Point $A$ has coordinates $(3, -1)$. A circle has equation $(x - 4)^2 + (y + 3)^2 = 5$. (a) Show that $A$ lies on the circumference of the circle. (b) Given that $AB$ is a diameter of the circle, find the coordinates of $B$. (c) Find the equation of the tangent to the circle at $A$.

(a) (3 - 4)^2 + (-1 + 3)^2 = 5, so A lies on the circumference · (b) (5, -5) · (c) y + 1 = 0.5(x - 3)

(a) Solve the equation $x^{\frac{1}{3}} - x^{\frac{1}{6}} = 2$. (b) Solve the simultaneous equations $$ \begin{aligned} \lg(x + 2y) &= 0 \\ x^2 + 4xy + y &= 1. \end{aligned} $$

(a) x = 64 · (b) x = 1, y = 0 and x = 1/2, y = 1/4

(a) Write down the set of values of $x$ for which $\log_5(12x - 4)$ exists. (b) Solve the equation $$ \log_5(12x - 4) = \frac{6}{\log_x 125} + 1. $$

(a) x > 1/3 · (b) x = 0.4 or x = 2

An arithmetic progression has common difference $d$. The 3rd term of this progression is 10. (a) Write down expressions for the 1st term and the 2nd term of this progression. Give your answers in terms of $d$ only. (b) When each of the first 3 terms is squared, the sum of these squares is 140 . There are two possible values for $d$. Using your answer to part (a), find the sum of the first 200 terms of the progression with the smaller value of $d$.

(a) 1st term: 10 - 2d, 2nd term: 10 - d · (b) 80000

4037/11 May/June 2025 — Additional Mathematics Solved Past Paper

Q1Medium-EasyVectors in two dimensions

(a)

Given that $\overrightarrow{PQ} = \begin{pmatrix} -3 \\ 7 \end{pmatrix}$ and $4\overrightarrow{PR} = \begin{pmatrix} -2 \\ 8 \end{pmatrix}$ , find $\overrightarrow{RQ}$ .

2M Similar

DifficultyEasy

Worked solution

Approach

Use the triangle relation $\overrightarrow{PR} + \overrightarrow{RQ} = \overrightarrow{PQ}$ , so $\overrightarrow{RQ} = \overrightarrow{PQ} - \overrightarrow{PR}$ .

Working

First find $\overrightarrow{PR}$ from the given scalar multiple:

\overrightarrow{PR} = \frac{1}{4}\begin{pmatrix} -2 \\ 8 \end{pmatrix} = \begin{pmatrix} -0.5 \\ 2 \end{pmatrix}

Then

\overrightarrow{RQ} = \overrightarrow{PQ} - \overrightarrow{PR} = \begin{pmatrix} -3 \\ 7 \end{pmatrix} - \begin{pmatrix} -0.5 \\ 2 \end{pmatrix} = \begin{pmatrix} -2.5 \\ 5 \end{pmatrix}

Answer

\overrightarrow{RQ} = \begin{pmatrix} -2.5 \\ 5 \end{pmatrix}

Final answer

(-2.5, 5) as a column vector

Detailed explanation

Walkthrough

The three points $P$ , $Q$ and $R$ form a triangle of vectors: going from $P$ to $R$ and then from $R$ to $Q$ is the same as going directly from $P$ to $Q$ . So $\overrightarrow{PR} + \overrightarrow{RQ} = \overrightarrow{PQ}$ , which rearranges to $\overrightarrow{RQ} = \overrightarrow{PQ} - \overrightarrow{PR}$ .

We are given $4\overrightarrow{PR}$ , not $\overrightarrow{PR}$ itself, so first divide every component by $4$ : this gives $\overrightarrow{PR} = \begin{pmatrix} -0.5 \\ 2 \end{pmatrix}$ .

Subtracting component by component gives $-3 - (-0.5) = -2.5$ for the top entry and $7 - 2 = 5$ for the bottom entry.

Key Takeaways

The triangle law lets you express any one of $\overrightarrow{PQ}$ , $\overrightarrow{QR}$ , $\overrightarrow{RP}$ in terms of the other two.
Scalar multiplication of a vector multiplies each component; undo it by dividing each component.
Vector subtraction is component-wise subtraction.

Common Mistakes

Forgetting to divide by $4$ and using $\begin{pmatrix} -2 \\ 8 \end{pmatrix}$ directly as $\overrightarrow{PR}$ .
Subtracting in the wrong order (computing $\overrightarrow{PR} - \overrightarrow{PQ}$ instead of $\overrightarrow{PQ} - \overrightarrow{PR}$ ), which gives the negative vector.
Sign slips when subtracting a negative component: $-3 - (-0.5) = -2.5$ , not $-3.5$ .

Things to Be Careful About

The mark scheme awards B2 outright for the correct answer or B1 for the correct unsimplified expression $-\frac{1}{4}\begin{pmatrix} -2 \\ 8 \end{pmatrix} + \begin{pmatrix} -3 \\ 7 \end{pmatrix}$ , so showing the subtraction step protects your method mark even if an arithmetic slip follows.

Techniques used

scale a vector by a scalaruse the triangle law of vector additionsubtract column vectors

(b)

The vectors $\mathbf{a}$ , $\mathbf{b}$ and $\mathbf{c}$ are such that $\mathbf{a} = \alpha \mathbf{i} + 6\mathbf{j}$ , $\mathbf{b} = 4\mathbf{i} + \beta \mathbf{j}$ and $\mathbf{c} = (2\alpha + 5\beta)\mathbf{i} + 20\mathbf{j}$ , where $\alpha$ and $\beta$ are scalars.

Given that $\mathbf{c} = 3\mathbf{a} - 2\mathbf{b}$ , find the values of $\alpha$ and $\beta$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write $3\mathbf{a} - 2\mathbf{b}$ in terms of $\alpha$ and $\beta$ , then equate the $\mathbf{i}$ and $\mathbf{j}$ components with those of $\mathbf{c}$ to get two equations.

Working

3\mathbf{a} - 2\mathbf{b} = 3(\alpha\mathbf{i} + 6\mathbf{j}) - 2(4\mathbf{i} + \beta\mathbf{j}) = (3\alpha - 8)\mathbf{i} + (18 - 2\beta)\mathbf{j}

Equating with $\mathbf{c} = (2\alpha + 5\beta)\mathbf{i} + 20\mathbf{j}$ :

3\alpha - 8 = 2\alpha + 5\beta \qquad \text{and} \qquad 18 - 2\beta = 20

From the $\mathbf{j}$ equation:

-2\beta = 2 \implies \beta = -1

Substituting into the $\mathbf{i}$ equation:

3\alpha - 8 = 2\alpha - 5 \implies \alpha = 3

Answer

\alpha = 3, \qquad \beta = -1

Final answer

alpha = 3, beta = -1

Detailed explanation

Walkthrough

Two vectors are equal only when their corresponding components are equal. First expand the combination $3\mathbf{a} - 2\mathbf{b}$ : multiplying $\mathbf{a}$ by $3$ gives $3\alpha\mathbf{i} + 18\mathbf{j}$ , and multiplying $\mathbf{b}$ by $2$ gives $8\mathbf{i} + 2\beta\mathbf{j}$ ; subtracting component-wise yields $(3\alpha - 8)\mathbf{i} + (18 - 2\beta)\mathbf{j}$ .

Setting this equal to $\mathbf{c}$ produces one equation from the $\mathbf{i}$ components ( $3\alpha - 8 = 2\alpha + 5\beta$ ) and one from the $\mathbf{j}$ components ( $18 - 2\beta = 20$ ).

The $\mathbf{j}$ equation contains only $\beta$ , so solve it first: $\beta = -1$ . Substituting into the $\mathbf{i}$ equation then gives $3\alpha - 8 = 2\alpha - 5$ , so $\alpha = 3$ .

Key Takeaways

Equality of vectors means equality of each component — this converts one vector equation into two scalar equations.
Solve the equation containing only one unknown first, then substitute.

Common Mistakes

Sign errors when expanding $-2(4\mathbf{i} + \beta\mathbf{j})$ , e.g. writing $+8$ instead of $-8$ .
Equating unlike components (matching the $\mathbf{i}$ part of one side with the $\mathbf{j}$ part of the other).
Solving only one of the two equations and leaving the other unknown unfound.

Things to Be Careful About

The mark scheme requires equating like vectors at least once for M1, then A1 for either value and full credit for both — so always show the component equations explicitly rather than jumping to the answers.

Techniques used

equate like components of equal vectorsform simultaneous equations in scalarssolve linear equations

Q24MMedium-EasyQuadratic functions

Solve the inequality $(3 - x)(5x + 8) \geq 9 - 3x$ .

DifficultyMedium-Easy

Worked solution

Approach

Expand the left-hand side, bring everything to one side to form a three-term quadratic inequality, factorise it, and use the critical values to write down the solution interval.

Working

Expand the brackets:

(3 - x)(5x + 8) = 15x + 24 - 5x^2 - 8x = -5x^2 + 7x + 24

So the inequality becomes

-5x^2 + 7x + 24 \geq 9 - 3x

Move everything to one side (adding $5x^2 + 3x - 24$ to both sides):

0 \geq 5x^2 - 10x - 15

Divide through by $5$ :

x^2 - 2x - 3 \leq 0

Factorise:

(x - 3)(x + 1) \leq 0

The critical values are $x = -1$ and $x = 3$ . The quadratic $x^2 - 2x - 3$ is negative between its roots, so

-1 \leq x \leq 3

Answer

-1 \leq x \leq 3

Final answer

-1 <= x <= 3

Detailed explanation

Walkthrough

The inequality compares a product of two brackets, $(3 - x)(5x + 8)$ , with the linear expression $9 - 3x$ . The first step is to expand the product: multiply each term in the first bracket by each term in the second to get $-5x^2 + 7x + 24$ . Now the inequality reads $-5x^2 + 7x + 24 \geq 9 - 3x$ . To solve a quadratic inequality we collect everything onto one side so that one side is zero. Moving all terms to the right (equivalently multiplying through by $-1$ ) gives $5x^2 - 10x - 15 \leq 0$ , which is exactly the three-term quadratic the mark scheme expects. Dividing by the common factor $5$ simplifies it to $x^2 - 2x - 3 \leq 0$ , which factorises neatly as $(x - 3)(x + 1) \leq 0$ .

The roots $x = -1$ and $x = 3$ are the critical values: they are where the quadratic equals zero, so they are always included in the solution of a $\leq$ inequality. Because the parabola $y = x^2 - 2x - 3$ opens upwards, it is below or on the $x$ -axis exactly between its two roots, giving $-1 \leq x \leq 3$ . A quick check: at $x = 0$ the original inequality reads $3 \times 8 = 24 \geq 9$ , which is true, and $0$ lies in the interval — consistent.

Key Takeaways

To solve a quadratic inequality, rearrange to a three-term quadratic on one side with zero on the other, factorise, and use the roots as critical values.
For an upwards-opening quadratic, the expression is negative between the roots and positive outside them; the inequality sign tells you which region to keep.
Critical values are included when the inequality is $\leq$ or $\geq$ (closed interval), excluded for $<$ or $>$ .
Dividing by a positive common factor (here $5$ ) does not change the inequality direction.

Common Mistakes

Sign errors when expanding $(3 - x)(5x + 8)$ — the $-x \times 5x$ term gives $-5x^2$ , and forgetting this sign wrecks the whole question.
Moving terms to the wrong side and forgetting to flip the inequality when multiplying/dividing by a negative number; the mark scheme accepts any consistent rearrangement ("oe"), but the final interval must follow correctly.
Solving $(x-3)(x+1) \leq 0$ as $x \leq -1$ or $x \geq 3$ — the wrong region; the upwards parabola is negative between the roots.
Writing the answer as two separate inequalities without connecting them with "and" — the mark scheme explicitly rejects this ("Do not accept separate inequalities unless connected with 'and'").
Excluding the endpoints $-1$ and $3$ : since the inequality is $\geq$ , the critical values must be included with $\leq$ signs.

Things to Be Careful About

The final answer must be a single connected statement, $-1 \leq x \leq 3$ (or " $x \geq -1$ and $x \leq 3$ "); separate inequalities score A0.
Check the rearrangement direction: the mark scheme's form is $5x^2 - 10x - 15 \leq 0$ ; if you keep the negative quadratic instead, the inequality direction flips and the same interval must result — make sure your working is consistent throughout.
Substituting a test value (e.g. $x = 0$ ) into the original inequality is a quick way to confirm you have chosen the correct region.
All working must be shown: the expansion, the rearrangement to the three-term quadratic, the factorisation, and the critical values each carry marks (M1, M1, A1, A1).

Techniques used

expand the product of two linear bracketsrearrange to a three-term quadratic inequalityfactorise the quadraticidentify critical values and select the correct solution region

Similar questions

Q3Medium-EasyCoordinate geometry of the circleStraight-line graphs

Point $A$ has coordinates $(3, -1)$ .
A circle has equation $(x - 4)^2 + (y + 3)^2 = 5$ .

(a)

Show that $A$ lies on the circumference of the circle.

1M Similar

(b)

Given that $AB$ is a diameter of the circle, find the coordinates of $B$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The centre of the circle is $C(4, -3)$ . Since $AB$ is a diameter, $C$ is the midpoint of $AB$ , so $B$ is the reflection of $A$ through $C$ : $B = 2C - A$ .

Working

The vector from $A$ to the centre is

\begin{pmatrix} 4 \\ -3 \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}

Continuing the same displacement from the centre:

\begin{pmatrix} x_B \\ y_B \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \end{pmatrix} + \begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ -5 \end{pmatrix}

Equivalently, using the midpoint equations $\frac{3 + x}{2} = 4$ and $\frac{-1 + y}{2} = -3$ gives $x = 5$ , $y = -5$ .

Answer

B(5, -5)

Final answer

(5, -5)

Detailed explanation

Walkthrough

From the equation $(x-4)^2 + (y+3)^2 = 5$ the centre is $C(4, -3)$ and the radius is $\sqrt{5}$ . A diameter passes through the centre, so the centre is the midpoint of $AB$ . Moving from $A(3,-1)$ to $C(4,-3)$ is a displacement of $(1, -2)$ ; continuing the same displacement past the centre lands on $B(5, -5)$ . The same result follows from the midpoint formula: $\frac{3+x}{2}=4$ gives $x=5$ , and $\frac{-1+y}{2}=-3$ gives $y=-5$ .

Key Takeaways

For a circle $(x-a)^2 + (y-b)^2 = r^2$ , the centre is $(a, b)$ . If $AB$ is a diameter, the centre is the midpoint of $AB$ , so each endpoint is found by reflecting the other through the centre.

Common Mistakes

Reading the centre as $(-4, 3)$ by forgetting the sign flip from $(x-4)$ and $(y+3)$ ; or subtracting instead of adding the displacement to get the wrong side of the centre. The mark scheme notes the M1 can also be earned by solving the line $y = -2x + 5$ with the circle to reach $x^2 - 8x + 15 = 0$ .

Things to Be Careful About

Both coordinates must be correct for the A1; the answer is cao at $(5, -5)$ . Show the midpoint reasoning (or equivalent working) so the M1 is clearly earned.

Techniques used

identify the centre and radius from the circle equationuse the midpoint property of a diameter

(c)

Find the equation of the tangent to the circle at $A$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The tangent at $A$ is perpendicular to the radius joining the centre $C(4, -3)$ to $A(3, -1)$ . Find the radius gradient, take its negative reciprocal, then form the line through $A$ .

Working

Gradient of the radius $CA$ :

m_{CA} = \frac{-3 - (-1)}{4 - 3} = \frac{-2}{1} = -2

Since the tangent is perpendicular to the radius:

m_{\text{tangent}} = \frac{-1}{-2} = \frac{1}{2}

Using the point $A(3, -1)$ :

y - (-1) = \frac{1}{2}(x - 3)

y + 1 = \frac{1}{2}(x - 3)

Answer

y + 1 = \frac{1}{2}(x - 3) \quad \text{oe, e.g. } y = \frac{1}{2}x - \frac{5}{2}

Final answer

y + 1 = 0.5(x - 3)

Detailed explanation

Walkthrough

A key geometric fact is that the tangent to a circle at a point is perpendicular to the radius drawn to that point. The radius runs from the centre $C(4, -3)$ to $A(3, -1)$ , so its gradient is $\frac{-3+1}{4-3} = -2$ . Perpendicular gradients multiply to $-1$ , so the tangent's gradient is $\frac{-1}{-2} = \frac{1}{2}$ . Finally, using the straight-line form $y - y_1 = m(x - x_1)$ with $m = \frac{1}{2}$ and $(x_1, y_1) = (3, -1)$ gives $y + 1 = \frac{1}{2}(x - 3)$ .

Key Takeaways

Tangent ⊥ radius at the point of contact; perpendicular gradients satisfy $m_1 m_2 = -1$ ; the point–gradient form builds the line once a point and gradient are known.

Common Mistakes

Taking the reciprocal without changing the sign (giving $-\frac{1}{2}$ ); using the centre instead of $A$ as the point on the line; forgetting the sign when computing $-3 - (-1)$ . The mark scheme allows differentiation routes too, but the radius method is simplest. ISW applies after a correct unsimplified answer.

Things to Be Careful About

The final A1 accepts any equivalent form ('oe'), e.g. $x - 2y - 5 = 0$ or $y = \frac{1}{2}x - \frac{5}{2}$ . Follow-through applies to the perpendicular step using your radius gradient, but the radius gradient itself must be correct.

Techniques used

find the gradient of the radiususe the perpendicular gradient relationform the tangent equation from a point and a gradient

Q4Medium[Legacy] Indices and surdsSimultaneous equationsLogarithmic and exponential functions

(a)

Solve the equation $x^{\frac{1}{3}} - x^{\frac{1}{6}} = 2$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Let $u = x^{\frac{1}{6}}$ . Then $x^{\frac{1}{3}} = \left(x^{\frac{1}{6}}\right)^2 = u^2$ , so the equation becomes a quadratic in $u$ .

Working

u^2 - u - 2 = 0

Factorise:

(u - 2)(u + 1) = 0

so $u = 2$ or $u = -1$ , i.e.

x^{\frac{1}{6}} = 2 \quad \text{or} \quad x^{\frac{1}{6}} = -1

The sixth root of a real number is never negative, so $x^{\frac{1}{6}} = -1$ must be rejected.

From $x^{\frac{1}{6}} = 2$ :

x = 2^6 = 64

Check: $64^{\frac{1}{3}} - 64^{\frac{1}{6}} = 4 - 2 = 2$ . ✓

Answer

x = 64

Final answer

x = 64

Detailed explanation

Walkthrough

The equation mixes two different fractional powers of $x$ . The key observation is that $x^{\frac{1}{3}} = \left(x^{\frac{1}{6}}\right)^2$ , because multiplying indices gives $\frac{1}{6} \times 2 = \frac{1}{3}$ . So if we write $u = x^{\frac{1}{6}}$ , the equation $x^{\frac{1}{3}} - x^{\frac{1}{6}} = 2$ becomes $u^2 - u - 2 = 0$ , an ordinary quadratic.

Factorising: we need two numbers with product $-2$ and sum $-1$ : these are $-2$ and $+1$ , giving $(u-2)(u+1) = 0$ , so $u = 2$ or $u = -1$ .

Converting back: $x^{\frac{1}{6}} = 2$ means raising both sides to the power 6, so $x = 2^6 = 64$ . But $x^{\frac{1}{6}} = -1$ is impossible — the sixth root of any real number is non-negative — so this root is rejected. A quick check confirms $4 - 2 = 2$ when $x = 64$ .

Key Takeaways

Equations in different powers of $x$ often hide a quadratic once you spot one power as the square of the other.
Always convert back carefully using index laws: $x^{\frac{1}{6}} = 2 \Rightarrow x = 2^6$ .
Even-index roots are never negative, so negative candidate values of $x^{\frac{1}{6}}$ must be discarded.

Common Mistakes

Raising every term to the sixth power directly: $(x^{\frac{1}{3}})^6 - (x^{\frac{1}{6}})^6 = 2^6$ scores zero marks — it does not preserve the equation.
Trying to combine the powers by subtracting indices ( $x^{\log_x 2}$ style manipulation) also scores zero.
Keeping the rejected root and answering $x = 64$ and $x = 1$ — the mark scheme explicitly requires rejecting $x^{\frac{1}{6}} = -1$ (which would give $x = 1$ ).
Sign slips when factorising $u^2 - u - 2$ .

Things to Be Careful About

The mark scheme awards a specific B1 for rejecting $x^{\frac{1}{6}} = -1$ (or ignoring $x = 1$ ), so state the rejection explicitly.
Do not raise both sides to the sixth power term by term — that method earns nothing.
Verify your answer by substitution where possible.

Techniques used

substitute to reduce the equation to a quadraticfactorise the quadraticraise both sides to a power to recover xreject an invalid root

(b)

Solve the simultaneous equations

\begin{aligned} \lg(x + 2y) &= 0 \\ x^2 + 4xy + y &= 1. \end{aligned}

5M Similar

DifficultyMedium

Worked solution

Approach

Use $\lg 1 = 0$ to turn the first equation into a linear relation between $x$ and $y$ , substitute into the second equation, and solve the resulting quadratic.

Working

Since $\lg(x + 2y) = 0$ :

x + 2y = 10^0 = 1 \quad \Rightarrow \quad x = 1 - 2y

Substitute into $x^2 + 4xy + y = 1$ :

(1 - 2y)^2 + 4(1 - 2y)y + y = 1

1 - 4y + 4y^2 + 4y - 8y^2 + y = 1

1 + y - 4y^2 = 1

y - 4y^2 = 0

y(1 - 4y) = 0

So $y = 0$ or $y = \frac{1}{4}$ .

When $y = 0$ : $x = 1 - 2(0) = 1$ .

When $y = \frac{1}{4}$ : $x = 1 - 2\left(\frac{1}{4}\right) = \frac{1}{2}$ .

Check: $(1, 0)$ : $\lg 1 = 0$ ✓ and $1 + 0 + 0 = 1$ ✓. $\left(\tfrac{1}{2}, \tfrac{1}{4}\right)$ : $\lg 1 = 0$ ✓ and $\tfrac{1}{4} + 4\cdot\tfrac{1}{2}\cdot\tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{4} + \tfrac{1}{2} + \tfrac{1}{4} = 1$ ✓.

Answer

x = 1,\ y = 0 \quad \text{and} \quad x = \frac{1}{2},\ y = \frac{1}{4}

Final answer

x = 1, y = 0 and x = 1/2, y = 1/4

Detailed explanation

Walkthrough

The first equation involves a base-10 logarithm. Since $\lg t = 0$ means $t = 10^0 = 1$ , the equation $\lg(x + 2y) = 0$ simply says $x + 2y = 1$ — a straight line.

Rearranging gives $x = 1 - 2y$ , which we substitute into the second (nonlinear) equation $x^2 + 4xy + y = 1$ . Expanding $(1-2y)^2 = 1 - 4y + 4y^2$ and $4(1-2y)y = 4y - 8y^2$ , everything collects neatly: the $-4y$ and $+4y$ cancel, leaving $1 + y - 4y^2 = 1$ , i.e. $y - 4y^2 = 0$ .

This factors as $y(1 - 4y) = 0$ , giving $y = 0$ or $y = \frac{1}{4}$ . Substituting each back into $x = 1 - 2y$ gives the pairs $(x, y) = (1, 0)$ and $\left(\frac{1}{2}, \frac{1}{4}\right)$ .

Key Takeaways

$\lg t = 0 \iff t = 1$ converts a logarithmic equation into a linear one instantly.
Substitution turns a line–curve system into a single quadratic in one variable.
Nonlinear simultaneous equations can have more than one solution pair — find them all.

Common Mistakes

Forgetting that $\lg 1 = 0$ and being unable to start.
Expanding $(1-2y)^2$ incorrectly (e.g. writing $1 - 4y^2$ instead of $1 - 4y + 4y^2$ ).
Losing a factor when simplifying: the mark scheme expects the correct quadratic $y - 4y^2 = 0$ (or equivalently $2x^2 - 3x + 1 = 0$ if eliminating $y$ instead).
Finding only one solution pair instead of both.
Dividing through by $y$ instead of factorising, which loses the $y = 0$ solution.

Things to Be Careful About

The elimination M-marks depend on having first obtained $x + 2y = 1$ correctly (B1), so get the log step right before anything else.
Both solution pairs are required for full marks; check each pair satisfies the original logarithmic equation (the argument $x + 2y$ must equal 1, which it does here, so no solutions need discarding on domain grounds).
Exact fractions are expected: $y = \frac{1}{4}$ , $x = \frac{1}{2}$ .

Techniques used

convert a logarithmic equation to linear formsubstitute to eliminate one unknownform and solve a quadraticgive both solution pairs

Q5Medium-EasyEquations, inequalities and graphsTrigonometry

(a)

Solve the equation $5|2x - 1| + 8 = 23$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Isolate $\left|2x - 1\right|$ , then split into the two cases $2x - 1 = \pm 3$ and solve each linear equation.

Working

5\left|2x - 1\right| + 8 = 23 \implies 5\left|2x - 1\right| = 15 \implies \left|2x - 1\right| = 3

Case 1:

2x - 1 = 3 \implies 2x = 4 \implies x = 2

Case 2:

2x - 1 = -3 \implies 2x = -2 \implies x = -1

Answer

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

Final answer

x = 2 or x = -1

Detailed explanation

Walkthrough

The equation contains a modulus, so we first isolate it: subtracting 8 from both sides gives $5\left|2x - 1\right| = 15$ , and dividing by 5 gives $\left|2x - 1\right| = 3$ . The modulus of an expression equals 3 when the expression itself is either 3 or $-3$ , so we split into two linear equations: $2x - 1 = 3$ gives $x = 2$ , and $2x - 1 = -3$ gives $x = -1$ . Both values check: at $x = 2$ , $5|4-1|+8 = 23$ ; at $x = -1$ , $5|-2-1|+8 = 23$ .

Key Takeaways

Always isolate the modulus before splitting into cases.
$|u| = k$ (with $k > 0$ ) means $u = k$ or $u = -k$ — two solutions.
Substituting back verifies both answers.

Common Mistakes

Forgetting the negative case and giving only one solution (the mark scheme awards A1 for either correct, A2 for both).
Dividing by 5 incorrectly or forgetting to divide at all.
Squaring both sides to form a quadratic is an accepted alternative route, but sign slips in $4x^2 - 4x - 8 = 0$ lose marks.

Things to Be Careful About

Both solutions are required for full credit — the scheme explicitly allows A1 for just one.
Answers must come from correct working (nfww applies to the alternative quadratic route).
Check each root in the original equation as a quick error check.

Techniques used

isolate the modulus expressionsplit the modulus equation into positive and negative casessolve each linear equation

(b)

On the axes, sketch the graph of $y = 5\sin x - 2$ for $0° \leq x \leq 360°$ .

Fig. 1

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The graph of $y = 5\sin x - 2$ is the ordinary sine wave stretched vertically by a factor of 5 and translated 2 units down. One full cycle on $0° \leq x \leq 360°$ : start at $(0, -2)$ , peak at $(90°, 3)$ , midline crossing at $(180°, -2)$ , trough at $(270°, -7)$ , end at $(360°, -2)$ .

Working

Amplitude $= 5$ , midline $y = -2$ , so the curve oscillates between $-2 + 5 = 3$ and $-2 - 5 = -7$ .

Key points:

(0, -2), \quad (90°, 3), \quad (180°, -2), \quad (270°, -7), \quad (360°, -2)

It crosses the $x$ -axis where $\sin x = \frac{2}{5}$ , i.e. at approximately $23.6°$ and $156.4°$ .

Answer

One full cycle of a sine wave with midline $y = -2$ , amplitude 5: starting at $(0, -2)$ , maximum 3 at $90°$ , minimum $-7$ at $270°$ , returning to $(360°, -2)$ .

Final answer

Sine curve over 0 to 360 degrees with midline y = -2, maximum 3 at 90 degrees, minimum -7 at 270 degrees, passing through (0, -2) and (360, -2)

Detailed explanation

Walkthrough

Multiplying $\sin x$ by 5 stretches the wave so its values run from $-5$ to $5$ instead of $-1$ to $1$ ; subtracting 2 then slides everything down by 2, giving a midline of $y = -2$ and a range from $-7$ to $3$ . The sine shape starts at its midline value at $x = 0$ , rises to its maximum at $90°$ , returns to the midline at $180°$ , dips to its minimum at $270°$ , and returns to the midline at $360°$ — exactly one cycle on the given interval.

Key Takeaways

In $y = a\sin x + c$ , the amplitude is $|a|$ and the midline is $y = c$ .
The maxima/minima of a shifted sine occur at the same $x$ -values as the unshifted curve ( $90°$ and $270°$ ).
A sketch needs correct shape, correct turning points, and correct start/end values.

Common Mistakes

Drawing the wave oscillating about $y = 0$ instead of $y = -2$ (the mark scheme pays B1 specifically for the correct midline).
Using amplitude 5 but centring wrongly, giving maxima 5 and minima $-5$ instead of 3 and $-7$ (B1 requires the correct amplitudes 3 and $-7$ ).
Drawing more than one cycle or an inverted (cosine-like) shape.
Marking the maximum at $x = 0$ instead of $90°$ .

Things to Be Careful About

The mark scheme awards two separate B1s: one for the correct amplitude values ( $3$ and $-7$ ) and one for the correct midline $y = -2$ — both features must be visible.
Exactly one cycle between $0°$ and $360°$ is required.
The curve must pass through $(0, -2)$ and $(360°, -2)$ , not through the origin.

Techniques used

identify amplitude and midline from a transformed sine functionmark maximum, minimum and interceptssketch one full cycle of the sine curve

Q6MediumCalculus

A curve has equation

y = \left(\frac{x^2 - 1}{x^2 + 1}\right)^4.

(a)

Show that $\frac{\mathrm{d}y}{\mathrm{d}x}$ can be written as $\frac{Ax(x^2 - 1)^3}{(x^2 + 1)^5}$ , where $A$ is a positive integer to be found.

5M Similar

DifficultyMedium

Worked solution

Approach

Use the chain rule on $y = u^4$ , where $u = \frac{x^2 - 1}{x^2 + 1}$ , and differentiate the inner function $u$ with respect to $x$ using the quotient rule.

Working

Let $u = \frac{x^2 - 1}{x^2 + 1}$ . By the quotient rule,

\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{(x^2 + 1)(2x) - (x^2 - 1)(2x)}{(x^2 + 1)^2}

\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2x(x^2 + 1 - x^2 + 1)}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2}

By the chain rule,

\frac{\mathrm{d}y}{\mathrm{d}x} = 4u^3 \frac{\mathrm{d}u}{\mathrm{d}x}

\frac{\mathrm{d}y}{\mathrm{d}x} = 4\left(\frac{x^2 - 1}{x^2 + 1}\right)^3 \left(\frac{4x}{(x^2 + 1)^2}\right)

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4(x^2 - 1)^3}{(x^2 + 1)^3} \times \frac{4x}{(x^2 + 1)^2} = \frac{16x(x^2 - 1)^3}{(x^2 + 1)^5}

This matches the required form with $A = 16$ .

Answer

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{16x(x^2 - 1)^3}{(x^2 + 1)^5} \quad (A = 16)

Final answer

16

Detailed explanation

Walkthrough

To find $\frac{\mathrm{d}y}{\mathrm{d}x}$ for the given function $y = \left(\frac{x^2 - 1}{x^2 + 1}\right)^4$ , we recognise it as a composite function of the form $y = [u(x)]^4$ .

First, we differentiate the inside function $u(x) = \frac{x^2 - 1}{x^2 + 1}$ using the quotient rule $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}$ :

\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{(x^2 + 1)(2x) - (x^2 - 1)(2x)}{(x^2 + 1)^2} = \frac{2x(x^2 + 1 - x^2 + 1)}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2}

Next, we apply the chain rule $\frac{\mathrm{d}y}{\mathrm{d}x} = 4u^3 \frac{\mathrm{d}u}{\mathrm{d}x}$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = 4\left(\frac{x^2 - 1}{x^2 + 1}\right)^3 \left(\frac{4x}{(x^2 + 1)^2}\right)

Multiplying the numerators and denominators together:
- Numerator: $4 \times (x^2 - 1)^3 \times 4x = 16x(x^2 - 1)^3$
- Denominator: $(x^2 + 1)^3 \times (x^2 + 1)^2 = (x^2 + 1)^5$

This gives the required expression with $A = 16$ .

Key Takeaways

When differentiating powers of a fraction, combining the chain rule with the quotient rule provides a systematic approach.
Simplify the derivative of the inner function first to avoid cumbersome algebra when multiplying terms out.

Common Mistakes

Sign errors in the numerator of the quotient rule, e.g. writing $(x^2+1)(2x) + (x^2-1)(2x)$ instead of subtracting.
Forgetting to multiply by the derivative of the inner function (omitting the chain rule step entirely).
Incorrectly combining powers of $(x^2 + 1)$ in the denominator (e.g. adding exponents incorrectly).

Things to Be Careful About

Ensure all steps of the algebraic simplification are clearly shown, as this is a "Show that" question where the target form is given.

Techniques used

differentiate using the chain ruledifferentiate the inner function using the quotient rulesimplify rational expressions by combining powers

(b)

3M

(i)

Show that the curve has stationary points where $x = -1$ , $x = 0$ and $x = 1$ .

1M Similar

DifficultyEasy

Worked solution

Approach

Stationary points occur where $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ . Equate the derivative found in part (a) to zero and solve for $x$ .

Working

\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies \frac{16x(x^2 - 1)^3}{(x^2 + 1)^5} = 0

Since the denominator $(x^2 + 1)^5 > 0$ for all real $x$ , this simplifies to:

16x(x^2 - 1)^3 = 0

16x = 0 \implies x = 0

(x^2 - 1)^3 = 0 \implies x^2 - 1 = 0 \implies x = \pm 1

Thus, the stationary points occur at $x = -1$ , $x = 0$ , and $x = 1$ .

Answer

x = -1, \; x = 0, \; x = 1

Final answer

x = -1, x = 0 and x = 1

Detailed explanation

Walkthrough

A stationary point on a curve is a point where the gradient is zero, i.e., $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ .

From part (a), the derivative is:

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{16x(x^2 - 1)^3}{(x^2 + 1)^5}

Setting $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ , a rational expression equals zero when its numerator is zero and its denominator is non-zero.

The denominator $(x^2 + 1)^5$ is never zero for real $x$ .
Setting the numerator to zero: $16x(x^2 - 1)^3 = 0$ .

This gives two factors:

$x = 0$
$(x^2 - 1)^3 = 0 \implies x^2 - 1 = 0 \implies x = \pm 1$

Hence, the curve has stationary points at $x = -1$ , $x = 0$ , and $x = 1$ .

Key Takeaways

For rational functions, setting $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ corresponds to setting the numerator equal to zero (provided the denominator is non-zero).

Common Mistakes

Forgetting the negative root when solving $x^2 - 1 = 0$ , leading to only $x = 1$ and $x = 0$ .

Things to Be Careful About

Ensure you clearly state each factor giving its corresponding root to justify all three stationary values.

Techniques used

set the derivative to zerosolve polynomial factors for stationary points

(ii)

Use the first derivative test to determine which two stationary points have the same nature and state whether they are maximum or minimum points.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the first derivative test by determining the sign of $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{16x(x^2 - 1)^3}{(x^2 + 1)^5}$ in the intervals around the stationary points $x = -1$ , $x = 0$ , and $x = 1$ .

Note that the denominator $(x^2 + 1)^5$ is strictly positive for all $x$ , so the sign of $\frac{\mathrm{d}y}{\mathrm{d}x}$ is determined entirely by the numerator $16x(x^2 - 1)^3$ .

Working

Test the sign of $\frac{\mathrm{d}y}{\mathrm{d}x}$ in each region:

For $x < -1$ (e.g. $x = -2$ ): $x < 0$ and $x^2 - 1 > 0$ , so $\frac{\mathrm{d}y}{\mathrm{d}x} < 0$ (negative).
For $-1 < x < 0$ (e.g. $x = -0.5$ ): $x < 0$ and $x^2 - 1 < 0$ , so $\frac{\mathrm{d}y}{\mathrm{d}x} > 0$ (positive).
For $0 < x < 1$ (e.g. $x = 0.5$ ): $x > 0$ and $x^2 - 1 < 0$ , so $\frac{\mathrm{d}y}{\mathrm{d}x} < 0$ (negative).
For $x > 1$ (e.g. $x = 2$ ): $x > 0$ and $x^2 - 1 > 0$ , so $\frac{\mathrm{d}y}{\mathrm{d}x} > 0$ (positive).

Summarising the sign changes across each stationary point:

Around $x = -1$ : the sign changes from negative to positive $\implies$ minimum point.
Around $x = 0$ : the sign changes from positive to negative $\implies$ maximum point.
Around $x = 1$ : the sign changes from negative to positive $\implies$ minimum point.

Therefore, the two stationary points with the same nature are $x = -1$ and $x = 1$ , and they are both minimum points.

Answer

x = -1 \text{ and } x = 1 \text{ are minimum points}

Final answer

x = -1 and x = 1 are minimum points

Detailed explanation

Walkthrough

The question explicitly requires the use of the first derivative test.

Recall the first derivative from part (a):

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{16x(x^2 - 1)^3}{(x^2 + 1)^5}

Since $(x^2 + 1)^5 > 0$ for all real $x$ , the sign of $\frac{\mathrm{d}y}{\mathrm{d}x}$ depends only on the sign of $x(x^2 - 1)^3$ , which has the same sign as $x(x^2 - 1)$ :
- In $(-\infty, -1)$ : $x$ is negative and $(x^2 - 1)$ is positive $\implies \frac{\mathrm{d}y}{\mathrm{d}x} < 0$ .
- In $(-1, 0)$ : $x$ is negative and $(x^2 - 1)$ is negative $\implies \frac{\mathrm{d}y}{\mathrm{d}x} > 0$ .
- In $(0, 1)$ : $x$ is positive and $(x^2 - 1)$ is negative $\implies \frac{\mathrm{d}y}{\mathrm{d}x} < 0$ .
- In $(1, \infty)$ : $x$ is positive and $(x^2 - 1)$ is positive $\implies \frac{\mathrm{d}y}{\mathrm{d}x} > 0$ .
Interpreting the behaviour:
- At $x = -1$ , the gradient changes from negative (decreasing) to positive (increasing), meaning $x = -1$ is a local minimum.
- At $x = 0$ , the gradient changes from positive to negative, meaning $x = 0$ is a local maximum.
- At $x = 1$ , the gradient changes from negative to positive, meaning $x = 1$ is a local minimum.
Thus, $x = -1$ and $x = 1$ share the same nature: they are both minimum points.

Key Takeaways

In the first derivative test:
- A change from negative to positive indicates a local minimum.
- A change from positive to negative indicates a local maximum.
- No change in sign indicates a point of inflection.
Testing the sign of factors separately makes sign determination much simpler than calculating exact values.

Common Mistakes

Using the second derivative test instead of the first derivative test; the mark scheme specifies M0 if only the second derivative is used because the question explicitly requested the first derivative test.
Picking test points incorrectly or confusing the signs of $(x^2 - 1)$ for values between $-1$ and $1$ .

Things to Be Careful About

Ensure you clearly state which two stationary points share the same nature ( $x = -1$ and $x = 1$ ) and specify that they are minimum points.

Techniques used

apply the first derivative testevaluate the sign of the gradient in adjacent intervalsclassify stationary points as maxima or minima

Q76MMediumFactors of polynomials

Solutions to this question by accurate drawing will not be accepted.

Find the $x$ -coordinates of the points where the curve $y = (2x - 9)(x^2 + 5) + 42$ cuts the $x$ -axis.

DifficultyMedium

Worked solution

Approach

The curve cuts the $x$ -axis where $y = 0$ . Expand $(2x - 9)(x^2 + 5) + 42$ into a cubic, find one linear factor using the factor theorem, divide out to get a quadratic, then factorise the quadratic.

Working

Set $y = 0$ :

(2x - 9)(x^2 + 5) + 42 = 0

Expand:

2x^3 - 9x^2 + 10x - 45 + 42 = 0 \quad\Rightarrow\quad 2x^3 - 9x^2 + 10x - 3 = 0

Test $x = 1$ : $2 - 9 + 10 - 3 = 0$ , so $x - 1$ is a factor. Dividing:

2x^3 - 9x^2 + 10x - 3 = (x - 1)(2x^2 - 7x + 3)

Factorise the quadratic:

2x^2 - 7x + 3 = (2x - 1)(x - 3)

So

(x - 1)(2x - 1)(x - 3) = 0

Answer

x = 1, \quad x = \frac{1}{2}, \quad x = 3

Final answer

x = 1/2, 1, 3

Detailed explanation

Walkthrough

The curve meets the $x$ -axis exactly where $y = 0$ , so we need to solve $(2x - 9)(x^2 + 5) + 42 = 0$ . First expand fully: multiplying out gives $2x^3 - 9x^2 + 10x - 45$ , and adding $42$ gives $2x^3 - 9x^2 + 10x - 3 = 0$ .

To factorise a cubic, look for an integer root by testing small values with the factor theorem. Trying $x = 1$ : $2(1)^3 - 9(1)^2 + 10(1) - 3 = 2 - 9 + 10 - 3 = 0$ , so $(x - 1)$ is a factor. Dividing the cubic by $(x - 1)$ (by inspection or long division) gives the quadratic $2x^2 - 7x + 3$ , which factorises as $(2x - 1)(x - 3)$ since $2x^2 - 6x - x + 3 = 2x^2 - 7x + 3$ .

So the equation becomes $(x - 1)(2x - 1)(x - 3) = 0$ , giving $x = 1$ , $x = \frac{1}{2}$ and $x = 3$ .

Key Takeaways

Setting $y = 0$ converts a curve-intercept question into an equation-solving question.
The factor theorem: if substituting $x = k$ gives zero, then $(x - k)$ is a factor.
A cubic with three real roots splits completely into three linear factors; each factor gives one $x$ -coordinate of an intercept.

Common Mistakes

Forgetting to add the $+42$ when expanding, leaving $-45$ instead of $-3$ — every later step then fails.
Testing only $x = 1$ style values carelessly; arithmetic slips in evaluating $p(k)$ give a wrong 'factor'.
Sign errors when dividing the cubic by $(x - 1)$ ; check by re-multiplying the factors back.
Factorising $2x^2 - 7x + 3$ incorrectly — the factors must multiply to give first term $2x^2$ and constant $+3$ .
Giving only some of the three roots; all three intercepts are required for full marks.

Things to Be Careful About

The final answers must come from correct working ('nfww' applies): expanding correctly, showing the factor found, the quadratic factor, its factorisation, and the roots.
The mark scheme accepts any of the three linear factors first ( $x - 1$ , $x - 3$ or $2x - 1$ ), but the corresponding quadratic must match.
State all three values clearly: $x = \frac{1}{2}$ , $1$ , $3$ .

Techniques used

expand the product to form a cubic equationuse the factor theorem to find a linear factordivide by inspection to obtain the quadratic factorfactorise the quadratic and solve for x

Similar questions

Q8MediumLogarithmic and exponential functions

(a)

Write down the set of values of $x$ for which $\log_5(12x - 4)$ exists.

1M Similar

(b)

Solve the equation

\log_5(12x - 4) = \frac{6}{\log_x 125} + 1.

6M Similar

DifficultyMedium

Worked solution

Approach

Use the change-of-base formula to rewrite $\log_x 125$ in base 5, so that every term shares one base. Then combine the right-hand side into a single logarithm, drop the logs to get a quadratic, solve it, and keep only roots satisfying part (a).

Working

By change of base,

\log_x 125 = \frac{\log_5 125}{\log_5 x} = \frac{3}{\log_5 x}

since $\log_5 125 = 3$ . Hence

\frac{6}{\log_x 125} = \frac{6}{\frac{3}{\log_5 x}} = 2\log_5 x.

The equation becomes

\log_5(12x - 4) = 2\log_5 x + 1.

Write $1 = \log_5 5$ and use the power and product laws:

\log_5(12x - 4) = \log_5 x^2 + \log_5 5 = \log_5(5x^2).

Equating arguments:

12x - 4 = 5x^2 \implies 5x^2 - 12x + 4 = 0.

Factorise:

(5x - 2)(x - 2) = 0 \implies x = \frac{2}{5} \text{ or } x = 2.

Both roots satisfy the domain condition $x > \frac{1}{3}$ from part (a), so both are valid.

Answer

x = 0.4 \quad \text{or} \quad x = 2

Final answer

x = 0.4 or x = 2

Detailed explanation

Walkthrough

The equation mixes two different bases ( $5$ and $x$ ), so the first job is to put everything into one base. The change-of-base formula gives $\log_x 125 = \frac{\log_5 125}{\log_5 x}$ , and since $5^3 = 125$ , the numerator is $3$ , so $\log_x 125 = \frac{3}{\log_5 x}$ . Dividing $6$ by this gives $6 \times \frac{\log_5 x}{3} = 2\log_5 x$ .

The equation is now entirely in base 5: $\log_5(12x - 4) = 2\log_5 x + 1$ . Using the power law, $2\log_5 x = \log_5 x^2$ , and writing $1$ as $\log_5 5$ , the product law combines the right-hand side into $\log_5(5x^2)$ .

With a single logarithm on each side and the same base, the arguments must be equal: $12x - 4 = 5x^2$ . Rearranging gives the quadratic $5x^2 - 12x + 4 = 0$ , which factorises as $(5x - 2)(x - 2) = 0$ , giving $x = \frac{2}{5} = 0.4$ or $x = 2$ .

Finally, both candidates are checked against the domain from part (a): $0.4 > \frac{1}{3}$ and $2 > \frac{1}{3}$ , so both are accepted.

Key Takeaways

Change of base converts an unfamiliar base into a workable one: $\log_a b = \frac{\log_c b}{\log_c a}$ .
$\frac{k}{\left(\frac{m}{t}\right)} = \frac{kt}{m}$ — dividing by a fraction multiplies by its reciprocal.
Combining logs with the product/power laws lets you drop the logarithms and solve a polynomial equation.
Every root of a log equation must be checked against the domain conditions (argument positive, base positive and not equal to 1).

Common Mistakes

Misapplying change of base, e.g. writing $\log_x 125 = \frac{\log_5 x}{\log_5 125}$ (inverted).
Treating $\frac{6}{\log_x 125}$ as $6\log_{125} x$ incorrectly without handling the factor correctly.
Forgetting that $1 = \log_5 5$ when combining terms on the right-hand side.
Equating coefficients instead of arguments after combining logs.
Sign errors forming $5x^2 - 12x + 4 = 0$ from $12x - 4 = 5x^2$ .
Accepting a root that violates the domain $x > \frac{1}{3}$ (here both survive, but the check must be made).

Things to Be Careful About

The mark scheme awards B1 marks dependent on a correct change of base being seen — show it explicitly.
Both answers here are exact decimals/fractions; give $x = 0.4$ (i.e. $\frac{2}{5}$ ) and $x = 2$ .
The base $x$ must also be positive and $x \neq 1$ ; both solutions satisfy this automatically since they exceed $\frac{1}{3}$ and neither equals 1.
"Dep" notes mean later marks are lost if the change of base was wrong, so the first step must be correct.

Techniques used

change the base of a logarithm using the change-of-base formulaevaluate log base 5 of 125 as 3combine logarithms using the product lawconvert a logarithmic equation to a quadratic by equating argumentsfactorise the quadratic and reject invalid roots against the domain

Q910MMediumCalculus

Fig. 1

The point $A$ with $x$ -coordinate $2$ lies on the curve $y = \sqrt{4x + 1}$ .
The diagram shows part of this curve and the tangent to the curve at $A$ .

Find the area of the shaded region enclosed by the curve, the tangent and the $x$ -axis.

DifficultyMedium

Worked solution

Approach

The shaded region is bounded above by the tangent at $A$ and below by the curve, sitting on the $x$ -axis. Its area equals the area of the large triangle formed by the tangent and the $x$ -axis, minus the area under the curve from its $x$ -intercept to $x = 2$ .

Working

At $A$ , where $x = 2$ :

y = \sqrt{4(2) + 1} = \sqrt{9} = 3

so $A = (2, 3)$ .

Differentiate using the chain rule:

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \times 4 = \frac{2}{\sqrt{4x+1}}

Gradient of the tangent at $A$ :

\frac{2}{\sqrt{9}} = \frac{2}{3}

Equation of the tangent:

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

Its $x$ -intercept ( $y = 0$ ):

0 - 3 = \frac{2}{3}(x - 2) \implies x - 2 = -\frac{9}{2} \implies x = -\frac{5}{2}

Area of the triangle under the tangent from $x = -\frac{5}{2}$ to $x = 2$ :

\frac{1}{2} \times \left(\frac{5}{2} + 2\right) \times 3 = \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}

The curve meets the $x$ -axis where $y = 0$ :

\sqrt{4x+1} = 0 \implies x = -\frac{1}{4}

Area under the curve from $x = -\frac{1}{4}$ to $x = 2$ :

\int_{-\frac{1}{4}}^{2} (4x+1)^{\frac{1}{2}}\,\mathrm{d}x = \left[ \frac{(4x+1)^{\frac{3}{2}}}{\frac{3}{2} \times 4} \right]_{-\frac{1}{4}}^{2} = \left[ \frac{(4x+1)^{\frac{3}{2}}}{6} \right]_{-\frac{1}{4}}^{2}

= \frac{9^{\frac{3}{2}}}{6} - \frac{0^{\frac{3}{2}}}{6} = \frac{27}{6} = \frac{9}{2}

Shaded area:

\frac{27}{4} - \frac{9}{2} = \frac{27}{4} - \frac{18}{4} = \frac{9}{4}

Answer

\frac{9}{4}

Final answer

9/4 square units

Detailed explanation

Walkthrough

First find the coordinates of $A$ : substituting $x = 2$ into $y = \sqrt{4x+1}$ gives $y = \sqrt{9} = 3$ , so $A = (2, 3)$ .

To draw the tangent we need its gradient, so differentiate. Since $y$ is a square root of a linear expression, the chain rule applies: differentiate $(4x+1)^{\frac{1}{2}}$ to get $\frac{1}{2}(4x+1)^{-\frac{1}{2}}$ , then multiply by the derivative of the inside, $4$ . This gives $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{\sqrt{4x+1}}$ , which at $x = 2$ is $\frac{2}{3}$ .

Using the point–slope form, the tangent is $y - 3 = \frac{2}{3}(x-2)$ . Setting $y = 0$ gives its $x$ -intercept: $x = 2 - \frac{9}{2} = -\frac{5}{2}$ .

The shaded region sits between the tangent (above) and the curve (below), both meeting the $x$ -axis. So:

\text{shaded area} = \text{area of triangle under tangent} - \text{area under curve}

The triangle has base from $x = -\frac{5}{2}$ to $x = 2$ (length $\frac{9}{2}$ ) and height $3$ , giving area $\frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}$ .

The curve hits the $x$ -axis where $4x + 1 = 0$ , i.e. $x = -\frac{1}{4}$ . The area under it from there to $x = 2$ uses the power rule for integration: raise the power from $\frac{1}{2}$ to $\frac{3}{2}$ and divide by $\frac{3}{2} \times 4 = 6$ . Evaluating gives $\frac{9^{\frac{3}{2}}}{6} - 0 = \frac{27}{6} = \frac{9}{2}$ .

Subtracting: $\frac{27}{4} - \frac{18}{4} = \frac{9}{4}$ .

Key Takeaways

The chain rule is needed whenever differentiating a function of a linear expression such as $(4x+1)^{\frac{1}{2}}$ — do not forget the inner factor of $4$ .
Areas between a curve and a line can be found by decomposition: here a triangle minus an integral.
Integrating $(ax+b)^n$ means dividing by $a(n+1)$ as well as increasing the power.
Intercepts anchor the limits: the tangent's intercept sets the triangle's base; the curve's intercept sets the integral's lower limit.

Common Mistakes

Forgetting the inner derivative $4$ when applying the chain rule, giving gradient $\frac{1}{6}$ instead of $\frac{2}{3}$ — the mark scheme requires the form $k(4x+1)^{-\frac{1}{2}}$ with $k \neq 0$ .
Using the wrong lower limit in the integral (e.g. $0$ instead of $-\frac{1}{4}$ ); the mark scheme allows wrong limits only if correct substitution into the integrated form is shown.
Dividing by $\frac{3}{2}$ but forgetting the extra factor $4$ when integrating, giving $\frac{(4x+1)^{\frac{3}{2}}}{\frac{3}{2}}$ instead of $\frac{(4x+1)^{\frac{3}{2}}}{6}$ .
Taking the base of the triangle as $\frac{5}{2}$ rather than $\frac{5}{2} + 2 = \frac{9}{2}$ — the base runs from $x = -\frac{5}{2}$ to $x = 2$ .
Subtracting the areas the wrong way round, or computing area under the curve from $0$ to $2$ only.
Arithmetic slips in $9^{\frac{3}{2}} = 27$ or in combining quarters and halves.

Things to Be Careful About

The final answer must be exactly $\frac{9}{4}$ (or an equivalent exact form); decimals are acceptable on this calculator paper but the exact fraction is safest.
The tangent's $x$ -intercept mark requires the correct straight-line equation — follow-through on your own gradient and point is allowed, but the intercept must come from a valid line.
Show every step: the scheme awards separate marks for the derivative structure, the substitution at $x=2$ , the line equation, each intercept, the triangle area, the integrated form, and the limit evaluation.
Check that the shaded region really is triangle-minus-curve-area by looking at the figure: the tangent lies above the curve throughout the interval, so no splitting into multiple integrals is needed.

Techniques used

differentiate a function of a linear function using the chain rulefind the equation of the tangent at a point on a curvefind x-intercepts of the tangent and of the curveintegrate (4x+1)^(1/2) by raising the powercompute the shaded area as triangle minus area under the curve

Similar questions

Q10MediumTrigonometry

(a)

Given that $0 \leq \theta < \frac{\pi}{2}$ , show that

\frac{\sin\theta}{\sqrt{\operatorname{cosec}^2\theta - 1}} + \frac{1}{\sqrt{1 + \tan^2\theta}}

can be written as $\sec\theta$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use $\operatorname{cosec}^2\theta - 1 = \cot^2\theta$ on the first denominator and $1 + \tan^2\theta = \sec^2\theta$ on the second. Since $0 \leq \theta < \frac{\pi}{2}$ , all of $\sin\theta$ , $\cos\theta$ , $\tan\theta$ are non-negative, so each square root simplifies without a modulus sign. Then combine into a single fraction.

Working

\frac{\sin\theta}{\sqrt{\cot^2\theta}} + \frac{1}{\sqrt{\sec^2\theta}}

Since $\cot\theta \geq 0$ and $\sec\theta \geq 0$ for $0 \leq \theta < \frac{\pi}{2}$ :

= \frac{\sin\theta}{\cot\theta} + \frac{1}{\sec\theta} = \sin\theta\tan\theta + \cos\theta

Write in terms of sines and cosines:

= \sin\theta \times \frac{\sin\theta}{\cos\theta} + \cos\theta

Combine over a common denominator:

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

which is the required result (AG).

Answer

The expression equals $\sec\theta$ , as required.

Final answer

Shown: the expression simplifies to sec(theta)

Detailed explanation

Walkthrough

The expression contains two square roots of expressions built from trig functions, so the natural first move is to convert what is under each root into a single squared function. The Pythagorean identity $1 + \cot^2\theta = \operatorname{cosec}^2\theta$ rearranges to $\operatorname{cosec}^2\theta - 1 = \cot^2\theta$ , so the first root is $\sqrt{\cot^2\theta}$ ; similarly $1 + \tan^2\theta = \sec^2\theta$ , so the second root is $\sqrt{\sec^2\theta}$ .

Taking a square root normally introduces a modulus sign, but here we are told $0 \leq \theta < \frac{\pi}{2}$ — the first quadrant — where every trig function is positive. So $\sqrt{\cot^2\theta} = \cot\theta$ and $\sqrt{\sec^2\theta} = \sec\theta$ with no ambiguity. The first fraction becomes $\frac{\sin\theta}{\cot\theta} = \sin\theta\tan\theta$ (since dividing by $\cot\theta$ multiplies by $\tan\theta$ ), and the second becomes $\frac{1}{\sec\theta} = \cos\theta$ .

Now write everything over one denominator: $\sin\theta\tan\theta = \frac{\sin^2\theta}{\cos\theta}$ , so adding $\cos\theta = \frac{\cos^2\theta}{\cos\theta}$ gives $\frac{\sin^2\theta + \cos^2\theta}{\cos\theta}$ . The fundamental identity $\sin^2\theta + \cos^2\theta = 1$ collapses the numerator to $1$ , leaving $\frac{1}{\cos\theta} = \sec\theta$ , exactly the printed target.

Key Takeaways

The three Pythagorean identities ( $\sin^2 + \cos^2 = 1$ , $1 + \tan^2 = \sec^2$ , $1 + \cot^2 = \operatorname{cosec}^2$ ) are the toolkit for simplifying anything involving $\operatorname{cosec}$ or $\sec$ .
$\sqrt{f^2(\theta)} = |f(\theta)|$ ; you may drop the modulus only when the question's interval guarantees positivity.
Writing everything in terms of $\sin\theta$ and $\cos\theta$ before combining fractions avoids most algebra errors.

Common Mistakes

Dropping the modulus sign without quoting or using the interval $0 \leq \theta < \frac{\pi}{2}$ — the mark scheme requires the range to justify this.
Omitting the argument $\theta$ throughout the working: the scheme states the last mark is not available if $\theta$ is persistently missing.
Misremembering the identity as $\operatorname{cosec}^2\theta - 1 = \tan^2\theta$ instead of $\cot^2\theta$ .
Sign slips when combining $\sin\theta\tan\theta + \cos\theta$ over the common denominator $\cos\theta$ .
Working backwards from $\sec\theta$ : for an AG part you must derive forward to the target, never start from it.

Things to Be Careful About

This is a "show that" (AG) item: every intermediate line must be shown — the scheme awards B1 for each stage and the final B1 depends on all three previous marks being earned.
Keep $\theta$ attached to every trig function at every line; persistent omission forfeits the final mark.
State explicitly why the roots simplify (first-quadrant positivity) rather than silently writing $\sqrt{\cot^2\theta} = \cot\theta$ .

Techniques used

apply the identity cosec^2 θ - 1 = cot^2 θapply the identity 1 + tan^2 θ = sec^2 θsimplify square roots of squared trig functions using the given range of θcombine fractions over a common denominator and use sin^2 + cos^2 = 1

(b)

Given that $\sec x = \alpha$ , where $\frac{3\pi}{2} < x \leq 2\pi$ , find $\sin x$ in terms of $\alpha$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Convert $\sec x = \alpha$ to $\cos x = \frac{1}{\alpha}$ , then use $\sin^2 x + \cos^2 x = 1$ . Since $\frac{3\pi}{2} < x \leq 2\pi$ is the fourth quadrant, $\sin x < 0$ , so take the negative root.

Working

\cos x = \frac{1}{\alpha}

Using $\sin^2 x + \cos^2 x = 1$ :

\sin^2 x = 1 - \left(\frac{1}{\alpha}\right)^2 = 1 - \frac{1}{\alpha^2}

so

\sin x = \pm\sqrt{1 - \frac{1}{\alpha^2}}

In the fourth quadrant ( $\frac{3\pi}{2} < x \leq 2\pi$ ), sine is negative, hence

\sin x = -\sqrt{1 - \frac{1}{\alpha^2}}

Answer

\sin x = -\sqrt{1 - \frac{1}{\alpha^2}}

Final answer

-√(1 - 1/α²)

Detailed explanation

Walkthrough

We are given $\sec x = \alpha$ and want $\sin x$ . The link between secant and sine runs through cosine: since $\sec x = \frac{1}{\cos x}$ , we get $\cos x = \frac{1}{\alpha}$ . Then the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ gives $\sin^2 x = 1 - \frac{1}{\alpha^2}$ , so $\sin x = \pm\sqrt{1 - \frac{1}{\alpha^2}}$ .

Choosing the sign is where the interval matters. The interval $\frac{3\pi}{2} < x \leq 2\pi$ is the fourth quadrant, where cosine is positive but sine is negative. So we must take the negative root: $\sin x = -\sqrt{1 - \frac{1}{\alpha^2}}$ .

(Equivalently, one may picture a right triangle with hypotenuse $\alpha$ and adjacent side $1$ , giving opposite $= \sqrt{\alpha^2 - 1}$ and $\sin x = -\frac{\sqrt{\alpha^2 - 1}}{\alpha}$ , which is the same answer.)

Key Takeaways

Reciprocal functions convert cleanly: $\sec x = \alpha \iff \cos x = \frac{1}{\alpha}$ .
The Pythagorean identity lets you move between any two of $\sin$ , $\cos$ , $\tan$ up to a sign.
The quadrant stated in the question exists precisely to fix that sign — always use it.

Common Mistakes

Giving $+\sqrt{1 - \frac{1}{\alpha^2}}$ by ignoring the fourth quadrant, where sine is negative — the A1 requires the negative sign.
Forgetting the $\pm$ at the squaring stage and only later realising a sign choice was needed.
Writing $\cos x = \alpha$ instead of $\frac{1}{\alpha}$ (confusing secant with cosine).
Algebra slips such as writing $1 - \frac{1}{\alpha^2}$ as $\frac{\alpha^2 - 1}{\alpha}$ instead of $\frac{\alpha^2 - 1}{\alpha^2}$ .

Things to Be Careful About

The final answer must carry the negative sign; the equivalent form $-\frac{\sqrt{\alpha^2 - 1}}{\alpha}$ is also accepted (oe).
Leave the answer in exact surd form in terms of $\alpha$ — no decimal evaluation is wanted.
Check consistency: in the fourth quadrant $\cos x > 0$ , which matches $\frac{1}{\alpha} > 0$ , i.e. $\alpha > 0$ ; the given interval already encodes this.

Techniques used

convert sec x to cos x via the reciprocal relationuse sin^2 x + cos^2 x = 1 to express sin x through cos xselect the correct sign from the given quadrant

Q11Medium-HardSeries

An arithmetic progression has common difference $d$ .
The 3rd term of this progression is 10.

(a)

Write down expressions for the 1st term and the 2nd term of this progression.
Give your answers in terms of $d$ only.

2M Similar

(b)

When each of the first 3 terms is squared, the sum of these squares is 140 .
There are two possible values for $d$ .

Using your answer to part (a), find the sum of the first 200 terms of the progression with the smaller value of $d$ .

7M Similar

DifficultyMedium-Hard

Worked solution

Approach

Using part (a), the first three terms are $10 - 2d$ , $10 - d$ and $10$ . Square each, set the sum equal to $140$ , solve the resulting quadratic for $d$ , take the smaller value ( $d = 4$ ), recover the first term, then use the sum formula $S_n = \frac{n}{2}\{2a + (n-1)d\}$ with $n = 200$ .

Working

Squaring each of the three terms and summing:

(10 - 2d)^2 + (10 - d)^2 + 10^2 = 140

Expanding:

(100 - 40d + 4d^2) + (100 - 20d + d^2) + 100 = 140

Collecting terms:

5d^2 - 60d + 300 = 140

5d^2 - 60d + 160 = 0

Dividing by 5 and factorising:

d^2 - 12d + 32 = 0 \quad\Rightarrow\quad (d - 4)(d - 8) = 0

d = 4 \quad \text{or} \quad d = 8

The smaller value is $d = 4$ . Then the first term is

a = 10 - 2(4) = 2

Sum of the first 200 terms:

S_{200} = \frac{200}{2}\left\{2(2) + 199 \times 4\right\} = 100\{4 + 796\} = 100 \times 800 = 80\,000

Answer

S_{200} = 80\,000

Final answer

80000

Detailed explanation

Walkthrough

First we carry forward the results from part (a): the three terms are $10 - 2d$ , $10 - d$ and $10$ . Squaring each term gives $(10-2d)^2$ , $(10-d)^2$ and $100$ , and their sum is stated to be $140$ . Expanding the squares using $(x-y)^2 = x^2 - 2xy + y^2$ produces a quadratic in $d$ alone. Simplifying to $5d^2 - 60d + 160 = 0$ and dividing by 5 gives $d^2 - 12d + 32 = 0$ , which factorises neatly as $(d-4)(d-8) = 0$ , so $d = 4$ or $d = 8$ . The question asks for the progression with the smaller value of $d$ , so we take $d = 4$ . Substituting back into the expression from part (a), the first term is $a = 10 - 2(4) = 2$ . Finally, the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}\{2a + (n-1)d\}$ ; with $n = 200$ , $a = 2$ and $d = 4$ this gives $S_{200} = 100\{4 + 796\} = 80\,000$ .

Key Takeaways

Squaring AP terms turns a linear condition into a quadratic equation in $d$ .
When a quadratic yields two valid roots, re-read the question to decide which one applies.
The sum formula $S_n = \frac{n}{2}\{2a + (n-1)d\}$ needs both the first term and the common difference — recover $a$ from the earlier part rather than guessing.

Common Mistakes

Forgetting to square the third term (the constant $100$ ), giving $5d^2 - 60d + 200 = 140$ and wrong roots.
Sign errors when expanding $(10 - 2d)^2$ or $(10 - d)^2$ .
Taking $d = 8$ instead of the smaller value $d = 4$ .
Using $a = 10$ (the third term) instead of the actual first term $2$ in the sum formula.
Writing $S_{200} = \frac{200}{2}\{2(2) + 200 \times 4\}$ — the last term correction uses $(n-1)d = 199 \times 4$ , not $200 \times 4$ .
Arithmetic slips in the final evaluation; the mark scheme requires the fully correct $80\,000$ .

Things to Be Careful About

Follow-through applies at several stages (the scheme says FT from their terms in (a), their quadratic, their $a$ and their $d$ ), but the final answer $80\,000$ must come from correct working.
The equation must be reduced to a solvable three-term quadratic form before factorising to earn the method marks.
Check the final answer's size: $S_{200}$ with positive terms should be large — $80\,000$ is consistent with an average term of about $400$ .

Techniques used

square each term and form an equationsolve a quadratic by factorisingselect the smaller rootapply the sum formula for an arithmetic progression

Q124MMediumPermutations and combinations

In this question $n \geq 6$ .

Use an algebraic method to show that $^{n}\mathrm{C}_{5} - {}^{n-1}\mathrm{C}_{5}$ can be written as $^{n-1}\mathrm{C}_{4}$ .

DifficultyMedium

Worked solution

Approach

Write both combinations in terms of factorials, take out the common factor $\dfrac{(n-1)!}{(n-6)!\,5!}$ , simplify what remains, and show the result equals $^{n-1}\mathrm{C}_{4}$ .

Working

{}^{n}\mathrm{C}_{5} - {}^{n-1}\mathrm{C}_{5} = \frac{n!}{(n-5)!\,5!} - \frac{(n-1)!}{(n-6)!\,5!}

Since $n! = n(n-1)!$ and $(n-5)! = (n-5)(n-6)!$ , factor out $\dfrac{(n-1)!}{(n-6)!\,5!}$ :

= \frac{(n-1)!}{(n-6)!\,5!}\left(\frac{n}{n-5} - 1\right)

Simplify the bracket:

\frac{n}{n-5} - 1 = \frac{n - (n-5)}{n-5} = \frac{5}{n-5}

So

{}^{n}\mathrm{C}_{5} - {}^{n-1}\mathrm{C}_{5} = \frac{(n-1)!}{(n-6)!\,5!} \times \frac{5}{n-5} = \frac{(n-1)!}{(n-6)!\,(n-5)\,4!}

Since $(n-6)!(n-5) = (n-5)!$ :

= \frac{(n-1)!}{(n-5)!\,4!} = {}^{n-1}\mathrm{C}_{4}

Answer

{}^{n}\mathrm{C}_{5} - {}^{n-1}\mathrm{C}_{5} = {}^{n-1}\mathrm{C}_{4}

Final answer

Shown: nC5 - (n-1)C5 = (n-1)C4

Detailed explanation

Walkthrough

The question asks for an algebraic proof that choosing 5 objects from $n$ and subtracting those choices that exclude one particular object leaves exactly the number of ways to choose 4 from the remaining $n-1$ — but we must prove it with factorials, not by argument alone.

First convert each combination to its factorial definition:

{}^{n}\mathrm{C}_{5} = \frac{n!}{(n-5)!\,5!}, \qquad {}^{n-1}\mathrm{C}_{5} = \frac{(n-1)!}{(n-6)!\,5!}

Both fractions share the building blocks $(n-1)!$ on top and $(n-6)!$ below, because $n! = n(n-1)!$ and $(n-5)! = (n-5)(n-6)!$ . Factoring out the common piece gives

\frac{(n-1)!}{(n-6)!\,5!}\left(\frac{n}{n-5} - 1\right)

The bracket simplifies over the common denominator $n-5$ :

\frac{n - (n-5)}{n-5} = \frac{5}{n-5}

Multiplying through, the extra factor of $5$ turns the $5!$ into $4!$ (since $5!/5 = 4!$ ), and the factor of $n-5$ combines with $(n-6)!$ back into $(n-5)!$ :

\frac{(n-1)!}{(n-6)!\,(n-5)\,4!} = \frac{(n-1)!}{(n-5)!\,4!}

This is precisely the factorial form of ${}^{n-1}\mathrm{C}_{4}$ , completing the argument. The condition $n \geq 6$ guarantees every factorial involved is defined.

Key Takeaways

The factorial definition ${}^{n}\mathrm{C}_{r} = \dfrac{n!}{(n-r)!\,r!}$ is the tool for proving identities between combinations algebraically.
Factorials telescope: $n! = n(n-1)!$ and $(k+1)! = (k+1)k!$ , so common factors can always be extracted.
The identity ${}^{n}\mathrm{C}_{r} - {}^{n-1}\mathrm{C}_{r} = {}^{n-1}\mathrm{C}_{r-1}$ is worth remembering; this proof establishes it for $r = 5$ .

Common Mistakes

Cancelling factorials incorrectly, e.g. treating $(n-5)!$ as if it were unrelated to $(n-6)!$ instead of using $(n-5)! = (n-5)(n-6)!$ .
Using the identity itself (i.e. quoting ${}^{n-1}\mathrm{C}_{4}$ properties) to simplify the left-hand side — the mark scheme explicitly disallows this; the proof must be self-contained via factorials.
Sign slips in the bracket: $\frac{n}{n-5} - 1 = \frac{5}{n-5}$ , not $\frac{n+5}{n-5}$ or similar.
Forgetting that $5!/5 = 4!$ when tidying the final fraction.
Not completing the final line: the mark scheme requires the last expression to be recognised as ${}^{n-1}\mathrm{C}_{4}$ , not left as an unevaluated fraction.

Things to Be Careful About

This is an "AG"-style show-that part: derive forward from the given left-hand side to the printed target; never start from ${}^{n-1}\mathrm{C}_{4}$ and work backwards.
Every intermediate algebraic line must be shown — the scheme awards M1 for substituting the factorial forms, M1 for factorising, M1 for simplifying the fraction, and A1 only for the completed argument ending at ${}^{n-1}\mathrm{C}_{4}$ .
The second and third M marks are dependent on earlier ones, so a wrong substitution early loses the whole chain even if later algebra is correct.
Keep all working exact in factorials; no decimals or numerical values of $n$ may be used since $n$ is general ( $n \geq 6$ ).

Techniques used

write combinations in factorial formfactorise the common factorialsimplify the resulting fractionrecognise the result as a combination

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