $y=-\frac{3}{5}x+\frac{76}{5}$ .

Key Takeaways

The tangent to a circle is perpendicular to the radius at the point of contact. This fact lets you find the tangent gradient from the radius gradient with $m_1m_2=-1$ .

Common Mistakes

Using the radius gradient $5/3$ as the tangent gradient instead of taking its negative reciprocal.
Forgetting the negative sign when applying $m_1m_2=-1$ .
Substituting the coordinates of the centre $(4,6)$ instead of the point of tangency $(7,11)$ into the line equation.
A sign error when simplifying $\frac{21}{5}+11$ to $\frac{76}{5}$ .

Things to Be Careful About

The mark scheme accepts the tangent equation in any equivalent form, such as $\frac{y-11}{x-7}=-\frac{3}{5}$ or $y=-\frac{3}{5}x+\frac{76}{5}$ . If you use implicit differentiation of the circle equation, substitute $x=7$ and $y=11$ into

2(x-4)+2(y-6)\frac{dy}{dx}=0

and solve for $\frac{dy}{dx}$ , which also gives $-\frac{3}{5}$ .

Techniques used

compute the gradient of a line segmentapply the perpendicular gradient conditionform the equation of a line in point-slope form

(c)

The other point on the circle with $x$ -coordinate 7 is $R$ .

Find the coordinates of the point of intersection of the tangent at $Q$ with the tangent at $R$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Substitute $x=7$ into the circle equation to find the two points on the circle with that $x$ -coordinate. One is $Q$ and the other is $R$ . Then find the tangent at $R$ , and solve it simultaneously with the tangent at $Q$ from part (b).

Working

Substitute $x=7$ into $x^2+y^2-8x-12y+18=0$ :

49 + y^2 - 56 - 12y + 18 = 0

y^2 - 12y + 11 = 0

(y-1)(y-11) = 0

So $y=1$ or $y=11$ . Since $Q$ is $(7,11)$ , the other point is

R = (7,1)

The radius to $R$ has gradient

m_{CR} = \frac{1-6}{7-4} = -\frac{5}{3}

so the tangent at $R$ has gradient $\frac{3}{5}$ . Its equation is

y-1 = \frac{3}{5}(x-7)

or

y = \frac{3}{5}x - \frac{16}{5}

The tangent at $Q$ is $y = -\frac{3}{5}x + \frac{76}{5}$ . Equating the two tangents:

\frac{3}{5}x - \frac{16}{5} = -\frac{3}{5}x + \frac{76}{5}

\frac{6}{5}x = \frac{92}{5}

x = \frac{46}{3}

Substitute into the tangent at $R$ :

y = \frac{3}{5}\cdot\frac{46}{3} - \frac{16}{5} = \frac{46}{5} - \frac{16}{5} = 6

Answer

The point of intersection is

\left( \frac{46}{3}, 6 \right)

or $x = \frac{46}{3}$ , $y=6$ .

Final answer

(46/3, 6)

Detailed explanation

Walkthrough

The circle has equation $(x-4)^2+(y-6)^2=34$ , or expanded as $x^2+y^2-8x-12y+18=0$ . To find points with $x$ -coordinate $7$ , substitute $x=7$ into the expanded form:

49+y^2-56-12y+18=0

This simplifies to $y^2-12y+11=0$ , which factorises as $(y-1)(y-11)=0$ . Hence the two points with $x=7$ are $(7,11)$ , which is $Q$ , and $(7,1)$ , which is $R$ .

The tangent at $R$ is perpendicular to the radius $CR$ , where $C=(4,6)$ . The gradient of $CR$ is

\frac{1-6}{7-4} = -\frac{5}{3}

so the tangent at $R$ has gradient $3/5$ . Its equation through $(7,1)$ is

y-1=\frac{3}{5}(x-7)

or $y=\frac{3}{5}x-\frac{16}{5}$ .

Now solve the two tangent equations together. Since the point of intersection lies on both lines, set the two expressions for $y$ equal:

\frac{3}{5}x-\frac{16}{5} = -\frac{3}{5}x+\frac{76}{5}

This gives $\frac{6}{5}x=\frac{92}{5}$ , so $x=\frac{46}{3}$ . Substitute back into either tangent to get $y=6$ .

Key Takeaways

When a line meets a circle at two points, substituting one coordinate gives a quadratic whose two roots are the two possible values of the other coordinate. Tangents to a circle are perpendicular to the radius at the point of contact. The intersection of two tangent lines is found by solving their line equations simultaneously.

Common Mistakes

Choosing $(7,11)$ as $R$ instead of $(7,1)$ ; $R$ is the other point with $x=7$ .
Using the same gradient for both tangents. The tangent at $Q$ has gradient $-3/5$ and the tangent at $R$ has gradient $3/5$ .
Forgetting to substitute the $x$ -value back into one of the tangent equations to find $y$ .
Simplifying $\frac{3}{5}\cdot\frac{46}{3}$ incorrectly; it equals $\frac{46}{5}$ .

Things to Be Careful About

The point $Q$ is $(7,11)$ , so when the quadratic gives $y=1$ and $y=11$ , the other point is $(7,1)$ . The tangent at $R$ must be perpendicular to the radius through $R$ , not to $PQ$ . When solving the two tangent equations, both gradients must be different for the lines to meet in a single point. The final coordinates may be left as $\left(\frac{46}{3},6\right)$ or as $x=\frac{46}{3}$ , $y=6$ ; $15.3$ is acceptable as an approximate check but the exact form is preferred.

Techniques used

substitute a coordinate into a circle equationfactorise a quadratic to find the second intersection pointform a tangent line using the perpendicular gradient conditionsolve simultaneous linear equations

Q8MediumSeriesQuadratics

The first three terms of a geometric progression are $a$ , $b$ and $c$ respectively, where $a$ , $b$ and $c$ are positive constants. The first three terms of an arithmetic progression are $a$ , $b$ and $-3c$ respectively.

(a)

Show that $a^2 - 10ac + 9c^2 = 0$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the fact that the three terms belong to a geometric progression to write an equation connecting $a$ , $b$ and $c$ . Then use the fact that the same three letters form the first three terms of an arithmetic progression to write a second equation. Eliminate $b$ to obtain the required equation in $a$ and $c$ .

Working

Let the common ratio of the geometric progression be $r$ . Since $a$ , $b$ , $c$ are consecutive terms,

r = \frac{b}{a} = \frac{c}{b},

so

b^2 = ac.

For the arithmetic progression, the common difference is constant:

b - a = -3c - b.

Rearrange:

2b = a - 3c, \qquad b = \frac{a - 3c}{2}.

Substitute this expression for $b$ into $b^2 = ac$ :

\left(\frac{a - 3c}{2}\right)^2 = ac.

Multiply out:

(a - 3c)^2 = 4ac.

Expand and simplify:

\begin{aligned} a^2 - 6ac + 9c^2 &= 4ac \\ a^2 - 10ac + 9c^2 &= 0. \end{aligned}

Answer

a^2 - 10ac + 9c^2 = 0

Final answer

a^2 - 10ac + 9c^2 = 0

Detailed explanation

Walkthrough

The geometric progression condition gives $b^2 = ac$ because the ratio between consecutive terms is constant. The arithmetic progression condition gives $b - a = -3c - b$ , because the common difference is constant. Rearranging the second equation expresses $b$ in terms of $a$ and $c$ . Substituting this into $b^2 = ac$ eliminates $b$ , leaving an equation involving only $a$ and $c$ . Expanding $(a - 3c)^2$ and simplifying gives the required result.

Key Takeaways

Recognising the defining property of a geometric progression (common ratio) and an arithmetic progression (common difference).
Eliminating a variable by substitution to obtain a relationship between the remaining variables.
Expanding squared binomials carefully.

Common Mistakes

Using only one progression and forgetting to combine both conditions.
Sign errors when rearranging $b - a = -3c - b$ ; the correct rearrangement is $2b = a - 3c$ .
Not showing the expansion of $(a - 3c)^2$ ; the mark scheme requires a convincing proof (AG.

Things to Be Careful About

Since $a$ , $b$ , $c$ are positive, the common ratio of the GP is positive, though this is not needed in part (a.
The mark scheme awards B1 for one correct equation connecting $a$ , $b$ , $c$ only, M1 for forming an equation in $a$ and $c$ only, and A1 for the final convincing proof.

Techniques used

equate common ratios in a geometric progressionequate common differences in an arithmetic progressioneliminate a variable by substitutionexpand and simplify a quadratic expression

(b)

It is now given that $a = 9$ and $c$ takes the smaller of its two possible values.

8M

(i)

Find the sum to infinity of the geometric progression.

5M Similar

DifficultyMedium

Worked solution

Approach

Substitute $a = 9$ into the equation from part (a, solve the resulting quadratic for $c$ , and choose the smaller positive value. Use the geometric progression property to find the common ratio $r$ , then apply the sum-to-infinity formula for a convergent geometric progression.

Working

With $a = 9$ , the equation becomes

81 - 90c + 9c^2 = ​0.

Divide by $9$ :

c^2 - 10c + 9 =​ 0.

Factorise:

(c - 1)(c - 9) =​ 0,

so $c = 1$ or $c = 9$ . The smaller value is $c = 1$ .

Since $b^2 = ac$ , with $a = 9$ and $c = 1$ ,

b^2 = 9,

and $b > 0$ , so $b = 3$ . Therefore the common ratio of the GP is

r = \frac{b}{a} = \frac{3}{9} = \frac{1}{3}.

(Alternatively, $r^2 = \frac{c}{a} = \frac{1}{9}$ , and $r > 0$ gives $r = \frac{1}{3}$ .

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

S_\infty = \frac{a}{1 - r} = \frac{9}{1 - \frac{1}{3}} = \frac{9}{\frac{2}{3}} = \frac{27}{2}.

Answer

S_\infty = \frac{27}{2}

Final answer

27/2

Detailed explanation

Walkthrough

Substitute $a = 9$ into the equation from part (a. This gives a quadratic in $c$ . Factorising gives two possible values, $c = 1$ and $c = 9$ ; the question specifies that $c$ takes the smaller value, so $c = 1$ . Then use $b^2 = ac$ to find $b = 3$ , because $b$ is positive. The common ratio is $r = b/a = 1/3$ . Since $|r|< 1$ , the geometric progression converges, and the sum to infinity is $a/(1-r) = 27/2$ .

Key Takeaways

Solving a quadratic equation by factorisation.
Using the positivity of terms to choose the correct sign of the common ratio.
Applying the sum-to-infinity formula $S_\infty = \frac{a}{1-r}$ only when $|r|< 1$ .

Common Mistakes

Forgetting that $c$ must take the smaller of its two values; including $c = 9$ as an additional solution is condoned, but the final answer must use $c = 1$ .
Using $r = -\frac{1}{3}$ ; since $a$ , $b$ , $c$ are positive, the common ratio must be positive.
Applying the sum-to-infinity formula without checking that $|r|< 1$ .

Things to Be Careful About

The mark scheme allows the extra solution $c = 9$ for the B1 mark, but any error that leads to $c = 1$ costs that B1.
The mark for finding $r$ may be implied by writing $r = \frac{1}{3}$ , $r = \pm\frac{1}{3}$ , or $r = -\frac{1}{3}$ .
The sum-to-infinity mark requires use of the correct formula with $a = 9$ and their $|r|< 1$ ; extra working with a negative $r$ is ignored.
Do not give an extra final answer; the A1 mark does not allow an additional incorrect answer.

Techniques used

solve a quadratic equation by factorisationselect the smaller positive rootdetermine the common ratio using positivity of termsapply the sum-to-infinity formula for a geometric progression

(ii)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the values $a = 9$ , $b = 3$ and $c = 1$ to write down the first three terms of the arithmetic progression. Find its common difference, then apply the formula for the sum of the first $n$ terms.

Working

The first three terms of the AP are

9,\quad 3,\quad -3(1) = -3.

Hence the common difference is

d = 3 - 9 = -6.

Using $S_n = \frac{n}{2}\left(2a + (n-1)d\right)$ with $n = 20$ , $a = 9$ and $d = -6$ :

\begin{aligned} S_{20} &= \frac{20}{2}\left(2(9) + (20-1)(-6)\right) \\ &= 10\left(18 - 114\right) \\ &= 10(-96) = -960. \end{aligned}

Answer

S_{20} = -960

Final answer

-960

Detailed explanation

Walkthrough

With $c = 1$ , the first three terms of the arithmetic progression are $9$ , $3$ , and $-3$ . The common difference is $d = 3 - 9 = -6$ . Then substitute $n = 20$ , $a = 9$ , $d = -6$ into the sum formula $S_n = \frac{n}{2}(2a + (n-1)d))$ . This gives $S_{20} = -960$ .

Key Takeaways

The formula for the sum of the first $n$ terms of an arithmetic progression: $S_n = \frac{n}{2}(2a + (n-1)d))$ .
Finding the common difference from consecutive terms.
Substituting carefully, including handling a negative common difference.

Common Mistakes

Using the wrong sign for the common difference; here $d = -6$ , not $6$ .
Forgetting to multiply by $\frac{n}{2}$ .
Using $n = 19$ instead of $n = 20$ .

Things to Be Careful About

The mark scheme awards B1 for stating $d = -6$ (or equivalent), M1 for using a correct sum formula with $n = 20$ , $a = 9$ and their non-zero $d$ , and A1 for the final answer $-960$ .
The common difference can be implied by quoting the correct formula then substituting; it does not have to be stated separately.

Techniques used

find the common difference of an arithmetic progressionapply the arithmetic series sum formula

Q9MediumDifferentiationFunctions

The function $f$ is defined by $f(x) = \frac{4}{(3x - 6)^2} + \frac{1}{(3x - 6)^3}$ for $x > 2$ .

(a)

Find an expression for $f'(x)$ and hence determine whether $f$ is an increasing function, a decreasing function or neither.

4M Similar

DifficultyMedium

Worked solution

Approach

Rewrite the terms as powers of $(3x-6)$ , differentiate using the chain rule, then use $x>2$ to determine the sign of $f'(x)$ .

Working

Let $u=3x-6$ , so $f(x)=4u^{-2}+u^{-3}$ and $\frac{du}{dx}=3$ .

f'(x)=4(-2)(3x-6)^{-3}(3)+1(-3)(3x-6)^{-4}(3)

f'(x)=-\frac{24}{(3x-6)^3}-\frac{9}{(3x-6)^4}.

For $x>2$ , $3x-6>0$ , so both terms are negative; hence $f'(x)<0$ for every $x$ in the domain. A function with a negative derivative on its whole domain is decreasing.

Answer

f'(x)=-\frac{24}{(3x-6)^3}-\frac{9}{(3x-6)^4}

and $f$ is decreasing.

Final answer

f'(x) = -24/(3x-6)^3 - 9/(3x-6)^4; f is decreasing.

Detailed explanation

Walkthrough

The expression is written with powers of $(3x-6)$ : $4(3x-6)^{-2}$ and $(3x-6)^{-3}$ . Each term is a composite function of the form $(3x-6)^n$ , so differentiating uses the chain rule: differentiate the outer power and multiply by the derivative of $3x-6$ , which is $3$ . Thus the first term contributes $4 \times (-2) \times 3(3x-6)^{-3}$ and the second contributes $1 \times (-3) \times 3(3x-6)^{-4}$ . Both contributions are negative for $x>2$ because $3x-6>0$ . A negative derivative on the whole domain means the function is decreasing.

Key Takeaways

The chain rule is needed for powers of a linear expression. The sign of $f'(x)$ on the domain tells whether $f$ is increasing, decreasing, or neither.

Common Mistakes

Forgetting the factor $3$ from the derivative of $3x-6$ is the most common error. Sign errors in the exponents also occur. The decreasing mark is only awarded if the derivative is written in the form $(-p)(3x-6)^{-3}+(-q)(3x-6)^{-4}$ with positive $p$ and $q$ , so keep the negative coefficients visible.

Things to Be Careful About

Use the domain $x>2$ : it makes $3x-6$ positive, so both terms of $f'(x)$ are negative. Do not try to find stationary points; the question only asks for the sign of the derivative.

Techniques used

rewrite terms as negative powersapply the chain rule to composite powersdetermine the sign of the derivative on the domain

(b)

State whether $f^{-1}$ exists. Give a reason for your answer.

1M Similar

(c)

The function $g$ is defined by $g(x) = 4x - 3$ for $x > a$ .

Find the range of $g$ in terms of the constant $a$ .

1M Similar

(d)

Find the set of values of $a$ for which the composite function $fg$ exists.

2M Similar

DifficultyMedium

Worked solution

Approach

For $fg$ to exist, every value taken by $g$ must lie in the domain of $f$ , so $g(x)>2$ for all $x>a$ .

Working

Since $g(x)=4x-3$ and $x>a$ , the range of $g$ is $(4a-3,\infty)$ . We need this range to be contained in the domain $(2,\infty)$ :

(4a-3,\infty) \subseteq (2,\infty)

so $4a-3 \geq 2$ . Hence $a \geq \frac{5}{4}$ . If $a=\frac{5}{4}$ , then $g(x)>2$ for every $x>a$ , so the endpoint is allowed.

Answer

a \geq \frac{5}{4}

or equivalently $\left[\frac{5}{4},\infty\right)$ .

Final answer

a ≥ 5/4, i.e. [5/4, ∞)

Detailed explanation

Walkthrough

For $fg$ to exist, every output of $g$ must be an allowed input of $f$ . The domain of $f$ is $x>2$ , so we need $g(x)>2$ for every $x>a$ . Since $g$ is increasing and $x>a$ , the smallest values of $g$ approach $4a-3$ but never reach it; the range is $(4a-3,\infty)$ . This range must lie inside $(2,\infty)$ , so $4a-3 \geq 2$ . Solving gives $a \geq \frac{5}{4}$ . If $a=\frac{5}{4}$ , then $g(x)>2$ for all $x>a$ , so the endpoint is allowed. Conversely, if $a<\frac{5}{4}$ , values of $x$ just above $a$ give $g(x)<2$ , so the composite would not be defined for those $x$ .

Key Takeaways

A composite function $fg$ exists exactly when the range of $g$ is a subset of the domain of $f$ . Strict inequalities must be handled carefully at endpoints.

Common Mistakes

Using $4a-3>2$ and excluding $a=\frac{5}{4}$ is a common endpoint error. Using $=$ , $\neq$ , or $<$ in the initial inequality is not accepted unless a correct inequality is later reached. Forgetting that the condition must hold for all $x>a$ , not just at one value, is also a mistake.

Things to Be Careful About

The mark scheme accepts both $a>\frac{5}{4}$ and $a \geq \frac{5}{4}$ , but the exact set of values is $a \geq \frac{5}{4}$ because the range of $g$ is open at its lower end.

Techniques used

require the range of g to be a subset of the domain of fsolve the resulting inequality for ahandle the strict endpoint correctly

Q10Medium-HardTrigonometryCircular Measure

Fig. 3

The diagram shows a circle with centre $A$ and radius $r$ passing through points $B$ , $C$ and $D$ . A larger circle of radius $s$ has centre $C$ and passes through $B$ and $D$ . The length $BD$ is also $s$ .

(a)

Show that $s = \sqrt{3}r$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $CB = CD = BD = s$ , triangle $BCD$ is equilateral, so $\angle BCD = \frac{\pi}{3}$ . Because $C$ lies on the smaller circle with centre $A$ , the inscribed angle theorem gives the central angle $\angle BAD = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}$ . Apply the cosine rule in the isoceles triangle $ABD$ (where $AB = AD = r$ ) to find $BD = s$ in terms of $r$ .

Working

Since $CB = CD = BD = s$ , triangle $BCD$ is equilateral, so

\angle BCD = \frac{\pi}{3}.

$C$ lies on the smaller circle, so $\angle BCD$ is the inscribed angle subtending chord $BD$ . By the inscribed angle theorem, the central angle subtending the same arc is twice the inscribed angle:

\angle BAD = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}.

In triangle $ABD$ , $AB = AD = r$ (both radii of the smaller circle) and $\angle BAD = \frac{2\pi}{3}$ . Apply the cosine rule:

\begin{aligned} s^2 &= AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos\left(\frac{2\pi}{3}\right) \\ &= r^2 + r^2 - 2r^2 \cdot \cos\left(\frac{2\pi}{3}\right) \\ &= 2r^2 - 2r^2 \cdot \left(-\frac{1}{2}\right) \\ &= 2r^2 + r^2 \\ &= 3r^2. \end{aligned}

Therefore

s = \sqrt{3}\, r.

Answer

s = \sqrt{3}\, r

Final answer

s = √3 r

Detailed explanation

Walkthrough

The first thing to notice is that $CB = CD = BD = s$ , so triangle $BCD$ is equilateral. This means each interior angle in the triangle is $\frac{\pi}{3}$ (or $60°$ ).

The next step uses the fact that $C$ lies on the smaller circle with centre $A$ . The angle $\angle BCD$ is an inscribed angle in this circle, subtending the chord $BD$ . The inscribed angle theorem tells us that the central angle subtending the same arc (the arc that does not contain $C$ ) is twice the inscribed angle. So $\angle BAD = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}$ .

Now we have a triangle $ABD$ with $AB = AD = r$ and $\angle BAD = \frac{2\pi}{3}$ . We want to find $BD = s$ . The cosine rule is the right tool here:

s^2 = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos(\angle BAD).

Substituting the values and using the exact trigonometric value $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$ , we get:

s^2 = r^2 + r^2 - 2r^2 \cdot \left(-\frac{1}{2}\right) = 2r^2 + r^2 = 3r^2.

Taking the positive square root (since $s$ is a length), $s = \sqrt{3}\, r$ .

Key Takeaways

Recognise when a triangle is equilateral (all three sides equal).
The inscribed angle theorem: the central angle subtending an arc is twice the inscribed angle subtending the same arc.
The cosine rule is a powerful tool for finding an unknown side of a triangle when you know two sides and the included angle (or any other combination).
Memorise exact values of $\sin$ and $\cos$ for special angles like $\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \pi$ .

Common Mistakes

Forgetting that the inscribed angle theorem relates the inscribed angle to the central angle subtending the same arc (the arc that does not contain the inscribed angle's vertex).
Sign errors: $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$ , not $+\frac{1}{2}$ .
The cosine rule gives $s^2$ , so we must take the positive square root to find $s$ .
Using degrees instead of radians in the cosine rule, or vice versa.

Things to Be Careful About

The "show that" question requires a clear, complete derivation. State the inscribed angle theorem and the cosine rule explicitly.
Make sure to identify the correct angle: the central angle, not the inscribed angle.
Keep calculations in exact form throughout to avoid losing accuracy.

Techniques used

recognise equilateral triangle BCDapply inscribed angle theorem to find central angle BADapply cosine rule in isoceles triangle ABDsubstitute exact value of cos(2π/3) and simplify

(b)

Find an expression for the area of the shaded region. Give your answer in the form $(a + b\pi)r^2$ , where $a$ and $b$ are constants to be found.

7M Similar

DifficultyMedium-Hard

Worked solution

Approach

The shaded region lies between the left arc of the larger circle and the left arc of the smaller circle, on the same side of the chord $BD$ . Compute the area of each segment (sector minus triangle), then subtract to obtain the shaded area.

Working

From part (a), $s = \sqrt{3}\, r$ , so $s^2 = 3r^2$ .

Central angles:

$\angle BAD = \frac{2\pi}{3}$ (from part a).
$\angle BCD = \frac{\pi}{3}$ (equilateral triangle $BCD$ ).

Sector areas:

\text{Sector } BAD = \frac{1}{2} r^2 \cdot \frac{2\pi}{3} = \frac{\pi r^2}{3}

\text{Sector } BCD = \frac{1}{2} s^2 \cdot \frac{\pi}{3} = \frac{1}{2} (3r^2) \cdot \frac{\pi}{3} = \frac{\pi r^2}{2}

Triangle areas:

\text{Triangle } ABD = \frac{1}{2} r^2 \sin\left(\frac{2\pi}{3}\right) = \frac{1}{2} r^2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}\, r^2}{4}

\text{Triangle } BCD = \frac{1}{2} s^2 \sin\left(\frac{\pi}{3}\right) = \frac{1}{2} (3r^2) \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}\, r^2}{4}

Segment areas (left of $BD$ ):

Each segment is bounded by the chord $BD$ and the corresponding left arc. The smaller circle's left arc bulges further to the left (away from the chord $BD$ ) than the larger circle's left arc, so the smaller circle's left segment contains the larger circle's left segment. The shaded region is the difference:

\text{Smaller left segment} = \text{Sector } BAD - \text{Triangle } ABD = \frac{\pi r^2}{3} - \frac{\sqrt{3}\, r^2}{4}

\text{Larger left segment} = \text{Sector } BCD - \text{Triangle } BCD = \frac{\pi r^2}{2} - \frac{3\sqrt{3}\, r^2}{4}

\begin{aligned} \text{Shaded area} &= \left(\frac{\pi r^2}{3} - \frac{\sqrt{3}\, r^2}{4}\right) - \left(\frac{\pi r^2}{2} - \frac{3\sqrt{3}\, r^2}{4}\right) \\ &= \frac{\pi r^2}{3} - \frac{\sqrt{3}\, r^2}{4} - \frac{\pi r^2}{2} + \frac{3\sqrt{3}\, r^2}{4} \\ &= -\frac{\pi r^2}{6} + \frac{2\sqrt{3}\, r^2}{4} \\ &= \frac{\sqrt{3}\, r^2}{2} - \frac{\pi r^2}{6} \\ &= \left(\frac{\sqrt{3}}{2} - \frac{\pi}{6}\right) r^2. \end{aligned}

Answer

\left(\frac{\sqrt{3}}{2} - \frac{\pi}{6}\right) r^2

with $a = \frac{\sqrt{3}}{2}$ and $b = -\frac{\pi}{6}$ .

Final answer

(√3/2 − π/6) r²

Detailed explanation

Walkthrough

The shaded region is a crescent between the two left arcs of the circles (both arcs lie on the same side of the chord $BD$ , away from the centres $A$ and $C$ ). To find its area, we compute the area of each segment (the region between a chord and its arc) and subtract.

Step 1: Identify the central angles.

$\angle BAD = \frac{2\pi}{3}$ — central angle of the smaller circle subtending chord $BD$ (from part a).
$\angle BCD = \frac{\pi}{3}$ — central angle of the larger circle, equal to the angle of the equilateral triangle $BCD$ at vertex $C$ .

Step 2: Compute the sector areas.

The sector area formula is $\frac{1}{2} r^2 \theta$ , where $\theta$ is the central angle in radians.

\text{Sector } BAD = \frac{1}{2} r^2 \cdot \frac{2\pi}{3} = \frac{\pi r^2}{3}

\text{Sector } BCD = \frac{1}{2} s^2 \cdot \frac{\pi}{3} = \frac{1}{2} (3r^2) \cdot \frac{\pi}{3} = \frac{\pi r^2}{2}.

(We substituted $s^2 = 3r^2$ from part a.)

Step 3: Compute the triangle areas.

The triangle area formula $\frac{1}{2} ab \sin C$ uses the included angle $C$ .

For isoceles triangle $ABD$ with two sides of length $r$ and included angle $\frac{2\pi}{3}$ :

\text{Triangle } ABD = \frac{1}{2} r^2 \sin\left(\frac{2\pi}{3}\right) = \frac{1}{2} r^2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}\, r^2}{4}.

For equilateral triangle $BCD$ with side $s$ :

\text{Triangle } BCD = \frac{1}{2} s^2 \sin\left(\frac{\pi}{3}\right) = \frac{1}{2} (3r^2) \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}\, r^2}{4}.

Step 4: Recognise the shaded region as a difference of segments.

This is the key step. The shaded region is the crescent between the two left arcs. A segment of a circle is the region between a chord and an arc. Both circles have a left segment (the region between chord $BD$ and the left arc).

In this configuration, the smaller circle's left arc bulges further to the left than the larger circle's left arc. You can verify this with a quick calculation: the chord $BD$ is at perpendicular distance $\frac{r}{2}$ from the centre $A$ of the smaller circle, and at perpendicular distance $\frac{s\sqrt{3}}{2} = \frac{3r}{2}$ from the centre $C$ of the larger circle. Since the smaller circle has radius $r$ and the larger has radius $s = \sqrt{3} r$ , the chord sits at half the smaller radius but at a much greater fraction $\frac{3r/2}{\sqrt{3} r} = \frac{\sqrt{3}}{2}$ of the larger radius. This means the larger circle's left arc is more "flat" near the chord, while the smaller circle's left arc extends out further to the left.

The shaded region is bounded by the smaller circle's arc (further to the left) and the larger circle's arc (closer to the chord). So the shaded region is the part of the smaller circle's left segment that is NOT in the larger circle's left segment:

\text{Shaded area} = \text{Smaller left segment} - \text{Larger left segment}.

Step 5: Compute the difference.

\text{Smaller left segment} = \frac{\pi r^2}{3} - \frac{\sqrt{3}\, r^2}{4}.

\text{Larger left segment} = \frac{\pi r^2}{2} - \frac{3\sqrt{3}\, r^2}{4}.

Subtracting:

\begin{aligned} \text{Shaded area} &= \left(\frac{\pi r^2}{3} - \frac{\sqrt{3}\, r^2}{4}\right) - \left(\frac{\pi r^2}{2} - \frac{3\sqrt{3}\, r^2}{4}\right) \\ &= \frac{\pi r^2}{3} - \frac{\sqrt{3}\, r^2}{4} - \frac{\pi r^2}{2} + \frac{3\sqrt{3}\, r^2}{4} \\ &= -\frac{\pi r^2}{6} + \frac{2\sqrt{3}\, r^2}{4} \\ &= \frac{\sqrt{3}\, r^2}{2} - \frac{\pi r^2}{6} \\ &= \left(\frac{\sqrt{3}}{2} - \frac{\pi}{6}\right) r^2. \end{aligned}

This is in the form $(a + b\pi) r^2$ with $a = \frac{\sqrt{3}}{2}$ and $b = -\frac{\pi}{6}$ .

Key Takeaways

Sector area = $\frac{1}{2} r^2 \theta$ , where $\theta$ is the central angle in radians.
Triangle area (using two sides and the included angle) = $\frac{1}{2} ab \sin C$ .
Segment area = sector area − triangle area (for a minor segment).
When a region is bounded by two arcs of different circles on the same side of a chord, it is the difference of two segments. Identify which arc is further from the chord (this segment contains the other).
Exact trigonometric values for special angles (e.g. $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ ) are essential.

Common Mistakes

Forgetting to use radians. The sector area formula $\frac{1}{2} r^2 \theta$ requires the angle in radians. Mixing degrees and radians will give the wrong answer.
Sign errors with trig values. $\sin\left(\frac{2\pi}{3}\right) = +\frac{\sqrt{3}}{2}$ (positive in the second quadrant), but $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$ (negative in the second quadrant).
Subtracting the segments in the wrong order. Make sure to identify which segment is "outer" (further from the chord) and which is "inner". The shaded region is outer minus inner.
Not substituting $s = \sqrt{3}\, r$ . The mark scheme explicitly requires simplification in terms of $r$ at intermediate steps.

Things to Be Careful About

The question requires the answer in the form $(a + b\pi) r^2$ . Make sure to write it in this exact form, with the constants $a$ and $b$ clearly identified.
The angle $\angle BCD$ in the larger circle is the central angle of the arc on the left of $BD$ (the arc not containing $C$ ). It equals $\frac{\pi}{3}$ , the angle of the equilateral triangle $BCD$ at $C$ .
Verify the final answer is positive: $\frac{\sqrt{3}}{2} \approx 0.866$ and $\frac{\pi}{6} \approx 0.524$ , so the answer is positive. ✓

Techniques used

identify central angles BAD and BCDcompute sector areas using (1/2)r²θcompute triangle areas using (1/2)ab sin Crecognise shaded region as difference of two segmentssubstitute s = √3 r and simplify to required form

Mathematics 9709/12 — October/November 2025

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working