(a) A 4-digit number is to be formed using the digits 0, 2, 4, 5, 6 and 8. The 4-digit number must not start with 0. Any digit may be used at most once in the 4-digit number. (i) Find how many 4-digit numbers can be formed. (ii) Find how many even 4-digit numbers can be formed. (iii) Find how many 4-digit numbers that are divisible by 5 can be formed. (b) Solve the equation $(n + 1) \times {}^{n+1}\mathrm{C}_{12} = 33(n - 10) \times {}^n\mathrm{C}_{10}$.

(a)(i) 300 · (a)(ii) 252 · (a)(iii) 108 · (b) n = 65

(a) Solve the equation $3\sec 3x = \sqrt{3}\operatorname{cosec} 3x$ for $-120° \leq x \leq 120°$. (b) Solve the equation $2\cos\left(y + \frac{\pi}{3}\right)\sin\left(y + \frac{\pi}{3}\right) = \sin\left(y + \frac{\pi}{3}\right)$ for $0 \leq y < 2\pi$.

(a) x = -110, -50, 10, 70 (degrees) · (b) y = 0, 2π/3, 4π/3, 5π/3

4037/22 May/June 2025 — Additional Mathematics Solved Past Paper

Q14MMedium-EasyEquations, inequalities and graphs

Solve the inequality $|5x + 2| \geq 3$ .

DifficultyMedium-Easy

Worked solution

Approach

The inequality $\left|5x + 2\right| \geq 3$ means that $5x + 2$ is at least $3$ or at most $-3$ . Split into two linear inequalities and solve each.

Working

5x + 2 \geq 3 \quad \text{or} \quad 5x + 2 \leq -3

Solving the first:

5x \geq 1 \implies x \geq \frac{1}{5}

Solving the second:

5x \leq -5 \implies x \leq -1

Answer

x \leq -1 \quad \text{or} \quad x \geq \frac{1}{5}

Final answer

x <= -1 or x >= 1/5

Detailed explanation

Walkthrough

The modulus $\left|5x + 2\right|$ measures the distance of $5x + 2$ from $0$ , so $\left|5x + 2\right| \geq 3$ says that $5x + 2$ is at least $3$ units away from zero on either side. This gives two cases: $5x + 2 \geq 3$ (to the right) or $5x + 2 \leq -3$ (to the left). Each case is a simple linear inequality: subtracting $2$ gives $5x \geq 1$ , so $x \geq \frac{1}{5}$ , and $5x \leq -5$ , so $x \leq -1$ . Because the original inequality is 'greater than or equal to', both branches keep their inequality directions and the answer is the union of the two outer regions — the middle region $-1 < x < \frac{1}{5}$ is exactly where the modulus is less than $3$ .

Key Takeaways

$|u| \geq k$ splits into $u \geq k$ or $u \leq -k$ ; the 'greater than' case points outwards.
Critical values come from solving the boundary equations $5x + 2 = 3$ and $5x + 2 = -3$ .
An alternative valid route is squaring both sides: $(5x+2)^2 \geq 9$ , giving $25x^2 + 20x - 5 \geq 0$ , with the same critical values.

Common Mistakes

Reversing an inequality when dividing by a negative number — here we divide by positive $5$ , so no reversal is needed, but candidates often flip signs unnecessarily.
Writing the answer as $-1 \leq x \leq \frac{1}{5}$ (the inner region), which is the solution of $|5x+2| \leq 3$ , not $\geq 3$ .
Dropping the equality: the boundaries $x = -1$ and $x = \frac{1}{5}$ must be included since the inequality is $\geq$ .
Forgetting one of the two branches entirely.

Things to Be Careful About

The mark scheme awards M1 for setting up the two inequalities (or the squared form), A2 for both critical values $\frac{1}{5}$ and $-1$ , and A1 for the final combined answer in the correct directions — so the final line must show $x \leq -1$ or $x \geq \frac{1}{5}$ , not just the critical values.
Equivalent forms ('oe') are accepted, e.g. $x \in (-\infty, -1] \cup [\tfrac{1}{5}, \infty)$ .

Techniques used

split the modulus inequality into two linear inequalitiessolve each linear inequalitycombine the solution regions

Similar questions

Q2Medium-EasyCalculus

In this question, all lengths are in metres and time is in seconds.

A particle $P$ moves in a straight line such that its displacement $s$ from a fixed point $O$ at time $t$ is given by $s = (t - 4)^2(t - 1)$ for $t \geq 0$ .

(a)

On the axes, sketch the displacement–time graph of $P$ , stating the intercepts with the axes.

Fig. 1

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Expand $s = (t-4)^2(t-1)$ to find the $s$ -intercept, and read the roots from the factors to place the curve. The repeated factor $(t-4)^2$ means the curve touches the $t$ -axis at $t = 4$ .

Working

Expand the displacement:

s = (t^2 - 8t + 16)(t - 1) = t^3 - 9t^2 + 24t - 16

Intercepts:

$s$ -intercept: $t = 0$ gives $s = -16$ , so the curve meets the $s$ -axis at $(0, -16)$ .
$t$ -intercepts: $s = 0$ gives $(t-4)^2(t-1) = 0$ , so $t = 1$ (crossing) and $t = 4$ (tangency, since the factor is squared).

Shape: for $t \geq 0$ the cubic starts at $(0, -16)$ , rises through $(1, 0)$ to a local maximum in the first quadrant, falls back to touch the $t$ -axis at the minimum $(4, 0)$ , then rises into the first quadrant.

Answer

A cubic curve from $(0, -16)$ crossing the $t$ -axis at $t = 1$ , with a maximum in the first quadrant, then touching the $t$ -axis at $(4, 0)$ before rising. Intercepts: $s$ -axis at $(0, -16)$ ; $t$ -axis at $(1, 0)$ and $(4, 0)$ (tangent).

Final answer

Cubic from (0, -16) crossing the t-axis at t = 1, maximum in the first quadrant, touching the t-axis at (4, 0); intercepts (0, -16), (1, 0), (4, 0)

Detailed explanation

Walkthrough

The displacement is given in factorised form, $s = (t-4)^2(t-1)$ , which makes the roots easy to read: $s = 0$ when $t = 4$ or $t = 1$ . Crucially, $(t-4)^2$ is a squared factor, so at $t = 4$ the curve does not cross the axis — it touches it and turns back, giving a point of tangency (which is also a local minimum).

To find where the curve meets the $s$ -axis we set $t = 0$ : $s = (0-4)^2(0-1) = 16 \times (-1) = -16$ . To be sure of the overall shape we can expand: $s = t^3 - 9t^2 + 24t - 16$ , a positive cubic, so it rises steeply for large $t$ . Between the roots the curve is positive (both factors $(t-4)^2$ and $(t-1)$ have the same sign for $1 < t < 4$ ), so it bulges above the axis between $t = 1$ and $t = 4$ , forming a maximum in the first quadrant. Since $t \geq 0$ , only the part of the curve from $t = 0$ onwards is drawn.

Key Takeaways

A repeated root of even multiplicity means the curve touches the axis (tangency) rather than crossing it.
The $s$ -intercept is found by substituting $t = 0$ into the displacement.
For $t \geq 0$ , only the right-hand portion of the cubic is sketched.

Common Mistakes

Drawing the curve crossing the $t$ -axis at $t = 4$ instead of touching it — the mark scheme explicitly requires "a point of tangency to the $t$ -axis at the minimum point".
Forgetting the negative $s$ -intercept $-16$ and starting the curve at the origin.
Omitting the local maximum between $t = 1$ and $t = 4$ ; the scheme requires "maximum in the first quadrant".
Sketching the curve for negative $t$ when the domain is $t \geq 0$ .

Things to Be Careful About

Both marks depend on attempting the correct cubic shape first; intercepts alone score nothing without the shape ("must have attempted the correct shape").
State the intercepts explicitly as coordinates or values: $(0, -16)$ , $(1, 0)$ , $(4, 0)$ .
The curve must meet the $s$ -axis — start the sketch at $t = 0$ , not to the right of it.

Techniques used

expand the cubic displacement expressionfind the s-intercept by setting t = 0identify a repeated root giving tangency to the t-axissketch the cubic shape for t >= 0

(b)

Find an expression for the velocity, $v$ , of $P$ .
Give your answer in a factorised form.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Velocity is the derivative of displacement. Differentiate $s = (t-4)^2(t-1)$ by the product rule, then factorise.

Working

Using the product rule with $u = (t-4)^2$ and $w = t - 1$ :

v = \frac{\mathrm{d}s}{\mathrm{d}t} = 2(t-4)(t-1) + (t-4)^2(1)

Factorise by taking out the common factor $(t-4)$ :

v = (t-4)\left[2(t-1) + (t-4)\right] = (t-4)(3t - 6) = 3(t-4)(t-2)

(Equivalently, expanding $s = t^3 - 9t^2 + 24t - 16$ and differentiating gives $v = 3t^2 - 18t + 24 = 3(t-4)(t-2)$ .)

Answer

v = 3(t-4)(t-2)

Final answer

v = 3(t - 4)(t - 2)

Detailed explanation

Walkthrough

Velocity is defined as the rate of change of displacement, so $v = \frac{\mathrm{d}s}{\mathrm{d}t}$ . The displacement is a product of two factors of $t$ , so the product rule applies: differentiate $(t-4)^2$ to get $2(t-4)$ and multiply by the untouched $(t-1)$ , then add $(t-4)^2$ times the derivative of $(t-1)$ , which is $1$ .

The result $2(t-4)(t-1) + (t-4)^2$ has a common factor of $(t-4)$ ; extracting it leaves $2(t-1) + (t-4) = 2t - 2 + t - 4 = 3t - 6 = 3(t-2)$ . So the velocity factorises neatly as $3(t-4)(t-2)$ .

Alternatively, expanding the cubic first ( $s = t^3 - 9t^2 + 24t - 16$ ) and differentiating term by term gives $v = 3t^2 - 18t + 24$ , which factorises to the same answer. The question asks for factorised form, so the final answer must be left as $3(t-4)(t-2)$ .

Key Takeaways

$v = \frac{\mathrm{d}s}{\mathrm{d}t}$ : velocity is the derivative of displacement with respect to time.
The product rule: $(uw)' = u'w + uw'$ .
Factorising before or after differentiating both work; the common-factor route is often quicker.

Common Mistakes

Forgetting the second term of the product rule (writing only $2(t-4)(t-1)$ ).
Leaving the answer unexpanded/unfactorised when the question demands factorised form.
Sign slips when differentiating $(t-4)^2$ via the chain rule.

Things to Be Careful About

The question explicitly says "Give your answer in a factorised form", so $3t^2 - 18t + 24$ alone would not earn the final accuracy mark.
Both the product-rule route and the expand-then-differentiate route are accepted by the mark scheme; either earns the method mark.

Techniques used

apply the product rule to differentiate the displacementexpand the cubic and differentiate term by termfactorise the resulting quadratic

(c)

On the axes, sketch the velocity–time graph of $P$ , stating the intercepts with the axes.

Fig. 2

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The velocity $v = 3(t-4)(t-2)$ is an upward-opening parabola with roots at $t = 2$ and $t = 4$ ; find the $v$ -intercept and sketch for $t \geq 0$ .

Working

$v$ -intercept: $t = 0$ gives $v = 3(-4)(-2) = 24$ , so the curve meets the $v$ -axis at $(0, 24)$ .
$t$ -intercepts: $v = 0$ gives $t = 2$ and $t = 4$ .

Since the coefficient of $t^2$ is positive ( $3$ ), the parabola opens upwards. Between $t = 2$ and $t = 4$ it dips below the $t$ -axis, so the minimum lies in the fourth quadrant (at $t = 3$ , $v = 3(-1)(1) = -3$ ).

Shape: from $(0, 24)$ the curve falls, crosses the $t$ -axis at $t = 2$ , reaches its minimum in the fourth quadrant, crosses again at $t = 4$ , and rises into the first quadrant.

Answer

An upward parabola from $(0, 24)$ crossing the $t$ -axis at $t = 2$ and $t = 4$ with its minimum in the fourth quadrant. Intercepts: $v$ -axis at $(0, 24)$ ; $t$ -axis at $(2, 0)$ and $(4, 0)$ .

Final answer

Upward parabola from (0, 24) crossing the t-axis at t = 2 and t = 4, minimum in the fourth quadrant; intercepts (0, 24), (2, 0), (4, 0)

Detailed explanation

Walkthrough

From part (b) the velocity is $v = 3(t-4)(t-2)$ , a quadratic with a positive leading coefficient, so its graph is a parabola opening upwards. The factors give the $t$ -intercepts immediately: $v = 0$ at $t = 2$ and $t = 4$ . The $v$ -intercept comes from setting $t = 0$ : $v = 3(-4)(-2) = 24$ .

Because the parabola opens upwards and both roots are positive, the section of the curve between the roots (for $2 < t < 4$ ) lies below the $t$ -axis — this is where the minimum sits, in the fourth quadrant (at $t = 3$ , $v = -3$ ). For $t \geq 0$ the curve starts high on the $v$ -axis at $24$ , descends through $(2, 0)$ , bottoms out below the axis, and climbs back through $(4, 0)$ into positive velocity.

Key Takeaways

The sign of the leading coefficient determines whether a parabola opens up or down.
Factorised form gives the roots directly; substituting $t = 0$ gives the vertical intercept.
A positive quadratic with two positive roots dips below the axis between them, placing the minimum in the fourth quadrant.

Common Mistakes

Drawing a downward parabola (misreading the sign of $3t^2$ ).
Giving the wrong $v$ -intercept by a sign slip in $3(-4)(-2)$ .
Placing the minimum above the axis or between the wrong roots.
Sketching for negative $t$ when the domain is $t \geq 0$ .

Things to Be Careful About

The mark scheme requires the correct quadratic shape first — intercepts alone do not score.
State all intercepts: $(0, 24)$ on the $v$ -axis, $(2, 0)$ and $(4, 0)$ on the $t$ -axis.
The curve must meet the $v$ -axis, so begin the sketch at $t = 0$ .

Techniques used

identify the quadratic from the factorised velocityfind the v-intercept by setting t = 0locate the t-intercepts from the factorssketch an upward parabola with minimum below the t-axis

(d)

Find an expression for the acceleration, $a$ , of $P$ .

1M Similar

(e)

On the axes, sketch the acceleration–time graph of $P$ , stating the intercepts with the axes.

Fig. 3

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The acceleration $a = 6t - 18$ is linear, so its graph is a straight line. Find the two intercepts and draw the line for $t \geq 0$ .

Working

$a$ -intercept: $t = 0$ gives $a = -18$ , so the line meets the $a$ -axis at $(0, -18)$ .
$t$ -intercept: $a = 0$ gives $6t = 18$ , so $t = 3$ .

The gradient is $6 > 0$ , so the line rises from $(0, -18)$ , crosses the $t$ -axis at $(3, 0)$ , and continues into the first quadrant.

Answer

A ruled straight line of positive gradient from $(0, -18)$ crossing the $t$ -axis at $(3, 0)$ and rising into the first quadrant. Intercepts: $a$ -axis at $(0, -18)$ ; $t$ -axis at $(3, 0)$ .

Final answer

Ruled straight line from (0, -18) crossing the t-axis at t = 3 with positive gradient; intercepts (0, -18), (3, 0)

Detailed explanation

Walkthrough

From part (d) the acceleration is $a = 6t - 18$ , which is of the form $mt + c$ with gradient $6$ and intercept $-18$ . Its graph is therefore a straight line. Setting $t = 0$ gives the $a$ -intercept $-18$ ; setting $a = 0$ gives $t = 3$ . Since the gradient is positive, the line climbs from $(0, -18)$ below the axis, crosses at $t = 3$ , and continues upward into the first quadrant. Only the part for $t \geq 0$ is drawn, and the mark scheme requires the line to be ruled (drawn with a straight edge).

Key Takeaways

A linear function of $t$ graphs as a straight line whose gradient and intercept are read directly from $mt + c$ .
The $t$ -intercept of $a = mt + c$ is $-c/m$ .

Common Mistakes

Plotting the $t$ -intercept at $t = 18$ or $t = -3$ instead of $t = 3$ (sign or division slip).
Drawing a freehand curve instead of a ruled straight line.
Omitting the negative $a$ -intercept $-18$ .

Things to Be Careful About

The mark scheme awards follow-through marks from the candidate's own linear acceleration: the $t$ -intercept must be $-(-18) \div 6 = 3$ and the $a$ -intercept $-18$ .
The line must meet the $a$ -axis and cross the $t$ -axis, and both intercepts must be stated.
This part is worth 3 marks — shape (ruled line, positive gradient), $t$ -intercept, and $a$ -intercept each carry a mark.

Techniques used

recognise a linear acceleration functionfind the a-intercept by setting t = 0find the t-intercept by solving a = 0sketch a ruled straight line

Q34MMedium-EasyFunctions

Functions $\mathrm{f}$ and $\mathrm{g}$ are such that

\mathrm{f}(x) = \frac{3x}{x + 4} \quad \text{for } x > 0

\mathrm{g}(x) = \sqrt{x + 2} \quad \text{for } x > -2

Solve the equation $\mathrm{fg}(x) = 1$ .

DifficultyMedium-Easy

Worked solution

Approach

The composite function $\mathrm{fg}(x)$ means apply $\mathrm{g}$ first, then $\mathrm{f}$ : substitute $\sqrt{x+2}$ into $\mathrm{f}$ . Set the result equal to $1$ and solve.

Working

\mathrm{fg}(x) = \frac{3\sqrt{x+2}}{\sqrt{x+2} + 4}

Setting this equal to $1$ :

\frac{3\sqrt{x+2}}{\sqrt{x+2} + 4} = 1

Multiply both sides by $\sqrt{x+2} + 4$ :

3\sqrt{x+2} = \sqrt{x+2} + 4

2\sqrt{x+2} = 4 \quad \Rightarrow \quad \sqrt{x+2} = 2

Squaring both sides:

x + 2 = 4 \quad \Rightarrow \quad x = 2

This satisfies the domain condition $x > 0$ , so it is valid.

Answer

x = 2

Final answer

x = 2

Detailed explanation

Walkthrough

The notation $\mathrm{fg}(x)$ means $\mathrm{f}\big(\mathrm{g}(x)\big)$ : first apply $\mathrm{g}$ , then apply $\mathrm{f}$ to the result. Since $\mathrm{g}(x) = \sqrt{x+2}$ , we feed $\sqrt{x+2}$ into $\mathrm{f}$ in place of $x$ , giving $\dfrac{3\sqrt{x+2}}{\sqrt{x+2}+4}$ . Setting this equal to $1$ and multiplying both sides by the denominator removes the fraction: $3\sqrt{x+2} = \sqrt{x+2} + 4$ . Collecting like terms gives $2\sqrt{x+2} = 4$ , so $\sqrt{x+2} = 2$ . Squaring gives $x + 2 = 4$ , hence $x = 2$ . Finally we check that $x = 2$ lies in the domain $x > 0$ , which it does, so the solution is valid. (An alternative accepted route is to find $\mathrm{f}^{-1}(x) = \dfrac{4x}{3-x}$ , set $\mathrm{g}(x) = \mathrm{f}^{-1}(1) = 2$ , and solve $\sqrt{x+2} = 2$ .)

Key Takeaways

$\mathrm{fg}(x)$ means apply $\mathrm{g}$ first, then $\mathrm{f}$ — order matters in composites.
An equation with an unknown inside an algebraic fraction can be solved by multiplying through by the denominator.
Squaring a square-root equation can introduce extraneous roots, so always check the answer against the domain of the original functions.

Common Mistakes

Computing $\mathrm{gf}(x)$ instead of $\mathrm{fg}(x)$ — the order of composition is reversed.
Forgetting to substitute $\sqrt{x+2}$ for every occurrence of $x$ in $\mathrm{f}$ (e.g. leaving the denominator as $x+4$ ).
Squaring incorrectly or forgetting to square at all after reaching $\sqrt{x+2} = 2$ .
Not checking the solution against the domain restriction $x > 0$ .
The mark scheme awards the final mark "nfww" — no marks without correct working — so each step must be shown.

Things to Be Careful About

The final answer must come from valid working ("nfww"): show the composite being formed and the clearing of the fraction.
Keep the domain conditions in mind: $\mathrm{f}$ requires $x > 0$ and $\mathrm{g}$ requires $x > -2$ ; the solution $x = 2$ satisfies both.
Give the exact value $x = 2$ ; no rounding issues arise here.

Techniques used

form the composite function fg(x)solve the resulting equation by clearing the algebraic fractionsquare to remove the square root and check the domain

Similar questions

Q4MediumCalculusTrigonometryStraight-line graphs

(a)

Given that $y = 4\sin 2x \cos 2x$ , find the value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ when $x = \frac{\pi}{6}$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $y = 4\sin 2x \cos 2x$ by the product rule (with chain rule for each factor), simplify with the double-angle identity $\sin 4x = 2\sin 2x\cos 2x$ , then substitute $x = \frac{\pi}{6}$ .

Working

\frac{\mathrm{d}y}{\mathrm{d}x} = 4\left(\cos 2x \cdot 2\cos 2x + \sin 2x \cdot (-2\sin 2x)\right)

= 8\cos^2 2x - 8\sin^2 2x = 8\left(\cos^2 2x - \sin^2 2x\right) = 8\cos 4x

At $x = \frac{\pi}{6}$ :

\left[\frac{\mathrm{d}y}{\mathrm{d}x}\right]_{x=\frac{\pi}{6}} = 8\cos\left(4 \times \frac{\pi}{6}\right) = 8\cos\frac{2\pi}{3} = 8 \times \left(-\frac{1}{2}\right) = -4

Answer

-4

Final answer

-4

Detailed explanation

Walkthrough

The function is a product of two factors, $4\sin 2x$ and $\cos 2x$ (or equivalently $\sin 2x$ and $4\cos 2x$ ), so the product rule applies: differentiate one factor while holding the other fixed, then swap, and add. Each factor itself contains $2x$ , so the chain rule gives an extra factor of 2 in each derivative — this is where marks are commonly lost if the inner derivative is forgotten.

The result $8\cos^2 2x - 8\sin^2 2x$ simplifies neatly because $\cos^2 A - \sin^2 A = \cos 2A$ ; here $A = 2x$ , so the derivative becomes $8\cos 4x$ . This simplification is not compulsory but makes the substitution trivial.

Substituting $x = \frac{\pi}{6}$ gives angle $4x = \frac{2\pi}{3}$ , whose cosine is $-\frac{1}{2}$ (second-quadrant angle), so the derivative equals $-4$ .

Key Takeaways

The product rule combined with the chain rule handles products like $\sin kx \cos kx$ .
Recognising $\cos^2 A - \sin^2 A = \cos 2A$ collapses the answer to a single term.
Exact values such as $\cos \frac{2\pi}{3} = -\frac{1}{2}$ must be known; no calculator rounding is needed.

Common Mistakes

Forgetting the inner derivative of $2x$ (writing $\frac{\mathrm{d}}{\mathrm{d}x}\sin 2x = \cos 2x$ ), which loses the structure mark.
Sign slips when differentiating $\cos 2x$ : its derivative is $-2\sin 2x$ .
Mis-evaluating $\cos \frac{2\pi}{3}$ as positive $\frac{1}{2}$ instead of $-\frac{1}{2}$ .
Leaving the answer as $8\cos \frac{2\pi}{3}$ without evaluating it.

Things to Be Careful About

The mark scheme requires the full product-rule structure before any credit for the final value, so show the expanded form explicitly.
Work in radians throughout — the question gives $x$ in radians.
The scheme allows follow-through only on derivatives of the form $a\cos^2 2x - b\sin^2 2x$ with positive integers $a, b$ , so keep coefficients integral.

Techniques used

apply the product rule to differentiate a product of trigonometric functionsdifferentiate sin of a multiple angle using the chain rulesimplify using the double-angle identityevaluate the derivative at an exact radian value

(b)

A curve has equation $y = 4\sin 2x \cos 2x$ .

The normal to the curve at the point where $x = \frac{\pi}{6}$ meets the $x$ -axis at the point $P$ .

Find the exact coordinates of $P$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use the derivative $\frac{\mathrm{d}y}{\mathrm{d}x} = -4$ from part (a). The normal's gradient is the negative reciprocal, $\frac{1}{4}$ . Find the point on the curve at $x = \frac{\pi}{6}$ , write the normal's equation, and set $y = 0$ to locate $P$ .

Working

At $x = \frac{\pi}{6}$ :

y = 4\sin\frac{\pi}{3}\cos\frac{\pi}{3} = 4 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} = \sqrt{3}

Gradient of normal:

m_{\text{normal}} = \frac{-1}{-4} = \frac{1}{4}

Equation of the normal through $\left(\frac{\pi}{6}, \sqrt{3}\right)$ :

y - \sqrt{3} = \frac{1}{4}\left(x - \frac{\pi}{6}\right)

At $P$ , $y = 0$ :

0 - \sqrt{3} = \frac{1}{4}\left(x - \frac{\pi}{6}\right)

x - \frac{\pi}{6} = -4\sqrt{3} \quad\Rightarrow\quad x = \frac{\pi}{6} - 4\sqrt{3}

Answer

P = \left(\frac{\pi}{6} - 4\sqrt{3},\; 0\right)

Final answer

(π/6 − 4√3, 0)

Detailed explanation

Walkthrough

This part builds directly on part (a): the tangent gradient there was $-4$ , so the normal — being perpendicular to the tangent — has gradient $\frac{-1}{-4} = \frac{1}{4}$ .

Next we need the actual point on the curve. Substituting $x = \frac{\pi}{6}$ into $y = 4\sin 2x\cos 2x$ uses the special values $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{3} = \frac{1}{2}$ , giving $y = \sqrt{3}$ exactly.

With a point and a gradient, the point–slope form gives the normal's equation: $y - \sqrt{3} = \frac{1}{4}(x - \frac{\pi}{6})$ . Since $P$ lies on the $x$ -axis, set $y = 0$ and solve: multiplying both sides by 4 gives $x - \frac{\pi}{6} = -4\sqrt{3}$ , so $x = \frac{\pi}{6} - 4\sqrt{3}$ .

Key Takeaways

A normal's gradient is the negative reciprocal of the tangent's gradient: $m_n = -\frac{1}{m_t}$ .
Point–slope form $y - y_1 = m(x - x_1)$ is the quickest route to a line's equation.
An $x$ -intercept is found by setting $y = 0$ ; a $y$ -intercept by setting $x = 0$ .
Exact answers mixing $\pi$ and surds are expected on this paper.

Common Mistakes

Using the tangent gradient $-4$ instead of the normal gradient $\frac{1}{4}$ .
Sign error in the negative reciprocal, e.g. writing $-\frac{1}{4}$ instead of $+\frac{1}{4}$ .
Evaluating $y$ incorrectly, e.g. forgetting that $2x = \frac{\pi}{3}$ when substituting.
Giving a decimal approximation instead of the exact form $\left(\frac{\pi}{6} - 4\sqrt{3}, 0\right)$ .
Dropping the $\frac{\pi}{6}$ term and answering $(-4\sqrt{3}, 0)$ .

Things to Be Careful About

The final accuracy mark depends on part (a) being fully correct (or a correct derivative stated here), so carry forward $-4$ carefully.
The question demands exact coordinates — leave $\pi$ and $\sqrt{3}$ unevaluated; decimals score nothing for the final mark.
Keep radians throughout; the coordinates mix $\pi$ and surds precisely because the working stays exact.

Techniques used

evaluate the curve's y-coordinate at a given xuse the negative reciprocal to obtain the normal gradientform the equation of the normal from a point and gradientfind the x-intercept by setting y = 0

Q5MediumPermutations and combinations

(a)

A 4-digit number is to be formed using the digits 0, 2, 4, 5, 6 and 8.
The 4-digit number must not start with 0.
Any digit may be used at most once in the 4-digit number.

5M

(i)

Find how many 4-digit numbers can be formed.

1M Similar

(ii)

Find how many even 4-digit numbers can be formed.

2M Similar

DifficultyMedium

Worked solution

Approach

An even number must end in $0, 2, 4, 6$ or $8$ . Split into two cases: ending in $0$ (which frees up all five non-zero digits for the front) and ending in $2, 4, 6$ or $8$ (where the front digit cannot be $0$ or the chosen units digit).

Working

Case 1: ends in $0$ . The first three places are filled from the five non-zero digits:

5 \times 4 \times 3 = 60

Case 2: ends in $2, 4, 6$ or $8$ . Choose the units digit ( $4$ ways). The first digit cannot be $0$ or the units digit, so $4$ ways; then $4$ and $3$ ways remain for the middle two places:

4 \times 4 \times 4 \times 3 = 192

Total:

60 + 192 = 252

Answer

252

Final answer

252

Detailed explanation

Walkthrough

Evenness depends only on the last digit, so we fix the units digit first. If it is $0$ , the leading-digit problem disappears entirely: any of the five non-zero digits can lead, giving $5 \times 4 \times 3 = 60$ numbers. If instead the units digit is one of $2, 4, 6, 8$ (four choices), the first digit now has two forbidden values — $0$ and the digit already used at the end — leaving $4$ choices, then $4$ and $3$ for the middle positions, giving $4 \times 4 \times 4 \times 3 = 192$ . The two cases cannot overlap, so adding them is valid.

Key Takeaways

When a condition interacts with the "no leading zero" rule, split into cases so each case has a clean count.
Disjoint cases may be added directly.

Common Mistakes

Counting $5 \times 5 \times 4 \times 4$ by treating every even units digit like the $0$ case — this double-counts numbers starting with $0$ .
Forgetting that when the units digit is fixed, the first digit loses two options (zero and the used digit).
Taking only one of the two cases.

Things to Be Careful About

The scheme's alternative route (all possible minus those ending in $5$ : $300 - 48 = 252$ ) also earns full marks.
The final answer $252$ is cao.

Techniques used

split into disjoint cases by the final digitcount arrangements avoiding repeated digitsadd the case counts

(iii)

Find how many 4-digit numbers that are divisible by 5 can be formed.

2M Similar

(b)

Solve the equation $(n + 1) \times {}^{n+1}\mathrm{C}_{12} = 33(n - 10) \times {}^n\mathrm{C}_{10}$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Write both binomial coefficients in factorial form, cancel the common factors of $n!$ , and simplify until the equation reduces to $(n+1)^2 = 33 \times 12 \times 11$ .

Working

(n + 1) \times {}^{n+1}\mathrm{C}_{12} = 33(n - 10) \times {}^{n}\mathrm{C}_{10}

(n + 1) \cdot \frac{(n+1)!}{12!(n-11)!} = 33(n - 10) \cdot \frac{n!}{10!(n-10)!}

Divide both sides by $n!$ (using $(n+1)! = (n+1)n!$ ):

\frac{(n+1)^2}{12!(n-11)!} = \frac{33(n-10)}{10!(n-10)!}

Since $12! = 12 \times 11 \times 10!$ and $(n-10)! = (n-10)(n-11)!$ :

\frac{(n+1)^2}{12 \times 11 \times 10! \,(n-11)!} = \frac{33(n-10)}{10!\,(n-10)(n-11)!}

Cancel $10!$ and $(n-11)!$ from both sides:

\frac{(n+1)^2}{12 \times 11} = \frac{33(n-10)}{n-10} = 33

so

(n+1)^2 = 33 \times 12 \times 11 = 4356

Taking the positive square root (since $n + 1 > 0$ for the combinations to be defined):

n + 1 = 66 \quad \Rightarrow \quad n = 65

Answer

n = 65

Final answer

n = 65

Detailed explanation

Walkthrough

The equation mixes binomial coefficients with algebraic factors, so the first job is to rewrite ${}^{n+1}\mathrm{C}_{12}$ and ${}^{n}\mathrm{C}_{10}$ as factorials. Expanding $(n+1)!$ as $(n+1) \times n!$ lets us cancel $n!$ across the equation, and expanding $12! = 12 \times 11 \times 10!$ together with $(n-10)! = (n-10)(n-11)!$ lets us cancel $10!$ and $(n-11)!$ too. What survives is remarkably clean: $(n+1)^2 = 33 \times 12 \times 11 = 4356$ . Since $n$ must be large enough for the combinations to exist, $n+1$ is positive, so we take the positive root $n + 1 = 66$ , giving $n = 65$ . Note the negative root would be rejected anyway because $n - 10$ appears on both sides and the original coefficients require $n \geq 12$ .

Key Takeaways

Converting ${}^{n}\mathrm{C}_r$ to factorials is the standard way to solve equations involving binomial coefficients.
Systematic cancellation of shared factorial blocks turns a messy equation into a simple one.
Domain restrictions ( $n \geq 12$ here) select the valid root.

Common Mistakes

Writing ${}^{n+1}\mathrm{C}_{12}$ with the wrong factorial pattern, e.g. misplacing the $(n-11)!$ term.
Failing to spot that $(n-10)$ cancels, and trying to expand everything instead.
Keeping the negative root $n = -67$ , which is invalid since binomial coefficients need $n \geq 12$ .
Arithmetic slips in $33 \times 12 \times 11$ ; it equals $4356$ , and $\sqrt{4356} = 66$ .

Things to Be Careful About

The scheme marks the simplified equation $(n+1)^2 = 33 \times 12 \times 11$ as B2 nfww — it must come from correct working, not backwards substitution.
The final B1 for $n = 65$ is dependent on earning the B2, and $65$ must be the only solution given.
Check the domain: $n = 65$ comfortably satisfies $n \geq 12$ .

Techniques used

express binomial coefficients in factorialssimplify using cancellation of common factorial factorssolve the resulting quadratic by square-rooting

Q66MMediumCalculus

The volume, $V$ , of a sphere is increasing at the constant rate of $2\pi\,\text{cm}^3\,\text{s}^{-1}$ .
Find the rate of change of the surface area, $S$ , of this sphere when the volume of the sphere is $36\pi\,\text{cm}^3$ .

DifficultyMedium

Worked solution

Approach

The volume $V$ , radius $r$ and surface area $S$ of the sphere are all functions of time $t$ . Use the chain rule to connect $\frac{\mathrm{d}V}{\mathrm{d}t}$ to $\frac{\mathrm{d}r}{\mathrm{d}t}$ , then $\frac{\mathrm{d}S}{\mathrm{d}t}$ to $\frac{\mathrm{d}r}{\mathrm{d}t}$ .

Working

First find the radius when $V = 36\pi$ :

V = \frac{4}{3}\pi r^3 \implies \frac{4}{3}\pi r^3 = 36\pi \implies r^3 = 27 \implies r = 3

Differentiate both formulas with respect to $r$ :

\frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2, \qquad \frac{\mathrm{d}S}{\mathrm{d}r} = 8\pi r

At $r = 3$ :

\frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi(9) = 36\pi, \qquad \frac{\mathrm{d}S}{\mathrm{d}r} = 8\pi(3) = 24\pi

Using the chain rule with $\frac{\mathrm{d}V}{\mathrm{d}t} = 2\pi$ :

\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \times \frac{\mathrm{d}r}{\mathrm{d}V} = 2\pi \times \frac{1}{36\pi} = \frac{1}{18}

Then:

\frac{\mathrm{d}S}{\mathrm{d}t} = \frac{\mathrm{d}S}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t} = 24\pi \times \frac{1}{18} = \frac{4\pi}{3}

Answer

\frac{\mathrm{d}S}{\mathrm{d}t} = \frac{4\pi}{3} \approx 4.19\,\text{cm}^2\,\text{s}^{-1}

Final answer

4.19 cm^2 s^-1 (exactly 4π/3)

Detailed explanation

Walkthrough

The question gives a rate ( $\frac{\mathrm{d}V}{\mathrm{d}t} = 2\pi$ ) and asks for another rate ( $\frac{\mathrm{d}S}{\mathrm{d}t}$ ), but there is no direct formula linking $S$ to $V$ . The bridge is the radius: both $V$ and $S$ are simple functions of $r$ , so we go through $\frac{\mathrm{d}r}{\mathrm{d}t}$ using the chain rule.

First we need the actual value of $r$ at the moment in question. Setting $\frac{4}{3}\pi r^3 = 36\pi$ gives $r^3 = 27$ , so $r = 3$ cm. This value is needed because all the derivatives below depend on it.

Next differentiate the two standard formulas:

Volume: $V = \frac{4}{3}\pi r^3$ gives $\frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2$ , which at $r=3$ is $36\pi$ .
Surface area: $S = 4\pi r^2$ gives $\frac{\mathrm{d}S}{\mathrm{d}r} = 8\pi r$ , which at $r=3$ is $24\pi$ .

The chain rule then links everything. Since $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t}$ , we get $\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{2\pi}{36\pi} = \frac{1}{18}$ cm per second — the radius is growing slowly because the sphere is already large.

Finally, $\frac{\mathrm{d}S}{\mathrm{d}t} = \frac{\mathrm{d}S}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t} = 24\pi \times \frac{1}{18} = \frac{4\pi}{3} \approx 4.19\,\text{cm}^2\,\text{s}^{-1}$ .

Key Takeaways

Related-rates problems are solved by chaining derivatives through an intermediate variable (here $r$ ).
You must evaluate every derivative at the specific instant given — here found from the volume condition.
Knowing the standard results $V = \frac{4}{3}\pi r^3$ and $S = 4\pi r^2$ and their derivatives is essential.

Common Mistakes

Forgetting to find $r = 3$ first and leaving answers in terms of $r$ — the mark scheme requires the numerical value "stated or very clearly implied" (nfww).
Confusing $\frac{\mathrm{d}V}{\mathrm{d}r}$ with $\frac{\mathrm{d}r}{\mathrm{d}V}$ when applying the chain rule — the reciprocals must be handled correctly.
Using $S = 4\pi r^2$ but differentiating incorrectly, or mixing up which derivative multiplies which.
Giving a decimal without showing the exact substitution $24\pi \times \frac{1}{18}$ — method marks require the working to be visible.
Rounding too early; the accepted decimal forms are $4.19$ or $4.188$ to $4.1888$ .

Things to Be Careful About

The answer may be left as the exact form $\frac{4\pi}{3}$ (isw applies after that), or as a decimal rounding to $4.19$ .
Every mark-scheme step must appear: the value $r=3$ , both derivatives with their values, the correct chain-rule statement including $\frac{\mathrm{d}S}{\mathrm{d}t}$ , and the final multiplication.
Units: the rate of change of surface area is in $\text{cm}^2\,\text{s}^{-1}$ since $S$ is an area.
Answers reached from wrong working score nothing (nfww on $r = 3$ ).

Techniques used

find the radius from the volume of a spheredifferentiate the sphere volume and surface area formulas with respect to rapply the chain rule to link dV/dt, dr/dt and dS/dtevaluate the derivatives at the given instant

Similar questions

Q7MediumSeriesLogarithmic and exponential functions

The first three terms of an arithmetic progression can be written as

2\ln(x^3), \quad 5\ln(x^2), \quad 2\ln(x^7).

(a)

Given that $x > 1$ , find the least number of terms for the sum of this progression to be greater than $43\ln(x^{24})$ .

6M Similar

DifficultyMedium

Worked solution

Approach

Use $\ln(x^k) = k\ln x$ to write every term as a multiple of $\ln x$ , identify $a$ and $d$ of the AP, then set up the sum-to- $n$ -terms inequality.

Working

Using the laws of logarithms:

2\ln(x^3) = 6\ln x, \qquad 5\ln(x^2) = 10\ln x, \qquad 2\ln(x^7) = 14\ln x

So the AP has first term $a = 6\ln x$ and common difference $d = 4\ln x$ .

The sum of $n$ terms must exceed $43\ln(x^{24}) = 43 \times 24 \ln x = 1032\ln x$ :

\frac{n}{2}\left(12\ln x + (n-1) \times 4\ln x\right) > 1032\ln x

Since $x > 1$ , $\ln x > 0$ , so divide both sides by $\ln x$ :

n\left(6 + 2(n-1)\right) > 1032

2n^2 + 4n - 1032 > 0

Divide by 2 and solve the quadratic equation $n^2 + 2n - 516 = 0$ :

n = \frac{-2 + \sqrt{4 + 2064}}{2} = \frac{-2 + \sqrt{2068}}{2} \approx \frac{-2 + 45.48}{2} \approx 21.74

The quadratic is positive for $n > 21.74$ , so the least whole number of terms is $n = 22$ .

Answer

n = 22

Final answer

22

Detailed explanation

Walkthrough

Each term is a logarithm of a power of $x$ , so the power law $\ln(x^k) = k\ln x$ converts all three terms into multiples of one quantity, $\ln x$ : they become $6\ln x$ , $10\ln x$ , $14\ln x$ . The differences are both $4\ln x$ , confirming this really is an arithmetic progression with first term $a = 6\ln x$ and common difference $d = 4\ln x$ .

The target is $43\ln(x^{24}) = 1032\ln x$ . Substituting into $S_n = \frac{n}{2}(2a + (n-1)d)$ gives an inequality in which every term contains $\ln x$ . Because $x > 1$ we know $\ln x > 0$ , so dividing through by $\ln x$ is safe and leaves a pure quadratic inequality: $2n^2 + 4n - 1032 > 0$ .

Solving the corresponding equation gives a positive root of about $21.74$ , so the inequality holds once $n \geq 22$ . Since $n$ counts terms, it must be a whole number, giving the least value $22$ .

Key Takeaways

Logarithm laws can turn a 'log-flavoured' progression into an ordinary numerical AP whose first term and difference are multiples of $\ln x$ .
Dividing an inequality by $\ln x$ is only valid because $x > 1$ guarantees $\ln x > 0$ ; if $0 < x < 1$ the sign would flip.
A quadratic inequality in $n$ needs the positive root found, then rounding up to the next integer when the inequality is strict.

Common Mistakes

Mis-simplifying the terms, e.g. writing $2\ln(x^3)$ as $\ln(2x^3)$ instead of $6\ln x$ — the mark scheme requires consistent powers of $\ln x$ throughout.
Forgetting that $x > 1$ justifies dividing by $\ln x$ without reversing the inequality.
Taking the positive root $21.74$ itself, or rounding down to $21$ : the answer must be $22$ exactly ('cao, nfww').
Sign slips forming $2n^2 + 4n - 1032$ , e.g. dropping the factor of 2 from $\frac{n}{2}$ .
Using the wrong target, e.g. $43\ln(x^{24})$ read as $(43\ln x)^{24}$ .

Things to Be Careful About

The final answer is cao and nfww: $n = 22$ scores nothing if reached from incorrect working.
Show the substituted sum formula before simplifying — the M marks are for identifying $a$ and $d$ correctly and using them in $S_n$ .
Keep exact working where possible; the root $\approx 21.74$ is only used to decide the integer, not as the answer.
All solutions/conditions matter: state why $\ln x > 0$ before cancelling it.

Techniques used

apply log laws to simplify each term to a multiple of ln xidentify the first term and common difference of the APform and solve a quadratic inequality in n from the sum formulaselect the least integer solution

(b)

Given that the 25th term of this progression is equal to 408, find the exact value of $x$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Carry forward $a = 6\ln x$ and $d = 4\ln x$ from part (a), form the 25th term, equate it to 408 and solve for $x$ .

Working

Using $u_n = a + (n-1)d$ with $a = 6\ln x$ and $d = 4\ln x$ :

u_{25} = 6\ln x + 24 \times 4\ln x = 6\ln x + 96\ln x = 102\ln x

Given $u_{25} = 408$ :

102\ln x = 408 \quad\Rightarrow\quad \ln x = 4

Converting to exponential form:

x = \mathrm{e}^4

Answer

x = \mathrm{e}^4

Final answer

x = e^4

Detailed explanation

Walkthrough

This part reuses the results from part (a): the progression has first term $a = 6\ln x$ and common difference $d = 4\ln x$ . The 25th term follows from $u_n = a + (n-1)d$ , giving $6\ln x + 24 \times 4\ln x = 102\ln x$ .

Setting this equal to 408 gives a simple linear equation in $\ln x$ : dividing by 102 yields $\ln x = 4$ . Exponentiating both sides converts this to $x = \mathrm{e}^4$ , which is the exact value required.

Key Takeaways

The nth-term formula $u_n = a + (n-1)d$ applied with symbolic first term and difference still produces a clean linear equation.
$\ln x = k$ rearranges to $x = \mathrm{e}^k$ — keep answers in exact exponential form unless told otherwise.

Common Mistakes

Using $n$ instead of $n - 1$ in the nth-term formula (writing $6\ln x + 25 \times 4\ln x$ ).
Giving a decimal approximation such as $54.6$ instead of the exact $\mathrm{e}^4$ — the mark scheme demands the exact form.
Arithmetic slips collecting $6 + 96 = 102$ or dividing $408 \div 102$ .
Confusing $\ln x = 4$ with $x = 10^4$ .

Things to Be Careful About

The B1 is for a correct expression for $u_{25}$ as a genuine AP term; show the substitution before simplifying.
The final answer must be the exact value $\mathrm{e}^4$ (A1), not a rounded decimal.
This part depends on the values of $a$ and $d$ established in part (a), so any error there carries through (the M1 is follow-through on a valid AP expression).

Techniques used

write the nth term of the AP using the first term and common differencesolve a linear logarithmic equationconvert a logarithmic equation to exponential form

Q811MMedium-HardCalculusLogarithmic and exponential functions

Fig. 1

The diagram shows part of the curve $y = \frac{15}{x} - \frac{5}{x^2}$ .

The curve meets the $x$ -axis at the point $A$ .
The curve has a maximum at the point $B$ .

Find the area of the shaded region enclosed by the line $AB$ and the curve.

Give your answer in exact form.

DifficultyMedium-Hard

Worked solution

Approach

Find the $x$ -coordinate of $A$ by setting $y = 0$ . Find the maximum point $B$ by differentiating and solving $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ . The shaded area equals the area under the curve from $A$ to $B$ minus the area of the triangle formed by the chord $AB$ and the $x$ -axis.

Working

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

\frac{15}{x} - \frac{5}{x^2} = 0 \quad\Rightarrow\quad \frac{15x - 5}{x^2} = 0 \quad\Rightarrow\quad x = \frac{5}{15} = \frac{1}{3}

Point $B$ : differentiate:

\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{15}{x^2} + \frac{10}{x^3}

At the maximum $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ :

-\frac{15}{x^2} + \frac{10}{x^3} = 0 \quad\Rightarrow\quad -15x + 10 = 0 \quad\Rightarrow\quad x = \frac{2}{3}

Then

y = \frac{15}{2/3} - \frac{5}{(2/3)^2} = \frac{45}{2} - \frac{45}{4} = \frac{45}{4}

so $B = \left(\frac{2}{3}, \frac{45}{4}\right)$ .

Area under the curve from $x = \frac{1}{3}$ to $x = \frac{2}{3}$ :

\int \left(\frac{15}{x} - \frac{5}{x^2}\right)\mathrm{d}x = 15\ln x + \frac{5}{x}

\left[15\ln x + \frac{5}{x}\right]_{\frac{1}{3}}^{\frac{2}{3}} = \left(15\ln\frac{2}{3} + \frac{15}{2}\right) - \left(15\ln\frac{1}{3} + 15\right)

Area of the triangle under the chord $AB$ :

\frac{1}{2} \times \frac{45}{4} \times \left(\frac{2}{3} - \frac{1}{3}\right) = \frac{1}{2} \times \frac{45}{4} \times \frac{1}{3} = \frac{15}{8}

Shaded area:

15\ln\frac{2}{3} - 15\ln\frac{1}{3} + \frac{15}{2} - 15 - \frac{15}{8} = 15\ln 2 - \frac{75}{8}

Answer

15\ln 2 - \frac{75}{8}

Final answer

15 ln 2 - 75/8

Detailed explanation

Walkthrough

First we locate the two points that define the shaded region. Point $A$ is where the curve crosses the $x$ -axis, so we set $y = 0$ : multiplying through by $x^2$ gives $15x - 5 = 0$ , hence $x = \frac{1}{3}$ .

Point $B$ is the maximum, so we differentiate. Writing $y = 15x^{-1} - 5x^{-2}$ and applying the power rule gives $\frac{\mathrm{d}y}{\mathrm{d}x} = -15x^{-2} + 10x^{-3}$ . Setting this to zero and multiplying by $x^3$ gives $-15x + 10 = 0$ , so $x = \frac{2}{3}$ . Substituting back gives $y = \frac{45}{4}$ , so $B = \left(\frac{2}{3}, \frac{45}{4}\right)$ .

The shaded region sits between the curve and the straight chord $AB$ . The standard strategy is: area under the curve minus area of the triangle formed by the chord and the $x$ -axis. The chord runs from $A\left(\frac{1}{3}, 0\right)$ up to $B\left(\frac{2}{3}, \frac{45}{4}\right)$ , so the triangle has base $\frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ and height $\frac{45}{4}$ , giving area $\frac{1}{2} \times \frac{45}{4} \times \frac{1}{3} = \frac{15}{8}$ .

The area under the curve uses the two standard integrals: $\int \frac{15}{x}\,\mathrm{d}x = 15\ln x$ and $\int -5x^{-2}\,\mathrm{d}x = \frac{5}{x}$ . Evaluating between the limits:

\left(15\ln\frac{2}{3} + \frac{15}{2}\right) - \left(15\ln\frac{1}{3} + 15\right)

Subtracting the triangle's area and simplifying the logarithms with the quotient law, $15\ln\frac{2}{3} - 15\ln\frac{1}{3} = 15\ln 2$ , and the rational terms give $\frac{15}{2} - 15 - \frac{15}{8} = -\frac{75}{8}$ . The exact answer is $15\ln 2 - \frac{75}{8}$ .

Key Takeaways

To find where a curve meets the $x$ -axis, set $y = 0$ ; for rational functions, multiply through by the denominator (noting $x \neq 0$ ).
Stationary points come from solving $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ ; negative powers differentiate by the power rule.
Area between a curve and a chord = area under the curve (definite integral) minus the triangle under the chord, $\frac{1}{2} \times \text{base} \times \text{height}$ .
$\int \frac{k}{x}\,\mathrm{d}x = k\ln x$ and $\int x^{-2}\,\mathrm{d}x = -x^{-1}$ .
Laws of logarithms ( $\ln a - \ln b = \ln\frac{a}{b}$ ) convert $15\ln\frac{2}{3} - 15\ln\frac{1}{3}$ into the clean exact form $15\ln 2$ .

Common Mistakes

Forgetting to subtract the triangle under the chord, giving just the area under the curve.
Getting the sign wrong on $\int -5x^{-2}\,\mathrm{d}x$ : it is $+\frac{5}{x}$ , not $-\frac{5}{x}$ .
Writing $\int \frac{15}{x}\,\mathrm{d}x = 15\ln|x|$ and mishandling the modulus, or dropping the $\ln$ entirely.
Arithmetic slips when evaluating $\frac{5}{x}$ at the limits: $\frac{5}{2/3} = \frac{15}{2}$ and $\frac{5}{1/3} = 15$ .
Giving a decimal answer (e.g. $10.99$ ) when the question demands exact form — the mark scheme requires $15\ln 2 - \frac{75}{8}$ .
Sign errors in the final combination: the rational terms sum to $-\frac{75}{8}$ , not $+\frac{75}{8}$ .
The final A1 is dependent on all previous marks being awarded, so an error anywhere upstream loses it.

Things to Be Careful About

The answer must be in exact form: $15\ln 2 - \frac{75}{8}$ (equivalently $15\ln 2 - 9.375$ is accepted as an exact equivalent since $\frac{75}{8}$ terminates, but the logarithm must remain as $\ln$ ).
The mark scheme marks the derivative with B2 (one mark per correct term), then M1 for solving $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ , then A2 for the coordinates of $B$ — show every step.
The integration step carries B2 (one mark for $15\ln x$ , one for $\frac{5}{x}$ ), and the substitution of limits is a separate M1 — write the bracketed evaluation explicitly.
The final A1 is dependent on all previous marks, so accuracy throughout matters, not just at the end.
Keep fractions exact ( $\frac{45}{4}$ , $\frac{15}{8}$ , $\frac{75}{8}$ ) rather than decimals during working to avoid rounding errors.
Note $x > 0$ throughout, so $\ln x$ is well defined on the interval of integration.

Techniques used

find the x-intercept of the curvedifferentiate a rational function and solve dy/dx = 0 to locate the maximumintegrate 1/x and 1/x^2 termssubtract the area of the triangle under chord AB from the area under the curvecombine logarithms using the laws of logarithms and give an exact answer

Similar questions

Q9MediumTrigonometry

(a)

Solve the equation $3\sec 3x = \sqrt{3}\operatorname{cosec} 3x$ for $-120° \leq x \leq 120°$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Rewrite the equation in sines and cosines, rearrange to $an 3x = \frac{1}{\sqrt{3}}$ , find all values of $3x$ in $-360° \leq 3x \leq 360°$ , then divide by 3.

Working

\frac{3}{\cos 3x} = \frac{\sqrt{3}}{\sin 3x}

Cross-multiplying:

3\sin 3x = \sqrt{3}\cos 3x \quad\Rightarrow\quad \tan 3x = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}

The basic angle is $30°$ . Since $-120° \leq x \leq 120°$ , we need $-360° \leq 3x \leq 360°$ . The tangent is positive in the first and third quadrants, giving:

3x = 30°, \; 210°, \; -150°, \; -330°

Dividing each by 3:

x = 10°, \; 70°, \; -50°, \; -110°

Answer

x = -110°, \; -50°, \; 10°, \; 70°

Final answer

x = -110, -50, 10, 70 (degrees)

Detailed explanation

Walkthrough

The equation mixes secant and cosecant, so the first move is to rewrite both in terms of sine and cosine: $\sec 3x = \frac{1}{\cos 3x}$ and $\operatorname{cosec} 3x = \frac{1}{\sin 3x}$ . That gives $\frac{3}{\cos 3x} = \frac{\sqrt{3}}{\sin 3x}$ , which this mark scheme awards B1 for on its own.

Cross-multiplying brings everything into one ratio: $3\sin 3x = \sqrt{3}\cos 3x$ . Dividing by $\cos 3x$ (valid because $\cos 3x = 0$ would make the original equation undefined) turns it into a single tangent equation, $\tan 3x = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$ , worth B2.

The reference angle with tangent $\frac{1}{\sqrt{3}}$ is $30°$ . Because the argument is $3x$ and $x$ runs from $-120°$ to $120°$ , the argument $3x$ must run over a full sweep of $-360°$ to $360°$ — twice the usual circle. Tangent is positive in the first and third quadrants, so within that sweep we get $3x = 30°, 210°, -150°, -330°$ (M1).

Finally divide every value by 3 to undo the triple angle: $x = 10°, 70°, -50°, -110°$ , all of which lie inside the required interval, and there are no extras (A2).

Key Takeaways

Convert sec and cosec to sine and cosine before solving — most trig equations become manageable once written in sin and cos only.
When the argument is a multiple like $3x$ , extend the solution interval by that factor: $x \in [-120°, 120°]$ means $3x \in [-360°, 360°]$ .
Divide by the multiple only at the very end, after all angles have been found.

Common Mistakes

Solving only for $0° \leq 3x \leq 360°$ and missing the negative-angle solutions $-150°$ and $-330°$ .
Dividing by 3 too early, e.g. writing $\tan x = \frac{1}{3\sqrt{3}}$ , which is wrong.
Dividing the equation by $\cos 3x$ without noting why no solutions are lost.
Including extra angles such as $3x = 390°$ whose thirds fall outside the interval.
The A marks are 'no extras in range': listing a wrong extra angle loses accuracy marks even if the correct ones are present.

Things to Be Careful About

The interval check must be done on $3x$ , not on $x$ : generate all valid triple angles first.
Answers are exact here ( $10°, 70°, -50°, -110°$ ); no rounding issues arise, but all four values are required for full credit.
The mark scheme allows radians for the M1 step, but degrees keep the arithmetic cleanest since the final answers are printed in degrees.

Techniques used

rewrite sec and cosec in terms of sine and cosinerearrange to a tangent equationextend solutions across the full interval for the triple angledivide by three to recover x

(b)

Solve the equation $2\cos\left(y + \frac{\pi}{3}\right)\sin\left(y + \frac{\pi}{3}\right) = \sin\left(y + \frac{\pi}{3}\right)$ for $0 \leq y < 2\pi$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Factorise out $\sin\left(y + \frac{\pi}{3}\right)$ to split the equation into two simpler equations, solve each for the compound angle, then subtract $\frac{\pi}{3}$ .

Working

2\cos\left(y + \tfrac{\pi}{3}\right)\sin\left(y + \tfrac{\pi}{3}\right) - \sin\left(y + \tfrac{\pi}{3}\right) = 0

\sin\left(y + \tfrac{\pi}{3}\right)\left(2\cos\left(y + \tfrac{\pi}{3}\right) - 1\right) = 0

Case 1:

\sin\left(y + \tfrac{\pi}{3}\right) = 0

Since $0 \leq y < 2\pi$ , we have $\frac{\pi}{3} \leq y + \frac{\pi}{3} < \frac{7\pi}{3}$ , so:

y + \tfrac{\pi}{3} = \pi \text{ or } 2\pi \quad\Rightarrow\quad y = \tfrac{2\pi}{3}, \; \tfrac{5\pi}{3}

Case 2:

\cos\left(y + \tfrac{\pi}{3}\right) = \tfrac{1}{2}

Within $\frac{\pi}{3} \leq y + \frac{\pi}{3} < \frac{7\pi}{3}$ :

y + \tfrac{\pi}{3} = \tfrac{\pi}{3} \text{ or } \tfrac{5\pi}{3} \quad\Rightarrow\quad y = 0, \; \tfrac{4\pi}{3}

Answer

y = 0, \; \frac{2\pi}{3}, \; \frac{4\pi}{3}, \; \frac{5\pi}{3}

Final answer

y = 0, 2π/3, 4π/3, 5π/3

Detailed explanation

Walkthrough

Both terms contain the factor $\sin\left(y + \frac{\pi}{3}\right)$ , so instead of dividing by it (which would lose solutions), bring everything to one side and factorise:

\sin\left(y + \tfrac{\pi}{3}\right)\left(2\cos\left(y + \tfrac{\pi}{3}\right) - 1\right) = 0

A product is zero when either factor is zero, giving two cases.

Case 1: $\sin\left(y + \frac{\pi}{3}\right) = 0$ (B1). Sine vanishes at multiples of $\pi$ . The compound angle ranges over $\frac{\pi}{3} \leq y + \frac{\pi}{3} < \frac{7\pi}{3}$ because $y$ starts at $0$ , not at $-\frac{\pi}{3}$ — so the admissible zeros are $\pi$ and $2\pi$ (M1). Subtracting $\frac{\pi}{3}$ gives $y = \frac{2\pi}{3}$ and $y = \frac{5\pi}{3}$ .

Case 2: $\cos\left(y + \frac{\pi}{3}\right) = \frac{1}{2}$ (B1). Cosine equals $\frac{1}{2}$ at $\pm\frac{\pi}{3}$ plus multiples of $2\pi$ ; within the shifted interval the valid values are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ . Subtracting $\frac{\pi}{3}$ gives $y = 0$ and $y = \frac{4\pi}{3}$ .

All four values lie in $0 \leq y < 2\pi$ (A2, nfww).

Key Takeaways

Never divide an equation by a trig factor — factorise instead, or you silently discard solutions.
When the argument is $y + c$ , shift the interval endpoints by $c$ before listing candidate angles.
Solve each factor of a product as its own equation and combine all results.

Common Mistakes

Cancelling $\sin\left(y + \frac{\pi}{3}\right)$ from both sides, losing the two roots $y = \frac{2\pi}{3}$ and $y = \frac{5\pi}{3}$ entirely.
Forgetting that $y + \frac{\pi}{3}$ starts at $\frac{\pi}{3}$ , not $0$ , and wrongly including $y + \frac{\pi}{3} = 0$ (which would give negative $y$ ).
Missing the second cosine solution $\frac{5\pi}{3}$ (i.e. taking only the principal value).
Subtracting $\frac{\pi}{3}$ incorrectly, e.g. computing $2\pi - \frac{\pi}{3}$ as $\frac{5\pi}{6}$ instead of $\frac{5\pi}{3}$ .
Giving decimal approximations instead of exact multiples of $\pi$ .

Things to Be Careful About

The answer must be exact ('oe' accepted but nfww — nothing scores from wrong working).
Check every final value against $0 \leq y < 2\pi$ ; note $2\pi$ itself is excluded.
Both factors must be solved — the mark scheme awards separate B1s for each factor set equal to zero, so omitting either case caps the score.

Techniques used

factorise out the common sine factorsolve each factor separatelyhandle the compound angle shift when recovering y

Q109MMedium-HardSeries

The first three terms, in descending powers of $x$ , in the expansion of $\left(3x^2 - a\right)^n \left(1 + \frac{1}{x^2}\right)^2$ can be written as $729x^{12} + 972x^{10} + bx^8$ , where $a$ , $b$ and $n$ are constants.

Find the values of $a$ , $b$ and $n$ .

DifficultyMedium-Hard

Worked solution

Approach

Expand $\left(3x^2 - a\right)^n$ as far as the third term using the binomial theorem, expand $\left(1 + \frac{1}{x^2}\right)^2$ fully, multiply matching powers of $x$ , and equate coefficients with $729x^{12} + 972x^{10} + bx^8$ .

Working

First term: the only way to get $x^{12}$ is $(3x^2)^n \times 1$ :

(3x^2)^n = 729x^{12} \implies 3^n = 729 = 3^6 \implies n = 6

Expansion of $\left(3x^2 - a\right)^6$ (first three terms):

(3x^2)^6 + \binom{6}{1}(3x^2)^5(-a) + \binom{6}{2}(3x^2)^4(-a)^2

= 729x^{12} - 1458ax^{10} + 1215a^2x^8

Expansion of $\left(1 + \frac{1}{x^2}\right)^2$ :

1 + \frac{2}{x^2} + \frac{1}{x^4}

Multiply and collect by power of $x$ .

Coefficient of $x^{12}$ : $729$ ✓.

Coefficient of $x^{10}$ :

-1458a + 2 \times 729 = -1458a + 1458 = 972

1458a = 486 \implies a = \frac{486}{1458} = \frac{1}{3}

Coefficient of $x^8$ :

b = 1215a^2 + 6(3x^2)^5(-a)\times\frac{2}{x^2}\text{-coefficient} + 729

More explicitly, the $x^8$ terms come from $1215a^2x^8 \times 1$ , from $-1458ax^{10} \times \frac{2}{x^2}$ , and from $729x^{12} \times \frac{1}{x^4}$ :

b = 1215a^2 - 2916a + 729

With $a = \frac{1}{3}$ :

b = 1215 \times \frac{1}{9} - 2916 \times \frac{1}{3} + 729 = 135 - 972 + 729 = -108

Answer

n = 6, \qquad a = \frac{1}{3}, \qquad b = -108

Final answer

n = 6, a = 1/3, b = -108

Detailed explanation

Walkthrough

The expression is a product of two binomial powers. To reach $x^{12}$ in the final answer, we need the highest power of the first factor times the constant of the second: $(3x^2)^n \times 1$ . Setting this equal to $729x^{12}$ gives $3^n = 729$ , and since $729 = 3^6$ , we get $n = 6$ .

Next we expand $\left(3x^2 - a\right)^6$ far enough to capture all contributions down to $x^8$ . The binomial theorem gives the first three terms:

(3x^2)^6 + \binom{6}{1}(3x^2)^5(-a) + \binom{6}{2}(3x^2)^4(-a)^2 = 729x^{12} - 1458ax^{10} + 1215a^2x^8

The second factor is a simple square: $\left(1 + \frac{1}{x^2}\right)^2 = 1 + \frac{2}{x^2} + \frac{1}{x^4}$ .

Now multiply. The coefficient of $x^{10}$ receives two contributions: $-1458ax^{10}$ (from the second term of the first expansion times the constant 1) and $729x^{12} \times \frac{2}{x^2} = 1458x^{10}$ . Equating their sum to 972 gives $-1458a + 1458 = 972$ , so $1458a = 486$ and $a = \frac{1}{3}$ .

The coefficient of $x^8$ receives three contributions: $1215a^2x^8 \times 1$ , $-1458ax^{10} \times \frac{2}{x^2} = -2916ax^8$ , and $729x^{12} \times \frac{1}{x^4} = 729x^8$ . So $b = 1215a^2 - 2916a + 729$ . Substituting $a = \frac{1}{3}$ : $b = 135 - 972 + 729 = -108$ .

Key Takeaways

Matching the highest-degree term of a product of expansions pins down an unknown index $n$ immediately.
When multiplying two expansions, each coefficient of the product collects contributions from several pairs of terms — track them by power of $x$ systematically.
The binomial coefficient formula $\binom{n}{r} = \frac{n(n-1)\cdots}{r!}$ with signs carried inside $(-a)^r$ is essential for the general-term approach.
Unknowns must be found in order: $n$ first, then $a$ from the next coefficient, then $b$ — each depends on the previous one.

Common Mistakes

Forgetting the sign alternation: writing $+1458ax^{10}$ instead of $-1458ax^{10}$ because $(-a)$ was treated as positive; the scheme's answers are cao/nfww, so wrong working scores nothing even if numbers look close.
Omitting the contribution of the second factor's $\frac{2}{x^2}$ and $\frac{1}{x^4}$ terms to the $x^{10}$ and $x^8$ coefficients — many candidates equate only the first expansion's terms.
Using $\binom{6}{2} = 15$ incorrectly or computing $\frac{6 \times 5}{2}$ wrongly when forming the third term.
Solving for $b$ before finding $a$ , or substituting the wrong value of $a$ into $1215a^2 - 2916a + 729$ .
Writing $3^n = 729$ but mis-evaluating the power (e.g. taking $n = 5$ ).

Things to Be Careful About

The mark scheme states nfww (no follow-through from wrong working): every value must come from correct algebra, not adjusted guesses.
Keep $(-a)^2 = a^2$ but $(-a)^1 = -a$ — sign discipline throughout.
All three values ( $n = 6$ , $a = \frac{1}{3}$ , $b = -108$ ) are required; partial answers score only partially (B2/B2/B2 split across the parts).
Check the final $b$ by substituting back: with $a = \frac{1}{3}$ the full coefficient equation $1215a^2 - 2916a + 729 = -108$ should hold exactly.

Techniques used

match the first term of the product to find nexpand each binomial factor to the required termscollect coefficients of like powers of xform and solve equations for a and b

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