Solve the following equations. (a) $$\log_2 x^2 + \log_{16} x = 18$$ (b) $$\mathrm{e}^{2x+1} - 10\mathrm{e}^{-2x-1} = 3$$

(a) x = 256 · (b) x = (-1 + ln 5)/2 = 0.305

A class of 10 students includes Abby and Ben. (a) A group of 5 students is to be selected from the class. Find the number of possible groups in the following cases. (i) There are no restrictions. (ii) The group includes both Abby and Ben. (iii) The group includes either Abby or Ben, but not both. (b) All 10 students are arranged in a line. How many arrangements are possible if there are exactly three students between Abby and Ben?

(a)(i) 252 · (a)(ii) 56 · (a)(iii) 140 · (b) 483840

4037/22 Oct/Nov 2024 — Additional Mathematics Solved Past Paper

Q13MMedium-EasySimultaneous equations

Solve the following simultaneous equations.

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

\frac{y^4}{x^5} = \frac{27}{16}

DifficultyMedium-Easy

Worked solution

Approach

From the first equation express $y$ in terms of $x$ , substitute into the second equation so that only $x$ remains, then solve.

Working

From $\dfrac{y}{x} = \dfrac{3}{2}$ :

y = \frac{3x}{2}

Substituting into $\dfrac{y^4}{x^5} = \dfrac{27}{16}$ :

\frac{\left(\frac{3x}{2}\right)^4}{x^5} = \frac{27}{16}

\frac{\frac{81}{16}x^4}{x^5} = \frac{27}{16} \quad\Longrightarrow\quad \frac{81}{16x} = \frac{27}{16}

Multiplying both sides by $16x$ :

81 = 27x \quad\Longrightarrow\quad x = 3

Then

y = \frac{3 \times 3}{2} = \frac{9}{2}

Answer

x = 3, \qquad y = \frac{9}{2}

Final answer

x = 3, y = 9/2

Detailed explanation

Walkthrough

The first equation is linear in form: dividing $y$ by $x$ gives $\frac{3}{2}$ , so $y = \frac{3x}{2}$ . This is the natural substitution to make, because it removes $y$ entirely from the second equation.

Substituting $y = \frac{3x}{2}$ into $\frac{y^4}{x^5} = \frac{27}{16}$ gives $\frac{(3x/2)^4}{x^5}$ . Raising the fraction to the fourth power gives $\frac{81}{16}x^4$ , so the left-hand side becomes $\frac{81x^4}{16x^5} = \frac{81}{16x}$ — the powers of $x$ cancel to leave a single $x$ in the denominator.

Setting $\frac{81}{16x} = \frac{27}{16}$ and multiplying through by $16x$ gives $81 = 27x$ , so $x = 3$ . Back-substituting into $y = \frac{3x}{2}$ gives $y = \frac{9}{2}$ .

Key Takeaways

With simultaneous equations where one is a simple ratio, express one variable in terms of the other and substitute.
Simplify powers carefully: $(3x/2)^4 = \frac{81}{16}x^4$ , and $x^4 \div x^5 = \frac{1}{x}$ .
Always back-substitute to find the second variable.

Common Mistakes

Mishandling the fourth power of the fraction, e.g. writing $(3x/2)^4 = \frac{3x^4}{2}$ or $\frac{81x^4}{8}$ .
Cancelling $x^4$ with $x^5$ the wrong way round, leaving $x$ in the numerator instead of the denominator.
Forgetting to back-substitute and giving only $x = 3$ (this loses an accuracy mark — the scheme awards A1 for one value, A2 for both).
Giving extra spurious solutions; the scheme requires exactly $x = 3$ and $y = \frac{9}{2}$ with no others.

Things to Be Careful About

Both values are required for full marks; one correct value alone earns only A1.
The answer may be given in equivalent forms ("oe"), e.g. $y = 4.5$ , but no other solutions are accepted.
Show the elimination step explicitly — the M1 is for a correct elimination of one unknown such as $\left(\frac{3}{2}\right)^4 \times \frac{1}{x} = \frac{27}{16}$ .

Techniques used

substitute one equation into the other to eliminate an unknownsimplify powers of a fractionsolve the resulting linear equation

Similar questions

Q2Medium-EasyCalculus

Variables $x$ and $y$ are related by the equation $y = x\sqrt{1 + 2x}$ .

(a)

Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write $y = x(1+2x)^{\frac{1}{2}}$ and differentiate using the product rule, with the chain rule for the square-root factor.

Working

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

By the product rule:

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

Combining over a common denominator:

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x + (1+2x)}{\sqrt{1+2x}} = \frac{1+3x}{\sqrt{1+2x}}

Answer

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1+3x}{\sqrt{1+2x}}

Final answer

(1 + 3x)/sqrt(1 + 2x)

Detailed explanation

Walkthrough

The function is a product of $x$ and $\sqrt{1+2x}$ , so the product rule applies: differentiate one factor at a time while the other stays unchanged, then add. The square root is rewritten as a power, $(1+2x)^{\frac{1}{2}}$ , and differentiated by the chain rule: bring down the $\frac{1}{2}$ , reduce the power by one, then multiply by the derivative of the bracket, which is $2$ . The two $2$ s cancel, leaving $(1+2x)^{-\frac{1}{2}}$ . Combining the two product-rule terms over the common denominator $\sqrt{1+2x}$ gives the compact form $\frac{1+3x}{\sqrt{1+2x}}$ , which is used in the later parts.

Key Takeaways

The product rule: $\frac{\mathrm{d}}{\mathrm{d}x}(uv) = u'v + uv'$ .
The chain rule for $(ax+b)^n$ : multiply by the derivative of the inner bracket.
Writing a derivative as a single fraction often makes later substitution and equation-solving much easier.

Common Mistakes

Forgetting the factor of $2$ from the chain rule when differentiating $(1+2x)^{\frac{1}{2}}$ .
Dropping the second term of the product rule (treating it as a chain-rule problem only).
Sign or power slips when writing the square root as a negative half power.

Things to Be Careful About

The mark scheme awards B2 for the correct derivative of $\sqrt{1+2x}$ and B1 for the completed product rule, so both stages must be visible.
The simplified single-fraction form is not required, but it is the most convenient form for parts (b) and (c); later simplification errors are ignored (isw) once the derivative is correct.

Techniques used

differentiate a square root using the chain ruleapply the product rule to x times a function of xsimplify the derivative to a single fraction

(b)

It is given that when $y = 12$ , $x = 4$ . Find the approximate change in $x$ when $y$ increases from 12 by the small amount 0.06.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the small-change relation $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ with the derivative from part (a) evaluated at $x = 4$ .

Working

At $x = 4$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1+3(4)}{\sqrt{1+2(4)}} = \frac{13}{\sqrt{9}} = \frac{13}{3}

Using $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ with $\delta y = 0.06$ :

0.06 = \frac{13}{3}\,\delta x

\delta x = \frac{0.06 \times 3}{13} = \frac{0.18}{13} = 0.01384\ldots

Answer

\delta x \approx 0.0138

Final answer

0.0138

Detailed explanation

Walkthrough

The derivative measures how fast $y$ changes per unit change in $x$ , so a small increase $\delta y$ in $y$ corresponds to a small change $\delta x$ given by $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ . Substituting $x = 4$ into the derivative from part (a) gives $\frac{13}{3}$ , since $\sqrt{1+8} = 3$ . Setting $\delta y = 0.06$ and rearranging gives $\delta x = 0.06 \div \frac{13}{3} = \frac{0.18}{13} \approx 0.0138$ . Since $y$ increases, $x$ also increases, so the change is positive.

Key Takeaways

The small-change approximation $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ links tiny changes in related variables.
The derivative must be evaluated at the given point before the approximation is applied.

Common Mistakes

Substituting $y = 12$ instead of $x = 4$ into the derivative.
Inverting the relation and computing $\frac{13}{3} \times 0.06$ instead of dividing.
Giving a negative answer, forgetting that an increase in $y$ gives an increase in $x$ here.

Things to Be Careful About

The answer is marked nfww, so the correct method must be visible; a bare correct number from wrong working scores nothing.
Give the answer to 3 significant figures: $0.0138$ (the exact value is $\frac{9}{650}$ ).

Techniques used

substitute into the derivative to find the rate of changeapply the small-change approximation delta y = (dy/dx) delta xrearrange to find the approximate change in x

(c)

Find the $x$ -coordinate of the stationary point on the curve $y = x\sqrt{1 + 2x}$ .

2M Similar

Q3Medium-EasyFactors of polynomials

DO NOT USE A CALCULATOR IN THIS QUESTION.

The polynomial $\mathrm{p}$ is defined by $\mathrm{p}(x) = ax^3 - 3x^2 - 3x + b$ , where $a$ and $b$ are constants.

(a)

Given that $x = 2$ and $x = -1$ are roots of the equation $\mathrm{p}(x) = 0$ , find $a$ and $b$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $x = 2$ and $x = -1$ are roots of $\mathrm{p}(x) = 0$ , the factor theorem gives $\mathrm{p}(2) = 0$ and $\mathrm{p}(-1) = 0$ . Substituting each root produces two linear equations in $a$ and $b$ , which are then solved simultaneously.

Working

Substituting $x = 2$ :

\mathrm{p}(2) = a(8) - 3(4) - 3(2) + b = 8a - 12 - 6 + b = 0

so

8a + b = 18

Substituting $x = -1$ :

\mathrm{p}(-1) = a(-1) - 3(1) + 3 + b = -a - 3 + 3 + b = 0

so

b = a

Substituting $b = a$ into $8a + b = 18$ :

9a = 18 \quad\Rightarrow\quad a = 2, \quad b = 2

Answer

a = 2, \quad b = 2

Final answer

a = 2, b = 2

Detailed explanation

Walkthrough

The factor theorem says that if $x = k$ is a root of $\mathrm{p}(x) = 0$ , then $\mathrm{p}(k) = 0$ . We are told both $x = 2$ and $x = -1$ are roots, so evaluating the polynomial at each of these values must give zero.

At $x = 2$ : $\mathrm{p}(2) = 8a - 12 - 6 + b = 8a + b - 18$ , so $8a + b = 18$ .

At $x = -1$ : $\mathrm{p}(-1) = -a - 3 + 3 + b = b - a$ , so $b = a$ .

Solving these together: putting $b = a$ into the first equation gives $9a = 18$ , hence $a = 2$ and $b = 2$ .

Key Takeaways

The factor theorem converts information about roots directly into equations on the coefficients.
Two unknown constants require two independent conditions — here the two given roots supply exactly those.
Solving simultaneous linear equations is often the final step of such problems.

Common Mistakes

Arithmetic slips when substituting: forgetting that $(-1)^3 = -1$ or that $(-1)^2 = +1$ .
Sign errors collecting the constant terms ( $-12 - 6 = -18$ , not $+18$ ).
Stopping after writing the two equations without actually solving them.

Things to Be Careful About

The mark scheme awards one mark per correct equation and one for the pair $a = 2$ , $b = 2$ — so both substituted equations must be shown explicitly.
No calculator is allowed here; all arithmetic is simple integer work.

Techniques used

apply the factor theorem to substitute each rootform two simultaneous linear equations in a and bsolve the simultaneous equations

(b)

Solve the equation $\mathrm{p}(x) = 0$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Carry forward from part (a): $\mathrm{p}(x) = 2x^3 - 3x^2 - 3x + 2$ , and since $x = 2$ and $x = -1$ are roots, $(x - 2)$ and $(x + 1)$ are factors. The cubic must factorise as $(x-2)(x+1)(2x - c)$ for some constant $c$ ; comparing coefficients finds $c$ , giving the third root.

Working

With $a = b = 2$ :

\mathrm{p}(x) = 2x^3 - 3x^2 - 3x + 2

Write

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

Comparing the coefficient of $x^2$ : expanding $(x-2)(x+1) = x^2 - x - 2$ , so

(x^2 - x - 2)(2x - c) = 2x^3 + (-c - 2)x^2 + \ldots

Matching $-c - 2 = -3$ gives $c = 1$ , hence the third factor is $(2x - 1)$ :

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

Setting each factor to zero:

x - 2 = 0, \qquad x + 1 = 0, \qquad 2x - 1 = 0

x = 2, \quad x = -1, \quad x = \tfrac{1}{2}

Answer

x = 2, \; \tfrac{1}{2}, \; -1

Final answer

x = 2, 1/2, -1

Detailed explanation

Walkthrough

From part (a) we know $\mathrm{p}(x) = 2x^3 - 3x^2 - 3x + 2$ and that $x = 2$ and $x = -1$ are roots, so $(x - 2)$ and $(x + 1)$ are factors. A cubic has three linear factors, and since the leading coefficient is $2$ , the third factor must be of the form $(2x - c)$ .

Expanding $(x-2)(x+1) = x^2 - x - 2$ and multiplying by $(2x - c)$ , the $x^2$ coefficient is $-c - 2$ . Setting this equal to $-3$ gives $c = 1$ , so the full factorisation is $(x-2)(x+1)(2x-1)$ .

Each factor set to zero gives a root: $x = 2$ , $x = -1$ , and $x = \frac{1}{2}$ .

(Equivalently, the sum of the roots equals $\frac{3}{2}$ , so $2 + (-1) + x = \frac{3}{2}$ gives $x = \frac{1}{2}$ directly.)

Key Takeaways

Known roots give known factors immediately via the factor theorem.
Comparing coefficients (or using the sum/product of roots of a cubic) recovers an unknown factor quickly.
A fully factorised cubic yields all its roots at once by setting each factor to zero.

Common Mistakes

Forgetting the leading coefficient $2$ and writing the third factor as $(x - c)$ instead of $(2x - c)$ .
Sign errors when matching the $x^2$ coefficient.
Giving only the two roots already found in part (a) and missing the new root $\frac{1}{2}$ .
Writing $x = \frac{1}{2}$ as $x = 2$ from misreading $2x - 1 = 0$ .

Things to Be Careful About

This part says "Solve", so all three roots are required, including the repeated-use roots from part (a).
No calculator is needed: the factorisation and root-finding are exact integer/fraction work.
The mark scheme accepts several routes (full factorisation, quadratic division, or sum/product of roots), but the final set $x = 2, \frac{1}{2}, -1$ must be complete.

Techniques used

write down the known factors from part (a)find the third factor by comparing coefficients or using the sum/product of rootssolve the remaining linear factor

Q45MMedium-EasyEquations, inequalities and graphs

Use a graphical method to solve the inequality $|2x - 8| > 4$ .

Fig. 1

DifficultyMedium-Easy

Worked solution

Approach

Draw $y = |2x - 8|$ as a V-shaped graph on the given grid, draw the line $y = 4$ , and read off where the V lies above the line.

Working

The vertex of $y = |2x - 8|$ is where $2x - 8 = 0$ , i.e. at $(4, 0)$ .

For $x < 4$ : $y = -(2x - 8) = 8 - 2x$ , giving $(0, 8)$ , $(2, 4)$ and $(8, 8)$ on this branch when extended to $x = 8$ via $y = 2x - 8$ .

So plot the two straight sections:

(0, 8) \to (4, 0) \to (8, 8)

Draw the horizontal line $y = 4$ . It meets the V-shaped graph where

|2x - 8| = 4 \implies 2x - 8 = \pm 4 \implies x = 2 \text{ or } x = 6

giving intersections at $(2, 4)$ and $(6, 4)$ — these are the critical values.

$|2x - 8| > 4$ holds where the V-graph is above the line $y = 4$ , which is to the left of $x = 2$ and to the right of $x = 6$ .

Answer

x < 2 \quad \text{or} \quad x > 6

Final answer

x < 2 or x > 6

Detailed explanation

Walkthrough

To solve $|2x - 8| > 4$ graphically, we compare two graphs: the modulus function $y = |2x - 8|$ and the constant line $y = 4$ . The inequality asks for the $x$ -values where the first graph sits strictly above the second.

First find the shape of $y = |2x - 8|$ . The expression inside the modulus is zero when $2x - 8 = 0$ , so $x = 4$ ; the vertex of the V is therefore at $(4, 0)$ . To the left of $x = 4$ the graph is the line $y = 8 - 2x$ (passing through $(0, 8)$ ), and to the right it is $y = 2x - 8$ (passing through $(8, 8)$ ). Drawing these two straight sections gives the characteristic V-shape.

Next draw the horizontal line $y = 4$ . It cuts the V at exactly two points. Algebraically these satisfy $|2x - 8| = 4$ , so $2x - 8 = 4$ giving $x = 6$ , or $2x - 8 = -4$ giving $x = 2$ . These are the critical values: $(2, 4)$ and $(6, 4)$ .

Between $x = 2$ and $x = 6$ the V dips below the line, so there the modulus is less than 4. Outside that interval — to the left of $x = 2$ and to the right of $x = 6$ — the V is above the line, which is precisely where $|2x - 8| > 4$ .

Key Takeaways

A modulus graph $y = |ax + b|$ is always a V with vertex where $ax + b = 0$ ; plotting the vertex plus one point on each branch fixes the sketch.
Solving $|f(x)| > k$ graphically means finding where the modulus curve lies above the line $y = k$ ; the intersection $x$ -values are the critical values.
Strict inequalities ( $>$ ) give strict answer intervals: the boundary values themselves are excluded.

Common Mistakes

Drawing only one branch of the V, or placing the vertex anywhere other than $(4, 0)$ — the mark scheme requires both sections through $(0,8)$ – $(2,4)$ and $(6,4)$ – $(8,8)$ , or at least a correct V with vertex $(4, 0)$ .
Forgetting to draw the line $y = 4$ (a separate B1 mark).
Giving the answer as $2 < x < 6$ — that is where the graph is below the line, the opposite region.
Writing $x \leq 2$ , $x \geq 6$ with equality signs; the inequality is strict, so the critical values are not included.
Reading the critical values off inaccurately from the grid instead of checking them against $|2x - 8| = 4$ .

Things to Be Careful About

The final A1 depends on the three earlier marks (correct V-graph, correct line, correct critical values from the intersecting graphs), so an accurate drawing is essential — the final answer scores nothing if reached from wrong working.
Both branches must be shown over the drawn range; the scheme credits 'at least one side accurate' only at M1 level, full accuracy is needed for A1.
State the answer as two separate regions joined by 'or': $x < 2$ or $x > 6$ .

Techniques used

sketch the V-shaped modulus graph with its vertex and end pointsdraw the horizontal line y = 4 on the same axesread off the intersection points as critical valuesidentify the x-values where the modulus graph lies above the line

Similar questions

Q5MediumLogarithmic and exponential functions

Solve the following equations.

(a)

$\log_2 x^2 + \log_{16} x = 18$

4M Similar

DifficultyMedium

Worked solution

Approach

Both logarithms can be written in base $2$ : use $\log_{16} x = \frac{\log_2 x}{\log_2 16} = \frac{\log_2 x}{4}$ , and $\log_2 x^2 = 2\log_2 x$ . Then solve for $\log_2 x$ .

Working

\log_{16} x = \frac{\log_2 x}{4}

So the equation becomes

2\log_2 x + \frac{1}{4}\log_2 x = 18

\frac{9}{4}\log_2 x = 18

\log_2 x = 18 \times \frac{4}{9} = 8

Converting to index form:

x = 2^8 = 256

Answer

x = 256

Final answer

x = 256

Detailed explanation

Walkthrough

The equation mixes base 2 and base 16, so the first job is to put everything in one base. Since $16 = 2^4$ , the change-of-base law gives $\log_{16} x = \frac{\log_2 x}{4}$ . The power law turns $\log_2 x^2$ into $2\log_2 x$ (valid because $x > 0$ , so no sign issue arises). The whole equation is then just a linear equation in the single unknown $\log_2 x$ : collecting terms gives $\frac{9}{4}\log_2 x = 18$ , so $\log_2 x = 8$ . Finally, 'the power of 2 that gives $x$ is 8' means $x = 2^8 = 256$ .

Key Takeaways

Change of base: $\log_b a = \frac{\log_c a}{\log_c b}$ ; when $b = c^n$ , this simplifies to $\log_{c^n} a = \frac{1}{n}\log_c a$ .
Power law: $\log_a x^k = k\log_a x$ .
Once all logs share a base, treat $\log_2 x$ as one unknown and solve linearly, then undo with index form.

Common Mistakes

Writing $\log_2 x^2 = (\log_2 x)^2$ instead of $2\log_2 x$ — the power law multiplies the log, it does not square it.
Forgetting that $\log_{16} x = \frac{1}{4}\log_2 x$ and treating it as equal to $\log_2 x$ .
Giving $x = 8$ (forgetting to convert back through the index form).
The mark scheme awards A1 only 'nfww' — the answer must come from correct working, not luck.

Things to Be Careful About

The answer must be exactly $256$ ; the scheme also accepts equivalent routes such as $x^9 = 2^{72}$ leading to $x = 2^8$ .
Show every log-law step: the M3 is awarded for correctly reaching an equation free of logarithms, so intermediate lines must be visible.
Check the solution satisfies the domain of the original equation ( $x > 0$ ) — $256$ does.

Techniques used

change all logarithms to a common base using the change-of-base lawapply the power law of logarithmscombine logarithmic terms into a single equationconvert from logarithmic to index form

(b)

$\mathrm{e}^{2x+1} - 10\mathrm{e}^{-2x-1} = 3$

4M Similar

DifficultyMedium

Worked solution

Approach

Multiply the whole equation by $\mathrm{e}^{2x+1}$ to remove the negative exponent, then recognise a quadratic in $\mathrm{e}^{2x+1}$ .

Working

Multiplying both sides by $\mathrm{e}^{2x+1}$ :

\mathrm{e}^{4x+2} - 10 = 3\mathrm{e}^{2x+1}

\left(\mathrm{e}^{2x+1}\right)^2 - 3\mathrm{e}^{2x+1} - 10 = 0

Factorising:

\left(\mathrm{e}^{2x+1} - 5\right)\left(\mathrm{e}^{2x+1} + 2\right) = 0

Since $\mathrm{e}^{2x+1} > 0$ always, only $\mathrm{e}^{2x+1} = 5$ is valid. Taking natural logarithms:

2x + 1 = \ln 5

x = \frac{-1 + \ln 5}{2} \approx 0.305

Answer

x = \frac{-1 + \ln 5}{2} \approx 0.305

Final answer

x = (-1 + ln 5)/2 = 0.305

Detailed explanation

Walkthrough

The term $\mathrm{e}^{-2x-1}$ is the reciprocal of $\mathrm{e}^{2x+1}$ , which hints at a hidden quadratic. Multiplying the entire equation by $\mathrm{e}^{2x+1}$ clears the fraction: $\mathrm{e}^{2x+1}\cdot\mathrm{e}^{2x+1} = \mathrm{e}^{4x+2} = \left(\mathrm{e}^{2x+1}\right)^2$ , giving a standard three-term quadratic in $u = \mathrm{e}^{2x+1}$ : $u^2 - 3u - 10 = 0$ . This factorises as $(u-5)(u+2)=0$ . The root $u = -2$ must be rejected because an exponential is always positive, leaving $\mathrm{e}^{2x+1} = 5$ . Taking $\ln$ of both sides gives $2x+1 = \ln 5$ , so $x = \frac{\ln 5 - 1}{2} \approx 0.305$ .

Key Takeaways

Equations mixing $\mathrm{e}^{kx}$ and $\mathrm{e}^{-kx}$ become quadratics after multiplying by $\mathrm{e}^{kx}$ .
Exponential functions are strictly positive, so negative roots for the exponential must be discarded.
Undoing an exponential requires natural logarithms: $\mathrm{e}^{y} = a \iff y = \ln a$ .

Common Mistakes

Forgetting to multiply EVERY term by $\mathrm{e}^{2x+1}$ (e.g. missing the $-10$ or the $3$ ).
Keeping the invalid root $\mathrm{e}^{2x+1} = -2$ and trying to take its logarithm — the scheme explicitly requires 'no other solution'.
Sign slips when rearranging: $2x + 1 = \ln 5$ gives $x = \frac{\ln 5 - 1}{2}$ , not $\frac{\ln 5 + 1}{2}$ .
Writing $\ln(\mathrm{e}^{2x+1})$ as something other than $2x+1$ .

Things to Be Careful About

The exact form $\frac{-1+\ln 5}{2}$ is accepted, as is the decimal $0.305$ (or $0.3047\ldots$ ); later mis-simplification is ignored (isw).
Only ONE solution exists — the negative root of the quadratic must be rejected because $\mathrm{e}^{2x+1} > 0$ for all real $x$ .
Show the quadratic stage explicitly ( $\mathrm{B1}$ ) before factorising ( $\mathrm{M1}$ ).

Techniques used

multiply through by an exponential factor to clear negative indicesrecognise a hidden quadratic in e^(2x+1)factorise the quadratictake natural logarithms to solve for x

Q65MMedium[Legacy] Indices and surds

DO NOT USE A CALCULATOR IN THIS QUESTION.

Write $(5 - \sqrt{3})(\sqrt{6} + \sqrt{2})^{-2}$ in the form $a + b\sqrt{3}$ , where $a$ and $b$ are constants.

DifficultyMedium

Worked solution

Approach

Rewrite the negative power as a fraction, expand $(\sqrt{6}+\sqrt{2})^2$ , then rationalise the denominator by multiplying top and bottom by its conjugate.

Working

(5 - \sqrt{3})(\sqrt{6} + \sqrt{2})^{-2} = \frac{5 - \sqrt{3}}{(\sqrt{6} + \sqrt{2})^2}

Expand the denominator:

(\sqrt{6} + \sqrt{2})^2 = 6 + 2\sqrt{12} + 2 = 8 + 2\sqrt{12} = 8 + 4\sqrt{3}

Rationalise by multiplying numerator and denominator by $8 - 4\sqrt{3}$ :

\frac{5 - \sqrt{3}}{8 + 4\sqrt{3}} \times \frac{8 - 4\sqrt{3}}{8 - 4\sqrt{3}}

Expand the numerator and use the difference of two squares on the denominator:

\frac{40 - 20\sqrt{3} - 8\sqrt{3} + 12}{8^2 - (4\sqrt{3})^2}

= \frac{52 - 28\sqrt{3}}{64 - 48} = \frac{52 - 28\sqrt{3}}{16}

Divide each term by 16:

= \frac{52}{16} - \frac{28}{16}\sqrt{3} = \frac{13}{4} - \frac{7}{4}\sqrt{3}

Answer

\frac{13}{4} - \frac{7}{4}\sqrt{3}

Final answer

13/4 - (7/4)√3

Detailed explanation

Walkthrough

The negative index means we take a reciprocal, so the expression becomes a fraction with $(\sqrt{6}+\sqrt{2})^2$ in the denominator. Expanding that square gives $6 + 2\sqrt{12} + 2$ , and since $\sqrt{12} = 2\sqrt{3}$ , the denominator is $8 + 4\sqrt{3}$ .

To remove the surd from the denominator we multiply top and bottom by the conjugate $8 - 4\sqrt{3}$ . The denominator then becomes a difference of two squares: $64 - 48 = 16$ . The numerator expands to $40 - 20\sqrt{3} - 8\sqrt{3} + 12 = 52 - 28\sqrt{3}$ .

Finally, dividing both terms of the numerator by 16 gives $\frac{13}{4} - \frac{7}{4}\sqrt{3}$ , which is exactly the form $a + b\sqrt{3}$ with $a = \frac{13}{4}$ and $b = -\frac{7}{4}$ .

Key Takeaways

A negative index turns an expression into a reciprocal — always rewrite it as a fraction first.
$(\sqrt{x}+\sqrt{y})^2 = x + y + 2\sqrt{xy}$ ; simplify the inner surd before proceeding.
Multiplying by the conjugate over itself removes surds from a denominator via the difference of two squares.
Answers must be collected into the exact requested form, here $a + b\sqrt{3}$ .

Common Mistakes

Writing $2\sqrt{12}$ as $\sqrt{24}$ or failing to simplify it to $4\sqrt{3}$ .
Sign slips when multiplying out $(5-\sqrt{3})(8-4\sqrt{3})$ — all four products must appear.
Computing $(4\sqrt{3})^2$ as $4 \times 3 = 12$ instead of $16 \times 3 = 48$ .
Stopping at $\frac{52 - 28\sqrt{3}}{16}$ instead of simplifying to the required $a + b\sqrt{3}$ form.
The mark scheme awards the final accuracy mark only if reached from correct working (nfww).

Things to Be Careful About

This is a no-calculator question: every step must be exact, with fractions and surds kept in exact form throughout.
Show every line of expansion — the scheme pays separate marks for rewriting as a fraction, for the expanded denominator, for the conjugate multiplication, and for the expanded numerator over the difference-of-squares denominator.
The final answer must be fully simplified: $\frac{13}{4} - \frac{7}{4}\sqrt{3}$ (or equivalent), not an unsimplified fraction.

Techniques used

rewrite a negative power as a reciprocalexpand a bracket of surds using the difference of two squares patternrationalise the denominator by multiplying by the conjugatesimplify the resulting surd expression into the required form

Similar questions

Q7MediumPermutations and combinations

A class of 10 students includes Abby and Ben.

(a)

A group of 5 students is to be selected from the class. Find the number of possible groups in the following cases.

5M

(i)

There are no restrictions.

1M Similar

(ii)

The group includes both Abby and Ben.

2M Similar

(iii)

The group includes either Abby or Ben, but not both.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

"Either Abby or Ben, but not both" splits into two disjoint cases: Abby in without Ben, or Ben in without Abby. Count each and add.

Working

Case 1 — Abby in, Ben out: choose the other 4 from the remaining 8:

{}^{8}\mathrm{C}_{4} = 70

Case 2 — Ben in, Abby out: again choose 4 from 8:

{}^{8}\mathrm{C}_{4} = 70

Total:

70 + 70 = {}^{8}\mathrm{C}_{4} \times 2 = 140

(Equivalently, ${}^{10}\mathrm{C}_{5} - {}^{8}\mathrm{C}_{5} - {}^{8}\mathrm{C}_{3} = 252 - 56 - 56 = 140$ .)

Answer

140

Final answer

140

Detailed explanation

Walkthrough

The condition excludes two situations: groups containing neither, and groups containing both. The cleanest route is to split into two disjoint cases:

Abby in, Ben out: Abby fills one place, so choose the remaining 4 from the other 8 students: ${}^{8}\mathrm{C}_{4} = 70$ .
Ben in, Abby out: by symmetry, also ${}^{8}\mathrm{C}_{4} = 70$ .

Adding the disjoint cases gives $70 + 70 = 140$ . An equally valid check is subtraction: all groups ( $252$ ) minus those with both ( $56$ ) minus those with neither ( ${}^{8}\mathrm{C}_{5} = 56$ ) equals $140$ .

Key Takeaways

"Either ... but not both" means multiply the single-inclusion count by 2 when the two people are symmetric.
Complementary counting (total minus forbidden cases) is often quicker than direct casework.

Common Mistakes

Forgetting the factor of 2 and answering 70.
Subtracting only one of the forbidden counts in the complementary method.
Including groups with both Abby and Ben by mistake.

Things to Be Careful About

B1 is given for either accepted route ( ${}^{8}\mathrm{C}_{4} \times 2$ or the subtraction form); the final answer must be exactly 140.

Techniques used

split into disjoint casescount each case with combinationsadd the cases

(b)

All 10 students are arranged in a line. How many arrangements are possible if there are exactly three students between Abby and Ben?

3M Similar

DifficultyMedium

Worked solution

Approach

Treat Abby and Ben as a pair with exactly three students between them. First count where such a pair can sit in the line of 10, then arrange the other 8 students, and multiply.

Working

If Abby sits in position $k$ , Ben must sit in position $k + 4$ (three students between them). The possible values of $k$ run from 1 to 6, giving 6 placements for the ordered pair (Abby before Ben).

The same 6 placements exist with Ben before Abby, so there are $6 \times 2 = 12$ ways to place the pair.

The remaining 8 students can be arranged in the remaining 8 positions in

8! = 40\,320

ways. Total arrangements:

12 \times 8! = 12 \times 40\,320 = 483\,840

Answer

483\,840

Final answer

483840

Detailed explanation

Walkthrough

We need lines of all 10 students in which exactly three students separate Abby and Ben.

Step 1 — position the pair. If Abby occupies position $k$ , Ben must occupy position $k+4$ , since positions $k+1$ , $k+2$ , $k+3$ hold the three students between them. Position $k+4$ must exist within the line of 10, so $k$ can be $1, 2, 3, 4, 5, 6$ — that is 6 placements with Abby on the left. Repeating with Ben on the left doubles this: $6 \times 2 = 12$ ways to place the pair.

Step 2 — arrange everyone else. Once the pair's positions are fixed, the other 8 students fill the remaining 8 slots in any order:

8! = 40\,320

Step 3 — multiply. The placement of the pair and the arrangement of the rest are independent:

12 \times 40\,320 = 483\,840

Key Takeaways

"Exactly $m$ between A and B" means their positions differ by $m + 1$ .
Count valid starting positions systematically, remember the pair can occur in either order, then multiply by the factorial of the remaining people.

Common Mistakes

Forgetting the factor of 2 for the order of Abby and Ben (giving $241\,920$ ).
Miscounting the number of valid gaps (using 5 or 7 instead of 6).
Treating it as a simple $10!$ arrangement or using combinations instead of permutations.

Things to Be Careful About

The mark scheme awards B1 for $8!$ alone, B2 for $8! \times 6$ or $8! \times 2$ , and full marks only for the complete product $483\,840$ — every factor must appear.

Techniques used

treat Abby and Ben plus the gap as a block structurecount positions for the pairarrange the remaining studentsmultiply the independent choices

Q86MMediumTrigonometry

Solve the equation $\cot^2 2\theta + 3\operatorname{cosec} 2\theta = 9$ for $-90° \leq \theta \leq 90°$ .

DifficultyMedium

Worked solution

Approach

Use the identity $\cot^2 A = \operatorname{cosec}^2 A - 1$ to rewrite the equation as a quadratic in $\operatorname{cosec} 2\theta$ , solve it, then convert each root back to angles.

Working

\cot^2 2\theta + 3\operatorname{cosec} 2\theta = 9

Using $\cot^2 2\theta = \operatorname{cosec}^2 2\theta - 1$ :

\operatorname{cosec}^2 2\theta - 1 + 3\operatorname{cosec} 2\theta - 9 = 0

\operatorname{cosec}^2 2\theta + 3\operatorname{cosec} 2\theta - 10 = 0

Factorise:

\left(\operatorname{cosec} 2\theta - 2\right)\left(\operatorname{cosec} 2\theta + 5\right) = 0

So $\operatorname{cosec} 2\theta = 2$ , giving $\sin 2\theta = \frac{1}{2}$ ; or $\operatorname{cosec} 2\theta = -5$ , giving $\sin 2\theta = -\frac{1}{5}$ .

Since $-90° \leq \theta \leq 90°$ , we need $-180° \leq 2\theta \leq 180°$ .

For $\sin 2\theta = \frac{1}{2}$ :

2\theta = 30° \quad \text{or} \quad 150° \quad \Rightarrow \quad \theta = 15°,\ 75°

For $\sin 2\theta = -\frac{1}{5}$ :

2\theta = -11.537° \quad \text{or} \quad -(180° - 11.537°) = -168.463°

\theta = -5.769° \approx -5.77°, \qquad \theta = -84.232° \approx -84.2°

All four values lie in range and there are no others.

Answer

\theta = 15°,\ 75°,\ -5.77°,\ -84.2°

Final answer

θ = 15°, 75°, -5.77°, -84.2°

Detailed explanation

Walkthrough

The equation mixes $\cot^2 2\theta$ and $\operatorname{cosec} 2\theta$ . The identity $\cot^2 A = \operatorname{cosec}^2 A - 1$ (the Pythagorean identity divided by $\sin^2 A$ ) converts everything into one function, $\operatorname{cosec} 2\theta$ . Substituting and collecting terms gives the quadratic $\operatorname{cosec}^2 2\theta + 3\operatorname{cosec} 2\theta - 10 = 0$ , which factorises neatly as $(\operatorname{cosec} 2\theta - 2)(\operatorname{cosec} 2\theta + 5) = 0$ .

Each root is converted via $\sin 2\theta = 1/\operatorname{cosec} 2\theta$ : so $\sin 2\theta = \frac{1}{2}$ or $-\frac{1}{5}$ .

Because the argument is the double angle $2\theta$ , the interval $-90° \leq \theta \leq 90°$ doubles to $-180° \leq 2\theta \leq 180°$ . For $\sin 2\theta = \frac{1}{2}$ the solutions are $2\theta = 30°$ and $150°$ , giving $\theta = 15°$ and $75°$ . For $\sin 2\theta = -\frac{1}{5}$ the principal value is $2\theta = -11.537°$ , and the second solution in the interval comes from the symmetry $\sin(-x) = -\sin x$ : $2\theta = -168.463°$ , giving $\theta = -5.77°$ and $-84.2°$ .

Key Takeaways

The identity $\cot^2 A = \operatorname{cosec}^2 A - 1$ turns mixed cot/cosec equations into quadratics in one function.
A quadratic in a reciprocal trig function must be converted back through $\sin = 1/\operatorname{cosec}$ before finding angles.
With a double angle, always double the given interval before listing solutions, then halve at the end.
The sine graph's symmetries ( $\sin(180° - x) = \sin x$ , $\sin(-x) = -\sin x$ ) generate all solutions in an interval.

Common Mistakes

Forgetting to double the interval: solving only for $2\theta$ in $[-90°, 90°]$ loses the $150°$ and $-168.463°$ solutions (the scheme penalises extras and missing values — "no other angles in range").
Stopping at $2\theta = 30°, 150°, -11.5°, -168.5°$ without halving to get $\theta$ .
Discarding $\operatorname{cosec} 2\theta = -5$ because it looks awkward — it yields two valid answers.
Sign slips when forming the quadratic: $\cot^2 - 1$ becomes $\operatorname{cosec}^2 - 10$ after moving the 9 across.
Rounding too early: keep $-11.537°$ unrounded until halving, then give angles to 1 decimal place ( $-5.77°$ may be quoted as $-5.8°$ ).
Dividing by $\sin 2\theta$ instead of using the identity would lose solutions.

Things to Be Careful About

The mark scheme awards B2 for the correct single-function quadratic, M1 for solving/factorising it, and A3 for all four angles with no extras — "nfww" means no marks from wrong working.
Angles in degrees are required to 1 decimal place where not exact: $15°$ and $75°$ exact, the others $-5.77°$ (or $-5.8°$ ) and $-84.2°$ .
Check every candidate angle lies in $-90° \leq \theta \leq 90°$ after halving; here all four do, but e.g. $2\theta = 191.5°$ would be rejected.

Techniques used

apply the identity cot^2 A = cosec^2 A - 1form and factorise a quadratic in cosecconvert cosec roots to sine valuessolve for the double angle over the extended intervalhalve to find theta and discard out-of-range angles

Similar questions

Q9Medium-EasyCalculus

In this question time is measured in seconds.

(a)

A particle is moving in a straight line with constant velocity of $6\,\text{ms}^{-1}$ . At time $t = 0$ , it passes a fixed point $A$ . At time $t = 5$ it suddenly changes direction and moves with a different constant velocity along the same straight line. It passes the point $A$ again at time $t = 15$ . Sketch the velocity–time graph for the motion.

Fig. 2

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For constant velocity, displacement is given by $\Delta s = v \times t$ . The particle travels away from point $A$ for $5\,\text{s}$ at $6\,\text{ms}^{-1}$ , and then returns to $A$ during the interval $t = 5$ to $t = 15$ ( $10\,\text{s}$ ) with a new constant velocity $v_2$ . Find $v_2$ so that the net displacement over $15\,\text{s}$ is zero, and sketch the two horizontal segments on the velocity–time graph.

Working

First stage of motion ( $0 \leq t \leq 5$ ):

s_1 = 6 \times 5 = 30\,\text{m}

Second stage of motion ( $5 \leq t \leq 15$ , duration $\Delta t = 10\,\text{s}$ ):
To return to $A$ , the displacement must be $-30\,\text{m}$ :

s_2 = v_2 \times (15 - 5) = 10v_2

Setting $s_1 + s_2 = 0$ :

30 + 10v_2 = 0 \implies v_2 = -3\,\text{ms}^{-1}

The graph consists of two horizontal straight-line segments:

$v = 6$ for $0 \leq t \leq 5$
$v = -3$ for $5 \leq t \leq 15$

Answer

Horizontal line at $v = 6$ from $t = 0$ to $t = 5$ , and horizontal line at $v = -3$ from $t = 5$ to $t = 15$ .

Final answer

Horizontal line at v = 6 for 0 <= t <= 5, and horizontal line at v = -3 for 5 <= t <= 15

Detailed explanation

Walkthrough

For the first $5$ seconds, the particle moves at a constant velocity of $6\,\text{ms}^{-1}$ . The displacement during this stage is the area under the velocity–time graph:

\Delta s_1 = 6 \times 5 = 30\,\text{m}

At $t = 5$ , the particle is $30\,\text{m}$ away from $A$ . It returns to $A$ at $t = 15$ , which means it travels for an additional $10\,\text{s}$ ( $15 - 5 = 10\,\text{s}$ ) with constant velocity $v_2$ .
To return to $A$ , the displacement in the second stage must cancel the first stage:

\Delta s_2 = -30\,\text{m}

v_2 = \frac{-30}{10} = -3\,\text{ms}^{-1}

On the grid, sketch a horizontal line at $v = 6$ from $t = 0$ to $t = 5$ , and a horizontal line at $v = -3$ from $t = 5$ to $t = 15$ .

Key Takeaways

On a velocity–time graph, constant velocity is represented by a horizontal straight line.
Displacement is the signed area between the velocity–time graph and the time axis (areas above the axis represent positive displacement, while areas below represent negative displacement).
Returning to the starting point means total displacement is zero, so the positive and negative areas must be equal in magnitude.

Common Mistakes

Plotting a positive return velocity ( $v = 3$ ) instead of $v = -3$ , forgetting that returning requires opposite direction.
Connecting $(5, 6)$ to $(5, -3)$ with a sloped line instead of recognising that the velocity changes suddenly, meaning two separate horizontal segments are drawn.
Dividing by $15$ instead of the duration of the second stage ( $10\,\text{s}$ ).

Things to Be Careful About

Ensure the time interval for the second segment is accurately drawn from $t = 5$ to $t = 15$ .
The grid has major markings at $v = 6$ and $v = -6$ ; $v = -3$ is exactly halfway between $0$ and $-6$ .

Techniques used

calculate displacement from constant velocitydetermine return velocity to achieve net zero displacementsketch horizontal segments on a velocity-time graph

(b)

Another particle is moving in a straight line with constant acceleration. At time $t = 0$ it passes a fixed point $B$ with velocity $-8\,\text{ms}^{-1}$ . It passes the point $B$ again at time $t = 20$ . Sketch the velocity–time graph for the motion.

Fig. 3

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For motion with constant acceleration, the velocity–time graph is a single straight line of equation $v = u + at$ . The particle starts at $t = 0$ with $u = -8\,\text{ms}^{-1}$ and returns to $B$ at $t = 20$ , meaning the total displacement at $t = 20$ is $s = 0$ . Use the constant acceleration formula $s = ut + \frac{1}{2}at^2$ to find the acceleration $a$ , determine the velocity at $t = 20$ and the time when $v = 0$ , then sketch the line.

Working

Using $s = ut + \frac{1}{2}at^2$ with $u = -8$ , $t = 20$ , and $s = 0$ :

0 = (-8)(20) + \frac{1}{2}a(20)^2

0 = -160 + 200a \implies a = \frac{160}{200} = 0.8\,\text{ms}^{-2}

Find the key points on the velocity–time graph $v = -8 + 0.8t$ :

At $t = 0$ : $v = -8\,\text{ms}^{-1} \implies (0, -8)$
When $v = 0$ : $0 = -8 + 0.8t \implies t = 10\,\text{s} \implies (10, 0)$
At $t = 20$ : $v = -8 + 0.8(20) = 8\,\text{ms}^{-1} \implies (20, 8)$

Draw a single straight line passing through $(0, -8)$ , $(10, 0)$ , and $(20, 8)$ .

Answer

A single straight line with positive gradient starting at $(0, -8)$ , crossing the $t$ -axis at $(10, 0)$ , and ending at $(20, 8)$ .

Final answer

Single straight line passing through (0, -8), (10, 0) and (20, 8)

Detailed explanation

Walkthrough

Constant acceleration implies that the velocity–time graph is a single straight line.
The initial velocity is $u = -8\,\text{ms}^{-1}$ at $t = 0$ , giving the initial point $(0, -8)$ .
Returning to point $B$ at $t = 20$ means the total displacement over the interval $[0, 20]$ is zero ( $s = 0$ ).
Using $s = ut + \frac{1}{2}at^2$ :

0 = -8(20) + \frac{1}{2}a(20^2) = -160 + 200a \implies a = 0.8\,\text{ms}^{-2}

Alternatively, by symmetry of constant acceleration, if net displacement is zero over $[0, T]$ , the triangular area below the $t$ -axis must equal the triangular area above the $t$ -axis. Thus, the line must cross the $t$ -axis halfway through the motion at $t = 10\,\text{s}$ , and the final velocity must be equal in magnitude but opposite in sign to the initial velocity, giving $v(20) = 8\,\text{ms}^{-1}$ .
Draw a single straight line from $(0, -8)$ through $(10, 0)$ up to $(20, 8)$ .

Key Takeaways

Constant acceleration corresponds to a straight line with constant gradient $a$ on a velocity–time graph.
Zero net displacement over an interval $[0, T]$ for constant acceleration means the graph is symmetric about $t = \frac{T}{2}$ , so the $t$ -intercept is $\left(\frac{T}{2}, 0\right)$ and the final velocity is $-u$ .

Common Mistakes

Drawing two separate line segments (e.g., a V-shape) instead of a single continuous straight line for constant acceleration.
Drawing a line with negative gradient starting at $(0, 8)$ and ending at $(20, -8)$ .
Misidentifying the $t$ -intercept (not at $t = 10$ ).

Things to Be Careful About

Ensure the line is drawn straight with a ruler across the whole range $0 \leq t \leq 20$ .
Clearly mark the key coordinates $(0, -8)$ , $(10, 0)$ , and $(20, 8)$ on the grid.

Techniques used

use constant acceleration kinematics equationsdetermine final velocity and time of zero velocity from zero net displacementsketch a single linear velocity-time graph

Q10MediumQuadratic functionsCalculusTrigonometry

The diagram shows part of the curve $y = x - \frac{x^2}{4}$ and the line $y = -4$ . The curve and the line intersect at the point $A$ .

Fig. 4

(a)

The maximum point on the curve is at a perpendicular distance $h$ from the line $y = -4$ .
Find the value of $h$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Complete the square on $y = x - \frac{x^2}{4}$ to find the maximum point, then find the vertical distance from that point down to the line $y = -4$ .

Working

Complete the square:

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

The maximum occurs at $x = 2$ , where $y = 1$ , so the maximum point is $(2, 1)$ .

The perpendicular distance $h$ between this point and the line $y = -4$ is the difference in $y$ -values:

h = 1 - (-4) = 5

Answer

h = 5

Final answer

5

Detailed explanation

Walkthrough

The curve is a downward-opening parabola, so its maximum can be found by completing the square. Writing $y = x - \frac{x^2}{4} = -\frac{1}{4}(x^2 - 4x)$ and completing the square inside the bracket gives $-\frac{1}{4}(x-2)^2 + 1$ . Since $-\frac{1}{4}(x-2)^2 \leq 0$ , the largest value of $y$ is $1$ , occurring when $x = 2$ : the maximum point is $(2, 1)$ . The line $y = -4$ is horizontal, so the perpendicular distance from any point to it is simply the vertical gap: $h = 1 - (-4) = 5$ .

Key Takeaways

Completing the square immediately reveals the vertex (maximum or minimum) of a parabola.
For a horizontal line $y = c$ , the perpendicular distance from a point is just the difference in $y$ -coordinates.

Common Mistakes

Sign slips when completing the square (forgetting the factor $-\frac{1}{4}$ outside).
Computing $h$ as $-4 - 1 = -5$ instead of taking the positive distance $1 - (-4) = 5$ .
Finding only the $x$ -coordinate of the maximum ( $x = 2$ ) without evaluating $y$ — the scheme awards B2 for this but not full marks.
Using the wrong root of the factorised form $x(1 - \frac{x}{4}) = 0$ ; the roots $x = 0$ and $x = 4$ locate the axis of symmetry $x = 2$ by symmetry, not the maximum itself.

Things to Be Careful About

The mark scheme says "nfww": $h = 5$ scores nothing from wrong working, so the method for finding the maximum must be shown.
The answer must be positive — $h$ is a distance.

Techniques used

complete the square to locate the maximum pointread off the maximum y-valuesubtract the y-values of the maximum point and the line

(b)

Find the exact $x$ -coordinate of $A$ .

3M Similar

DifficultyMedium

Worked solution

Approach

At the intersection point $A$ the curve and the line have the same coordinates, so set $x - \frac{x^2}{4} = -4$ and solve the resulting quadratic. From the diagram, $A$ lies in the fourth quadrant, so take the larger root.

Working

Set the curve equal to the line:

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

Multiply through by $-4$ and rearrange into standard form:

x^2 - 4x - 16 = 0

Solve using the quadratic formula:

x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-16)}}{2} = \frac{4 \pm \sqrt{16 + 64}}{2} = \frac{4 \pm \sqrt{80}}{2}

x = \frac{4 \pm 4\sqrt{5}}{2} = 2 \pm 2\sqrt{5}

Since $A$ is in the fourth quadrant (positive $x$ ), take the positive root:

x = 2 + 2\sqrt{5}

Answer

x = 2 + 2\sqrt{5}

Final answer

2 + 2√5

Detailed explanation

Walkthrough

Point $A$ lies on both the curve and the line, so its coordinates satisfy both equations simultaneously. Setting $x - \frac{x^2}{4} = -4$ and multiplying by $-4$ gives the three-term quadratic $x^2 - 4x - 16 = 0$ . This does not factorise over the integers, so the quadratic formula is used: the discriminant is $16 + 64 = 80 = 16 \times 5$ , giving $\sqrt{80} = 4\sqrt{5}$ and hence $x = 2 \pm 2\sqrt{5}$ . The diagram shows $A$ in the fourth quadrant with a positive $x$ -coordinate, so the answer is $x = 2 + 2\sqrt{5}$ .

Key Takeaways

Intersections are found by equating the two expressions for $y$ .
When a quadratic has no rational factors, use the formula and simplify the surd exactly.
A sketch tells you which root to keep.

Common Mistakes

Taking the negative root $2 - 2\sqrt{5}$ , which contradicts the position of $A$ shown in the figure.
Sign errors forming the quadratic: forgetting to move everything to one side before applying the formula.
Leaving the answer as $\frac{4 + \sqrt{80}}{2}$ instead of simplifying to $2 + 2\sqrt{5}$ .
Rounding to a decimal when an exact surd is demanded.

Things to Be Careful About

The question asks for the exact $x$ -coordinate, so the surd form $2 + 2\sqrt{5}$ is required; a decimal such as $6.47$ would not earn the final accuracy mark.
The mark scheme also accepts completing the square: $y = 1 - \frac{1}{4}(x-2)^2 = -4$ leads to the same result.

Techniques used

set curve equal to line to form a quadraticsolve the quadratic using the formulaselect the correct root matching the diagram

(c)

Find the acute angle between the tangent to the curve at $A$ and the line $y = -4$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate the curve to obtain the gradient function, evaluate it at $x = 2 + 2\sqrt{5}$ to get the tangent's gradient, then use $\tan\theta = |m|$ since the line $y = -4$ is horizontal.

Working

Differentiate:

\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{x}{2}

Substitute $x = 2 + 2\sqrt{5}$ :

m = 1 - \frac{2 + 2\sqrt{5}}{2} = 1 - 1 - \sqrt{5} = -\sqrt{5}

The line $y = -4$ is horizontal, so the acute angle $\theta$ between the tangent and the line satisfies:

\tan\theta = \left|-\sqrt{5}\right| = \sqrt{5}

\theta = \tan^{-1}\sqrt{5} = 65.905...°

Answer

\theta = 65.9° \quad (1 \text{ d.p.})

Final answer

65.9 degrees

Detailed explanation

Walkthrough

The gradient of the tangent at any point comes from the derivative: differentiating $y = x - \frac{x^2}{4}$ gives $\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{x}{2}$ . Substituting the $x$ -coordinate of $A$ found in part (b), $x = 2 + 2\sqrt{5}$ , gives $m = 1 - (2+2\sqrt{5})/2 = -\sqrt{5}$ . Because the line $y = -4$ is horizontal, the angle between the tangent and the line equals the angle the tangent makes with the horizontal direction, whose tangent is the magnitude of the gradient: $\tan\theta = \sqrt{5}$ , so $\theta = \tan^{-1}\sqrt{5} \approx 65.9°$ .

Key Takeaways

The derivative evaluated at a point gives the gradient of the tangent there.
A gradient $m$ corresponds to an angle $\tan^{-1}|m|$ with the horizontal; a negative gradient means the tangent slopes downwards, so the acute angle uses the modulus.
Results from earlier parts are carried forward — here the exact $x$ -coordinate from part (b).

Common Mistakes

Dropping the minus sign and reporting $\tan^{-1}(-\sqrt{5}) \approx -65.9°$ instead of the acute angle $65.9°$ .
Arithmetic slips substituting the surd: $1 - \frac{2+2\sqrt{5}}{2}$ simplifies cleanly to $-\sqrt{5}$ , not $-\sqrt{5}/2$ or similar.
Giving the answer in radians ( $1.15$ ) when the question context expects degrees, or rounding incorrectly.
Using the gradient at $x = 2$ (the maximum, where the gradient is $0$ ) instead of at $A$ .

Things to Be Careful About

Angles in degrees should be given to 1 decimal place: $65.9°$ (the radian equivalent $1.15$ is accepted by the scheme if radians were used consistently).
The question asks for the acute angle, so the magnitude of the gradient must be used.
The final M1 depends on having both the correct derivative and the correct $x$ -coordinate of $A$ , so part (b) must be right (or correctly followed through).

Techniques used

differentiate the curve to get the gradient functionevaluate the gradient at the x-coordinate of Ause tan inverse to find the angle between the tangent and the horizontal

Q11MediumVectors in two dimensions

In this question $\mathbf{i}$ is a unit vector in the positive $x$ -direction and $\mathbf{j}$ is a unit vector in the positive $y$ -direction. Time is in seconds and distances are in metres.

The diagram shows the initial positions and velocities of two particles, $A$ and $B$ , that move in the $x$ - $y$ plane.

Fig. 5

Particle $A$ starts from the origin $O$ at time $t = 0$ . It moves with constant speed $10\,\text{ms}^{-1}$ in the direction $60°$ above the $x$ -axis.

(a)

Find the exact values of the components of the velocity of particle $A$ in the $x$ -direction and the $y$ -direction.

2M Similar

DifficultyEasy

Worked solution

Approach

Resolve the speed $10\,\text{ms}^{-1}$ at $60°$ above the $x$ -axis into components using cosine for the $x$ -component and sine for the $y$ -component.

Working

v_x = 10\cos 60° = 10 \times \frac{1}{2} = 5

v_y = 10\sin 60° = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}

Answer

The velocity of $A$ is $\begin{pmatrix}5 \\ 5\sqrt{3}\end{pmatrix}\,\text{ms}^{-1}$ .

Final answer

x-component 5 ms^-1, y-component 5√3 ms^-1

Detailed explanation

Walkthrough

Particle $A$ moves at constant speed $10\,\text{ms}^{-1}$ in a direction making $60°$ with the positive $x$ -axis. To split any vector of magnitude $v$ at angle $\theta$ to the $x$ -axis, the horizontal component is $v\cos\theta$ (the adjacent side) and the vertical component is $v\sin\theta$ (the opposite side). Using the exact special-angle values $\cos 60° = \frac{1}{2}$ and $\sin 60° = \frac{\sqrt{3}}{2}$ gives the components $5$ and $5\sqrt{3}$ exactly — no calculator needed.

Key Takeaways

Resolving a vector: $x$ -component $= v\cos\theta$ , $y$ -component $= v\sin\theta$ when $\theta$ is measured from the positive $x$ -axis.
Special angles ( $30°$ , $45°$ , $60°$ ) should give exact surd answers.

Common Mistakes

Swapping sine and cosine, giving $5\sqrt{3}$ horizontally and $5$ vertically.
Giving decimal approximations instead of the exact form $5\sqrt{3}$ .

Things to Be Careful About

The mark scheme awards B2 for both correct, B1 for either one correct — so even if unsure of one, state the other.
The question says "exact values", so leave $5\sqrt{3}$ as a surd; do not round.

Techniques used

resolve a velocity into horizontal and vertical componentsuse exact values of sin 60° and cos 60°

(b)

Find, in terms of $t$ , the position vector of particle $A$ at time $t$ .

1M Similar

(c)

Particle $B$ starts from the point $(2\sqrt{3}, 9)$ at time $t = 0$ . It moves with constant speed $\frac{5}{3}\,\text{ms}^{-1}$ parallel to the positive $x$ -axis.

Find, in terms of $t$ , the position vector of particle $B$ at time $t$ .

2M Similar

DifficultyEasy

Worked solution

Approach

$B$ starts at $(2\sqrt{3}, 9)$ and moves parallel to the positive $x$ -axis at $\frac{5}{3}\,\text{ms}^{-1}$ , so its velocity is $\begin{pmatrix}\frac{5}{3} \\ 0\end{pmatrix}$ .

Working

\mathbf{r}_B = \begin{pmatrix}2\sqrt{3} \\ 9\end{pmatrix} + t\begin{pmatrix}\frac{5}{3} \\ 0\end{pmatrix} = \begin{pmatrix}2\sqrt{3} + \frac{5}{3}t \\ 9\end{pmatrix}

Answer

\mathbf{r}_B = \begin{pmatrix}2\sqrt{3} + \frac{5}{3}t \\ 9\end{pmatrix}

Final answer

(2√3 + (5/3)t) i + 9 j

Detailed explanation

Walkthrough

Unlike particle $A$ , particle $B$ does not start at the origin: its initial position vector is $\begin{pmatrix}2\sqrt{3} \\ 9\end{pmatrix}$ . Its velocity is entirely horizontal because it moves parallel to the positive $x$ -axis, so the velocity vector is $\begin{pmatrix}\frac{5}{3} \\ 0\end{pmatrix}$ — the $y$ -component is zero, meaning $B$ stays on the line $y = 9$ forever. Adding velocity times time to the initial position gives the position vector at time $t$ . This expression is needed in part (d).

Key Takeaways

A general position vector has the form $\begin{pmatrix}a \\ b\end{pmatrix} + t\begin{pmatrix}c \\ d\end{pmatrix}$ where $(a, b)$ is the start point and $(c, d)$ the velocity.
Motion parallel to an axis means the perpendicular component of velocity is zero.

Common Mistakes

Omitting the initial position $\begin{pmatrix}2\sqrt{3} \\ 9\end{pmatrix}$ and writing only the velocity part.
Putting a non-zero $y$ -component in the velocity.

Things to Be Careful About

The mark scheme gives B1 for either the $x$ -component $2\sqrt{3} + \frac{5}{3}t$ or the $y$ -component $9$ seen, so partial credit exists — but the full B2 needs the complete correct vector including $t$ .

Techniques used

add an initial position vector to a velocity-times-time term

(d)

Hence show that the particles collide.

4M Similar

DifficultyMedium

Worked solution

Approach

Using the results of parts (b) and (c), set the position vectors equal. If the same value of $t$ satisfies both the $x$ -equation and the $y$ -equation, the particles are at the same point at the same time — they collide.

Working

Carrying forward:

\mathbf{r}_A = t\begin{pmatrix}5 \\ 5\sqrt{3}\end{pmatrix}, \qquad \mathbf{r}_B = \begin{pmatrix}2\sqrt{3} + \frac{5}{3}t \\ 9\end{pmatrix}

$y$ -components equal:

5\sqrt{3}\,t = 9 \quad\Rightarrow\quad t = \frac{9}{5\sqrt{3}} = \frac{9\sqrt{3}}{15} = \frac{3\sqrt{3}}{5}

$x$ -components equal:

5t = 2\sqrt{3} + \frac{5}{3}t \quad\Rightarrow\quad 5t - \frac{5}{3}t = 2\sqrt{3} \quad\Rightarrow\quad \frac{10}{3}t = 2\sqrt{3}

t = \frac{6\sqrt{3}}{10} = \frac{3\sqrt{3}}{5}

Both equations give the same time $t = \frac{3\sqrt{3}}{5}$ , so the particles occupy the same position at the same instant.

(At this time both particles are at $\left(\frac{9}{\sqrt{3}}, 9\right) = \left(3\sqrt{3}, 9\right)$ .)

Answer

Both component equations yield $t = \frac{3\sqrt{3}}{5}$ , so the particles collide.

Final answer

The particles collide at t = 3√3/5 s

Detailed explanation

Walkthrough

Two moving objects collide if there is a single time $t$ at which their position vectors are identical. Since we already have both position vectors in terms of $t$ , we equate them component by component. Equating $y$ -components gives $5\sqrt{3}\,t = 9$ , which solves to $t = \frac{9}{5\sqrt{3}}$ ; rationalising by multiplying numerator and denominator by $\sqrt{3}$ gives $t = \frac{3\sqrt{3}}{5}$ . Equating $x$ -components gives $5t = 2\sqrt{3} + \frac{5}{3}t$ ; collecting the $t$ -terms gives $\frac{10}{3}t = 2\sqrt{3}$ , so $t = \frac{6\sqrt{3}}{10} = \frac{3\sqrt{3}}{5}$ . Because both independent equations produce the same time, there really is an instant when the particles share a position — that is the collision. As a check, substituting back shows both are then at $\left(3\sqrt{3}, 9\right)$ .

Key Takeaways

Collision means equal position vectors at the same $t$ — you must check both components give the same time, not just one.
Rationalising denominators converts $\frac{9}{5\sqrt{3}}$ into the cleaner $\frac{3\sqrt{3}}{5}$ .

Common Mistakes

Solving only one component equation and claiming a collision without checking the other.
Setting the two position vectors' magnitudes equal instead of equating components.
Arithmetic slips when collecting $5t - \frac{5}{3}t = \frac{10}{3}t$ .
Leaving $t = \frac{9}{5\sqrt{3}}$ unrationalised — acceptable as "oe" but the tidy form is safer.

Things to Be Careful About

The word "Hence" requires you to use the position vectors from parts (b) and (c); re-deriving them from scratch risks losing method credit.
The mark scheme's M1 requires both position vectors in terms of $t$ , with at least one of the standard form $\begin{pmatrix}a \\ b\end{pmatrix} + t\begin{pmatrix}c \\ d\end{pmatrix}$ .
The A2 requires showing the exact times from both equations are the same (or showing both particles at the same point at that time) — one equation alone scores only partially.

Techniques used

equate corresponding components of two position vectorssolve linear equations in tshow both equations give the same time to prove collision

Q126MMediumCalculus

A metal tank is in the shape of a cuboid with a square base of side $x\,\text{m}$ and an open top. The tank has a volume of $5\,\text{m}^3$ . Given that $x$ can vary, and that the area of the metal used to make the tank is a minimum, find the dimensions of the tank.

DifficultyMedium

Worked solution

Approach

The tank is an open cuboid with square base of side $x$ m and height $h$ m. Use the volume $x^2h = 5$ to eliminate $h$ , write the metal area (base plus four sides) in terms of $x$ alone, differentiate, set to zero and solve.

Working

Volume constraint:

x^2h = 5 \quad\Rightarrow\quad h = \frac{5}{x^2}

Surface area (base $x^2$ plus four sides each $xh$ ):

A = x^2 + 4xh = x^2 + 4x\left(\frac{5}{x^2}\right) = x^2 + \frac{20}{x}

Differentiate:

\frac{\mathrm{d}A}{\mathrm{d}x} = 2x - \frac{20}{x^2}

Set equal to zero and solve:

2x - 20x^{-2} = 0 \quad\Rightarrow\quad 2x^3 = 20 \quad\Rightarrow\quad x^3 = 10

x = \sqrt[3]{10} = 2.154\ldots

Then

h = \frac{5}{x^2} = \frac{5}{(2.154\ldots)^2} = 1.077\ldots

Answer

x = 2.15\,\text{m}, \qquad h = 1.08\,\text{m}

Final answer

Square base side x = 2.15 m, height h = 1.08 m

Detailed explanation

Walkthrough

The tank has no lid, so the metal used is the square base plus four rectangular sides. With base side $x$ and height $h$ , that area is $x^2 + 4xh$ . The volume condition $x^2h = 5$ lets us write $h = \frac{5}{x^2}$ , so the area becomes a function of $x$ alone: $A = x^2 + \frac{20}{x}$ . Differentiating gives $\frac{\mathrm{d}A}{\mathrm{d}x} = 2x - 20x^{-2}$ ; setting this to zero finds the stationary point. Solving $2x^3 = 20$ gives $x = \sqrt[3]{10} = 2.154\ldots$ , and substituting back into $h = \frac{5}{x^2}$ gives $h = 1.077\ldots$ . Since $\frac{\mathrm{d}^2 A}{\mathrm{d}x^2} = 2 + 40x^{-3} > 0$ at this point, it is indeed a minimum.

Key Takeaways

Optimisation problems with two variables need a constraint equation to reduce to one variable before differentiating.
An open-top box's surface area is base plus four sides — do not include a top.
Setting a derivative to zero locates stationary points; checking the second derivative confirms minimum versus maximum.

Common Mistakes

Including a top face ( $6xh$ or $2x^2 + 4xh$ ) — the tank is open.
Forgetting to substitute $h = \frac{5}{x^2}$ correctly, giving a wrong surface-area expression.
Sign slips when differentiating $\frac{20}{x} = 20x^{-1}$ , which gives $-20x^{-2}$ .
Stopping after finding $x$ without computing $h$ — both dimensions are required.
Rounding too early: keep exact values until the final answers $x = 2.15$ , $h = 1.08$ (nfww applies).

Things to Be Careful About

The mark scheme requires the constraint $x^2h = 5$ to be seen (M1), then the surface area of form $ax^2 + \frac{b}{x}$ (B1), its derivative of form $ax + \frac{b}{x^2}$ (B1), equating to zero and solving (M1), and both final values correct from correct working (A1, dependent on all previous marks). Wrong working scores nothing for the final answer.
Give answers to 3 significant figures: $x = 2.15\,\text{m}$ , $h = 1.08\,\text{m}$ .

Techniques used

form the volume constraint equationeliminate a variable by substitutionexpress the surface area as a function of one variabledifferentiate the surface area functionset the derivative to zero and solveevaluate the remaining dimension

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