(a) Show that the equation $\frac{7 \tan \theta}{\cos \theta} + 12 = 0$ can be expressed as $$ 12 \sin^2 \theta - 7 \sin \theta - 12 = 0 $$ (b) Hence solve the equation $\frac{7 \tan \theta}{\cos \theta} + 12 = 0$ for $0^\circ \leq \theta \leq 360^\circ$.

(a) 12 sin^2 θ - 7 sin θ - 12 = 0 · (b) θ = 228.6°, 311.4°

The first and second terms of an arithmetic progression are $\tan \theta$ and $\sin \theta$ respectively, where $0 < \theta < \frac{1}{2}\pi$. (a) Given that $\theta = \frac{1}{4}\pi$, find the exact sum of the first 40 terms of the progression. (b) The first and second terms of a geometric progression are $\tan \theta$ and $\sin \theta$ respectively, where $0 < \theta < \frac{1}{2}\pi$. (i) Find the sum to infinity of the progression in terms of $\theta$. (ii) Given that…

(a) 390√2 - 740 · (b)(i) tanθ/(1 - cosθ) · (b)(ii) 3.46

The equation of a circle is $(x - 6)^2 + (y + a)^2 = 18$. The line with equation $y = 2a - x$ is a tangent to the circle. (a) Find the two possible values of the constant $a$. (b) For the greater value of $a$, find the equation of the diameter which is perpendicular to the given tangent.

(a) a = 0 or a = 4 · (b) y = x - 10

9709/12 May/June 2024 — Mathematics Solved Past Paper

Q13MMedium-EasySeries

The coefficient of $x^2$ in the expansion of $(1 - 4x)^6$ is 12 times the coefficient of $x^2$ in the expansion of $(2 + ax)^5$ .

Find the value of the positive constant $a$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the binomial theorem to extract the coefficient of $x^2$ in each expansion, then combine the given condition and solve for the positive constant $a$ .

Working

For $(1 - 4x)^6$ , the term in $x^2$ is

\binom{6}{2}(-4x)^2 = 15 \times 16x^2 = 240x^2

so the coefficient of $x^2$ is $240$ .

For $(2 + ax)^5$ , the term in $x^2$ is

\binom{5}{2}(2)^3(ax)^2 = 10 \times 8 \times a^2x^2 = 80a^2x^2

so the coefficient of $x^2$ is $80a^2$ .

The coefficient in $(1 - 4x)^6$ is 12 times the coefficient in $(2 + ax)^5$ :

240 = 12 \times 80a^2

240 = 960a^2

a^2 = \frac{240}{960} = \frac{1}{4}

Since $a$ is positive,

a = \frac{1}{2}

Answer

a = \frac{1}{2}

Final answer

a = 1/2

Detailed explanation

Walkthrough

We compare the coefficients of $x^2$ in two binomial expansions. First, expand $(1 - 4x)^6$ : the $x^2$ term is obtained by choosing the $-4x$ factor twice and the $1$ factor four times. The binomial coefficient is $\binom{6}{2}=15$ , and $(-4x)^2=16x^2$ , so the coefficient is $15 \times 16 = 240$ .

Second, expand $(2 + ax)^5$ : the $x^2$ term is obtained by choosing the $ax$ factor twice and the $2$ factor three times. The binomial coefficient is $\binom{5}{2}=10$ , the constant factor is $2^3=8$ , and the variable part gives $a^2x^2$ , so the coefficient is $10 \times 8 \times a^2 = 80a^2$ .

The condition states that the first coefficient is 12 times the second, so we set $240 = 12 \times 80a^2$ . This simplifies to $960a^2 = 240$ , giving $a^2 = \frac{1}{4}$ . Since $a$ is specified to be positive, we take the positive square root: $a = \frac{1}{2}$ .

Key Takeaways

For $(p+qx)^n$ , the term containing $x^r$ is $\binom{n}{r}p^{n-r}q^r x^r$ .
The coefficient of a power of $x$ is the numerical multiplier of that term, not the whole term including $x^r$ .
When a coefficient is a multiple of another, set up an equation between the coefficients before solving.

Common Mistakes

Forgetting the factor $2^3$ when finding the $x^2$ coefficient in $(2+ax)^5$ .
Forgetting to multiply the second coefficient by 12.
Taking the negative root $a = -\frac{1}{2}$ even though the question specifies a positive constant.

Things to Be Careful About

The sign in $(1 - 4x)^6$ : $(-4)^2$ is positive, so the $x^2$ coefficient is $240$ , not $-240$ .
The mark scheme condones $\pm 0.5$ , but the final answer should be $a = 0.5$ because $a$ is positive.
Always simplify the ratio $\frac{240}{960}$ fully to $\frac{1}{4}$ before taking the square root.

Techniques used

extract the coefficient of x^2 using the binomial theoremequate coefficients using the given multiplesolve the resulting equation for the positive constant

Approach

Compare the transformed equation $y = 4(x-3)^2 - 8$ with the original $y = x^2$ . The factor $4$ outside the square gives a vertical stretch, the term $(x-3)$ gives a horizontal translation, and the $-8$ gives a vertical translation. The vertical stretch must be applied before the translation in the $y$ -direction.

Working

Start from $y = x^2$ .

Apply a stretch with factor $4$ in the $y$ -direction:

y = 4x^2

Then translate by $\begin{pmatrix}3\\-8\end{pmatrix}$ , i.e. $3$ units in the positive $x$ -direction and $8$ units in the negative $y$ -direction:

y = 4(x-3)^2 - 8

This is a stretch and the two translations only. The stretch is applied before the translation in the $y$ -direction.

Answer

Stretch factor $4$ in the $y$ -direction, parallel to the $y$ -axis.
Translation by $\begin{pmatrix}3\\-8\end{pmatrix}$ , i.e. $3$ units to the right and $8$ units down.

Equivalently, after the stretch, translate by $\begin{pmatrix}3\\0\end{pmatrix}$ and then by $\begin{pmatrix}0\\-8\end{pmatrix}$ .

Final answer

Stretch factor 4 in the y-direction, then translation by vector (3, -8) (3 units right and 8 units down).

Detailed explanation

Walkthrough

The original curve is $y=x^2$ . After the transformation we want $y=4(x-3)^2-8$ . Break the target expression into separate changes from $y=x^2$ .

The coefficient $4$ on the square multiplies every $y$ -value by $4$ . This is a stretch with factor $4$ in the $y$ -direction.
Replacing $x$ by $x-3$ inside the square shifts the graph $3$ units to the right, i.e. a translation by $\begin{pmatrix}3\\0\end{pmatrix}$ .
Subtracting $8$ at the end shifts the graph $8$ units down, i.e. a translation by $\begin{pmatrix}0\\-8\end{pmatrix}$ .

Now check the order. If the vertical translation were done before the vertical stretch, we would get $y=4(x^2-8)=4x^2-32$ , which is not the target. Therefore the vertical stretch must be done before the vertical translation. The two translations can be combined as $\begin{pmatrix}3\\-8\end{pmatrix}$ and applied after the stretch.

Key Takeaways

For $y=f(x)$ : $y=af(x)$ is a vertical stretch with factor $a$ ; $y=f(x-a)$ is a horizontal translation of $a$ units; $y=f(x)+c$ is a vertical translation of $c$ units.
When transformations act on the same axis, their order matters. A vertical stretch and a vertical translation do not commute.
A translation vector $\begin{pmatrix}a\\b\end{pmatrix}$ means $a$ units horizontally and $b$ units vertically.

Common Mistakes

Saying the stretch is in the $x$ -direction. The factor $4$ multiplies the whole function value, so it is in the $y$ -direction.
Thinking that $x-3$ means a shift to the left. It is actually a shift to the right by $3$ units.
Translating down by $8$ before the vertical stretch, which would give $4x^2-32$ instead of $4(x-3)^2-8$ .
Omitting the order. The mark scheme requires the vertical stretch to be before the translation in the $y$ -direction.
Writing only the translation by $\begin{pmatrix}3\\-8\end{pmatrix}$ without mentioning the stretch, or vice versa.

Things to Be Careful About

Use “factor 4” or “scale factor 4” for the vertical stretch; both are accepted.
The translation $\begin{pmatrix}3\\-8\end{pmatrix}$ counts as two translations, one in each direction. If you separate them, give both vectors.
Numerical values must be correct. Loose words such as “shift”, “up”, “down” are accepted only if the intention is clear.
The $x$ -axis is invariant under a vertical stretch, so the stretch is parallel to the $y$ -axis.

Techniques used

identify vertical stretch factoridentify horizontal and vertical translationsdetermine the correct order of transformations

Approach

Rewrite $\tan\theta$ as $\frac{\sin\theta}{\cos\theta}$ and simplify the compound fraction. Then use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to replace $\cos^2\theta$ , and rearrange to obtain the required quadratic form.

Working

Start with the given equation:

\frac{7 \tan \theta}{\cos \theta} + 12 = 0

Use the identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$ :

\frac{7 \cdot \frac{\sin\theta}{\cos\theta}}{\cos\theta} + 12 = 0

Simplify the compound fraction:

\frac{7\sin\theta}{\cos^2\theta} + 12 = 0

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

7\sin\theta + 12\cos^2\theta = 0

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

7\sin\theta + 12(1 - \sin^2\theta) = 0

Expand and collect terms:

7\sin\theta + 12 - 12\sin^2\theta = 0

Rearrange to the required form:

12\sin^2\theta - 7\sin\theta - 12 = 0

Answer

12\sin^2\theta - 7\sin\theta - 12 = 0

Final answer

12 sin^2 θ - 7 sin θ - 12 = 0

Detailed explanation

Walkthrough

The aim of this part is to transform the given trigonometric equation into the stated quadratic form in $\sin\theta$ .

Step 1: Replace $\tan\theta$ with $\frac{\sin\theta}{\cos\theta}$ . This is the fundamental tangent identity and is the only way to bring everything into terms of $\sin$ and $\cos$ .

Step 2: Simplify the compound fraction. Dividing by $\cos\theta$ is the same as multiplying by $\frac{1}{\cos\theta}$ , so $\frac{\sin\theta}{\cos\theta} \div \cos\theta = \frac{\sin\theta}{\cos^2\theta}$ . The denominator becomes $\cos^2\theta$ , not $\cos\theta$ — this is a common trap.

Step 3: Multiply every term by $\cos^2\theta$ to clear the denominator. This gives $7\sin\theta + 12\cos^2\theta = 0$ . Note that the constant 12 must also be multiplied by $\cos^2\theta$ .

Step 4: Use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to write $\cos^2\theta = 1 - \sin^2\theta$ . Substituting this in eliminates the cosine term completely.

Step 5: Expand and rearrange. Expanding $12(1 - \sin^2\theta)$ gives $12 - 12\sin^2\theta$ , so the equation becomes $7\sin\theta + 12 - 12\sin^2\theta = 0$ . Multiplying through by $-1$ and reordering gives $12\sin^2\theta - 7\sin\theta - 12 = 0$ , which is exactly the required form.

Key Takeaways

This question tests the two core trigonometric identities: $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta = 1$ . It also tests algebraic manipulation of fractions — specifically, simplifying a compound fraction and correctly multiplying every term through by the denominator. The skill of expressing a trigonometric equation as a quadratic in a single trigonometric function is fundamental and appears again in part (b).

Common Mistakes

Incorrectly simplifying the compound fraction: $\frac{\tan\theta}{\cos\theta}$ is $\frac{\sin\theta}{\cos^2\theta}$ , not $\frac{\sin\theta}{\cos\theta}$ .
Forgetting to multiply the constant term 12 by $\cos^2\theta$ when clearing the denominator.
Sign errors when rearranging: the final form requires $12\sin^2\theta - 7\sin\theta - 12 = 0$ , so care is needed with the signs after expansion.
Using the identity incorrectly: $\cos^2\theta = 1 - \sin^2\theta$ , not $1 + \sin^2\theta$ .

Things to Be Careful About

The mark scheme awards M1 for the use of $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and DM1 for the use of $\sin^2 + \cos^2 = 1$ , so both identities must be shown explicitly. The final A1 requires the equation in exactly the stated form — the answer is given in the question (AG), so the working must lead to it convincingly. The mark scheme condones the use of abbreviations like $s$ , $c$ and $t$ , but the final answer must be correct.

Techniques used

substitute tan theta as sin theta over cos thetasimplify the compound fractionapply the Pythagorean identity to eliminate cos squared theta

(b)

Hence solve the equation $\frac{7 \tan \theta}{\cos \theta} + 12 = 0$ for $0^\circ \leq \theta \leq 360^\circ$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the quadratic from part (a) and factorise it to find possible values of $\sin\theta$ . Reject any value outside the range $-1 \leq \sin\theta \leq 1$ , then find all angles in $0^\circ \leq \theta \leq 360^\circ$ that satisfy the remaining equation.

Working

From part (a):

12\sin^2\theta - 7\sin\theta - 12 = 0

Factorise the quadratic:

(4\sin\theta + 3)(3\sin\theta - 4) = 0

Set each factor to zero:

4\sin\theta + 3 = 0 \quad \Rightarrow \quad \sin\theta = -\frac{3}{4}

3\sin\theta - 4 = 0 \quad \Rightarrow \quad \sin\theta = \frac{4}{3}

Since $\frac{4}{3} > 1$ , this has no solution because $-1 \leq \sin\theta \leq 1$ . So:

\sin\theta = -\frac{3}{4}

Find the principal value:

\theta = \sin^{-1}\left(-\frac{3}{4}\right) = -48.6^\circ

$\sin\theta$ is negative in the third and fourth quadrants. Therefore, in the range $0^\circ \leq \theta \leq 360^\circ$ :

\theta = 180^\circ + 48.6^\circ = 228.6^\circ

\theta = 360^\circ - 48.6^\circ = 311.4^\circ

Answer

\theta = 228.6^\circ, \ 311.4^\circ

Final answer

θ = 228.6°, 311.4°

Detailed explanation

Walkthrough

This part uses the result from part (a) directly — the word "Hence" tells us to build on the quadratic we just derived.

Step 1: Factorise $12\sin^2\theta - 7\sin\theta - 12$ . We look for two binomial factors: $(4\sin\theta + 3)(3\sin\theta - 4)$ multiplies out to $12\sin^2\theta - 16\sin\theta + 9\sin\theta - 12 = 12\sin^2\theta - 7\sin\theta - 12$ , which checks out.

Step 2: Set each factor to zero. This gives $\sin\theta = -\frac{3}{4}$ and $\sin\theta = \frac{4}{3}$ .

Step 3: Reject $\sin\theta = \frac{4}{3}$ . Since the sine function only takes values between $-1$ and $1$ , $\frac{4}{3}$ is impossible. This is a crucial step — many students miss it.

Step 4: Solve $\sin\theta = -\frac{3}{4}$ . The inverse sine of $-\frac{3}{4}$ is approximately $-48.6^\circ$ . This is the principal value, but it is not in our required range $0^\circ \leq \theta \leq 360^\circ$ .

Step 5: Find the angles in the range. Since $\sin\theta$ is negative in the third and fourth quadrants, we add $180^\circ$ to get the third-quadrant angle ( $180^\circ + 48.6^\circ = 228.6^\circ$ ) and subtract from $360^\circ$ to get the fourth-quadrant angle ( $360^\circ - 48.6^\circ = 311.4^\circ$ ).

Key Takeaways

This question combines two skills: solving a quadratic equation in a trigonometric function, and finding all solutions of a trigonometric equation within a specified interval. The key insight is that the quadratic may produce roots that are invalid for the sine function, and these must be rejected before finding angles. The quadrant method for finding all angles in a $360^\circ$ range is essential.

Common Mistakes

Forgetting to reject $\sin\theta = \frac{4}{3}$ — this would give no valid angles from that root, but it must be explicitly dismissed.
Only giving one solution instead of two. In the range $0^\circ$ to $360^\circ$ , $\sin\theta = -\frac{3}{4}$ has two solutions.
Giving $-48.6^\circ$ as a final answer — it is outside the required range.
Using the wrong quadrants: $\sin\theta$ is negative in quadrants III and IV, so the angles are $180^\circ + \alpha$ and $360^\circ - \alpha$ , where $\alpha = 48.6^\circ$ .

Things to Be Careful About

The mark scheme awards M1 for the factorisation, B1 for identifying $\sin\theta = -\frac{3}{4}$ (or $\frac{4}{3}$ ), and B1 for the two correct angles. The angles must be in the range $0^\circ \leq \theta \leq 360^\circ$ — answers outside this range are ignored. The mark scheme accepts answers within rounding tolerance (AWRT), so $228.6^\circ$ and $311.4^\circ$ are expected, but $229^\circ$ and $311^\circ$ are also condoned. No other angles in the range satisfy the equation.

Techniques used

factorise the quadratic in sin thetareject the root that exceeds the sine rangedetermine all angles in the required interval

Q4Medium-HardFunctions

The function $f$ is defined as follows:

f(x) = \sqrt{x} - 1 \text{ for } x > 1

(a)

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

1M Similar

DifficultyEasy

Worked solution

Approach

Let $y = f(x)$ , then interchange the roles of $x$ and $y$ and solve for $y$ . This gives the inverse function.

Working

y = \sqrt{x} - 1

Add $1$ to both sides and square:

\begin{aligned} y + 1 &= \sqrt{x} \\ (y + 1)^2 &= x \end{aligned}

Interchanging $x$ and $y$ :

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

Answer

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

Final answer

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

Detailed explanation

Walkthrough

To find the inverse of a function, we start by writing $y = f(x)$ . We then swap the variables $x$ and $y$ , which effectively reverses the input/output relationship, and finally solve the new equation for $y$ in terms of $x$ . Here, isolating $\sqrt{x}$ by adding $1$ to both sides gives $y + 1 = \sqrt{x}$ , and squaring removes the square root to give $(y+1)^2 = x$ . Swapping $x$ and $y$ then gives $f^{-1}(x) = (x+1)^2$ .

Key Takeaways

The inverse reverses the effect of the original function: if $f$ takes $x$ to $\sqrt{x} - 1$ , then $f^{-1}$ must take the output back, undoing the "minus $1$ " and the "square root".
A square root in $f$ becomes a square in $f^{-1}$ .
The domain restriction $x > 1$ on $f$ translates into the range $f(x) > 0$ for the inverse.

Common Mistakes

Forgetting to swap the variables and just solving for $x$ in terms of $y$ (this gives the original $f$ back, not the inverse).
Squaring before adding $1$ — that would give $y^2 = x - 2\sqrt{x}\cdot 0 + 1$ which is wrong because you cannot square $y$ alone; you must first isolate the square root.

Things to Be Careful About

Make sure the square root is isolated before squaring. Once isolated, squaring is the safe operation.
The domain $x > 1$ on $f$ is needed to keep the range of $f$ non-negative, which is required since $f^{-1}$ uses $(x+1)^2$ where the input represents $\sqrt{x} - 1 \geq 0$ .

Techniques used

swap x and y in y = f(x)solve for ystate inverse as a function of x

(b)

Fig. 4.1

The diagram shows the graph of $y = g(x)$ where $g(x) = \frac{1}{x^2 + 2}$ for $x \in \mathbb{R}$ .

State the range of $g$ and explain whether $g^{-1}$ exists.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

For $g(x) = \dfrac{1}{x^2 + 2}$ , find the largest and smallest values it can take, then check whether the function is one-to-one.

Working

The denominator $x^2 + 2$ is smallest when $x = 0$ , giving $x^2 + 2 = 2$ . Therefore $g$ attains its maximum at $x = 0$ :

g(0) = \frac{1}{0^2 + 2} = \frac{1}{2}

As $|x| \to \infty$ , $x^2 + 2 \to \infty$ and $g(x) \to 0$ , but $g(x)$ never equals $0$ . Since $g$ is positive for all real $x$ , the range is:

0 < g(x) \leq \frac{1}{2}

The graph is symmetric about the $y$ -axis, so a horizontal line such as $y = \tfrac{1}{3}$ cuts the curve in two points. Hence $g$ is many-to-one and $g^{-1}$ does not exist.

Answer

Range: $0 < g(x) \leq \dfrac{1}{2}$ . $g^{-1}$ does not exist because $g$ is not one-to-one.

Final answer

Range: 0 < g(x) ≤ 1/2. g^{-1} does not exist because g is not one-to-one.

Detailed explanation

Walkthrough

For a function of the form $\frac{1}{\text{positive expression}}$ , the maximum occurs where the denominator is smallest. Here the denominator $x^2 + 2$ has minimum $2$ at $x = 0$ , so the maximum of $g$ is $\frac{1}{2}$ . The function never reaches $0$ because the denominator is always at least $2$ , and the function value gets arbitrarily close to $0$ as $|x|$ grows. Putting these together gives the range $0 < g(x) \leq \frac{1}{2}$ .

An inverse function only exists when the original function is one-to-one. Looking at the graph, the bell shape is symmetric about the $y$ -axis: a horizontal line at, say, $y = 0.3$ crosses the curve at two different $x$ -values. This means each $y$ -value in the range comes from more than one $x$ -value, so the function is many-to-one and has no inverse.

Key Takeaways

For $y = \frac{1}{f(x)}$ where $f(x) > 0$ , the range is $\left(0, \frac{1}{\min f}\right]$ when $\min f$ is attained.
A function has an inverse only if it is one-to-one; the horizontal line test is the graphical check.
Even functions ( $g(-x) = g(x)$ ) are immediately non-injective unless their domain is restricted to one side of the axis.

Common Mistakes

Stating the range as $0 < g(x) < \frac{1}{2}$ — the upper bound is achieved (at $x = 0$ ), so the inequality is $\leq$ , not $<$ .
Saying " $g^{-1}$ does not exist" without giving a reason. The mark scheme requires a justification involving the function not being one-to-one.
Writing the range as two separate inequalities joined by a comma (the mark scheme explicitly forbids this form).

Things to Be Careful About

Always include the maximum value with $\leq$ when the maximum is actually attained.
The justification for the non-existence of the inverse must mention the failure of injectivity (one-to-one / horizontal line test) — that is what examiners want to see.

Techniques used

identify the maximum value of the rational functiondetermine the limiting behaviour as |x| growsstate the range using inequalitiesapply the horizontal line test to assess whether the function is one-to-one

(c)

The function $h$ is defined by $h(x) = \frac{1}{x^2 + 2}$ for $x \geq 0$ .

Solve the equation $hf(x) = f\left(\frac{25}{16}\right)$ . Give your answer in the form $a + b\sqrt{c}$ , where $a$ , $b$ and $c$ are integers.

4M Similar

DifficultyMedium-Hard

Worked solution

Approach

First evaluate $f\left(\tfrac{25}{16}\right)$ . Then form the composite $\mathrm{hf}(x) = h(f(x))$ and set it equal to that constant. Invert the equation, isolate $\sqrt{x}$ , and square. Finally discard any solution that violates the domain $x > 1$ .

Working

Evaluate the right-hand side:

f\left(\frac{25}{16}\right) = \sqrt{\frac{25}{16}} - 1 = \frac{5}{4} - 1 = \frac{1}{4}

Form the composite. With $h(u) = \dfrac{1}{u^2 + 2}$ and $f(x) = \sqrt{x} - 1$ :

\mathrm{hf}(x) = h\bigl(\sqrt{x} - 1\bigr) = \frac{1}{\bigl(\sqrt{x} - 1\bigr)^2 + 2}

Set this equal to $\dfrac{1}{4}$ and invert both sides:

\frac{1}{\bigl(\sqrt{x} - 1\bigr)^2 + 2} = \frac{1}{4}

\bigl(\sqrt{x} - 1\bigr)^2 + 2 = 4

Isolate the squared term:

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

Take square roots (noting the sign will be checked against the domain):

\sqrt{x} - 1 = \pm\sqrt{2}

So

\sqrt{x} = 1 \pm \sqrt{2}

and squaring gives the candidate solutions

x = (1 \pm \sqrt{2})^2

Check the domain $x > 1$ . The value $\sqrt{x} = 1 - \sqrt{2}$ is negative, so it cannot be a square root; this branch is rejected. Only $\sqrt{x} = 1 + \sqrt{2}$ is valid, giving

x = (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}

Answer

x = 3 + 2\sqrt{2}

Final answer

x = 3 + 2√2

Detailed explanation

Walkthrough

The equation $\mathrm{hf}(x) = f\bigl(\tfrac{25}{16}\bigr)$ has two parts: we need a number (the right-hand side) and a function of $x$ (the left-hand side). The right-hand side is straightforward — just plug $\tfrac{25}{16}$ into $f$ to get $\tfrac{1}{4}$ .

The left-hand side is a composite. To form $\mathrm{hf}(x)$ , we take the formula for $h$ , replace its input with $f(x)$ , and simplify. Here that means $u = \sqrt{x} - 1$ in $h(u) = \tfrac{1}{u^2 + 2}$ , producing $\mathrm{hf}(x) = \tfrac{1}{(\sqrt{x} - 1)^2 + 2}$ .

Setting the two sides equal gives a fraction equal to $\tfrac{1}{4}$ . Reciprocating both sides turns the equation into $(\sqrt{x} - 1)^2 + 2 = 4$ , i.e. $(\sqrt{x} - 1)^2 = 2$ . Taking square roots gives $\sqrt{x} - 1 = \pm\sqrt{2}$ , so $\sqrt{x} = 1 + \sqrt{2}$ or $\sqrt{x} = 1 - \sqrt{2}$ .

A square root is by definition non-negative, so $\sqrt{x} = 1 - \sqrt{2}$ is rejected. Squaring the surviving branch yields $x = (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}$ . The discarded alternative $x = 3 - 2\sqrt{2}$ does not satisfy the original domain restriction $x > 1$ (in fact it is about $0.17$ ), confirming why it is extraneous.

Key Takeaways

Composite function notation $\mathrm{hf}$ means "first apply $f$ , then apply $h$ ".
A fraction equalling a fraction can be cleared by reciprocating only when both sides are non-zero; here both equal $\tfrac{1}{4}$ and $\tfrac{1}{4}$ so this is safe.
When squaring introduces a $\pm$ , always re-check against the domain of the original function. Here the domain $x > 1$ implies $\sqrt{x} > 1$ , so the negative branch is automatically impossible.
The answer is in surd form $a + b\sqrt{c}$ with integers $a, b, c$ , confirming the instruction.

Common Mistakes

Forgetting to evaluate $f\bigl(\tfrac{25}{16}\bigr)$ first and using the wrong target value on the right.
Squaring only one side instead of isolating the square root before squaring — this introduces extra terms and is hard to undo.
Reporting both $(1 + \sqrt{2})^2$ and $(1 - \sqrt{2})^2$ without checking the domain — the mark scheme requires the negative branch to be explicitly discarded.
Writing $x = \pm 1\sqrt{2}$ as the final answer, which the mark scheme explicitly awards $A0$ for.

Things to Be Careful About

The domain $x > 1$ from $f$ is essential. Even though $h$ is defined for $x \geq 0$ , the original function $f$ only allows $x > 1$ , so any solution with $x \leq 1$ is invalid.
The composite $\mathrm{hf}$ also requires $\sqrt{x} - 1 \geq 0$ (so that the input to $h$ lies in its domain $x \geq 0$ ). This is the same condition $x \geq 1$ , and combined with the original $x > 1$ it gives $x > 1$ .
Squaring can introduce extraneous solutions, so the final answer must always be re-checked in the original equation.

Techniques used

evaluate f at a specific valueform the composite function hf by substituting f(x) into hequate the composite to a constant and invertsolve the resulting equation for √xsquare to find xdiscard the extraneous solution that violates the domain x > 1

Q5MediumSeriesTrigonometry

The first and second terms of an arithmetic progression are $\tan \theta$ and $\sin \theta$ respectively, where $0 < \theta < \frac{1}{2}\pi$ .

(a)

Given that $\theta = \frac{1}{4}\pi$ , find the exact sum of the first 40 terms of the progression.

4M Similar

DifficultyMedium

Worked solution

Approach

Use the first term $a = \tan\theta$ and the second term $\sin\theta$ to find the common difference $d$ . Substitute $\theta = \frac{\pi}{4}$ and use the formula for the sum of the first $n$ terms of an arithmetic progression.

Working

For an arithmetic progression with first term $a$ and common difference $d$ , the second term is $a + d$ . Hence

d = \sin\theta - \tan\theta

When $\theta = \frac{\pi}{4}$ ,

\tan\frac{\pi}{4} = 1, \qquad \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}

so

d = \frac{\sqrt{2}}{2} - 1

The sum of the first $n$ terms is

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

Substitute $n = 40$ , $a = 1$ and $d = \frac{\sqrt{2}}{2} - 1$ :

S_{40} = \frac{40}{2}\left(2(1) + 39\left(\frac{\sqrt{2}}{2} - 1\right)\right)

S_{40} = 20\left(2 + \frac{39\sqrt{2}}{2} - 39\right) = 20\left(\frac{39\sqrt{2}}{2} - 37\right)

S_{40} = 390\sqrt{2} - 740

Answer

S_{40} = 390\sqrt{2} - 740

Final answer

390√2 - 740

Detailed explanation

Walkthrough

We have an arithmetic progression whose first two terms are $\tan\theta$ and $\sin\theta$ . For any AP, the common difference is obtained by subtracting consecutive terms, so $d = \sin\theta - \tan\theta$ . With $\theta = \frac{\pi}{4}$ , we use the exact values $\tan\frac{\pi}{4}=1$ and $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$ , giving $d = \frac{\sqrt{2}}{2} - 1$ .

Then apply the AP sum formula $S_n = \frac{n}{2}(2a + (n-1)d)$ . Here $n=40$ , $a=1$ , and $d$ is the value above. Substituting and simplifying gives $S_{40}=390\sqrt{2}-740$ . The key is to keep the exact surd form throughout; a decimal answer such as $-188$ is not exact and would not satisfy the request.

Key Takeaways

This question tests the definition of an arithmetic progression, the use of exact trigonometric values at special angles, and the formula for the sum of the first $n$ terms. It also emphasises careful algebraic simplification with surds.

Common Mistakes

Using the wrong formula for the sum, such as $S_n = \frac{n}{2}(a+l)$ with $l$ as the second term instead of the last term.
Evaluating $\tan\frac{\pi}{4}$ or $\sin\frac{\pi}{4}$ incorrectly, or using degrees instead of radians.
Making a sign error when simplifying $2 + 39\left(\frac{\sqrt{2}}{2}-1\right)$ , especially the $2-39=-37$ step.
Giving a rounded decimal instead of the exact surd form.

Things to Be Careful About

The common difference is $\sin\theta - \tan\theta$ , not the other way round.
Use $n=40$ and remember the formula contains $2a$ , not just $a$ .
Keep exact values such as $\sqrt{2}$ until the final answer.
The mark scheme allows a decimal sight of $-0.29$ for the common difference, but the final sum must be exact.

Techniques used

find common difference of an arithmetic progressionsubstitute exact trigonometric valuesapply arithmetic series sum formulasimplify a radical expression

(b)

The first and second terms of a geometric progression are $\tan \theta$ and $\sin \theta$ respectively, where $0 < \theta < \frac{1}{2}\pi$ .

5M

(i)

Find the sum to infinity of the progression in terms of $\theta$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

For a geometric progression, divide the second term by the first term to find the common ratio $r$ . Simplify $r$ using $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , then apply the sum-to-infinity formula $S_\infty = \frac{a}{1-r}$ , valid because $0<\theta<\frac{\pi}{2}$ gives $0<\cos\theta<1$ .

Working

The first term is $a = \tan\theta$ and the second term is $\sin\theta$ . Therefore

r = \frac{\sin\theta}{\tan\theta} = \frac{\sin\theta}{\frac{\sin\theta}{\cos\theta}} = \cos\theta

Since $0<\theta<\frac{\pi}{2}$ , we have $0<\cos\theta<1$ , so the series converges. The sum to infinity is

S_\infty = \frac{a}{1-r} = \frac{\tan\theta}{1-\cos\theta}

Answer

S_\infty = \frac{\tan\theta}{1-\cos\theta}

Final answer

tanθ/(1 - cosθ)

Detailed explanation

Walkthrough

For a geometric progression, the common ratio is the second term divided by the first term. Since the first term is $\tan\theta$ and the second is $\sin\theta$ , $r = \frac{\sin\theta}{\tan\theta}$ . Using $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , this simplifies to $r = \cos\theta$ .

Because $0<\theta<\frac{\pi}{2}$ , we know $0<\cos\theta<1$ , so the geometric series converges. The sum to infinity is $S_\infty = \frac{a}{1-r}$ , where $a=\tan\theta$ and $r=\cos\theta$ . This gives $S_\infty = \frac{\tan\theta}{1-\cos\theta}$ .

Key Takeaways

This part links geometric progressions with trigonometry. The key skills are finding the common ratio, simplifying it using a trigonometric identity, checking the convergence condition $|r|<1$ , and applying the sum-to-infinity formula.

Common Mistakes

Using $r = \frac{\tan\theta}{\sin\theta}$ instead of $\frac{\sin\theta}{\tan\theta}$ .
Forgetting that the first term is $\tan\theta$ , not $\sin\theta$ .
Leaving the answer as a complex fraction such as $\frac{\tan\theta}{1-\frac{\sin\theta}{\tan\theta}}$ , which the mark scheme does not allow.
Omitting $\theta$ in the final expression.

Things to Be Careful About

The domain $0<\theta<\frac{\pi}{2}$ guarantees $0<\cos\theta<1$ , so the sum to infinity exists. No need to state it explicitly for the marks, but it justifies the formula.
Simplify $\frac{\sin\theta}{\tan\theta}$ to $\cos\theta$ before using the sum formula.
Keep the final answer as a single fraction in terms of $\theta$ .

Techniques used

find common ratio of a geometric progressionsimplify ratio using trigonometric identityapply sum to infinity formulaverify convergence condition

(ii)

Given that $\theta = \frac{1}{3}\pi$ , find the sum of the first 10 terms of the progression. Give your answer correct to 3 significant figures.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the geometric progression with first term $a = \tan\theta$ and common ratio $r = \cos\theta$ . Substitute $\theta = \frac{\pi}{3}$ , then apply the formula for the sum of the first $n$ terms, $S_n = \frac{a(1-r^n)}{1-r}$ , with $n=10$ .

Working

For $\theta = \frac{\pi}{3}$ :

a = \tan\frac{\pi}{3} = \sqrt{3}

and

r = \cos\frac{\pi}{3} = \frac{1}{2}

Using the finite GP sum formula with $n=10$ :

S_{10} = \sqrt{3}\left(\frac{1 - \left(\frac{1}{2}\right)^{10}}{1 - \frac{1}{2}}\right)

S_{10} = \sqrt{3}\left(\frac{1 - \frac{1}{1024}}{\frac{1}{2}}\right) = \sqrt{3} \cdot 2\left(1 - \frac{1}{1024}\right)

S_{10} = \frac{1023\sqrt{3}}{512} \approx 3.46

Answer

S_{10} = 3.46 \text{ (3 s.f.)}

Final answer

3.46

Detailed explanation

Walkthrough

For the geometric progression, the first term is $a=\tan\theta$ and the common ratio is $r=\cos\theta$ , as found in part (i). With $\theta=\frac{\pi}{3}$ , $a=\tan\frac{\pi}{3}=\sqrt{3}$ and $r=\cos\frac{\pi}{3}=\frac{1}{2}$ .

The sum of the first $10$ terms is given by $S_{10} = \frac{a(1-r^{10})}{1-r}$ . Substituting $a=\sqrt{3}$ and $r=\frac{1}{2}$ gives $\sqrt{3}\left(\frac{1-(\frac{1}{2})^{10}}{1-\frac{1}{2}}\right)$ . Since $(\frac{1}{2})^{10}=\frac{1}{1024}$ , this simplifies to $\frac{1023\sqrt{3}}{512}\approx 3.46$ , correct to 3 significant figures.

Key Takeaways

This part applies the finite geometric series formula with exact trigonometric values. It also checks the ability to evaluate powers of $\frac{1}{2}$ and round to a required degree of accuracy.

Common Mistakes

Using $n=9$ instead of $n=10$ . The question asks for the first 10 terms; although $S_9$ may round to the same value, it is not the correct sum.
Forgetting to subtract $r^{10}$ from 1 in the numerator.
Using $r=1$ or $r=\frac{\sqrt{3}}{2}$ incorrectly; the ratio is $\cos\theta$ , not $\tan\theta$ .
Giving the exact answer $\frac{1023\sqrt{3}}{512}$ without the required 3 significant figure approximation. (The exact form is accepted, but the question asks for 3 s.f.)

Things to Be Careful About

$\tan\frac{\pi}{3} = \sqrt{3}$ and $\cos\frac{\pi}{3} = \frac{1}{2}$ .
When dividing by $1 - \frac{1}{2} = \frac{1}{2}$ , multiply by 2.
Round $3.4607...$ to $3.46$ , not $3.47$ .
The mark scheme allows the exact form $\frac{1023\sqrt{3}}{512}$ but the final answer should be stated as $3.46$ to 3 significant figures.

Techniques used

substitute exact trigonometric valuesapply finite geometric series sum formulaevaluate and round to significant figures

Q6MediumDifferentiationQuadraticsIntegration

The curve with equation $y = 2x - 8x^{\frac{1}{2}}$ has a minimum point at $A$ and intersects the positive $x$ -axis at $B$ .

(a)

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

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

To find the minimum point $A$ , differentiate the curve equation and set $\frac{dy}{dx} = 0$ to find the $x$ -coordinate, then substitute back to find $y$ . To find the $x$ -intercept $B$ , set $y = 0$ and solve for $x$ .

Working

Given the curve equation:

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

Differentiate with respect to $x$ :

\frac{dy}{dx} = 2 - 8 \left( \frac{1}{2} x^{-\frac{1}{2}} \right) = 2 - 4x^{-\frac{1}{2}}

At the minimum point $A$ , $\frac{dy}{dx} = 0$ :

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

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

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

\frac{1}{\sqrt{x}} = \frac{1}{2} \implies \sqrt{x} = 2 \implies x = 4

Substitute $x = 4$ into the original equation to find the $y$ -coordinate:

y = 2(4) - 8(4)^{\frac{1}{2}} = 8 - 8(2) = 8 - 16 = -8

So the coordinates of $A$ are $(4, -8)$ .

To find $B$ , set $y = 0$ :

2x - 8x^{\frac{1}{2}} = 0

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

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

This gives $x^{\frac{1}{2}} = 0 \implies x = 0$ or $x^{\frac{1}{2}} = 4 \implies x = 16$ . Since $B$ is on the positive $x$ -axis, $x = 16$ .

So the coordinates of $B$ are $(16, 0)$ .

Answer

A = (4, -8), \quad B = (16, 0)

Final answer

A = (4, -8), B = (16, 0)

Detailed explanation

Walkthrough

First, we find the minimum point $A$ by differentiating the curve equation $y = 2x - 8x^{1/2}$ . The derivative is $\frac{dy}{dx} = 2 - 4x^{-1/2}$ . Setting this to zero gives $2 = 4/\sqrt{x}$ , so $\sqrt{x} = 2$ and $x = 4$ . Substituting $x = 4$ back into the curve equation gives $y = 8 - 16 = -8$ , so $A = (4, -8)$ .

Next, we find the $x$ -intercept $B$ by setting $y = 0$ . The equation $2x - 8x^{1/2} = 0$ can be factored as $2x^{1/2}(x^{1/2} - 4) = 0$ . The solutions are $x = 0$ and $x = 16$ . Since $B$ is on the positive $x$ -axis, we take $x = 16$ , giving $B = (16, 0)$ .

Key Takeaways

Differentiation is used to find stationary points (minima/maxima) by setting $\frac{dy}{dx} = 0$ .
Equations involving fractional powers like $x^{1/2}$ can be solved by factoring or substituting $u = x^{1/2}$ to form a quadratic equation.

Common Mistakes

Forgetting to substitute the $x$ -value back into the original curve equation to find the $y$ -coordinate.
Making algebraic errors when differentiating fractional powers, such as forgetting to multiply by the power or mishandling the negative exponent.
Selecting $x = 0$ instead of $x = 16$ for point $B$ when asked for the positive $x$ -axis intersection.

Things to Be Careful About

Ensure the derivative is simplified correctly: $8 \times \frac{1}{2} = 4$ , not $16$ .
When solving $2x - 8x^{1/2} = 0$ , remember that $x = 0$ is a valid solution but does not correspond to point $B$ as specified in the question.

Techniques used

differentiate the curve equationequate derivative to zero to find stationary pointsubstitute x-value to find y-coordinateset y = 0 and solve using substitution

(b)

Fig. 6.1

The diagram shows the curve with equation $y = 2x - 8x^{\frac{1}{2}}$ and the line $AB$ . It is given that the equation of $AB$ is $y = \frac{2x - 32}{3}$ .

Find the area of the shaded region between the curve and the line.

5M Similar

DifficultyMedium

Worked solution

Approach

The area of the shaded region between the line $AB$ and the curve is found by integrating the difference between the upper function (the line) and the lower function (the curve) from $x = 4$ to $x = 16$ . The limits are the $x$ -coordinates of points $A$ and $B$ found in part (a).

Working

The equation of the line $AB$ is:

y = \frac{2x - 32}{3} = \frac{2}{3}x - \frac{32}{3}

The equation of the curve is:

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

The area $A$ is given by:

A = \int_{4}^{16} \left[ \left( \frac{2}{3}x - \frac{32}{3} \right) - \left( 2x - 8x^{\frac{1}{2}} \right) \right] dx

Simplify the integrand:

\frac{2}{3}x - \frac{32}{3} - 2x + 8x^{\frac{1}{2}} = -\frac{4}{3}x + 8x^{\frac{1}{2}} - \frac{32}{3}

Integrate term by term:

\int \left( -\frac{4}{3}x + 8x^{\frac{1}{2}} - \frac{32}{3} \right) dx = -\frac{4}{3} \cdot \frac{x^2}{2} + 8 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{32}{3}x

= -\frac{2}{3}x^2 + \frac{16}{3}x^{\frac{3}{2}} - \frac{32}{3}x

Evaluate from $x = 4$ to $x = 16$ :

At $x = 16$ :

-\frac{2}{3}(16)^2 + \frac{16}{3}(16)^{\frac{3}{2}} - \frac{32}{3}(16) = -\frac{512}{3} + \frac{16}{3}(64) - \frac{512}{3}

= -\frac{512}{3} + \frac{1024}{3} - \frac{512}{3} = 0

At $x = 4$ :

-\frac{2}{3}(4)^2 + \frac{16}{3}(4)^{\frac{3}{2}} - \frac{32}{3}(4) = -\frac{32}{3} + \frac{16}{3}(8) - \frac{128}{3}

= -\frac{32}{3} + \frac{128}{3} - \frac{128}{3} = -\frac{32}{3}

Subtract the lower limit value from the upper limit value:

A = 0 - \left( -\frac{32}{3} \right) = \frac{32}{3}

Answer

\frac{32}{3} \text{ or } 10\frac{2}{3}

Final answer

32/3

Detailed explanation

Walkthrough

The shaded region is bounded above by the line $y = \frac{2x - 32}{3}$ and below by the curve $y = 2x - 8x^{1/2}$ , between $x = 4$ and $x = 16$ . To find the area, we integrate the difference (line minus curve) over this interval.

The integrand simplifies to $-\frac{4}{3}x + 8x^{1/2} - \frac{32}{3}$ . Integrating this gives $-\frac{2}{3}x^2 + \frac{16}{3}x^{3/2} - \frac{32}{3}x$ .

Evaluating at the upper limit $x = 16$ : $-\frac{2}{3}(256) + \frac{16}{3}(64) - \frac{32}{3}(16) = -\frac{512}{3} + \frac{1024}{3} - \frac{512}{3} = 0$ .

Evaluating at the lower limit $x = 4$ : $-\frac{2}{3}(16) + \frac{16}{3}(8) - \frac{32}{3}(4) = -\frac{32}{3} + \frac{128}{3} - \frac{128}{3} = -\frac{32}{3}$ .

The area is $0 - (-\frac{32}{3}) = \frac{32}{3}$ .

Key Takeaways

The area between two curves is found by integrating $\int (y_{\text{upper}} - y_{\text{lower}}) dx$ .
Careful algebraic simplification of the integrand before integrating reduces the chance of errors.
Evaluating definite integrals requires substituting both limits and subtracting the lower limit result from the upper limit result.

Common Mistakes

Integrating the curve and line separately and subtracting the results, but making sign errors when combining the integrals.
Forgetting to raise $x^{1/2}$ to $x^{3/2}$ and divide by $3/2$ during integration, leading to incorrect coefficients.
Evaluating $16^{3/2}$ incorrectly (it is $(\sqrt{16})^3 = 4^3 = 64$ , not $16^2 = 256$ ).
Subtracting the upper limit value from the lower limit value instead of the other way around, resulting in a negative area.

Things to Be Careful About

Ensure the line is above the curve in the interval $[4, 16]$ so that (line - curve) is positive. The diagram confirms this.
Use the exact $x$ -values from part (a) as the limits of integration; do not use the $y$ -values or any other numbers.
When integrating $8x^{1/2}$ , the result is $8 \cdot \frac{x^{3/2}}{3/2} = \frac{16}{3}x^{3/2}$ , not $\frac{8}{3}x^{3/2}$ .

Techniques used

integrate the line equationintegrate the curve equationsubtract the integrals to find the area between curvesevaluate the definite integral using limits from part (a)

Q7MediumCoordinate GeometryQuadratics

The equation of a circle is $(x - 6)^2 + (y + a)^2 = 18$ . The line with equation $y = 2a - x$ is a tangent to the circle.

(a)

Find the two possible values of the constant $a$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Substitute the tangent line into the circle equation to obtain a quadratic in $x$ . Since the line is tangent, this quadratic must have exactly one solution, so its discriminant is zero. Then solve the resulting equation for $a$ .

Working

Substitute $y = 2a - x$ into the circle equation:

(x-6)^2 + (2a - x + a)^2 = 18

Since $2a - x + a = 3a - x$ , expand:

(x-6)^2 + (3a - x)^2 = 18

x^2 - 12x + 36 + x^2 - 6ax + 9a^2 = 18

Collect all terms on one side:

2x^2 - 12x - 6ax + 9a^2 + 36 - 18 = 0

2x^2 - (12 + 6a)x + (9a^2 + 18) = 0

For a tangent, the discriminant of this quadratic in $x$ is zero:

(12 + 6a)^2 - 4(2)(9a^2 + 18) = 0

Expand:

(144 + 144a + 36a^2) - (72a^2 + 144) = 0

-36a^2 + 144a = 0

Factor:

36a(4 - a) = 0

Hence $a = 0$ or $a = 4$ .

Answer

a = 0 \quad \text{or} \quad a = 4

Final answer

a = 0 or a = 4

Detailed explanation

Walkthrough

The line is tangent to the circle, so it touches the circle at exactly one point. To find the intersection algebraically, substitute $y = 2a - x$ into the circle equation. This gives a quadratic equation in $x$ whose coefficients contain $a$ .

For the line to be tangent, the quadratic in $x$ must have exactly one solution. A quadratic $Ax^2 + Bx + C = 0$ has exactly one solution when its discriminant $B^2 - 4AC$ is zero. Setting the discriminant to zero gives an equation in $a$ , which is then solved to find the two possible values.

Key Takeaways

This question connects geometry with algebra: a line is tangent to a circle exactly when substituting the line into the circle gives a quadratic with one repeated root. The main technique is to use the discriminant condition. It also reviews expanding quadratics and solving a quadratic equation in a parameter.

Common Mistakes

Forgetting to move the 18 to the left-hand side when collecting terms.
Expanding $(3a - x)^2$ incorrectly, especially the sign of the $6ax$ term.
Using the discriminant of the equation in $a$ instead of the discriminant of the quadratic in $x$ .
Solving $36a(4-a)=0$ and forgetting that $a=0$ is also a valid value.

Things to Be Careful About

The coefficient of $x$ in the quadratic is $-(12+6a)$ , so $B^2 = (12+6a)^2$ .
The constant term is $9a^2 + 18$ after simplifying $9a^2 + 36 - 18$ .
The discriminant must be set equal to zero because a tangent gives exactly one point of intersection.
Both $a=0$ and $a=4$ are required by the question.

Techniques used

substitute the line equation into the circleexpand and collect terms into a quadratic in xapply the discriminant condition for tangencysolve the resulting quadratic in a

(b)

For the greater value of $a$ , find the equation of the diameter which is perpendicular to the given tangent.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the greater value $a = 4$ . Find the centre of the circle, then write the equation of the diameter through the centre with gradient perpendicular to the tangent.

Working

With $a = 4$ , the circle is

(x-6)^2 + (y+4)^2 = 18

so the centre is $(6, -4)$ .

The tangent line is

y = 2(4) - x = 8 - x

which has gradient $-1$ . A diameter perpendicular to this tangent therefore has gradient $1$ .

Using the centre $(6, -4)$ with gradient $1$ :

y - (-4) = 1(x - 6)

y + 4 = x - 6

y = x - 10

Answer

y = x - 10

Final answer

y = x - 10

Detailed explanation

Walkthrough

The greater value from part (a) is $a = 4$ . Substitute this into the circle equation to get $(x-6)^2 + (y+4)^2 = 18$ , so the centre is $(6, -4)$ .

The given tangent is $y = 2(4) - x = 8 - x$ , whose gradient is $-1$ . A diameter perpendicular to this tangent has gradient $1$ , because perpendicular gradients multiply to $-1$ .

A diameter of the circle is a line through the centre. Use the point-slope form of a line through $(6, -4)$ with gradient $1$ .

Key Takeaways

This part uses two facts: the centre of a circle is read directly from its equation, and a diameter perpendicular to a tangent has the reciprocal gradient of the tangent. The equation of a straight line through a known point with a known gradient is then straightforward.

Common Mistakes

Choosing $a = 0$ instead of $a = 4$ when the question asks for the greater value.
Using gradient $-1$ for the diameter, because the tangent has gradient $-1$ .
Using the tangent point instead of the centre, or using the wrong point.
Sign errors in point-slope form, such as writing $y + 4 = x + 6$ .

Things to Be Careful About

The centre is $(6, -4)$ , not $(6, 4)$ .
The tangent line $y = 8 - x$ has gradient $-1$ ; the perpendicular gradient is $1$ .
Simplify the line equation to $y = x - 10$ .
If using the tangent point $(9, -1)$ , the same equation is obtained; the mark scheme allows either the centre or the tangent point as long as the gradient is $1$ .

Techniques used

use the centre of the circlefind the gradient perpendicular to the tangentwrite the equation of the diameter through the centre

Q8MediumCircular MeasureTrigonometry

Fig. 8.1

The diagram shows a symmetrical plate $ABCDEF$ . The line $ABCD$ is straight and the length of $BC$ is $2\text{ cm}$ . Each of the two sectors $ABF$ and $DCE$ is of radius $r\text{ cm}$ and each of the angles $ABF$ and $DCE$ is equal to $\frac{1}{3}\pi$ radians.

(a)

It is given that $r = 0.4\text{ cm}$ .

6M

(i)

Show that the length $EF = 2.4\text{ cm}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Consider the right-angled triangle formed by the radius CE, the vertical through C, and the horizontal through E. Use the trig ratio in this triangle to find the horizontal offset XE, then add both offsets to BC using symmetry.

Working

Let X be the point directly below C at the same level as E. Then CX is vertical, XE is horizontal, and the right angle is at X. The line CD is horizontal, so CX is perpendicular to CD.

The angle of the sector at C is $\angle DCE = \frac{\pi}{3}$ . The angle between CD and the vertical CX is $\frac{\pi}{2}$ , so the angle between CX and CE is

\angle ECX = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}

In the right-angled triangle CEX (hypotenuse CE = 0.4):

\frac{XE}{CE} = \sin\frac{\pi}{6}

XE = 0.4 \times \sin\frac{\pi}{6} = 0.4 \times \frac{1}{2} = 0.2

By symmetry, the horizontal offset from B to F is also 0.2. Hence

EF = BC + 2 \times XE = 2 + 2 \times 0.2 = 2.4

Answer

EF = 2.4\text{ cm}

Final answer

EF = 2.4 cm

Detailed explanation

Walkthrough

We want the bottom edge EF of the plate. E is the endpoint of the radius CE of the sector centred at C, and F is the corresponding point of the sector centred at B. The line ABCD is straight, so we can think of E and F as being horizontally offset from C and B by some distance.

To find the horizontal offset, we form a right-angled triangle. Drop a perpendicular from E to a vertical line through C, calling the foot X. Then CX is vertical (perpendicular to the horizontal line CD), XE is horizontal, CE is the radius (length 0.4), and the right angle is at X.

The sector angle at C is $\angle DCE = \frac{\pi}{3}$ . Since CX is perpendicular to CD, the angle between CX and CE is $\frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$ .

In the right-angled triangle CEX, with the angle at C equal to $\frac{\pi}{6}$ and hypotenuse CE = 0.4, the side opposite this angle is XE. Using the sine ratio:

XE = CE \sin\frac{\pi}{6} = 0.4 \times \frac{1}{2} = 0.2

By symmetry, the horizontal offset from B to F is also 0.2. Therefore $EF = BC + 2 \times 0.2 = 2.4$ cm.

Key Takeaways

The bottom edge EF of this symmetric plate equals BC plus twice the horizontal offset of each sector.
The horizontal offset of one sector equals the radius times $\cos\frac{\pi}{3} = \frac{1}{2}$ (or equivalently, the radius times $\sin\frac{\pi}{6}$ ).
The right triangle inside a sector relates the radius, the horizontal offset, and the sector angle.

Common Mistakes

Forgetting to add the offset on the OTHER side too, giving EF = BC + 0.2 = 2.2 instead of 2.4.
Using the full sector angle $\frac{\pi}{3}$ as the angle at C in the right triangle, instead of its complement $\frac{\pi}{6}$ .
Confusing the horizontal offset (which is $r\cos\frac{\pi}{3} = \frac{r}{2}$ ) with the vertical drop (which is $r\sin\frac{\pi}{3} = \frac{r\sqrt{3}}{2}$ ).

Things to Be Careful About

The right angle is at X (the foot of the perpendicular), not at C.
The angle in the right triangle at C is the COMPLEMENT of the sector angle: $\frac{\pi}{6} = \frac{\pi}{2} - \frac{\pi}{3}$ .
All angles are in radians, and $\sin\frac{\pi}{6} = \frac{1}{2}$ is an exact value.

Techniques used

form a right-angled triangle from a sector radiusapply the sine ratio for a known angleuse symmetry to combine both horizontal offsets

(ii)

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

4M Similar

DifficultyMedium

Worked solution

Approach

Split the plate into a central trapezium with parallel sides BC (top) and FE (bottom), plus the two sectors at the ends. Find the perpendicular height of the trapezium using the right triangle from part (i), compute the sector areas, and add everything together.

Working

Height of the trapezium. The perpendicular distance from F (or E) to the line ABCD is found from the right-angled triangle CEX used in part (i):

CX = CE \cos\frac{\pi}{6} = 0.4 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{5}

Area of the trapezium. The parallel sides are BC = 2 (top) and FE = 2.4 (bottom), and the height is $\frac{\sqrt{3}}{5}$ :

A_{\text{trap}} = \frac{1}{2}(BC + FE) \times \text{height} = \frac{1}{2}(2 + 2.4) \times \frac{\sqrt{3}}{5}

A_{\text{trap}} = \frac{1}{2} \times 4.4 \times \frac{\sqrt{3}}{5} = \frac{11\sqrt{3}}{25}

Area of one sector. Using the formula $A = \frac{1}{2}r^2\theta$ with r = 0.4 and $\theta = \frac{\pi}{3}$ :

A_{\text{sector}} = \frac{1}{2} \times 0.4^2 \times \frac{\pi}{3} = \frac{1}{2} \times 0.16 \times \frac{\pi}{3} = \frac{2\pi}{75}

Total area. Adding the trapezium and the two equal sectors:

A = \frac{11\sqrt{3}}{25} + 2 \times \frac{2\pi}{75} = \frac{11\sqrt{3}}{25} + \frac{4\pi}{75}

A \approx 0.7621 + 0.1676 \approx 0.9297

Answer

A \approx 0.930\text{ cm}^2

Final answer

0.930 cm²

Detailed explanation

Walkthrough

The plate has an irregular shape, but it can be split into pieces whose areas we can compute.

Choosing the decomposition. The plate is bounded by the top line ABCD, two arcs (one at each end), and the bottom line FE. The natural split is:

A central trapezium with parallel top side BC = 2 and parallel bottom side FE = 2.4 (the result of part (i)).
Two circular sectors at the ends: one for sector ABF and one for sector DCE.

Trapezium height. The height of the trapezium is the perpendicular distance between BC and FE, which equals the vertical drop from C to E. In the right-angled triangle CEX from part (i), with CE = 0.4 and the angle at C equal to $\frac{\pi}{6}$ :

CX = 0.4 \cos\frac{\pi}{6} = 0.4 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{5}

Trapezium area. Using $A = \frac{1}{2}(a + b)h$ with a = 2, b = 2.4, and $h = \frac{\sqrt{3}}{5}$ :

A_{\text{trap}} = \frac{1}{2}(2 + 2.4) \times \frac{\sqrt{3}}{5} = \frac{11\sqrt{3}}{25} \approx 0.7621

Sector area. Each sector has radius r = 0.4 and angle $\theta = \frac{\pi}{3}$ :

A_{\text{sector}} = \frac{1}{2}r^2\theta = \frac{1}{2}(0.4)^2 \times \frac{\pi}{3} = \frac{2\pi}{75} \approx 0.0838

Two such sectors give a total of $\frac{4\pi}{75} \approx 0.1676$ .

Adding the pieces.

A = \frac{11\sqrt{3}}{25} + \frac{4\pi}{75} \approx 0.7621 + 0.1676 \approx 0.9297 \approx 0.930\text{ cm}^2

Key Takeaways

A non-standard shape can be split into standard pieces (trapeziums, sectors) whose areas are easy to compute.
Sector area formula: $A = \frac{1}{2}r^2\theta$ (angle in radians).
Trapezium area formula: $A = \frac{1}{2}(a + b)h$ , where a and b are the parallel sides and h is the perpendicular distance between them.
The trapezium height here is the vertical drop CX, not the radius r.

Common Mistakes

Forgetting to multiply the single sector area by 2 (there are two sectors).
Using BC's length as the top of the trapezium but then including the radii AB and CD in the area (they are not part of the trapezium, they are part of the sectors).
Mixing up the trapezium height: it is CX (the vertical drop), not the radius r = 0.4.

Things to Be Careful About

The top side of the trapezium is BC (length 2), not ABCD (length $2 + 2r$ ). The portions AB and CD are inside the sectors, not the trapezium.
The trapezium height is the perpendicular distance CX = $\frac{\sqrt{3}}{5}$ , not the radius r = 0.4.
Round only at the end of the calculation if you want the most accurate 3 s.f. result.
The exact form $\frac{11\sqrt{3}}{25} + \frac{4\pi}{75}$ is mathematically equal to 0.930 cm² to 3 s.f.

Techniques used

compute the perpendicular drop using trigonometrycompute the area of a sector using the formuladecompose the plate into a trapezium and two sectorssum the component areas

(b)

It is given instead that the perimeter of the plate is $6\text{ cm}$ .

Find the value of $r$ . Give your answer correct to 3 significant figures.

4M Similar

DifficultyMedium

Worked solution

Approach

The perimeter is the sum of all six boundary segments: the four straight parts (AB, BC, CD, EF) and the two arcs. Write each in terms of r, add them up, set the total equal to 6, and solve the resulting linear equation for r.

Working

Boundary components in terms of r.

$AB = r$ (radius of sector ABF)
$BC = 2$ (given)
$CD = r$ (radius of sector DCE)
arc $DE = r\theta = r \times \frac{\pi}{3} = \frac{\pi r}{3}$ (arc length formula)
$EF = 2 + r$ (from part (a)(i) with general r: $EF = 2 + 2r\cos\frac{\pi}{3} = 2 + r$ )
arc $FA = \frac{\pi r}{3}$

Perimeter equation.

P = r + 2 + r + \frac{\pi r}{3} + (2 + r) + \frac{\pi r}{3}

P = 3r + 4 + \frac{2\pi r}{3}

Setting P = 6 and solving.

3r + 4 + \frac{2\pi r}{3} = 6

3r + \frac{2\pi r}{3} = 2

r\left(3 + \frac{2\pi}{3}\right) = 2

r \cdot \frac{9 + 2\pi}{3} = 2

r = \frac{6}{9 + 2\pi}

Numerical value.

r = \frac{6}{9 + 2\pi} \approx \frac{6}{15.2832} \approx 0.3926

Answer

r \approx 0.393\text{ cm}

Final answer

0.393 cm

Detailed explanation

Walkthrough

The perimeter of the plate is the total length around its boundary, which consists of six pieces: the two radii at the top (AB and CD), the middle straight part BC, the bottom straight part EF, and the two arc pieces (the curved parts of the two sectors).

Expressing each piece in terms of r.

AB and CD are radii of the sectors, each equal to r.
BC = 2 is given.
EF is the bottom of the plate. From part (i), with general r, the horizontal offset of each sector is $r\cos\frac{\pi}{3} = \frac{r}{2}$ , so $EF = 2 + 2 \times \frac{r}{2} = 2 + r$ . The crucial point is to keep this in terms of r — we must not substitute r = 0.4 from part (a).
Each arc has length $r\theta = r \cdot \frac{\pi}{3}$ .

Adding them up.

P = r + 2 + r + \frac{\pi r}{3} + (2 + r) + \frac{\pi r}{3} = 3r + 4 + \frac{2\pi r}{3}

Solving P = 6.

3r + 4 + \frac{2\pi r}{3} = 6 \quad \Rightarrow \quad r\left(3 + \frac{2\pi}{3}\right) = 2

r = \frac{2}{3 + \frac{2\pi}{3}} = \frac{6}{9 + 2\pi} \approx 0.3926

Rounded to 3 significant figures, $r \approx 0.393$ cm.

Key Takeaways

The perimeter of a sector is the sum of two radii and one arc: $2r + r\theta$ .
When solving a problem with a parameter r, keep expressions in terms of r throughout — do not substitute numerical values from earlier parts.
The boundary of the plate here has 6 pieces; missing one gives a noticeably wrong answer.

Common Mistakes

Using EF = 2.4 (the value from part (a) with r = 0.4) instead of EF = 2 + r. This gives a wrong value of approximately 0.391, not 0.393.
Forgetting to include one of the six boundary pieces (e.g. one of the arcs).
Mixing degrees and radians (e.g. using 60° directly in the arc length formula instead of $\frac{\pi}{3}$ ).

Things to Be Careful About

$EF$ depends on r. With general r, $EF = 2 + 2r\cos\frac{\pi}{3} = 2 + r$ .
The answer 0.393 (3 s.f.) comes from the correct EF = 2 + r. Using EF = 2.4 (the constant value from part (a)) would give 0.391 instead — a tell-tale sign of the wrong substitution.
Make sure both arcs are counted: there are two sectors, hence two arcs.

Techniques used

write each boundary segment in terms of rform a linear equation in r from the perimetersolve the equation for r

Q9MediumDifferentiationQuadraticsIntegration

A function $f$ is such that $f'(x) = 6(2x - 3)^2 - 6x$ for $x \in \mathbb{R}$ .

(a)

Determine the set of values of $x$ for which $f(x)$ is decreasing.

4M Similar

DifficultyMedium

Worked solution

Approach

A function is decreasing where its derivative is negative. Set $f'(x) < 0$ , expand and simplify the quadratic, factorise it, and read off the interval between the roots.

Working

f'(x) = 6(2x-3)^2 - 6x < 0

Expand and simplify:

6(4x^2 - 12x + 9) - 6x = 24x^2 - 72x + 54 - 6x = 24x^2 - 78x + 54 < 0

Divide by 6:

4x^2 - 13x + 9 < 0

Factorise:

(4x-9)(x-1) < 0

The roots are $x = 1$ and $x = \frac{9}{4}$ . Since the quadratic has a positive leading coefficient, it is negative between its roots.

Answer

1 < x < \frac{9}{4}

Final answer

1 < x < 9/4

Detailed explanation

Walkthrough

A function is decreasing where its derivative is negative, so we start from $f'(x) < 0$ .

Substitute the given derivative:

6(2x-3)^2 - 6x < 0

Expand the square: $(2x-3)^2 = 4x^2 - 12x + 9$ . Multiplying by 6 gives $24x^2 - 72x + 54$ , and subtracting $6x$ gives $24x^2 - 78x + 54$ . Dividing by 6 simplifies the inequality to $4x^2 - 13x + 9 < 0$ .

Factorise the quadratic: $(4x-9)(x-1) < 0$ . The roots are $x = 1$ and $x = \frac{9}{4}$ . Because the coefficient of $x^2$ is positive, the graph of the quadratic is a U-shape, so the expression is negative between the two roots. Therefore the set of values is $1 < x < \frac{9}{4}$ .

Key Takeaways

The sign of $f'(x)$ tells whether $f$ is increasing or decreasing: negative derivative means decreasing.
Solving a quadratic inequality requires finding the roots and then using the sign of the leading coefficient to decide the interval.
For a positive quadratic, the expression is negative between the roots and positive outside them.

Common Mistakes

Using $f'(x) > 0$ instead of $f'(x) < 0$ .
Making sign errors when expanding $6(2x-3)^2$ or collecting like terms.
Giving $x > 1$ or $x < \frac{9}{4}$ separately instead of the combined interval $1 < x < \frac{9}{4}$ .
Forgetting to divide by 6 before factorising, although the same roots can still be found.

Things to Be Careful About

The mark scheme condones $\le 0$ or $\ge 0$ , but the mathematically correct condition for decreasing is $f'(x) < 0$ .
The final interval must link both inequalities, not be written as two disconnected conditions.
If you use the alternative method $2x - 3 < \sqrt{x}$ , be cautious about squaring and domain restrictions.

Techniques used

set derivative less than zeroexpand and simplify the quadratic expressionfactorise the quadraticdetermine the sign interval of a quadratic inequality

(b)

Given that $f(1) = -1$ , find $f(x)$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Integrate $f'(x)$ term by term. For the first term use the reverse chain rule for $(2x-3)^2$ . Add the constant of integration $C$ , then use the condition $f(1) = -1$ to determine $C$ .

Working

\begin{aligned} f(x) &= \int \left[6(2x-3)^2 - 6x\right]\,dx \\ &= (2x-3)^3 - 3x^2 + C \end{aligned}

Check the first integral: the derivative of $(2x-3)^3$ is $3(2x-3)^2 \cdot 2 = 6(2x-3)^2$ , so the integral is correct.

Use $f(1) = -1$ :

-1 = (2-3)^3 - 3(1)^2 + C = -1 - 3 + C = -4 + C

Hence $C = 3$ .

Answer

f(x) = (2x-3)^3 - 3x^2 + 3

Final answer

f(x) = (2x-3)^3 - 3x^2 + 3

Detailed explanation

Walkthrough

To recover $f(x)$ from $f'(x)$ , integrate the derivative.

For the first term, integrate $6(2x-3)^2$ . Since the derivative of $(2x-3)^3$ is $3(2x-3)^2 \cdot 2 = 6(2x-3)^2$ , the integral is $(2x-3)^3$ . Equivalently, substitute $u = 2x-3$ : then $du = 2\,dx$ , so

\int 6(2x-3)^2\,dx = \int 6u^2 \cdot \frac{1}{2}\,du = 3 \cdot \frac{u^3}{3} = u^3 = (2x-3)^3.

For the second term, $\int -6x\,dx = -3x^2$ . Add the constant of integration $C$ .

Now use $f(1) = -1$ :

-1 = (2-3)^3 - 3(1)^2 + C = -1 - 3 + C = -4 + C.

So $C = 3$ , giving

f(x) = (2x-3)^3 - 3x^2 + 3.

Key Takeaways

Integration is the reverse of differentiation; for a term like $(ax+b)^n$ , divide by $a(n+1)$ after increasing the power.
Always include the constant of integration before using an initial condition.
An initial condition such as $f(1) = -1$ determines the constant $C$ .

Common Mistakes

Forgetting to add $+C$ before substituting $x = 1$ .
Integrating $6(2x-3)^2$ incorrectly as $\frac{6(2x-3)^3}{3}$ and forgetting the factor from the chain rule.
Sign errors when integrating $-6x$ .
Stopping at $f(x) = (2x-3)^3 - 3x^2 + C$ without finding $C$ .

Things to Be Careful About

The mark scheme awards marks for each correct integral and for the correct use of $f(1) = -1$ .
The final answer must have simplified coefficients, so $C = 3$ should be substituted and the expression simplified.
If you expand the derivative first, you get the equivalent answer $f(x) = 8x^3 - 39x^2 + 54x - 24$ ; both forms are acceptable as long as the working is consistent.

Techniques used

integrate a composite power termintegrate a linear power termadd the constant of integrationsubstitute the initial condition to find the constant

Q10MediumDifferentiationCoordinate Geometry

The equation of a curve is $y = (5 - 2x)^{\frac{3}{2}} + 5$ for $x < \frac{5}{2}$ .

(a)

A point $P$ is moving along the curve in such a way that the $y$ -coordinate of point $P$ is decreasing at 5 units per second.

Find the rate at which the $x$ -coordinate of point $P$ is increasing when $y = 32$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate the curve to find $\frac{dy}{dx}$ using the chain rule. Find the $x$ -coordinate where $y = 32$ . Then use the related-rates chain rule $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ , with $\frac{dy}{dt} = -5$ (since the $y$ -coordinate is decreasing), to find $\frac{dx}{dt}$ .

Working

Given $y = (5-2x)^{\frac{3}{2}} + 5$ , differentiate with respect to $x$ :

\frac{dy}{dx} = \frac{3}{2}(5-2x)^{\frac{1}{2}} \cdot (-2) = -3(5-2x)^{\frac{1}{2}}

When $y = 32$ :

(5-2x)^{\frac{3}{2}} + 5 = 32

(5-2x)^{\frac{3}{2}} = 27

5 - 2x = 27^{\frac{2}{3}} = 9

x = -2

At $x = -2$ :

\frac{dy}{dx} = -3\left(5-2(-2)\right)^{\frac{1}{2}} = -3(9)^{\frac{1}{2}} = -9

The $y$ -coordinate is decreasing at 5 units per second, so $\frac{dy}{dt} = -5$ .

Using the chain rule for related rates:

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

-5 = -9 \cdot \frac{dx}{dt}

\frac{dx}{dt} = \frac{5}{9}

Answer

The $x$ -coordinate of $P$ is increasing at $\frac{5}{9}$ units per second.

Final answer

5/9 units per second

Detailed explanation

Walkthrough

We are told the $y$ -coordinate of point $P$ is decreasing at 5 units per second, so $\frac{dy}{dt} = -5$ . We need to find $\frac{dx}{dt}$ when $y = 32$ .

The chain rule connects these rates: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ . So we first need $\frac{dy}{dx}$ at the point where $y = 32$ .

Step 1: Differentiate $y = (5-2x)^{\frac{3}{2}} + 5$ . Using the chain rule, the derivative of $(5-2x)^{\frac{3}{2}}$ is $\frac{3}{2}(5-2x)^{\frac{1}{2}} \cdot (-2) = -3(5-2x)^{\frac{1}{2}}$ . The constant 5 differentiates to 0.

Step 2: Find the $x$ -coordinate where $y = 32$ . Set $(5-2x)^{\frac{3}{2}} + 5 = 32$ , so $(5-2x)^{\frac{3}{2}} = 27$ . Raising both sides to the power $\frac{2}{3}$ gives $5 - 2x = 27^{\frac{2}{3}} = (27^{\frac{1}{3}})^2 = 3^2 = 9$ , so $x = -2$ .

Step 3: Substitute $x = -2$ into the derivative: $\frac{dy}{dx} = -3(5-2(-2))^{\frac{1}{2}} = -3(9)^{\frac{1}{2}} = -3 \times 3 = -9$ .

Step 4: Apply the chain rule: $-5 = -9 \cdot \frac{dx}{dt}$ , so $\frac{dx}{dt} = \frac{5}{9}$ .

The positive sign of $\frac{dx}{dt}$ confirms the $x$ -coordinate is increasing, as stated in the question.

Key Takeaways

Related-rates problems use the chain rule $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ to connect rates of change.
A decreasing quantity has a negative rate of change; this determines the sign in the equation.
The chain rule for differentiating $(ax+b)^n$ requires multiplying by the derivative of the inside function, $a$ .

Common Mistakes

Forgetting that 'decreasing' means $\frac{dy}{dt}$ is negative — writing $\frac{dy}{dt} = +5$ gives the wrong sign for $\frac{dx}{dt}$ .
Forgetting the factor $-2$ (the derivative of $5-2x$ ) when applying the chain rule to $(5-2x)^{\frac{3}{2}}$ .
Miscomputing $27^{\frac{2}{3}}$ — it equals $(27^{\frac{1}{3}})^2 = 3^2 = 9$ , not a larger number.

Things to Be Careful About

The domain $x < \frac{5}{2}$ is satisfied by $x = -2$ , so the point is valid.
Include units in the final answer: units per second.
The mark scheme requires the differentiation step to be shown explicitly to earn the method mark.

Techniques used

differentiate a composite function using the chain rulefind the x-coordinate for a given y-valueapply the related-rates chain rulesolve for the rate of change of x

(b)

Point $A$ on the curve has $y$ -coordinate 32. Point $B$ on the curve is such that the gradient of the curve at $B$ is $-3$ .

Find the equation of the perpendicular bisector of $AB$ . Give your answer in the form $ax + by + c = 0$ , where $a$ , $b$ and $c$ are integers.

6M Similar

DifficultyMedium

Worked solution

Approach

Find the coordinates of $A$ (where $y = 32$ ) and $B$ (where the gradient is $-3$ ). Then compute the gradient of $AB$ , take its negative reciprocal to get the perpendicular gradient, find the midpoint of $AB$ , and use the point-slope form to write the equation of the perpendicular bisector.

Working

From part (a), when $y = 32$ , $x = -2$ . So $A = (-2, 32)$ .

For $B$ , the gradient is $-3$ :

\frac{dy}{dx} = -3(5-2x)^{\frac{1}{2}} = -3

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

5 - 2x = 1

x = 2

At $x = 2$ :

y = (5-4)^{\frac{3}{2}} + 5 = 1 + 5 = 6

So $B = (2, 6)$ .

Gradient of $AB$ :

m_{AB} = \frac{32 - 6}{-2 - 2} = \frac{26}{-4} = -\frac{13}{2}

Gradient of the perpendicular bisector:

m_{\perp} = -\frac{1}{m_{AB}} = \frac{2}{13}

Midpoint of $AB$ :

\left(\frac{-2+2}{2}, \frac{32+6}{2}\right) = (0, 19)

Equation of the perpendicular bisector using point-slope form:

y - 19 = \frac{2}{13}(x - 0)

Multiply through by 13:

13y - 247 = 2x

Rearrange:

2x - 13y + 247 = 0

Answer

2x - 13y + 247 = 0

Final answer

2x - 13y + 247 = 0

Detailed explanation

Walkthrough

We need the perpendicular bisector of segment $AB$ . A perpendicular bisector passes through the midpoint of $AB$ and has gradient equal to the negative reciprocal of the gradient of $AB$ .

Step 1: Find $A$ . From part (a), when $y = 32$ , $x = -2$ , so $A = (-2, 32)$ .

Step 2: Find $B$ . The gradient at $B$ is $-3$ , so set $\frac{dy}{dx} = -3(5-2x)^{\frac{1}{2}} = -3$ . This gives $(5-2x)^{\frac{1}{2}} = 1$ , so $5 - 2x = 1$ and $x = 2$ . Substituting into the curve equation gives $y = (5-4)^{\frac{3}{2}} + 5 = 1 + 5 = 6$ , so $B = (2, 6)$ .

Step 3: Compute the gradient of $AB$ : $m_{AB} = \frac{32 - 6}{-2 - 2} = \frac{26}{-4} = -\frac{13}{2}$ . The perpendicular gradient is the negative reciprocal: $m_{\perp} = -\frac{1}{m_{AB}} = \frac{2}{13}$ .

Step 4: Find the midpoint of $AB$ : $\left(\frac{-2+2}{2}, \frac{32+6}{2}\right) = (0, 19)$ .

Step 5: Use the point-slope form of a line through $(0, 19)$ with gradient $\frac{2}{13}$ : $y - 19 = \frac{2}{13}x$ . Multiply through by 13: $13y - 247 = 2x$ , then rearrange to $2x - 13y + 247 = 0$ .

Check: the midpoint $(0, 19)$ satisfies $2(0) - 13(19) + 247 = 0$ . ✓

Key Takeaways

The perpendicular bisector of a segment passes through its midpoint and has gradient equal to the negative reciprocal of the segment's gradient.
Setting the derivative equal to a given gradient locates the point on the curve where that gradient occurs.
The midpoint formula averages the coordinates of the two endpoints.
The general form of a line is $ax + by + c = 0$ with integer coefficients.

Common Mistakes

Forgetting to take the negative reciprocal of the gradient of $AB$ .
Using the wrong midpoint (e.g. averaging only one coordinate or using a single point).
Sign errors when rearranging to the form $ax + by + c = 0$ .
Not simplifying the perpendicular gradient (e.g. leaving $\frac{4}{26}$ instead of $\frac{2}{13}$ ).

Things to Be Careful About

The final answer must be in the form $ax + by + c = 0$ with integer coefficients; integer multiples of the answer are also accepted.
The perpendicular gradient is positive ( $\frac{2}{13}$ ) because the gradient of $AB$ is negative.
The mark scheme requires clear working when using the gradient and midpoint formulae — show the differences in coordinates explicitly.

Techniques used

differentiate and set the gradient equal to a given valuesolve for the coordinates of a point on a curvecompute the gradient of a line segment and its negative reciprocalfind the midpoint of two pointsform the equation of a perpendicular bisector

Mathematics 9709/12 — May/June 2024

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