The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = 1 - 4x - x^2$ for all real values of $x$. (a) Write $\mathrm{f}(x)$ in the form $a - (x+b)^2$, where $a$ and $b$ are constants. (b) Find the range of $\mathrm{f}$. (c) The function $\mathrm{g}$ is defined by $\mathrm{g}(x) = 1 - 4x - x^2$ for $x \geq k$, where $k$ is a constant. State the least possible value of $k$ such that $\mathrm{g}$ has an inverse. (d) Using your value of $k$, find $\mathrm{g}^{-1}(x)$, stating…

(a) f(x) = 5 - (x+2)^2 · (b) f(x) = -2

(a) Find $\int \frac{1}{\sqrt{3x+2}}\,\mathrm{d}x$. (b) Find, in terms of $a$, $\int_{0.5}^{a} \mathrm{e}^{(1-2x)}\,\mathrm{d}x$.

(a) (2/3)(3x + 2)^(1/2) + c · (b) 1/2 - (1/2)e^(1-2a)

(a) Solve the equation $\sin 4x = \frac{1}{2}$ for $0 \leq x \leq \frac{\pi}{4}$, giving your answers in terms of $\pi$. (b) The diagram shows parts of the graphs of $y = \sin 4x$ and $y = \frac{1}{2}$. Find the exact area of the shaded region enclosed by the curve and the line.

(a) x = pi/24 and x = 5pi/24 · (b) √3/4 − π/12

(a) There are 3 girls and 2 boys standing in a straight line. Find the number of possible orders in each of the following cases. (i) No girls are next to each other. (ii) The 2 boys are not next to each other. (b) 12 people, including Anjie and Bubay, are divided into 3 groups of 4 people. Anjie and Bubay must not be in the same group. Find the number of ways in which the 3 groups can be selected.

(a)(i) 12 · (a)(ii) 72 · (b) 4200

4037/23 Oct/Nov 2024 — Additional Mathematics Solved Past Paper

Q13MMedium-EasyEquations, inequalities and graphs

Fig. 1

The diagram shows the graph of $y = (x+1)(x-1)(x-2)$ . Use the graph to solve the inequality

(x+1)(x-1)(x-2) < 1

DifficultyMedium-Easy

Worked solution

Approach

The inequality $(x+1)(x-1)(x-2) < 1$ asks for the values of $x$ where the given cubic curve lies below the horizontal line $y = 1$ . Draw $y = 1$ on the graph, find where it meets the curve, then read off the intervals.

Working

Draw the horizontal line $y = 1$ across the grid. It cuts the cubic at three points; reading from the graph:

x \approx -0.8, \qquad x \approx 0.55, \qquad x \approx 2.25

The curve is below $y = 1$ :

to the left of the first crossing (the curve rises steeply from below), i.e. for $x < -0.8$ ;
between the second and third crossings (the curve dips through its local minimum near $(1.5, -0.63)$ before rising again), i.e. for $0.55 < x < 2.25$ ;
between $x = -0.8$ and $x = 0.55$ the curve is above the line (it passes through the local maximum near $y = 2.11$ ), so this interval is excluded.

Answer

x < -0.8 \quad \text{or} \quad 0.55 < x < 2.25

Final answer

x < -0.8 or 0.55 < x < 2.25

Detailed explanation

Walkthrough

The graph shows $y = (x+1)(x-1)(x-2)$ , a positive cubic with roots at $x = -1$ , $1$ , $2$ . To solve $(x+1)(x-1)(x-2) < 1$ we compare the curve with the constant function $y = 1$ : wherever the curve sits below height $1$ , the inequality holds.

Step 1: draw the horizontal line $y = 1$ on the same axes. This line crosses the cubic three times because the curve's local maximum (about $2.11$ ) is above $1$ and its local minimum (about $-0.63$ ) is below $1$ .

Step 2: read the $x$ -coordinates of the three intersection points from the grid: approximately $x = -0.8$ , $x = 0.55$ and $x = 2.25$ .

Step 3: decide on which side of each crossing the curve is below the line. Coming from far left, the cubic comes up from very negative values, so it is below $1$ until $x = -0.8$ . Between $-0.8$ and $0.55$ it bulges above the line (through the maximum). It dips back under at $x = 0.55$ , stays below through the minimum, and climbs above again at $x = 2.25$ . Hence the solution set is $x < -0.8$ together with $0.55 < x < 2.25$ .

Key Takeaways

A one-variable inequality can be solved graphically by comparing the curve with an appropriate horizontal line.
The number of solutions of $f(x) = k$ equals the number of times the line $y = k$ crosses the curve — here three, since $k = 1$ lies between the local maximum and minimum.
The solution set of a strict inequality uses strict inequalities ( $<$ ) and excludes the boundary points themselves.

Common Mistakes

Giving only one or two of the three intervals — the mark scheme awards B1 for either correct inequality but B2 needs both, so all three intersection readings must be used.
Including the boundary values with $\leq$ instead of $<$ : the question asks for strictly less than $1$ , so $x = -0.8$ , $0.55$ , $2.25$ are excluded.
Taking the middle interval the wrong way round (e.g. writing $-0.8 < x < 0.55$ ), which happens if you forget that between those crossings the curve is above the line near the local maximum.
Reading the intersection coordinates inaccurately from the grid — values should be read to about the nearest small division ( $-0.8$ , $0.55$ , $2.25$ ).

Things to Be Careful About

The answer must come from the graph as instructed — do not attempt to expand and solve the cubic algebraically.
Strict inequality: exclude the endpoints exactly as read.
Both pieces of the solution set are required for full credit (B2); a single correct piece earns B1.

Techniques used

draw the horizontal line y = 1 on the given graphread off the x-coordinates of the intersection pointsidentify where the cubic lies below the line to solve the inequality

Approach

Complete the square for $\mathrm{f}(x) = 1 - 4x - x^2$ , keeping the negative $x^2$ outside the bracket so the result has the form $a - (x+b)^2$ .

Working

Group the terms in $x$ :

\mathrm{f}(x) = 1 - \left(x^2 + 4x\right)

Complete the square inside the bracket: half of $4$ is $2$ , and $(x+2)^2 = x^2 + 4x + 4$ , so

x^2 + 4x = (x+2)^2 - 4

Substitute back:

\mathrm{f}(x) = 1 - \left((x+2)^2 - 4\right) = 1 - (x+2)^2 + 4

\mathrm{f}(x) = 5 - (x+2)^2

Answer

\mathrm{f}(x) = 5 - (x+2)^2, \quad \text{so } a = 5,\ b = 2

Final answer

f(x) = 5 - (x+2)^2

Detailed explanation

Walkthrough

The question asks us to rewrite $1 - 4x - x^2$ as $a - (x+b)^2$ . Because the coefficient of $x^2$ is negative, we first factor it out of the two terms containing $x$ : write $1 - (x^2 + 4x)$ . Then complete the square inside the bracket: take half the coefficient of $x$ ( $4 \div 2 = 2$ ), form $(x+2)^2$ , and correct for the extra constant introduced — since $(x+2)^2 = x^2 + 4x + 4$ , we have $x^2 + 4x = (x+2)^2 - 4$ . Substituting gives $1 - [(x+2)^2 - 4] = 1 - (x+2)^2 + 4 = 5 - (x+2)^2$ . So $a = 5$ and $b = 2$ .

Key Takeaways

Completing the square rewrites any quadratic in vertex form, revealing its maximum or minimum directly.
With a negative leading coefficient, keep the minus sign outside the bracket so the squared term is subtracted.
The correction constant inside the bracket is $\left(\frac{b}{2}\right)^2$ ; here that is $4$ .

Common Mistakes

Writing $-(x+2)^2$ without adjusting the constant, losing the $+4$ correction and getting $-3 - (x+2)^2$ instead of $5 - (x+2)^2$ .
Giving only part of the answer: the mark scheme awards B1 for just $-(x+2)^2$ or just $a=5,\ b=2$ , but both are needed for B2.
Sign slips when distributing the outer minus over the bracketed expression.

Things to Be Careful About

The answer must be exactly $5 - (x+2)^2$ for full marks (B2); partial forms earn B1 only.
Check by expanding: $5 - (x^2 + 4x + 4) = 1 - 4x - x^2$ , which matches the original.

Techniques used

complete the square on a quadratic with a negative leading coefficientidentify the constants in the completed-square form

(b)

Find the range of $\mathrm{f}$ .

1M Similar

(c)

The function $\mathrm{g}$ is defined by $\mathrm{g}(x) = 1 - 4x - x^2$ for $x \geq k$ , where $k$ is a constant.

State the least possible value of $k$ such that $\mathrm{g}$ has an inverse.

1M Similar

(d)

Using your value of $k$ , find $\mathrm{g}^{-1}(x)$ , stating its domain and range.

5M Similar

DifficultyMedium

Worked solution

Approach

Write $y = \mathrm{g}(x)$ using the completed-square form, swap the variables, and solve for $y$ . Choose the positive square root because the restricted domain $x \geq -2$ forces $\mathrm{g}^{-1}(x) \geq -2$ . The domain of $\mathrm{g}^{-1}$ is the range of $\mathrm{g}$ , and its range is the restricted domain of $\mathrm{g}$ .

Working

Let

y = 5 - (x+2)^2, \quad x \geq -2

Swap variables:

x = 5 - (y+2)^2

Rearrange:

(y+2)^2 = 5 - x

y + 2 = \pm\sqrt{5-x}

Since $x \geq -2$ in the domain of $\mathrm{g}$ , we need $y \geq -2$ , i.e. $y + 2 \geq 0$ , so take the positive root:

y = -2 + \sqrt{5-x}

Hence

\mathrm{g}^{-1}(x) = -2 + \sqrt{5-x}

Domain: the outputs of $\mathrm{g}$ on $x \geq -2$ run from $5$ downwards, so

x \leq 5

Range: the inputs to $\mathrm{g}^{-1}$ produce values from $-2$ upwards, so

\mathrm{g}^{-1}(x) \geq -2

Answer

\mathrm{g}^{-1}(x) = -2 + \sqrt{5-x}, \quad x \leq 5, \quad \mathrm{g}^{-1}(x) \geq -2

Final answer

g^-1(x) = -2 + sqrt(5-x), domain x <= 5, range g^-1(x) >= -2

Detailed explanation

Walkthrough

To invert $\mathrm{g}$ , start from $y = 5 - (x+2)^2$ (using the completed-square form from part (a)) and interchange $x$ and $y$ , giving $x = 5 - (y+2)^2$ . Rearranging isolates the square: $(y+2)^2 = 5 - x$ , so $y + 2 = \pm\sqrt{5-x}$ . The choice of sign matters: because the domain of $\mathrm{g}$ was restricted to $x \geq -2$ , the inverse must return values $≥ -2$ , so $y + 2 ≥ 0$ and we take the positive root, giving $\mathrm{g}^{-1}(x) = -2 + \sqrt{5-x}$ . The domain of the inverse is the range of the original restricted function: on $x ≥ -2$ , $\mathrm{g}$ takes every value up to and including $5$ , so the domain is $x ≤ 5$ . The range of the inverse mirrors the restricted domain: $\mathrm{g}^{-1}(x) ≥ -2$ .

Key Takeaways

To find an inverse algebraically: set $y = \mathrm{f}(x)$ , swap variables, solve for $y$ .
When taking a square root during inversion, the restricted domain tells you which branch to keep.
Domain of $\mathrm{f}^{-1}$ = range of $\mathrm{f}$ ; range of $\mathrm{f}^{-1}$ = (restricted) domain of $\mathrm{f}$ .

Common Mistakes

Keeping both roots $-2 \pm \sqrt{5-x}$ — this scores only A1, not A2, because the function would not be one-to-one.
Taking the negative root, giving values below $-2$ , contradicting the restricted domain.
Swapping the domain and range statements (writing $x ≥ -2$ as the domain of the inverse).
Forgetting to state the domain and range at all — each carries a separate mark.
Starting from the uncompleted form $1 - 4x - x^2$ and misapplying the quadratic formula signs.

Things to Be Careful About

The mark scheme requires the swap-and-rearrange method (M1) and the exact answer $-2 + \sqrt{5-x}$ (A2, with A1 for the $±$ version); later simplification errors are ignored (isw).
Domain and range each carry a B1 with follow-through from earlier parts, provided they are stated in the correct inequality form.
Check consistency: substituting $x = 5$ into $\mathrm{g}^{-1}$ gives $-2$ , matching $\mathrm{g}(-2) = 5$ .

Techniques used

swap variables and rearrange to find an inverse functiontake a square root choosing the branch consistent with the restricted domaindeduce the domain of the inverse from the range of the originaldeduce the range of the inverse from the restricted domain of the original

Q3MediumTrigonometry

(a)

Show that $(2\tan\theta + \sec\theta)(2\tan\theta - \sec\theta) = 3\tan^2\theta - 1$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Expand the left-hand side as a difference of squares, then use the identity $\sec^2\theta = 1 + \tan^2\theta$ to reduce everything to tangents.

Working

(2\tan\theta + \sec\theta)(2\tan\theta - \sec\theta) = (2\tan\theta)^2 - \sec^2\theta

= 4\tan^2\theta - \sec^2\theta

Using $\sec^2\theta = 1 + \tan^2\theta$ :

4\tan^2\theta - (1 + \tan^2\theta) = 4\tan^2\theta - 1 - \tan^2\theta

= 3\tan^2\theta - 1

which is the required result.

Answer

(2\tan\theta + \sec\theta)(2\tan\theta - \sec\theta) = 3\tan^2\theta - 1

Final answer

(2 tan theta + sec theta)(2 tan theta - sec theta) = 3 tan^2 theta - 1, shown by expanding and using sec^2 theta = 1 + tan^2 theta

Detailed explanation

Walkthrough

The left-hand side has the shape $(a+b)(a-b)$ with $a = 2\tan\theta$ and $b = \sec\theta$ , so it collapses immediately to a difference of squares: $(2\tan\theta)^2 - \sec^2\theta = 4\tan^2\theta - \sec^2\theta$ . The target answer contains only $\tan\theta$ , so the next job is to remove the $\sec^2\theta$ . The Pythagorean identity $\sec^2\theta = 1 + \tan^2\theta$ does exactly that. Substituting gives $4\tan^2\theta - 1 - \tan^2\theta$ , and collecting like terms leaves $3\tan^2\theta - 1$ , matching the printed result.

Key Takeaways

Recognising a difference of squares saves work — never multiply out four terms when $(a+b)(a-b)$ is visible.
The identity $\sec^2\theta = 1 + \tan^2\theta$ converts any expression in $\sec\theta$ into one purely in $\tan\theta$ .

Common Mistakes

Expanding all four products instead of spotting the difference of squares — slower and more error-prone.
Using $\sec^2\theta = 1 - \tan^2\theta$ by confusing it with $\cos^2\theta = 1 - \sin^2\theta$ ; the sign is wrong.
Forgetting to square the coefficient: $(2\tan\theta)^2 = 4\tan^2\theta$ , not $2\tan^2\theta$ .
In a "show that" question, starting from the given answer and working backwards scores nothing — every line must run forward from the left-hand side.

Things to Be Careful About

This is an AG (Answer Given) part: the mark scheme requires sufficient correct detail, so both the expansion step ( $4\tan^2\theta - \sec^2\theta$ ) and the justified use of the identity must appear explicitly. Do not skip straight from the product to $3\tan^2\theta - 1$ .

Techniques used

expand the product of two brackets as a difference of squaresapply the identity sec squared equals one plus tan squaredsimplify to the given target form

(b)

Hence solve the equation $(2\tan\theta + \sec\theta)(2\tan\theta - \sec\theta) = 1$ for $0° \leq \theta \leq 180°$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the result of part (a): the equation becomes $3\tan^2\theta - 1 = 1$ . Solve for $\tan\theta$ , then find both angles in $0° \leq \theta \leq 180°$ .

Working

By part (a):

3\tan^2\theta - 1 = 1

3\tan^2\theta = 2 \quad\Rightarrow\quad \tan^2\theta = \frac{2}{3}

Taking the square root:

\tan\theta = \pm\sqrt{\frac{2}{3}}

For $\tan\theta = \sqrt{\tfrac{2}{3}}$ :

\theta = 39.23°

For $\tan\theta = -\sqrt{\tfrac{2}{3}}$ , the angle lies in the second quadrant:

\theta = 180° - 39.23° = 140.77°

Both values lie in $0° \leq \theta \leq 180°$ and there are no others.

Answer

\theta = 39.2° \quad \text{or} \quad \theta = 140.8°

Final answer

theta = 39.2 degrees or 140.8 degrees

Detailed explanation

Walkthrough

The word "Hence" tells us to use the identity just proved: the left-hand side of the equation is exactly $3\tan^2\theta - 1$ . Setting this equal to $1$ gives $3\tan^2\theta = 2$ , so $\tan^2\theta = \frac{2}{3}$ . Because squaring loses sign information, taking the square root produces two branches, $\tan\theta = +\sqrt{\frac{2}{3}}$ and $\tan\theta = -\sqrt{\frac{2}{3}}$ ; keeping only the positive root would lose half the solutions. The positive branch gives an acute angle, $\arctan\sqrt{\frac{2}{3}} \approx 39.23°$ . Tangent is negative in the second quadrant, so the negative branch gives $180° - 39.23° = 140.77°$ . Both lie inside $0° \leq \theta \leq 180°$ , and since tangent has exactly one solution per branch in this interval, these are the only answers.

Key Takeaways

"Hence" questions are designed so the earlier result substitutes directly — always reuse it rather than re-expanding.
Solving $\tan^2\theta = k$ yields $\tan\theta = \pm\sqrt{k}$ , giving one angle in each of the two quadrants where the relevant sign occurs.
For tangent, the supplementary-angle relation is $\theta_2 = 180° - \theta_1$ (unlike sine's $180° - \theta$ and cosine's $360° - \theta$ patterns).

Common Mistakes

Discarding the negative root of $\tan^2\theta = \frac{2}{3}$ and reporting only $39.2°$ — the mark scheme awards A1 for one correct value but A2 requires both with no extras.
Using $360° - 39.23° = 320.8°$ instead of $180° - 39.23°$ : outside the interval and wrong quadrant logic for tangent.
Re-expanding $(2\tan\theta + \sec\theta)(2\tan\theta - \sec\theta)$ from scratch instead of quoting part (a).
Rounding too early: keep $39.232°$ unrounded until the final answer, then give $39.2°$ (angles in degrees to 1 decimal place).

Things to Be Careful About

The mark scheme demands both correct values and no extras in range (A2); one correct value ignoring extras earns only A1. Give angles to 1 decimal place: $39.2°$ and $140.8°$ . Both solutions in $0° \leq \theta \leq 180°$ are required, not just the principal one.

Techniques used

use the identity proved in part (a)solve a quadratic equation in tan thetatake the square root giving both signsfind all angles in the given interval including the second quadrant

Q4MediumCircular measureCalculus

The diagram shows a design for a logo. The logo is a sector of a circle, radius $r\text{ cm}$ , with angle $\alpha$ radians.

Fig. 1

The area of the logo is $9\text{ cm}^2$ .

(a)

Show that the perimeter, $P\text{ cm}$ , of the logo is given by

P = 2r + \frac{18}{r}

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The logo is a sector of radius $r$ and angle $\alpha$ radians. Use the sector area formula to eliminate $\alpha$ from the perimeter formula $P = 2r + r\alpha$ .

Working

The area of a sector is

\frac{1}{2}r^2\alpha = 9

so

\alpha = \frac{18}{r^2}

The perimeter of the sector is the two radii plus the arc:

P = 2r + r\alpha

Substituting $\alpha = \frac{18}{r^2}$ :

P = 2r + r \times \frac{18}{r^2} = 2r + \frac{18}{r}

as required.

Answer

P = 2r + \frac{18}{r}

Final answer

P = 2r + 18/r (shown)

Detailed explanation

Walkthrough

A sector's area is half the product of the two radii and the angle between them: $\frac{1}{2}r^2\alpha$ . Setting this equal to the given area of $9$ gives an equation linking $\alpha$ and $r$ , which we solve for $\alpha = \frac{18}{r^2}$ .

The perimeter consists of two straight sides, each of length $r$ , plus the curved arc. The arc length is $s = r\alpha$ , so $P = 2r + r\alpha$ . Substituting the expression for $\alpha$ gives $P = 2r + r \times \frac{18}{r^2} = 2r + \frac{18}{r}$ , which is exactly the printed target — so the derivation is complete.

Key Takeaways

Sector formulas only hold when the angle is in radians: $A = \frac{1}{2}r^2\alpha$ and $s = r\alpha$ .
When asked to show a given result, derive forward from the data; never start from the target.
Eliminating one variable using a second equation is the key move here.

Common Mistakes

Using degree-based sector formulas ( $\frac{\alpha}{360}\pi r^2$ ) instead of the radian forms.
Forgetting that the perimeter includes both straight radii as well as the arc.
Algebra slips when cancelling: $r \times \frac{18}{r^2} = \frac{18}{r}$ , not $\frac{18}{r^3}$ or similar.
Starting from the printed answer and working backwards — this scores nothing on a 'show that' part.

Things to Be Careful About

This is an AG (Answer Given) part: every algebraic line between the given information and the target must appear, with sufficient correct detail.
Keep the angle symbol $\alpha$ clearly in radians throughout.
The final line must match the printed form $2r + \frac{18}{r}$ exactly.

Techniques used

use the sector area formula in radiansrearrange to express the angle in terms of rsubstitute into the perimeter formula for a sector

(b)

Given that $r$ can vary, find the stationary value of $P$ and determine its nature.

5M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $P = 2r + \frac{18}{r}$ with respect to $r$ , set the derivative to zero to find the stationary value, then use the second derivative to determine its nature.

Working

Differentiate:

\frac{\mathrm{d}P}{\mathrm{d}r} = 2 - \frac{18}{r^2}

At a stationary point:

2 - \frac{18}{r^2} = 0 \implies r^2 = 9 \implies r = 3

(taking the positive root since $r > 0$ ).

Substitute into $P$ :

P = 2(3) + \frac{18}{3} = 6 + 6 = 12

Determine the nature using the second derivative:

\frac{\mathrm{d}^2P}{\mathrm{d}r^2} = \frac{36}{r^3}

At $r = 3$ :

\frac{36}{3^3} = \frac{36}{27} > 0

Since the second derivative is positive at $r = 3$ , the stationary value is a minimum.

Answer

P = 12 \text{ (a minimum), occurring when } r = 3

Final answer

Minimum stationary value P = 12, at r = 3

Detailed explanation

Walkthrough

We treat $P = 2r + \frac{18}{r}$ as a function of $r$ alone (the angle has already been eliminated). Writing $\frac{18}{r}$ as $18r^{-1}$ and differentiating gives $2 - 18r^{-2} = 2 - \frac{18}{r^2}$ .

A stationary point occurs where the gradient is zero, so we set $2 - \frac{18}{r^2} = 0$ . Rearranging gives $r^2 = 9$ , so $r = 3$ (we reject $r = -3$ because a radius must be positive). Substituting $r = 3$ back into the original formula gives $P = 6 + 6 = 12$ .

To decide whether this is a maximum or minimum, differentiate again: $\frac{\mathrm{d}^2P}{\mathrm{d}r^2} = \frac{36}{r^3}$ . At $r = 3$ this equals $\frac{36}{27} > 0$ , and a positive second derivative means the curve is concave up there — so $P = 12$ is a minimum.

Key Takeaways

To find stationary values, set the first derivative to zero, solve, then substitute back into the original function.
The second derivative test classifies the point: positive means minimum, negative means maximum.
Physical context matters: $r > 0$ selects which root is valid.

Common Mistakes

Keeping both roots $r = \pm 3$ — the negative radius is invalid in this context.
Forgetting to substitute back to find the stationary value of $P$ , not just where it occurs.
Differentiating $\frac{18}{r}$ incorrectly (e.g. getting $+\frac{18}{r^2}$ instead of $-\frac{18}{r^2}$ ).
Stating 'minimum' without justification — the mark scheme requires the second derivative evaluated (or a sign-change argument), dependent on having found $r = 3$ and no other value.

Things to Be Careful About

The mark scheme awards the final A1 only if it depends on $r = 3$ being the sole solution — check no extra roots are claimed.
The derivative answer is marked 'isw' (ignore subsequent working), but the classification step needs explicit evaluation such as $\frac{36}{3^3} > 0$ .
State the conclusion clearly: the stationary value is $P = 12$ and it is a minimum.

Techniques used

differentiate the expression for Pset the derivative to zero and solveevaluate P at the stationary pointuse the second derivative test to determine the nature

Q58MMediumCalculusStraight-line graphs

The tangent to the curve $y = \frac{\sqrt{x+1}}{x}$ at the point where $x = 3$ meets the line $y = x - 16$ at the point $A$ . Find the coordinates of $A$ .

DifficultyMedium

Worked solution

Approach

Find the $y$ -coordinate on the curve at $x = 3$ , differentiate the curve by the quotient rule to get the gradient there, form the equation of the tangent, then solve it simultaneously with $y = x - 16$ to locate $A$ .

Working

At $x = 3$ :

y = \frac{\sqrt{3+1}}{3} = \frac{2}{3}

Differentiate using the quotient rule:

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

Gradient at $x = 3$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3 \times \frac{1}{2}(4)^{-\frac{1}{2}} - \sqrt{4}}{3^2} = \frac{3 \times \frac{1}{4} - 2}{9} = \frac{-\frac{5}{4}}{9} = -\frac{5}{36}

Tangent through $\left(3, \frac{2}{3}\right)$ :

y - \frac{2}{3} = -\frac{5}{36}(x - 3)

y = -\frac{5}{36}x + \frac{15}{36} + \frac{24}{36} = -\frac{5}{36}x + \frac{13}{12}

Intersect with $y = x - 16$ :

x - 16 = -\frac{5}{36}x + \frac{13}{12}

x + \frac{5}{36}x = \frac{13}{12} + 16

\frac{41}{36}x = \frac{205}{12}

x = \frac{205}{12} \times \frac{36}{41} = \frac{205 \times 3}{41} = 15

Then $y = 15 - 16 = -1$ , so $A(15, -1)$ .

Answer

A(15, -1)

Final answer

A(15, -1)

Detailed explanation

Walkthrough

The question threads three ideas together. First we need the actual point of tangency: substituting $x = 3$ into $y = \frac{\sqrt{x+1}}{x}$ gives $y = \frac{\sqrt{4}}{3} = \frac{2}{3}$ .

Next we need the gradient of the curve at that point. The function is a quotient — a square root divided by $x$ — so the quotient rule is the right tool: differentiate top times bottom minus bottom times derivative of top, all over the bottom squared. Writing $u = \sqrt{x+1}$ and $v = x$ , we get $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{2}(x+1)^{-\frac{1}{2}}$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = 1$ , giving

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

Substituting $x = 3$ : the numerator is $3 \times \frac{1}{4} - 2 = -\frac{5}{4}$ , and dividing by $9$ gives gradient $-\frac{5}{36}$ .

With a point and a gradient we write the tangent as $y - \frac{2}{3} = -\frac{5}{36}(x-3)$ , which rearranges to $y = -\frac{5}{36}x + \frac{13}{12}$ .

Finally, the meeting point $A$ with the line $y = x - 16$ is found by setting the two expressions for $y$ equal: $x - 16 = -\frac{5}{36}x + \frac{13}{12}$ . Collecting terms gives $\frac{41}{36}x = \frac{205}{12}$ , so $x = 15$ and then $y = 15 - 16 = -1$ .

Key Takeaways

The quotient rule applies whenever one function is divided by another; here the numerator itself needed the chain rule for $\sqrt{x+1}$ .
A tangent needs two ingredients: the point on the curve (from substitution) and the gradient there (from the derivative).
The intersection of two lines is found by equating their $y$ -expressions and solving the resulting linear equation.

Common Mistakes

Forgetting to subtract $v\,\frac{\mathrm{d}u}{\mathrm{d}x}$ in the correct order in the quotient rule, or writing $x^2$ instead of $(x)^2$ underneath.
Mis-evaluating $(4)^{-\frac{1}{2}}$ — it equals $\frac{1}{2}$ , not $-2$ or $2$ .
Sign slips when expanding $y - \frac{2}{3} = -\frac{5}{36}(x-3)$ ; the constant term must come out as $+\frac{13}{12}$ .
Solving $x - 16 = -\frac{5}{36}x + \frac{13}{12}$ incorrectly by adding instead of subtracting $\frac{5}{36}x$ .
The mark scheme awards the final B2 only if all earlier marks are earned, so an arithmetic slip early loses marks twice over.

Things to Be Careful About

The mark scheme says "isw" after $-\frac{5}{36}$ : later mis-simplification of that value is ignored, but the value itself must be exact.
The final B2 depends on all previous marks being awarded, so every step matters — show the substituted derivative before evaluating it.
The tangent equation may be given in any equivalent form ("oe"), e.g. $y - \frac{2}{3} = -\frac{5}{36}(x-3)$ or $y = -\frac{5}{36}x + \frac{13}{12}$ .
Keep fractions exact throughout; decimals such as $-0.139$ for the gradient risk accuracy loss.

Techniques used

evaluate the curve at the given x-valuedifferentiate using the quotient ruleevaluate the gradient at the pointform the tangent equation from a point and a gradientsolve simultaneous linear equations to find the intersection

Approach

Rewrite $\dfrac{1}{\sqrt{3x+2}}$ as $(3x+2)^{-\frac{1}{2}}$ and integrate using the reverse chain rule: raise the power by one and divide by (new power $\times$ derivative of the inner function).

Working

\int (3x+2)^{-\frac{1}{2}}\,\mathrm{d}x = \frac{(3x+2)^{\frac{1}{2}}}{\frac{1}{2} \times 3} + c

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

Answer

\frac{2}{3}(3x+2)^{\frac{1}{2}} + c

Final answer

(2/3)(3x + 2)^(1/2) + c

Detailed explanation

Walkthrough

The integrand $\dfrac{1}{\sqrt{3x+2}}$ looks awkward with its square root, so the first move is to write it in index form: $\sqrt{3x+2} = (3x+2)^{\frac{1}{2}}$ , so the reciprocal is $(3x+2)^{-\frac{1}{2}}$ . Now it has the form (linear function) $^n$ , which integrates by the reverse chain rule: increase the power by one to get $\frac{1}{2}$ , then divide by the new power multiplied by the derivative of the inside function, which is $3$ . That gives $\dfrac{(3x+2)^{\frac{1}{2}}}{\frac{1}{2}\times 3} = \dfrac{2}{3}(3x+2)^{\frac{1}{2}}$ . Because this is an indefinite integral, the constant of integration $+c$ must be included — the mark scheme explicitly shows it.

Key Takeaways

Convert roots and reciprocals to fractional/negative indices before integrating.
The reverse chain rule: $\int (ax+b)^n\,\mathrm{d}x = \dfrac{(ax+b)^{n+1}}{a(n+1)} + c$ .
Always append $+c$ to an indefinite integral.

Common Mistakes

Forgetting to divide by the derivative of the inner function ( $3$ ), giving $2(3x+2)^{\frac{1}{2}}$ — the mark scheme only awards the B1 for $k(3x+2)^{\frac{1}{2}}$ and the final mark requires the correct $\frac{2}{3}$ .
Omitting the constant of integration.
Sign or index slips when converting $\dfrac{1}{\sqrt{3x+2}}$ to $(3x+2)^{-\frac{1}{2}}$ .

Things to Be Careful About

The answer is accepted "oe" (or equivalent), so forms like $\dfrac{2\sqrt{3x+2}}{3} + c$ are fine, but $+c$ is expected.
The B1 is for any non-zero constant multiple $k(3x+2)^{\frac{1}{2}}$ , so the method mark survives a wrong coefficient — but the final accuracy mark needs $\frac{2}{3}$ exactly.

Techniques used

rewrite the integrand as a power of a linear functionintegrate using the reverse chain ruleinclude the constant of integration

(b)

Find, in terms of $a$ , $\int_{0.5}^{a} \mathrm{e}^{(1-2x)}\,\mathrm{d}x$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Integrate $\mathrm{e}^{1-2x}$ by dividing by the derivative of the exponent, then evaluate between the limits $0.5$ and $a$ .

Working

\int \mathrm{e}^{1-2x}\,\mathrm{d}x = -\frac{1}{2}\mathrm{e}^{1-2x}

Evaluate from $0.5$ to $a$ :

\left[-\frac{1}{2}\mathrm{e}^{1-2x}\right]_{0.5}^{a} = -\frac{1}{2}\mathrm{e}^{1-2a} - \left(-\frac{1}{2}\mathrm{e}^{1-2(0.5)}\right)

Since $1 - 2(0.5) = 0$ and $\mathrm{e}^{0} = 1$ :

= -\frac{1}{2}\mathrm{e}^{1-2a} + \frac{1}{2}

Answer

\frac{1}{2} - \frac{1}{2}\mathrm{e}^{1-2a}

Final answer

1/2 - (1/2)e^(1-2a)

Detailed explanation

Walkthrough

The integrand is $\mathrm{e}^{1-2x}$ , an exponential whose index is a linear function of $x$ . Integrating it uses the reverse chain rule: since the derivative of the index $1-2x$ is $-2$ , the antiderivative is $\mathrm{e}^{1-2x}$ divided by $-2$ , i.e. $-\frac{1}{2}\mathrm{e}^{1-2x}$ . This earns the B2 in the mark scheme. Then the square-bracket evaluation form applies: substitute the upper limit $a$ and the lower limit $0.5$ and subtract. At the lower limit the index becomes $1 - 2(0.5) = 0$ , and $\mathrm{e}^{0} = 1$ , so the lower-limit term is $-\frac{1}{2}$ . Subtracting gives $-\frac{1}{2}\mathrm{e}^{1-2a} + \frac{1}{2}$ , an exact answer in terms of $a$ as the question requires.

Key Takeaways

$\int \mathrm{e}^{ax+b}\,\mathrm{d}x = \dfrac{1}{a}\mathrm{e}^{ax+b} + c$ — divide by the coefficient of $x$ in the index.
Definite integration: evaluate the antiderivative at the upper limit minus the lower limit.
$\mathrm{e}^{0} = 1$ makes the lower limit here collapse to a simple constant.

Common Mistakes

Forgetting the minus sign from the $-2$ in the index, giving $+\frac{1}{2}\mathrm{e}^{1-2x}$ — the mark scheme's B1 still credits $k\mathrm{e}^{(1-2x)}$ for any non-zero $k$ , but the final answer then has the wrong sign.
Subtracting the limits in the wrong order, giving $\frac{1}{2}\mathrm{e}^{1-2a} - \frac{1}{2}$ .
Mis-evaluating $1 - 2(0.5)$ or forgetting $\mathrm{e}^{0} = 1$ .
Leaving a $+c$ in a definite integral (harmless but unnecessary) or giving a decimal approximation instead of the exact form in terms of $a$ .

Things to Be Careful About

The answer must be exact and in terms of $a$ — do not substitute a numerical value for $a$ .
The mark scheme says "isw": later mis-simplification after the correct answer is ignored, but the correct exact form $-\frac{\mathrm{e}^{1-2a}}{2} + \frac{1}{2}$ (or equivalent) is required for all 3 marks.
Keep the bracket evaluation clearly shown: upper limit minus lower limit, so the sign of the $\frac{1}{2}$ term is visible.

Techniques used

integrate an exponential function with a linear exponentapply the square-bracket evaluation formsubstitute exact limits and simplify

Q7MediumSeries

(a)

In the expansion of $\left(x + x^2\right)^8$ in ascending powers of $x$ , the 3rd and 6th terms are equal.

Find the value of $x$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Write the general term of the expansion of $\left(x + x^2\right)^8$ using the binomial theorem, pick out the 3rd and 6th terms, set them equal and solve for $x$ .

Working

The general term is

\binom{8}{r}x^{8-r}\left(x^2\right)^r = \binom{8}{r}x^{8+r}

3rd term ( $r = 2$ ):

\binom{8}{2}x^{10} = 28x^{10}

6th term ( $r = 5$ ):

\binom{8}{5}x^{13} = 56x^{13}

Setting them equal:

28x^{10} = 56x^{13}

x^3 = \frac{28}{56} = \frac{1}{2}

x = \sqrt[3]{\frac{1}{2}}

Answer

x = \sqrt[3]{\frac{1}{2}}

Final answer

x = cube root of 1/2

Detailed explanation

Walkthrough

In any binomial expansion $\left(a+b\right)^n$ , the general $(r+1)$ -th term is $\binom{n}{r}a^{n-r}b^r$ . Here $a = x$ and $b = x^2$ , so each term carries a power $x^{8-r} \cdot x^{2r} = x^{8+r}$ — the powers increase by 1 per term because the second part of the binomial itself contains an $x$ .

The 3rd term corresponds to $r = 2$ : its coefficient is $\binom{8}{2} = 28$ , giving $28x^{10}$ . The 6th term corresponds to $r = 5$ : its coefficient is $\binom{8}{5} = 56$ , giving $56x^{13}$ .

Equating them gives $28x^{10} = 56x^{13}$ . Dividing both sides by $28x^{10}$ leaves $1 = 2x^3$ , so $x^3 = \frac{1}{2}$ and $x = \sqrt[3]{\frac{1}{2}}$ .

Key Takeaways

The general term formula lets you jump straight to any term without expanding fully.
When the binomial contains powers of $x$ in both parts, add the exponents carefully: here the power rises by 1 each term, not by 1 in one place and down elsewhere.
Equating two terms often reduces to a simple power equation after cancelling common factors.

Common Mistakes

Using $r$ instead of $r+1$ as the term number: the 3rd term has $r = 2$ , not $r = 3$ .
Getting the power wrong: forgetting that $\left(x^2\right)^r$ contributes $x^{2r}$ , so the total power is $8 + r$ , not $8 - r$ or $8 - 2r$ .
Cancelling incorrectly when equating $28x^{10} = 56x^{13}$ — dividing by $x^{10}$ must leave $x^3$ on one side only.
Giving a decimal approximation instead of the exact cube root; the mark scheme requires the exact form (nfww).

Things to Be Careful About

The answer must be exact: $x = \sqrt[3]{\frac{1}{2}}$ (or equivalent), not a rounded decimal.
Check coefficients: $\binom{8}{2} = 28$ and $\binom{8}{5} = 56$ ; a slip here makes the final equation unsolvable cleanly.
The mark scheme notes "isw" — later incorrect simplification is ignored once the correct value is reached.

Techniques used

write the general term of a binomial expansionidentify the 3rd and 6th termsequate the terms and solve for x

(b)

In the expansion of $\left(x + \frac{2}{x}\right)^n$ in decreasing powers of $x$ , the 6th term is a constant.

4M

(i)

Find the value of the positive integer $n$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write the 6th term of $\left(x + \frac{2}{x}\right)^n$ using the general term formula; it is constant exactly when the net power of $x$ is zero.

Working

The 6th term corresponds to $r = 5$ :

\binom{n}{5}x^{n-5}\left(\frac{2}{x}\right)^5 = \binom{n}{5}2^5 x^{n-10}

For this term to be constant, the power of $x$ must be zero:

n - 10 = 0 \quad\Rightarrow\quad n = 10

Answer

n = 10

Final answer

n = 10

Detailed explanation

Walkthrough

The 6th term of $\left(x + \frac{2}{x}\right)^n$ uses $r = 5$ in the general term $\binom{n}{r}a^{n-r}b^r$ with $a = x$ and $b = \frac{2}{x}$ . This gives $\binom{n}{5}x^{n-5}\left(\frac{2}{x}\right)^5$ . Simplifying the powers: $x^{n-5} \cdot x^{-5} = x^{n-10}$ .

A term is constant precisely when no $x$ remains, i.e. the exponent is zero: $n - 10 = 0$ , giving $n = 10$ .

Key Takeaways

A "constant term" means the power of $x$ is zero — set the exponent equal to zero.
In a binomial like $\left(x + \frac{k}{x}\right)^n$ , the power of $x$ decreases by 2 per term, so the middle term (when $n$ is even) is the constant one.

Common Mistakes

Using $r = 6$ for the 6th term instead of $r = 5$ .
Writing the power as $n - 5 - 5$ incorrectly, e.g. forgetting the $x^{-5}$ from $\left(\frac{2}{x}\right)^5$ .
Setting the coefficient rather than the power of $x$ to zero.

Things to Be Careful About

Counting terms: the $(r+1)$ -th term has index $r$ , so the 6th term always uses $r = 5$ .
The mark scheme accepts the correct term in any form, but the reasoning $n - 5 = 5$ (power of first part equals power taken from second part) must be visible.

Techniques used

write the general term of a binomial expansionset the power of x equal to zero for a constant term

(ii)

Find the value of the 6th term.

2M Similar

Q8MediumTrigonometryCalculus

(a)

Solve the equation $\sin 4x = \frac{1}{2}$ for $0 \leq x \leq \frac{\pi}{4}$ , giving your answers in terms of $\pi$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $u = 4x$ . The range $0 \leq x \leq \frac{\pi}{4}$ becomes $0 \leq u \leq \pi$ , in which $\sin u = \frac{1}{2}$ has exactly two solutions.

Working

\sin u = \frac{1}{2} \quad \Rightarrow \quad u = \frac{\pi}{6} \quad \text{or} \quad u = \pi - \frac{\pi}{6} = \frac{5\pi}{6}

Dividing by 4:

4x = \frac{\pi}{6} \Rightarrow x = \frac{\pi}{24}, \qquad 4x = \frac{5\pi}{6} \Rightarrow x = \frac{5\pi}{24}

Both lie in $0 \leq x \leq \frac{\pi}{4}$ , and no further solutions exist in this interval.

Answer

x = \frac{\pi}{24} \quad \text{and} \quad x = \frac{5\pi}{24}

Final answer

x = pi/24 and x = 5pi/24

Detailed explanation

Walkthrough

The equation involves $\sin 4x$ , so we first treat $4x$ as a single angle $u$ . Since $x$ runs from $0$ to $\frac{\pi}{4}$ , the angle $u = 4x$ runs from $0$ to $\pi$ — exactly one full positive half-cycle of the sine curve, in which $\sin u = \frac{1}{2}$ is attained twice. The first solution is the standard special angle $u = \frac{\pi}{6}$ ; the second uses the symmetry $\sin(\pi - \theta) = \sin\theta$ , giving $u = \frac{5\pi}{6}$ . Finally, dividing each by $4$ converts back to $x$ , giving $\frac{\pi}{24}$ and $\frac{5\pi}{24}$ .

Key Takeaways

For $\sin(kx) = c$ , solve for the angle $kx$ first, then divide by $k$ at the very end.
The second solution of $\sin u = k$ in a half-cycle is $\pi - \arcsin k$ .
Always transform the given interval for $x$ into an interval for the substituted angle so you know how many solutions to expect.

Common Mistakes

Dividing by 4 too early: writing $\sin x = \frac{1}{8}$ and solving for $x$ directly gives wrong angles.
Finding only $x = \frac{\pi}{24}$ and missing the second solution $\frac{5\pi}{24}$ .
Including extra solutions such as $x = \frac{13\pi}{24}$ , which lie outside the given range — the mark scheme penalises extras.

Things to Be Careful About

Answers must be exact, in terms of $\pi$ — no decimals.
The range is $0 \leq x \leq \frac{\pi}{4}$ , so $u = 4x$ only reaches $\pi$ : exactly two solutions, no more.
The mark scheme awards one mark for either correct value (or for $4x = \frac{\pi}{6}, \frac{5\pi}{6}$ seen) and the second only for both values with no extras in range.

Techniques used

solve sin of a multiple angle by substituting u = 4xuse the two angles in the principal cycle for sin u = 1/2divide by 4 and keep answers in terms of pi

(b)

Fig. 1

The diagram shows parts of the graphs of $y = \sin 4x$ and $y = \frac{1}{2}$ .

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

5M Similar

DifficultyMedium

Worked solution

Approach

The shaded region lies between the curve $y = \sin 4x$ (above) and the line $y = \frac{1}{2}$ (below), between the intersection points found in part (a): $x = \frac{\pi}{24}$ and $x = \frac{5\pi}{24}$ . Its area is the integral of the difference of the two functions between those limits.

Working

\text{Area} = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \left( \sin 4x - \frac{1}{2} \right) \mathrm{d}x

Integrating term by term:

\int \left( \sin 4x - \frac{1}{2} \right) \mathrm{d}x = -\frac{1}{4}\cos 4x - \frac{x}{2}

Substituting the exact limits:

\begin{aligned} \left[ -\frac{1}{4}\cos 4x - \frac{x}{2} \right]_{\frac{\pi}{24}}^{\frac{5\pi}{24}} &= \left( -\frac{1}{4}\cos\frac{5\pi}{6} - \frac{5\pi}{48} \right) - \left( -\frac{1}{4}\cos\frac{\pi}{6} - \frac{\pi}{48} \right) \\ &= \left( -\frac{1}{4} \cdot \left(-\frac{\sqrt{3}}{2}\right) - \frac{5\pi}{48} \right) - \left( -\frac{1}{4} \cdot \frac{\sqrt{3}}{2} - \frac{\pi}{48} \right) \\ &= \frac{\sqrt{3}}{8} - \frac{5\pi}{48} + \frac{\sqrt{3}}{8} + \frac{\pi}{48} \\ &= \frac{\sqrt{3}}{4} - \frac{\pi}{12} \end{aligned}

Answer

\text{Area} = \frac{\sqrt{3}}{4} - \frac{\pi}{12}

Final answer

√3/4 − π/12

Detailed explanation

Walkthrough

The shaded area sits between two graphs: the curve $y = \sin 4x$ on top and the horizontal line $y = \frac{1}{2}$ underneath. Wherever one curve lies above another, the area of the strip between them is found by integrating (top function) − (bottom function). The strip exists only between the two crossing points, which part (a) gave us as $x = \frac{\pi}{24}$ and $x = \frac{5\pi}{24}$ — this is why the parts are threaded together.

The integral of $\sin 4x$ needs the chain rule in reverse: differentiating $-\frac{1}{4}\cos 4x$ brings down the inner derivative $4$ , which cancels the $-\frac{1}{4}$ , giving back $\sin 4x$ . The line integrates to $-\frac{x}{2}$ .

At the upper limit, $4 \times \frac{5\pi}{24} = \frac{5\pi}{6}$ , and $\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ ; at the lower limit, $4 \times \frac{\pi}{24} = \frac{\pi}{6}$ , and $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ . Substituting and subtracting, the two cosine terms each contribute $\frac{\sqrt{3}}{8}$ (total $\frac{\sqrt{3}}{4}$ ), while the $x$ -terms give $-\frac{5\pi}{48} + \frac{\pi}{48} = -\frac{\pi}{12}$ .

Key Takeaways

Area between two curves $= \int (\text{upper} - \text{lower})\,\mathrm{d}x$ between their intersection points.
$\int \sin kx\,\mathrm{d}x = -\frac{1}{k}\cos kx$ — divide by the inner coefficient.
Exact limits and exact trig values of special angles give an exact answer in surds and $\pi$ .

Common Mistakes

Integrating $\sin 4x$ as $\cos 4x$ or $-\cos 4x$ , forgetting the factor $\frac{1}{4}$ (the mark scheme gives B1 only for $k\cos 4x$ with $k < 0$ or $k = \frac{1}{4}$ ).
Subtracting in the wrong order at the limits, or using the limits $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ (the $4x$ values) instead of the $x$ values from part (a).
Integrating $\sin 4x$ alone and forgetting to subtract the area under the line $y = \frac{1}{2}$ .
Using decimal approximations for $\cos\frac{\pi}{6}$ instead of the exact $\frac{\sqrt{3}}{2}$ , losing the exact form.
Sign slips when $\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ is negative.

Things to Be Careful About

The limits must be the $x$ -values $\frac{\pi}{24}$ and $\frac{5\pi}{24}$ from part (a), in radians, in the correct (increasing) order — the scheme awards the substitution mark only for exact limits in correct order.
The final answer must be exact: $\frac{\sqrt{3}}{4} - \frac{\pi}{12}$ or an exact equivalent; a decimal answer scores A0.
The plan mark requires the integral of the difference $\sin 4x - \frac{1}{2}$ (or the equivalent split $\int \sin 4x\,\mathrm{d}x - \frac{1}{2}(\text{width})$ ); integrating only the curve does not earn it.
Follow-through applies to the limits from part (a), but only if at least the integration B1 was earned.

Techniques used

set up the area between a curve and a line as an integral of the differenceintegrate sin of a multiple angleapply the square-bracket evaluation with exact limitssimplify to an exact surd and pi form

Q94MMedium-Easy[Legacy] Indices and surds

DO NOT USE A CALCULATOR IN THIS QUESTION.

Write $\frac{16+11\sqrt{10}}{2+\sqrt{10}} + 1$ in the form $p + q\sqrt{10}$ , where $p$ and $q$ are integers.

DifficultyMedium-Easy

Worked solution

Approach

First add the $1$ by writing it over the common denominator, then rationalise the denominator by multiplying top and bottom by the conjugate $2 - \sqrt{10}$ , and simplify.

Working

\frac{16+11\sqrt{10}}{2+\sqrt{10}} + 1 = \frac{16+11\sqrt{10}}{2+\sqrt{10}} + \frac{2+\sqrt{10}}{2+\sqrt{10}} = \frac{18+12\sqrt{10}}{2+\sqrt{10}}

Rationalise by multiplying by $\dfrac{2-\sqrt{10}}{2-\sqrt{10}}$ :

\frac{18+12\sqrt{10}}{2+\sqrt{10}} \times \frac{2-\sqrt{10}}{2-\sqrt{10}}

Expand the numerator and use $(2+\sqrt{10})(2-\sqrt{10}) = 4-10 = -6$ :

\frac{36 - 18\sqrt{10} + 24\sqrt{10} - 120}{-6} = \frac{-84 + 6\sqrt{10}}{-6}

= 14 - \sqrt{10}

This is of the form $p + q\sqrt{10}$ with integers $p = 14$ and $q = -1$ .

Answer

14 - \sqrt{10}

Final answer

14 - √10

Detailed explanation

Walkthrough

The expression is a surd fraction plus $1$ . The first move is to combine them into a single fraction: writing $1$ as $\dfrac{2+\sqrt{10}}{2+\sqrt{10}}$ gives $\dfrac{18+12\sqrt{10}}{2+\sqrt{10}}$ , since $16+2 = 18$ and $11\sqrt{10}+\sqrt{10} = 12\sqrt{10}$ .

The denominator $2+\sqrt{10}$ is irrational, so we rationalise it. Multiplying numerator and denominator by the conjugate $2-\sqrt{10}$ leaves the value unchanged but turns the denominator into a rational number: $(2+\sqrt{10})(2-\sqrt{10}) = 4 - 10 = -6$ .

Expanding the numerator with the distributive law: $18 \times 2 = 36$ , $18 \times (-\sqrt{10}) = -18\sqrt{10}$ , $12\sqrt{10} \times 2 = 24\sqrt{10}$ , and $12\sqrt{10} \times (-\sqrt{10}) = -120$ . Collecting terms gives $-84 + 6\sqrt{10}$ over $-6$ .

Dividing each term by $-6$ : $-84 \div (-6) = 14$ and $6\sqrt{10} \div (-6) = -\sqrt{10}$ , so the value is $14 - \sqrt{10}$ , which is of the required form $p + q\sqrt{10}$ with $p = 14$ , $q = -1$ .

Key Takeaways

To add a whole number to a surd fraction, put everything over the common denominator first.
Rationalising a denominator of the form $a + \sqrt{b}$ means multiplying top and bottom by the conjugate $a - \sqrt{b}$ , using the difference of two squares.
Expand surd products term by term, remembering that $\sqrt{10} \times \sqrt{10} = 10$ .
Divide every term of the numerator by the denominator separately at the end.

Common Mistakes

Adding the $1$ only to the numerator instead of forming a proper common denominator.
Sign slips in the conjugate: using $2+\sqrt{10}$ again instead of $2-\sqrt{10}$ , which changes nothing.
Arithmetic errors in the expansion, especially $12\sqrt{10} \times (-\sqrt{10}) = -120$ (not $+120$ ) and the denominator $-6$ (not $+6$ ).
Forgetting to divide the surd term by $-6$ , leaving $\sqrt{10}$ instead of $-\sqrt{10}$ .
The mark scheme awards the final mark only if all previous steps are correct (nfww), so a wrong expansion followed by a 'correct-looking' answer scores nothing.

Things to Be Careful About

The question is non-calculator: all arithmetic must be shown exactly, with no decimal approximations.
The final answer must be in the exact form $p + q\sqrt{10}$ with integers $p$ and $q$ — here $p = 14$ and $q = -1$ , so $-\sqrt{10}$ is a legitimate $q\sqrt{10}$ term with $q = -1$ .
Keep the denominator sign straight: $(2+\sqrt{10})(2-\sqrt{10}) = 4 - 10 = -6$ , and dividing by $-6$ flips both signs.
Show every expansion line; the scheme requires sufficient correct detail for each mark.

Techniques used

combine the fraction with 1 over a common denominatormultiply numerator and denominator by the conjugateexpand brackets involving surdssimplify the resulting surd expression

Approach

The distance run each week forms a geometric progression with first term $a = 10$ and common ratio $r = 1 + 0.10 = 1.1$ .
Set the formula for the sum of the first $n$ terms $S_n > 200$ , simplify the expression, and use logarithms to solve for the smallest integer $n$ .

Working

The sum of the first $n$ terms is given by:

S_n = \frac{a(r^n - 1)}{r - 1}

Substituting $a = 10$ and $r = 1.1$ :

\frac{10(1.1^n - 1)}{1.1 - 1} > 200

\frac{10(1.1^n - 1)}{0.1} > 200

100(1.1^n - 1) > 200

1.1^n - 1 > 2

1.1^n > 3

Take logarithms of both sides:

n \lg 1.1 > \lg 3

n > \frac{\lg 3}{\lg 1.1}

n > 11.5267\ldots

Since $n$ must be a whole number of weeks, the smallest possible integer value of $n$ is $12$ .

Answer

12

Final answer

12

Detailed explanation

Walkthrough

Identify the progression: Suzma starts with $10\text{ km}$ in week 1, so $a = 10$ . Increasing by $10\%$ each week means multiplying the previous week's distance by $1 + 0.10 = 1.1$ , so $r = 1.1$ .
Set up the sum inequality: We require the total distance over $n$ whole weeks to exceed $200\text{ km}$ , so $S_n > 200$ .
Using $S_n = \frac{a(r^n - 1)}{r - 1}$ , we have $\frac{10(1.1^n - 1)}{1.1 - 1} > 200$ .
Simplify the inequality:
- $\frac{10(1.1^n - 1)}{0.1} = 100(1.1^n - 1)$
- $100(1.1^n - 1) > 200 \implies 1.1^n - 1 > 2 \implies 1.1^n > 3$
Solve using logarithms: Taking the base-10 log (or natural log) of both sides gives $n \lg 1.1 > \lg 3$ , which yields $n > \frac{\lg 3}{\lg 1.1} \approx 11.53$ .
Find the integer answer: Since $n$ represents whole weeks, we round up to the next integer, giving $n = 12$ .

Key Takeaways

For a percentage increase of $p\%$ , the common ratio is $r = 1 + \frac{p}{100}$ .
When solving an inequality of the form $b^n > k$ where $b > 1$ , taking logarithms preserves the inequality direction because $\lg b > 0$ .
Read carefully whether the question asks for a single term $T_n$ or a cumulative total $S_n$ .

Common Mistakes

Using the $n$ th term formula $T_n = ar^{n-1}$ instead of the sum formula $S_n$ .
Incorrect common ratio, e.g. taking $r = 0.1$ instead of $r = 1.1$ .
Rounding $11.53$ down to $11$ instead of rounding up to $12$ to satisfy the strict inequality $> 200$ .

Things to Be Careful About

Ensure the inequality sign is maintained properly. Since $\lg(1.1) > 0$ , dividing by $\lg(1.1)$ does not flip the inequality sign.
The question specifically asks for whole weeks, requiring the smallest integer satisfying the condition.

Techniques used

identify geometric progression parametersset up the sum of a geometric progression inequalitysolve an exponential inequality using logarithmsround up to find the minimum integer value

(b)

A geometric progression has 1st term $a$ and common ratio $r$ , where $a \neq 0$ and $r \neq 1$ . The 1st, 2nd and 3rd terms of the geometric progression are the 1st, 3rd and 7th terms of an arithmetic progression. Find the value of $r$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Let the geometric progression have terms $T_1 = a$ , $T_2 = ar$ , and $T_3 = ar^2$ .
Let the arithmetic progression have first term $A$ and common difference $d$ . Since the 1st term of the GP is the 1st term of the AP, $A = a$ .
Express the 3rd and 7th terms of the AP in terms of $a$ and $d$ , equate them to the GP terms, eliminate $d$ , and solve for $r$ .

Working

From the arithmetic progression:

1st term: $a$
3rd term: $a + 2d = ar$
7th term: $a + 6d = ar^2$

From the equation for the 3rd term:

2d = ar - a = a(r - 1) \implies d = \frac{a(r - 1)}{2}

From the equation for the 7th term:

6d = ar^2 - a = a(r^2 - 1)

Substitute $d = \frac{a(r - 1)}{2}$ into the 7th term equation:

6\left(\frac{a(r - 1)}{2}\right) = a(r^2 - 1)

3a(r - 1) = a(r^2 - 1)

Since $a \neq 0$ , we can divide by $a$ :

3(r - 1) = r^2 - 1

3r - 3 = r^2 - 1

r^2 - 3r + 2 = 0

(r - 1)(r - 2) = 0

This gives $r = 1$ or $r = 2$ .

Since it is given that $r \neq 1$ , we have:

r = 2

Answer

2

Final answer

2

Detailed explanation

Walkthrough

Define the terms of both progressions:
- GP terms: $T_1 = a$ , $T_2 = ar$ , $T_3 = ar^2$ .
- AP terms: $U_1 = a$ , $U_3 = a + 2d$ , $U_7 = a + 6d$ .
Equate the corresponding terms:
- $ar = a + 2d \implies 2d = a(r - 1)$
- $ar^2 = a + 6d \implies 6d = a(r^2 - 1)$
Eliminate the variable $d$ :
- Multiply the first equation by $3$ : $6d = 3a(r - 1)$ .
- Equate with the second equation: $3a(r - 1) = a(r^2 - 1)$ .
Solve for $r$ :
- Since $a \neq 0$ , divide both sides by $a$ : $3(r - 1) = r^2 - 1$ .
- Rearrange into standard quadratic form: $r^2 - 3r + 2 = 0$ .
- Factorise: $(r - 1)(r - 2) = 0$ , giving $r = 1$ or $r = 2$ .
Apply restrictions:
- The question states $r \neq 1$ , so we discard $r = 1$ and conclude $r = 2$ .

Key Takeaways

Setting up simultaneous equations using standard term definitions ( $T_n = ar^{n-1}$ for GP, $U_n = a + (n-1)d$ for AP) is the standard method for linking progressions.
Eliminating nuisance variables (like $d$ ) allows you to isolate the desired parameter ( $r$ ).

Common Mistakes

Forgetting that the $n$ th term of an AP is $a + (n-1)d$ and writing $a + 3d$ or $a + 7d$ instead of $a + 2d$ and $a + 6d$ .
Dividing by $(r - 1)$ without acknowledging that $r \neq 1$ .
Leaving both $r = 1$ and $r = 2$ as the final answer, ignoring the given condition $r \neq 1$ .

Things to Be Careful About

Ensure you clearly reject $r = 1$ based on the problem statement $r \neq 1$ .
Take care with arithmetic when eliminating $d$ to avoid algebraic sign errors.

Techniques used

express terms of GP and AP using standard formulaeeliminate the common difference dfactorise and solve the resulting polynomial in rapply given domain restrictions to select the valid root

Q11MediumPermutations and combinations

(a)

There are 3 girls and 2 boys standing in a straight line. Find the number of possible orders in each of the following cases.

4M

(i)

No girls are next to each other.

2M Similar

(ii)

The 2 boys are not next to each other.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use complementary counting: subtract the number of arrangements where the 2 boys are next to each other from the total number of unrestricted arrangements of the 5 people.

Working

Total unrestricted arrangements of the 5 people:

5! = 120

Treat the 2 boys as a single unit $(\text{BB})$ . This creates 4 units to arrange (3 girls and 1 boy-pair):

4! = 24

The 2 boys can be arranged within their unit in $2!$ ways:

4! \times 2! = 24 \times 2 = 48

Subtract the arrangements where the boys are together from the total:

5! - (4! \times 2!) = 120 - 48 = 72

Answer

72

Final answer

72

Detailed explanation

Walkthrough

We want to find the number of ways to arrange 3 girls and 2 boys in a line such that the 2 boys are not next to each other.

Method: Complementary Counting (Subtraction)

First, find the total number of ways to arrange all 5 people without any restrictions:

5! = 120

Next, calculate the number of arrangements where the 2 boys are standing next to each other. Treat the two boys as a single "block" or unit. We now have 4 items to arrange: the 3 girls and the single boy-block. There are $4!$ ways to arrange these 4 items.
Inside the boy-block, the 2 boys can be ordered in $2!$ ways.
Therefore, the number of arrangements where the boys are together is:

4! \times 2! = 24 \times 2 = 48

Finally, subtract this from the total:

120 - 48 = 72

(Alternative approach using the gap method): Arrange the 3 girls first in $3! = 6$ ways. This creates 4 possible gaps (before the first girl, between girls, and after the last girl). Choose 2 distinct gaps for the 2 boys in $^{4}\mathrm{P}_{2} = 12$ ways. Total ways $= 6 \times 12 = 72$ .

Key Takeaways

When asked for items "not next to each other", complementary counting (Total minus Together) or the gap method are the two standard and reliable methods.

Common Mistakes

Forgetting to multiply by $2!$ for the internal arrangement of the two boys in the block.
Incorrectly computing $5! - 2!$ instead of treating the boys as a single entity among the 5 items.

Things to Be Careful About

Ensure the difference between the total permutations $5!$ and the bundled permutations $4! \times 2!$ is fully evaluated.

Techniques used

use complementary counting with the bundle methodapply the gap method as an alternative strategy

(b)

12 people, including Anjie and Bubay, are divided into 3 groups of 4 people. Anjie and Bubay must not be in the same group.

Find the number of ways in which the 3 groups can be selected.

2M Similar

DifficultyMedium

Worked solution

Approach

12 people are divided into 3 groups of 4. Since Anjie and Bubay must be in different groups, place Anjie into Group 1 and Bubay into Group 2. Group 3 contains neither of them.

Group 1 needs 3 more people chosen from the remaining 10.
Group 2 needs 3 more people chosen from the remaining 7.
Group 3 will automatically take the remaining 4 people.

Working

Number of ways to choose 3 additional people for Anjie's group from the 10 available people:

{}^{10}\mathrm{C}_{3} = 120

Number of ways to choose 3 additional people for Bubay's group from the remaining 7 people:

{}^{7}\mathrm{C}_{3} = 35

The remaining 4 people form the third group in ${}^{4}\mathrm{C}_{4} = 1$ way.

Multiply these combinations:

{}^{10}\mathrm{C}_{3} \times {}^{7}\mathrm{C}_{3} \times {}^{4}\mathrm{C}_{4} = 120 \times 35 \times 1 = 4200

Answer

4200

Final answer

4200

Detailed explanation

Walkthrough

We have 12 people to divide into 3 unnamed groups of 4 people each, with the restriction that Anjie (A) and Bubay (B) must not be in the same group.

Distinguish the groups by their special members:
Because A and B are in different groups, we can uniquely identify the three groups as:
- Anjie's group (contains A plus 3 others)
- Bubay's group (contains B plus 3 others)
- The third group (contains neither A nor B, so it has 4 people)
Select members for each group from the remaining 10 people:
- For Anjie's group, we need to choose 3 more people from the 10 remaining people: $^{10}\mathrm{C}_{3}$ .
- For Bubay's group, we need to choose 3 more people from the 7 remaining people: $^{7}\mathrm{C}_{3}$ .
- The third group is automatically filled with the final 4 people: $^{4}\mathrm{C}_{4} = 1$ .
Calculate the total:

{}^{10}\mathrm{C}_{3} \times {}^{7}\mathrm{C}_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 120 \times 35 = 4200

Note: We do not need to divide by any group factorial because the presence of A in one group, B in another, and neither in the third makes all three groups distinct.

Key Takeaways

Placing fixed individuals into groups naturally breaks the symmetry among otherwise identical/unlabelled groups.
When groups are rendered distinct by the members assigned to them, standard sequential selection via combinations can be multiplied directly.

Common Mistakes

Dividing by $3!$ erroneously after distinguishing the groups with Anjie and Bubay.
Forgetting that Anjie's and Bubay's groups only require 3 additional members each, not 4.

Things to Be Careful About

Ensure the pool of remaining people decreases appropriately at each step ( $10 \to 7 \to 4$ ).

Techniques used

place constrained elements into distinct groupsuse combinations to select remaining group membersaccount for unlabeled equal-sized groups

Q12Medium-HardCalculus

A particle moves in a straight line. Its velocity, $v\text{ ms}^{-1}$ , at time $t$ seconds is given by

v = \cos t - \sin t

(a)

Find the acceleration, $a\text{ ms}^{-2}$ , when $t = \frac{\pi}{3}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Acceleration is the derivative of velocity with respect to time, so differentiate $v = \cos t - \sin t$ and substitute $t = \frac{\pi}{3}$ .

Working

a = \frac{\mathrm{d}v}{\mathrm{d}t} = -\sin t - \cos t

At $t = \frac{\pi}{3}$ :

a = -\sin\frac{\pi}{3} - \cos\frac{\pi}{3} = -\frac{\sqrt{3}}{2} - \frac{1}{2}

a = -\frac{1 + \sqrt{3}}{2} \approx -1.37 \, \text{ms}^{-2}

Answer

a = -\frac{1 + \sqrt{3}}{2} \, \text{ms}^{-2} \approx -1.37 \, \text{ms}^{-2}

Final answer

-(1 + √3)/2 ms^-2 (≈ -1.37 ms^-2)

Detailed explanation

Walkthrough

The acceleration is the rate of change of velocity, so we differentiate $v$ with respect to $t$ . Differentiating $\cos t$ gives $-\sin t$ , and differentiating $-\sin t$ gives $-\cos t$ , so $a = -\sin t - \cos t$ . Substituting the special-angle values $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos\frac{\pi}{3} = \frac{1}{2}$ gives the exact answer, which may also be quoted as the decimal $-1.37$ to 3 significant figures.

Key Takeaways

Acceleration is $\frac{\mathrm{d}v}{\mathrm{d}t}$ in kinematics.
The derivatives of sine and cosine alternate signs: $\frac{\mathrm{d}}{\mathrm{d}t}(\cos t) = -\sin t$ , $\frac{\mathrm{d}}{\mathrm{d}t}(\sin t) = \cos t$ .

Common Mistakes

Dropping a sign when differentiating: writing $a = \sin t + \cos t$ instead of $-\sin t - \cos t$ loses the accuracy mark.
Mixing up $\sin\frac{\pi}{3}$ and $\cos\frac{\pi}{3}$ values.
Rounding too early or giving fewer than 3 significant figures if using decimals.

Things to Be Careful About

The mark scheme accepts $-\frac{1+\sqrt{3}}{2}$ or the decimal $-1.366...$ rounded to at least 3 decimal places; the final answer mark is for the value itself.
Keep angles in radians throughout — the question is stated in radians and no degree conversion should appear.

Techniques used

differentiate velocity to obtain accelerationevaluate the derivative at the given time

(b)

The displacement of the particle from a fixed point $O$ at time $t$ is $s$ metres. The particle passes through $O$ when $t = 0$ .

Find the displacement at the time when the particle first changes direction after passing through $O$ .

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

The particle first changes direction when its velocity first becomes zero after $t = 0$ . Solve $v = 0$ to find that time, integrate $v$ to get $s$ (using $s = 0$ at $t = 0$ to fix the constant), then substitute the time into $s$ .

Working

First find when the particle changes direction:

v = 0 \implies \cos t - \sin t = 0 \implies \tan t = 1

The first positive solution is

t = \frac{\pi}{4}

Now integrate the velocity to get displacement:

s = \int v \, \mathrm{d}t = \int (\cos t - \sin t)\,\mathrm{d}t = \sin t + \cos t + c

Use the condition that the particle passes through $O$ ( $s = 0$ ) when $t = 0$ :

0 = \sin 0 + \cos 0 + c = 0 + 1 + c \implies c = -1

So

s = \sin t + \cos t - 1

At $t = \frac{\pi}{4}$ :

s = \sin\frac{\pi}{4} + \cos\frac{\pi}{4} - 1 = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} - 1 = \sqrt{2} - 1

Answer

s = \sqrt{2} - 1 \approx 0.414 \, \text{m}

Final answer

√2 − 1 m (≈ 0.414 m)

Detailed explanation

Walkthrough

A particle changes direction exactly when its velocity passes through zero, so step one is to solve $\cos t - \sin t = 0$ . Rearranging gives $\tan t = 1$ , whose first positive solution is $t = \frac{\pi}{4}$ — this is the first time after passing through $O$ that the particle turns around.

Step two is to recover displacement from velocity by integration. Since $s$ is the integral of $v$ , integrating term by term gives $s = \sin t + \cos t + c$ . The constant $c$ is unknown until we use the given information that $s = 0$ when $t = 0$ : substituting gives $0 = 0 + 1 + c$ , so $c = -1$ and the displacement equation is $s = \sin t + \cos t - 1$ .

Finally, substitute $t = \frac{\pi}{4}$ : both sine and cosine equal $\frac{\sqrt{2}}{2}$ there, so $s = \sqrt{2} - 1 \approx 0.414$ metres. An equivalent route is to evaluate the definite integral $\int_{0}^{\pi/4}(\cos t - \sin t)\,\mathrm{d}t$ , which gives the same result directly.

Key Takeaways

Direction changes occur where $v = 0$ (and $v$ changes sign).
Displacement from velocity requires integration plus an initial condition to pin down the constant of integration.
Special-angle values such as $\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ give exact answers.

Common Mistakes

Forgetting the constant of integration $c$ — without it the final substitution gives $\sqrt{2}$ instead of $\sqrt{2} - 1$ .
Solving $\cos t - \sin t = 0$ incorrectly, e.g. dividing by $\cos t$ without recognising it becomes $\tan t = 1$ .
Working in degrees instead of radians when finding $t = \frac{\pi}{4}$ or evaluating the integral.
Confusing displacement with distance travelled — here they coincide on $[0, \frac{\pi}{4}]$ since $v > 0$ there, but the method must still be via $s$ .
Using $t = \frac{\pi}{4}$ degrees-equivalent ( $45°$ ) inside radian-mode expressions inconsistently.

Things to Be Careful About

The final accuracy mark depends on all previous marks being correct (dep on all previous marks awarded), so check each stage.
Angles must be in radians: the mark scheme explicitly requires this for the definite-integral alternative.
The answer may be left as the exact form $\sqrt{2} - 1$ or as $0.414[21...]$ ; later mis-simplification is ignored (isw).

Techniques used

solve v = 0 to find when direction changesintegrate velocity to obtain displacementuse initial condition s = 0 at t = 0 to find the constantsubstitute the turning time into the displacement equation

(c)

Find an expression for $a$ in terms of $s$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

From part (a), $a = -\sin t - \cos t$ . From part (b), $s = \sin t + \cos t - 1$ , so $\sin t + \cos t = s + 1$ . Substitute to eliminate $t$ .

Working

a = -(\sin t + \cos t)

Since $s = \sin t + \cos t - 1$ , we have $\sin t + \cos t = s + 1$ , hence

a = -(s + 1) = -s - 1

Answer

a = -s - 1

Final answer

a = −s − 1

Detailed explanation

Walkthrough

Part (a) gave acceleration as $a = -\sin t - \cos t$ , which is just the negative of $(\sin t + \cos t)$ . Part (b) gave the displacement as $s = \sin t + \cos t - 1$ , so rearranging shows $\sin t + \cos t = s + 1$ . Substituting this into the expression for $a$ removes all dependence on $t$ , leaving $a = -(s+1) = -s - 1$ . This makes sense physically: the acceleration is proportional to how far the particle is from the point $s = -1$ , pulling it back like a spring.

Key Takeaways

To express one kinematic quantity in terms of another, eliminate the parameter $t$ between their equations.
Recognising that $a$ is the negative of the bracket appearing in $s$ makes the elimination immediate.

Common Mistakes

Sign slips: writing $a = s + 1$ instead of $-(s+1)$ .
Attempting to re-differentiate or re-integrate rather than simply eliminating $t$ using earlier results.

Things to Be Careful About

This part relies on both earlier results — carry forward $a = -\sin t - \cos t$ and $s = \sin t + \cos t - 1$ exactly as established.
Equivalent forms such as $a = -(s+1)$ are accepted (oe), and later mis-simplification is ignored (isw).

Techniques used

relate acceleration to displacement by eliminating t

Additional Mathematics 4037/23 — October/November 2024

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