Additional Mathematics 4037 / 11 — May/June 2024

DifficultyMedium-Easy

Worked solution

Approach

The curve is a cubic with leading term $-\frac{1}{5}(x)(2x)(x) = -\frac{2}{5}x^{3}$ , so it has the negative-cubic shape: it comes down from the top left and goes down to the bottom right. The $x$ -intercepts come from the factors and the $y$ -intercept from substituting $x = 0$ .

Working

$x$ -intercepts: solve $-\frac{1}{5}(x+2)(2x-1)(x+5) = 0$ :

x = -5, \qquad x = -2, \qquad x = \frac{1}{2}

$y$ -intercept: put $x = 0$ :

y = -\frac{1}{5}(2)(-1)(5) = -\frac{1}{5}(-10) = 2

Since the leading coefficient is negative, the curve starts high on the left, falls to a local minimum between $x = -5$ and $x = -2$ (in the second quadrant), rises to a local maximum between $x = -2$ and $x = \frac{1}{2}$ (in the first quadrant), passing through $(0, 2)$ , then falls again after $x = \frac{1}{2}$ .

Answer

A negative cubic crossing the $x$ -axis at $x = -5$ , $-2$ and $\frac{1}{2}$ , crossing the $y$ -axis at $(0, 2)$ , with a local minimum in the second quadrant and a local maximum in the first quadrant.

Final answer

Negative cubic crossing the x-axis at -5, -2 and 1/2, the y-axis at 2, with a minimum in the second quadrant and a maximum in the first quadrant

Detailed explanation

Walkthrough

The factors of the cubic give the $x$ -intercepts immediately: each bracket equals zero at $x = -2$ , $x = \frac{1}{2}$ and $x = -5$ . To get the $y$ -intercept we substitute $x = 0$ into the whole expression, which gives $2$ . Finally, the overall shape is fixed by the leading term: multiplying the $x$ -terms of the brackets gives $-\frac{2}{5}x^{3}$ , a negative cubic, which falls from the top left, rises, then falls to the bottom right. That fixes where the local minimum (between $-5$ and $-2$ , below the axis) and local maximum (between $-2$ and $\frac{1}{2}$ , above the axis) sit.

Key Takeaways

The roots of a factorised polynomial are read straight from the brackets.
The sign of the leading coefficient determines the end behaviour of a cubic.
The $y$ -intercept is found by substituting $x = 0$ .

Common Mistakes

Drawing a positive cubic shape (up on the left) instead of a negative one — the mark scheme requires the cubic shape for the intercept marks.
Forgetting the factor $2x - 1$ gives the root $x = \frac{1}{2}$ , not $x = 1$ .
Sign errors when evaluating the $y$ -intercept: $(2)(-1)(5) = -10$ , and the minus sign in front makes it $+2$ .
Placing the maximum and minimum in the wrong quadrants.

Things to Be Careful About

The mark scheme awards the intercept marks only if the curve has a correct cubic shape, so get the shape right first.
The intercepts must actually be marked on the axes ( $-5$ , $-2$ , $\frac{1}{2}$ on the $x$ -axis and $2$ on the $y$ -axis), not just listed.

Techniques used

find x-intercepts from the factorsevaluate the y-intercept by substitutiondetermine the cubic shape from the leading coefficientsketch the curve with turning points in the correct quadrants

(b)

Hence solve the inequality $-\frac{1}{5}(x+2)(2x-1)(x+5) \geq 0$ .

DifficultyMedium-Easy

Worked solution

Approach

Using the sketch from part (a), $-\frac{1}{5}(x+2)(2x-1)(x+5) \geq 0$ holds where the curve is on or above the $x$ -axis. From the sketch this happens to the left of $x = -5$ and between $x = -2$ and $x = \frac{1}{2}$ , endpoints included.

Working

The curve is above the $x$ -axis for $x < -5$ and for $-2 < x < \frac{1}{2}$ , and touches zero at $x = -5$ , $-2$ , $\frac{1}{2}$ . Since the inequality is $\geq 0$ , the endpoints are included:

x \leq -5 \quad \text{or} \quad -2 \leq x \leq \frac{1}{2}

Answer

x \leq -5, \qquad -2 \leq x \leq \frac{1}{2}

Final answer

x <= -5, -2 <= x <= 1/2

Detailed explanation

Walkthrough

The inequality asks where the cubic takes values that are zero or positive, i.e. where the sketch sits on or above the $x$ -axis. From part (a) the curve is above the axis for $x < -5$ and for $-2 < x < \frac{1}{2}$ , and it equals zero exactly at $x = -5$ , $-2$ and $\frac{1}{2}$ . Because the inequality uses $\geq$ (not $>$ ), those three points must be included, giving $x \leq -5$ together with $-2 \leq x \leq \frac{1}{2}$ .

Key Takeaways

A polynomial inequality can be solved by reading the sign regions off its graph — this is exactly what 'Hence' is asking for.
The type of inequality sign ( $\geq$ versus $>$ ) decides whether endpoints are included.

Common Mistakes

Writing strict inequalities $x < -5$ , $-2 < x < \frac{1}{2}$ : the mark scheme gives this SC1 only, since the endpoints must be included for $\geq 0$ .
Reading the wrong regions (where the curve is below the axis) instead of above.
Omitting one of the two solution intervals.

Things to Be Careful About

Both intervals are required — a single interval scores only one of the two B1 marks.
'Hence' means the sketch from part (a) must actually be used; the answer follows from where that curve is non-negative.

Techniques used

read the solution of an inequality from the sketchidentify where the curve lies on or above the x-axisinclude the endpoints since equality is allowed

Q2Medium-EasyFactors of polynomials

DO NOT USE A CALCULATOR IN THIS QUESTION.

The polynomial $\mathrm{p}$ is such that $\mathrm{p}(x) = 6x^3 - 35x^2 + 34x + 45$ .

(a)

Find $\mathrm{p}(x)$ in the form $(2x-5)\mathrm{q}(x) + r$ , where $\mathrm{q}(x)$ is a polynomial and $r$ is a constant.

DifficultyMedium-Easy

Worked solution

Approach

Divide $\mathrm{p}(x) = 6x^3 - 35x^2 + 34x + 45$ by $(2x - 5)$ using algebraic long division; the quotient is $\mathrm{q}(x)$ and the constant left over is $r$ .

Working

\begin{array}{r} 3x^2 - 10x - 8 \\ 2x-5 \;\big)\; \overline{\;6x^3 - 35x^2 + 34x + 45} \\ \underline{-(6x^3 - 15x^2)} \qquad\qquad\quad \\ -20x^2 + 34x \qquad\quad \\ \underline{-(-20x^2 + 50x)} \qquad\quad \\ -16x + 45 \quad \\ \underline{-(-16x + 40)} \quad \\ 5 \end{array}

So

\mathrm{p}(x) = (2x - 5)(3x^2 - 10x - 8) + 5

Answer

\mathrm{q}(x) = 3x^2 - 10x - 8, \qquad r = 5

Final answer

q(x) = 3x^2 - 10x - 8, r = 5

Detailed explanation

Walkthrough

We are asked to write the cubic in the form $(2x-5)\mathrm{q}(x) + r$ . This is exactly what division by $(2x-5)$ produces: the quotient is $\mathrm{q}(x)$ and the remainder is the constant $r$ .

Long division proceeds term by term. First, divide the leading term $6x^3$ by $2x$ to get $3x^2$ ; multiply back to get $6x^3 - 15x^2$ and subtract, leaving $-20x^2$ . Next, $-20x^2 \div 2x = -10x$ ; multiply back to get $-20x^2 + 50x$ and subtract, leaving $-16x$ . Finally, $-16x \div 2x = -8$ ; multiply back to get $-16x + 40$ and subtract, leaving remainder $5$ .

So $\mathrm{p}(x) = (2x-5)(3x^2 - 10x - 8) + 5$ , giving both required quantities. As a check, the remainder theorem says the remainder on division by $(2x - 5)$ is $\mathrm{p}\!\left(\frac{5}{2}\right)$ , which indeed equals $5$ .

Key Takeaways

Dividing a polynomial by a linear factor gives a quotient plus a remainder, matching the form divisor × quotient + remainder.
The remainder theorem offers a quick check of the remainder value.

Common Mistakes

Sign slips when subtracting each product during long division — every subtraction must flip both signs.
Forgetting that the divisor has leading coefficient $2$ , so terms like $-20x^2 \div 2x$ must be divided by $2x$ , not $x$ .
Stopping at the quadratic without stating the remainder $r$ .

Things to Be Careful About

The mark scheme awards M1 for any valid method (long division, synthetic division, or forming an identity) and A1 for each of $\mathrm{q}(x)$ and $r$ — both values must be stated.
This paper is non-calculator, so all arithmetic here is exact integer work.

Techniques used

divide the cubic by a linear factor using algebraic long divisionidentify the quotient polynomialread off the remainder

(b)

Hence write the expression $\mathrm{p}(x) - 5$ as a product of linear factors.

DifficultyMedium-Easy

Worked solution

Approach

From part (a), $\mathrm{p}(x) - 5 = (2x-5)(3x^2 - 10x - 8)$ . Factorise the quadratic into two linear factors.

Working

Using part (a):

\mathrm{p}(x) - 5 = (2x - 5)(3x^2 - 10x - 8)

Factorise $3x^2 - 10x - 8$ : we need two numbers multiplying to $3 \times (-8) = -24$ and adding to $-10$ : these are $-12$ and $2$ .

3x^2 - 10x - 8 = 3x^2 - 12x + 2x - 8 = 3x(x - 4) + 2(x - 4) = (3x + 2)(x - 4)

Therefore

\mathrm{p}(x) - 5 = (3x + 2)(x - 4)(2x - 5)

Answer

\mathrm{p}(x) - 5 = (3x + 2)(x - 4)(2x - 5)

Final answer

(3x + 2)(x - 4)(2x - 5)

Detailed explanation

Walkthrough

The word 'Hence' tells us to use part (a): since $\mathrm{p}(x) = (2x-5)(3x^2 - 10x - 8) + 5$ , subtracting $5$ removes the remainder term entirely, giving $\mathrm{p}(x) - 5 = (2x-5)(3x^2 - 10x - 8)$ .

The quadratic $3x^2 - 10x - 8$ factorises by splitting the middle term: we seek two numbers with product $-24$ (that is $3 \times (-8)$ ) and sum $-10$ , namely $-12$ and $+2$ . Splitting gives $3x(x-4) + 2(x-4) = (3x+2)(x-4)$ .

So the whole expression becomes a product of three linear factors, as required.

Key Takeaways

'Hence' questions expect you to build on the earlier answer rather than start again.
Factoring a quadratic with non-unit leading coefficient via splitting the middle term (product–sum method).

Common Mistakes

Re-dividing the original cubic instead of simply subtracting $5$ from the part (a) form.
Choosing the wrong pair of numbers for the split (e.g. $-12$ and $+2$ versus $+12$ and $-2$ ).
Leaving one factor as a quadratic — the question demands linear factors only.

Things to Be Careful About

The mark scheme requires seeing $\mathrm{p}(x) - 5$ expressed fully as a product of three linear factors for the final A1.
Check by expanding mentally: the constant term should be $2 \times (-4) \times (-5) = 40$ , and $\mathrm{p}(0) - 5 = 45 - 5 = 40$ . ✓

Techniques used

subtract 5 from the polynomial using part (a)factorise the resulting quadratic into two linear factors

(c)

Hence write down the solutions of the equation $\mathrm{p}(x) = 5$ .

1M Similar

Q3MediumLogarithmic and exponential functions

(a)

Write $1 + \lg(x^2 - 1) - 2\lg(x-1)$ , where $x > 1$ , as a single logarithm to base 10. Give your answer in its simplest form.

DifficultyMedium-Easy

Worked solution

Approach

Write the constant 1 as $\lg 10$ , move the coefficient 2 into the logarithm as a power, then combine everything into one logarithm using the multiplication and division rules, and simplify the algebra inside.

Working

Since $1 = \lg 10$ :

1 + \lg(x^2 - 1) - 2\lg(x - 1) = \lg 10 + \lg(x^2 - 1) - \lg(x - 1)^2

Combining using the multiplication and division rules:

= \lg\frac{10(x^2 - 1)}{(x - 1)^2}

Factorise $x^2 - 1 = (x - 1)(x + 1)$ and cancel the common factor $x - 1$ (valid since $x > 1$ ):

\lg\frac{10(x - 1)(x + 1)}{(x - 1)^2} = \lg\frac{10(x + 1)}{x - 1}

Answer

\lg\frac{10(x + 1)}{x - 1}

Final answer

lg(10(x + 1)/(x - 1))

Detailed explanation

Walkthrough

The expression mixes a plain number with logarithms, so the first move is to write $1$ as $\lg 10$ , since $10^1 = 10$ . The term $2\lg(x - 1)$ has a coefficient in front of a logarithm, so the power rule turns it into $\lg(x-1)^2$ , putting everything inside logarithms. Then the multiplication rule $\lg A + \lg B = \lg(AB)$ and the division rule $\lg A - \lg B = \lg\frac{A}{B}$ collapse the three logarithms into one. Finally the algebra inside is simplified: $x^2 - 1$ factorises as $(x-1)(x+1)$ , and because $x > 1$ the factor $x-1$ is non-zero and cancels with one power of $(x-1)^2$ in the denominator.

Key Takeaways

The three laws of logarithms: power rule, product rule, quotient rule.
Any constant $k$ can be written as $\lg 10^k$ , so $1 = \lg 10$ .
Simplifying the argument of a logarithm (factorising and cancelling) is part of giving the answer "in its simplest form".

Common Mistakes

Forgetting to write $1$ as $\lg 10$ and leaving a stray constant.
Writing $2\lg(x-1)$ as $\lg 2(x-1)$ instead of $\lg(x-1)^2$ — the power rule multiplies the exponent, not the argument.
Cancelling $x - 1$ without noting $x > 1$ guarantees it is non-zero.
Leaving the answer as $\lg\frac{10(x^2-1)}{(x-1)^2}$ — the mark scheme requires the simplified form, so the factorisation and cancellation step earns its own mark.

Things to Be Careful About

The final answer must be the single simplified logarithm $\lg\frac{10(x+1)}{x-1}$ ; an unsimplified but otherwise correct form loses the final accuracy mark.
The condition $x > 1$ is what makes the cancellation legitimate — quote it or at least ensure the cancellation is shown.
"oe" in the mark scheme means an equivalent single-logarithm form is accepted, but it must be a single logarithm in simplest form.

Techniques used

rewrite 1 as lg 10apply the power rule to move the coefficient 2 inside the logarithmcombine logarithms using the multiplication and division rulesfactorise and cancel the common factor (x - 1)

(b)

Solve the equation

4\log_5(x+1) = 9\log_{(x+1)}5

giving your answers in the form $a + b\sqrt{c}$ , where $a$ , $b$ and $c$ are constants.

5M Similar

DifficultyMedium

Worked solution

Approach

Use the change-of-base rule to write both logarithms in the same base, giving a quadratic in $\log_5(x+1)$ . Solve it, then undo the logarithm by raising 5 to each side.

Working

By the change-of-base rule, $\log_{(x+1)}5 = \dfrac{1}{\log_5(x+1)}$ , so the equation becomes

4\log_5(x + 1) = \frac{9}{\log_5(x + 1)}

Multiplying both sides by $\log_5(x+1)$ :

\left(\log_5(x + 1)\right)^2 = \frac{9}{4}

Taking the square root (keeping both signs):

\log_5(x + 1) = \pm\frac{3}{2}

Undoing the logarithm:

x + 1 = 5^{\pm\frac{3}{2}}

Evaluating each root:

5^{\frac{3}{2}} = 5\sqrt{5}, \qquad 5^{-\frac{3}{2}} = \frac{1}{5\sqrt{5}} = \frac{\sqrt{5}}{25}

x = -1 + 5\sqrt{5} \quad \text{or} \quad x = -1 + \frac{\sqrt{5}}{25}

Answer

x = -1 + 5\sqrt{5} \quad \text{or} \quad x = -1 + \frac{\sqrt{5}}{25}

Final answer

x = -1 + 5√5 or x = -1 + √5/25

Detailed explanation

Walkthrough

The equation has logarithms with two different bases, $5$ and $(x+1)$ . The change-of-base rule $\log_a b = \frac{\log_c b}{\log_c a}$ shows that $\log_{(x+1)}5$ and $\log_5(x+1)$ are reciprocals, so the equation becomes $4u = \frac{9}{u}$ where $u = \log_5(x+1)$ . Multiplying through by $u$ gives the quadratic $u^2 = \frac{9}{4}$ , so $u = \pm\frac{3}{2}$ — both signs must be kept. Each value of $u$ is then converted back to $x$ by writing $\log_5(x+1) = u$ as $x + 1 = 5^u$ . Finally $5^{3/2} = 5\sqrt{5}$ and $5^{-3/2} = \frac{1}{5\sqrt{5}} = \frac{\sqrt{5}}{25}$ (rationalising by multiplying top and bottom by $\sqrt{5}$ ), and subtracting 1 gives both answers in the required form $a + b\sqrt{c}$ .

Key Takeaways

Change of base: $\log_a b = \frac{1}{\log_b a}$ , which turns mixed-base equations into one variable.
A quadratic in $\log_5(x+1)$ has two roots, and each must be converted back through the exponential.
Negative fractional indices: $5^{-3/2} = \frac{1}{5^{3/2}}$ , and surds are rationalised to match the form $a + b\sqrt{c}$ .

Common Mistakes

Dropping the negative root and giving only $x = -1 + 5\sqrt{5}$ — the mark scheme explicitly allows the $\pm$ to be carried through but both answers are needed for the accuracy mark.
Writing $\log_{(x+1)}5$ as $\frac{\log_5(x+1)}{\log_5 5}$ upside down — the change-of-base fraction must be $\frac{\log_5 5}{\log_5(x+1)} = \frac{1}{\log_5(x+1)}$ .
Forgetting to subtract the 1 when converting $x + 1 = 5^{\pm 3/2}$ back to $x$ .
Leaving $5^{-3/2}$ as $\frac{1}{5\sqrt{5}}$ without rationalising to $\frac{\sqrt{5}}{25}$ , which is not in the form $a + b\sqrt{c}$ .

Things to Be Careful About

Both answers are required; the mark scheme awards the final accuracy mark only when both $x = -1 + 5\sqrt{5}$ and $x = -1 + \frac{1}{25}\sqrt{5}$ appear.
Answers must be exact surd form $a + b\sqrt{c}$ — no decimals.
The method marks are dependent: the quadratic must come from a correct change of base, and the exponentiation step depends on the logarithms having been handled correctly, so each stage must be shown.

Techniques used

apply the change-of-base ruleform a quadratic equation in the logarithmsolve the quadratic by taking a square rootundo the logarithm by exponentiating base 5simplify surd answers

Q4MediumSeries

(a)

The first three terms, in ascending powers of $x$ , in the expansion of $(3+px)^n$ are $243 + 810x + qx^2$ , where $n$ , $p$ and $q$ are constants. Find the values of $n$ , $p$ and $q$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Expand $(3+px)^n$ using the binomial theorem and match each printed term with the corresponding coefficient.

Working

The first three terms are

3^n + \binom{n}{1}3^{n-1}(px) + \binom{n}{2}3^{n-2}(px)^2

Matching the constant term:

3^n = 243 \implies n = 5

Matching the coefficient of $x$ , using $n = 5$ :

\binom{5}{1}3^{4}p = 810 \implies 5 \times 81 \times p = 810 \implies 405p = 810 \implies p = 2

Matching the coefficient of $x^2$ , using $n = 5$ and $p = 2$ :

q = \binom{5}{2}3^{3}p^{2} = 10 \times 27 \times 4 = 1080

Answer

n = 5, \quad p = 2, \quad q = 1080

Final answer

n = 5, p = 2, q = 1080

Detailed explanation

Walkthrough

The binomial theorem gives the opening terms of $(3+px)^n$ as $3^n + n\cdot 3^{n-1}px + \frac{n(n-1)}{2}\cdot 3^{n-2}p^2x^2$ . The question hands us these three terms, so we compare them one at a time.

First, the constant term is $3^n$ , and this must equal $243$ . Since $243 = 3^5$ , we get $n = 5$ immediately.

Second, the coefficient of $x$ is $\binom{5}{1}3^4 p = 5 \times 81 \times p = 405p$ . Setting $405p = 810$ gives $p = 2$ .

Third, the coefficient of $x^2$ is $\binom{5}{2}3^3 p^2 = 10 \times 27 \times 4 = 1080$ , so $q = 1080$ .

Key Takeaways

The binomial expansion $(a+bx)^n = a^n + \binom{n}{1}a^{n-1}bx + \binom{n}{2}a^{n-2}b^2x^2 + \ldots$
Matching coefficients term by term lets you recover unknown constants one after another.
Each found value feeds into the next equation, so an early error propagates.

Common Mistakes

Writing the second term as $n(3p)x$ forgetting that the power of 3 drops to $n-1$ .
Using $\binom{n}{2}$ incorrectly, e.g. taking it as $n(n-1)$ without dividing by 2.
Squaring $p$ in the third term but forgetting to square it when solving for $q$ .
Misreading $243$ as a power of another base.

Things to Be Careful About

The mark scheme allows follow-through on their $n$ and their $p$ , so even if one value is wrong, later correct method still earns method marks.
Keep exact integer arithmetic throughout; all values here come out exactly.

Techniques used

write the first terms of the binomial expansionequate the constant term to find nform and solve an equation from the coefficient of xsubstitute back to find the coefficient of x squared

(b)

Find the term independent of $y$ in the expansion of $\left(2y - \frac{1}{3y^2}\right)^6$ . Give your answer in exact form.

DifficultyMedium-Easy

Worked solution

Approach

Write the general term of $\left(2y - \dfrac{1}{3y^2}\right)^6$ and choose the term in which the powers of $y$ cancel.

Working

The general term is

\binom{6}{r}(2y)^{6-r}\left(-\frac{1}{3y^2}\right)^r

The power of $y$ is $(6-r) - 2r = 6 - 3r$ . For independence from $y$ :

6 - 3r = 0 \implies r = 2

So the required term is

\binom{6}{2}(2y)^4\left(-\frac{1}{3y^2}\right)^2 = 15 \times 16y^4 \times \frac{1}{9y^4}

= \frac{240}{9} = \frac{80}{3}

Answer

\frac{80}{3}

Final answer

80/3

Detailed explanation

Walkthrough

In any binomial expansion, the general term combines powers of both parts of the bracket. Here the first part contributes $y^{6-r}$ and the second part contributes $y^{-2r}$ , so the total power of $y$ is $6 - 3r$ . We want the term with no $y$ at all, so set $6 - 3r = 0$ , giving $r = 2$ .

Substituting $r = 2$ : the coefficient is $\binom{6}{2} = 15$ , the first part gives $(2y)^4 = 16y^4$ , and the second gives $\left(-\frac{1}{3y^2}\right)^2 = \frac{1}{9y^4}$ (the sign disappears because the power is even). Multiplying: $15 \times 16 \times \frac{1}{9} = \frac{240}{9} = \frac{80}{3}$ .

Key Takeaways

The general term $\binom{n}{r}a^{n-r}b^r$ is the tool for picking out a specific term without expanding fully.
A negative power of $y$ arises from $\frac{1}{y^2}$ ; adding positive and negative exponents finds the independent term.
An even power makes the minus sign vanish.

Common Mistakes

Choosing $r = 4$ instead of $r = 2$ by counting from the wrong end of the expansion (though $\binom{6}{4}$ equals $\binom{6}{2}$ , mixing up which factor gets which exponent changes the answer).
Forgetting to square the denominator's 3, writing $\frac{1}{3y^4}$ instead of $\frac{1}{9y^4}$ .
Keeping a spurious minus sign despite the even power.
Giving a rounded decimal such as 26.7 instead of the exact fraction — the mark scheme demands the exact form.

Things to Be Careful About

The mark scheme condones small bracket or coefficient slips at M1 level but the final answer must be exact: $\frac{80}{3}$ , $\frac{240}{9}$ or $26\frac{2}{3}$ are accepted.
Check the exponent arithmetic carefully: $(6-r) - 2r = 0$ is the whole key to the question.

Techniques used

set up the general term of a binomial expansionsolve for the power that makes the y exponent zeroevaluate the chosen term exactly

Q5Medium-EasyTrigonometry

(a)

The diagram shows the graph of $y = a\cos bx + c$ , for $-360^\circ \leq x \leq 360^\circ$ , where $a$ , $b$ and $c$ are constants. Find the values of $a$ , $b$ and $c$ .

Fig. 1

DifficultyMedium-Easy

Worked solution

Approach

For $y = a\cos bx + c$ , the amplitude is $|a|$ , the midline is $y = c$ , and the period is $\frac{360°}{b}$ . Read each feature off the graph.

Working

Maximum value is $1$ at $x = 0°$ ; minimum value is $-9$ .

Amplitude:

a = \frac{1 - (-9)}{2} = 5

Midline:

c = \frac{1 + (-9)}{2} = -4

The curve completes one full cycle between the minima at $x = -240°$ and $x = 240°$ , so the period is $480°$ :

\frac{360°}{b} = 480° \quad \Rightarrow \quad b = \frac{360}{480} = \frac{3}{4}

Answer

a = 5, \quad b = \frac{3}{4}, \quad c = -4

Final answer

a = 5, b = 3/4, c = -4

Detailed explanation

Walkthrough

A cosine curve of the form $y = a\cos bx + c$ oscillates about the horizontal line $y = c$ with height $|a|$ above and below that line. The graph reaches a highest point of $1$ and a lowest point of $-9$ , so the middle of the oscillation sits halfway between them: $c = -4$ , and the distance from the middle to either extreme is $a = 5$ . Since the maximum occurs at $x = 0°$ with a positive peak (not inverted), $a$ itself is positive. Finally, consecutive minima occur at $x = -240°$ and $x = 240°$ , so one whole cycle spans $480°$ . Since the standard cosine has period $360°$ , the factor $b$ compresses or stretches the period to $\frac{360°}{b}$ , giving $b = \frac{360}{480} = \frac{3}{4}$ .

Key Takeaways

For $y = a\cos bx + c$ : amplitude $= |a|$ , midline $= c$ , period $= \frac{360°}{b}$ .
The sign of $a$ tells you whether the curve starts at a maximum ( $a > 0$ ) or minimum ( $a < 0$ ) at $x = 0$ .
Period can be measured between any two corresponding points one full cycle apart, such as successive minima.

Common Mistakes

Taking $a$ as the maximum value ( $1$ ) instead of half the difference between maximum and minimum.
Taking $c$ as the minimum instead of the midline.
Measuring the period incorrectly — e.g. reading $240°$ as the period when it is only the distance from the origin to a minimum (half of three-quarters of a cycle).
Forgetting that $b$ divides into $360°$ : writing $b = 480$ instead of $b = \frac{3}{4}$ .

Things to Be Careful About

Each of $a$ , $b$ , $c$ carries its own B1 mark — state all three clearly.
The mark scheme allows any equivalent form for $b$ (oe), e.g. $0.75$ .
Check consistency: substituting back, $5\cos\left(\frac{3}{4} \times 240°\right) - 4 = 5(-1) - 4 = -9$ , matching the printed minimum.

Techniques used

read the amplitude from the maximum and minimum valuesread the vertical shift (midline) from the graphdetermine the period and hence the coefficient b

(b)

The line $y = p$ is a tangent to the curve $y = 3 - 2\sin 6\theta$ . Write down the possible values of $p$ .

DifficultyMedium-Easy

Worked solution

Approach

Since $-1 \leq \sin 6\theta \leq 1$ , the curve $y = 3 - 2\sin 6\theta$ lies between fixed bounds. A horizontal line $y = p$ can be tangent to the curve only where it touches a maximum or minimum.

Working

The sine term satisfies

-1 \leq \sin 6\theta \leq 1

so

-2 \leq -2\sin 6\theta \leq 2

and therefore

1 \leq y \leq 5

The maximum value $y = 5$ occurs when $\sin 6\theta = -1$ , and the minimum value $y = 1$ occurs when $\sin 6\theta = 1$ . At these points the tangent to the curve is horizontal, i.e. the line $y = p$ with $p = 5$ or $p = 1$ .

Answer

p = 1 \quad \text{or} \quad p = 5

Final answer

p = 1 or p = 5

Detailed explanation

Walkthrough

The expression $3 - 2\sin 6\theta$ takes every value between $3 - 2(1) = 1$ and $3 - 2(-1) = 5$ . The sine wave turns around smoothly at these extremes, so the gradient there is zero — meaning the tangent line is horizontal, of the form $y = p$ . No other horizontal line can touch the curve, because everywhere else the curve is strictly climbing or falling through intermediate values. Hence the two possible tangents are $y = 1$ and $y = 5$ .

Key Takeaways

For $y = k + m\sin(n\theta)$ , the range runs from $k - |m|$ to $k + |m|$ .
Horizontal tangents to a sine or cosine curve occur exactly at its maximum and minimum points.

Common Mistakes

Giving an inequality such as $1 \leq p \leq 5$ — the mark scheme explicitly does not allow this; only the two boundary values are tangents.
Sign slips: forgetting that subtracting $2\sin 6\theta$ makes the maximum occur when sine equals $-1$ .
Reporting only one of the two values.

Things to Be Careful About

Both values earn separate B1 marks — give both.
Write equalities ( $p = 1$ , $p = 5$ ), not inequalities, as inequalities score zero per the mark scheme.

Techniques used

identify the range of a sine function under scaling and translationrecognise horizontal tangents at maxima and minima

Q64MMedium-EasyCalculus

Find

\int_2^4 \left(\frac{2}{2x-3} - \frac{3}{(3x-5)^2}\right)\,\mathrm{d}x

giving your answer in exact form.

DifficultyMedium-Easy

Worked solution

Approach

Integrate each term separately: the reciprocal of a linear function integrates to a logarithm, and the squared reciprocal integrates to a negative power. Then apply the limits $2$ and $4$ and simplify to the exact form.

Working

\int \frac{2}{2x-3}\,\mathrm{d}x = \ln(2x-3)

\int -\frac{3}{(3x-5)^2}\,\mathrm{d}x = -3 \times \frac{(3x-5)^{-1}}{-1 \times 3} = \frac{1}{3x-5}

So

\int_2^4 \left(\frac{2}{2x-3} - \frac{3}{(3x-5)^2}\right)\mathrm{d}x = \left[\ln(2x-3) + \frac{1}{3x-5}\right]_2^4

At $x = 4$ :

\ln 5 + \frac{1}{7}

At $x = 2$ :

\ln 1 + \frac{1}{1} = 1

Subtracting:

\ln 5 + \frac{1}{7} - 1 = \ln 5 - \frac{6}{7}

Answer

\ln 5 - \frac{6}{7}

Final answer

ln 5 - 6/7

Detailed explanation

Walkthrough

The integral splits into two standard pieces. The first term, $\frac{2}{2x-3}$ , is a constant over a linear function: its integral is $\ln(2x-3)$ , because the coefficient $2$ on top exactly matches the derivative of the inside, $2$ , so no adjustment factor is needed. The second term, $-\frac{3}{(3x-5)^2}$ , is a negative power: write it as $-3(3x-5)^{-2}$ and integrate using the rule that $\int (ax+b)^{-2}\,\mathrm{d}x = -\frac{1}{a}(ax+b)^{-1}$ , giving $+\frac{1}{3x-5}$ .

Applying the limits: at $x = 4$ , $\ln(8-3) + \frac{1}{12-5} = \ln 5 + \frac{1}{7}$ ; at $x = 2$ , $\ln 1 + \frac{1}{1} = 1$ . The difference is $\ln 5 + \frac{1}{7} - 1 = \ln 5 - \frac{6}{7}$ , which is exact as required.

Key Takeaways

$\int \frac{a}{ax+b}\,\mathrm{d}x = \ln(ax+b)$ when the numerator matches the derivative of the denominator.
$\int (ax+b)^{-2}\,\mathrm{d}x = -\frac{1}{a}(ax+b)^{-1}$ — divide by the coefficient of $x$ when integrating a function of a linear expression.
Definite integrals are evaluated with the square-bracket form: substitute the upper limit, subtract the lower limit.
$\ln 1 = 0$ , so the lower-limit logarithm vanishes here.

Common Mistakes

Writing $\ln(2x-3)$ without checking the numerator matches the derivative — here it does, but a wrong coefficient (e.g. $2\ln(2x-3)$ ) loses the B1.
Sign or coefficient slips in the second term: the integral of $-\frac{3}{(3x-5)^2}$ is $+\frac{1}{3x-5}$ , not $-\frac{1}{3x-5}$ or $\frac{3}{3x-5}$ .
Forgetting to subtract the lower-limit value, or evaluating $\ln 1$ as $1$ instead of $0$ .
Giving a decimal answer: the question demands exact form, cao.

Things to Be Careful About

The answer must be the exact form $\ln 5 - \frac{6}{7}$ (cao) — a decimal approximation scores nothing.
The mark scheme allows unsimplified intermediate forms, so $\ln 5 + \frac{1}{7} - 1$ earns the method mark, but the final line must be fully simplified.
Keep the antiderivative in the form $a\ln(2x-3) \pm \frac{b}{3x-5}$ before substituting limits so the method mark is visible.

Techniques used

integrate a reciprocal linear function to a logarithmintegrate a negative power of a linear functionapply limits using the square-bracket evaluation formcombine logarithmic and fractional terms into exact form

Q73MMediumTrigonometry

Given that $2 + \cot\theta = 3x$ and $\sin\theta = \sqrt{y}$ , find $y$ in terms of $x$ .

DifficultyMedium

Worked solution

Approach

Use the identity $\cot^2\theta + 1 = \operatorname{cosec}^2\theta$ together with $\operatorname{cosec}\theta = \frac{1}{\sin\theta}$ , then substitute $\cot\theta = 3x - 2$ and $\sin\theta = \sqrt{y}$ .

Working

From $2 + \cot\theta = 3x$ :

\cot\theta = 3x - 2

Using $\cot^2\theta + 1 = \operatorname{cosec}^2\theta$ and $\operatorname{cosec}\theta = \frac{1}{\sin\theta}$ :

(3x - 2)^2 + 1 = \frac{1}{\sin^2\theta}

Since $\sin\theta = \sqrt{y}$ , we have $\sin^2\theta = y$ :

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

Taking reciprocals:

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

Answer

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

Final answer

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

Detailed explanation

Walkthrough

The question gives two pieces of information involving the same angle: $2 + \cot\theta = 3x$ and $\sin\theta = \sqrt{y}$ . The goal is to eliminate $\theta$ entirely and produce an equation linking only $x$ and $y$ .

The key tool is the Pythagorean identity $\cot^2\theta + 1 = \operatorname{cosec}^2\theta$ , because it connects $\cot\theta$ (which we can write in terms of $x$ ) directly to $\operatorname{cosec}\theta$ , which is just the reciprocal of $\sin\theta$ — and $\sin\theta$ is given in terms of $y$ . This is why this identity is the right one here rather than, say, $\sin^2 + \cos^2 = 1$ : it links exactly the two quantities we know about.

First rearrange the first equation to isolate $\cot\theta = 3x - 2$ . Squaring gives $(3x-2)^2$ , so the identity yields $(3x-2)^2 + 1 = \operatorname{cosec}^2\theta$ . Since $\operatorname{cosec}\theta = \frac{1}{\sin\theta}$ , squaring gives $\operatorname{cosec}^2\theta = \frac{1}{\sin^2\theta}$ , and because $\sin\theta = \sqrt{y}$ we get $\sin^2\theta = y$ . So $(3x-2)^2 + 1 = \frac{1}{y}$ , and inverting both sides gives the final answer. Note that writing $4 + \cot^2\theta = 9x^2$ (squaring before isolating $\cot\theta$ ) loses the cross term and scores nothing unless recovered — always isolate $\cot\theta$ first.

An alternative route uses $\cot\theta = \frac{\cos\theta}{\sin\theta}$ to get $y = \frac{\cos^2\theta}{(3x-2)^2}$ , then $\cos^2\theta = 1 - \sin^2\theta = 1 - y$ , leading to the same result after solving $y(3x-2)^2 = 1 - y$ .

Key Takeaways

The identity $\cot^2\theta + 1 = \operatorname{cosec}^2\theta$ is the bridge between cotangent and cosecant; combined with $\operatorname{cosec}\theta = \frac{1}{\sin\theta}$ it converts a statement about $\cot\theta$ into one about $\sin\theta$ .
Eliminating a shared parameter ( $\theta$ ) from two equations is a recurring technique in Additional Mathematics.
Isolate the trig function before squaring or applying identities — squaring a sum like $2 + \cot\theta$ directly destroys information.

Common Mistakes

Squaring $2 + \cot\theta = 3x$ as $4 + \cot^2\theta = 9x^2$ : the mark scheme explicitly awards B0 M0 A0 for this unless recovered, since the cross term $4\cot\theta$ is lost.
Forgetting to square when converting $\operatorname{cosec}\theta = \frac{1}{\sin\theta}$ into $\operatorname{cosec}^2\theta = \frac{1}{\sin^2\theta}$ .
Writing $\sin^2\theta = \sqrt{y}$ instead of $\sin^2\theta = y$ .
Sign slips when rearranging $2 + \cot\theta = 3x$ to $\cot\theta = 3x - 2$ .
Leaving the answer as $\frac{1}{y} = (3x-2)^2 + 1$ without finishing by expressing $y$ in terms of $x$ , which is what was asked.

Things to Be Careful About

The final answer must have $y$ as the subject: $y = \frac{1}{(3x-2)^2+1}$ (equivalent forms are accepted, "oe").
Keep the exact fractional form — no decimals are involved or wanted here.
Check the algebra at each step: the identity applies to squares, so both sides must be squared consistently.

Techniques used

apply the Pythagorean identity cot squared plus one equals cosec squareduse the reciprocal relation between cosecant and sinerearrange to express y in terms of x

Q85MMediumTrigonometry

Solve the equation

4\sin^2\left(2\alpha - \frac{\pi}{3}\right) = 1

for $-\frac{\pi}{2} \leq \alpha \leq \frac{\pi}{2}$ . Give your answers in terms of $\pi$ .

DifficultyMedium

Worked solution

Approach

Take the square root of both sides to get $\sin\left(2\alpha - \frac{\pi}{3}\right) = \pm\frac{1}{2}$ . Transform the interval for $\alpha$ into an interval for the inner angle $\theta = 2\alpha - \frac{\pi}{3}$ , find every angle in that interval with sine $\pm\frac{1}{2}$ , then convert back to $\alpha$ .

Working

Taking the square root:

\sin\left(2\alpha - \frac{\pi}{3}\right) = \pm\frac{1}{2}

Let $\theta = 2\alpha - \frac{\pi}{3}$ . Since $-\frac{\pi}{2} \leq \alpha \leq \frac{\pi}{2}$ :

2\left(-\frac{\pi}{2}\right) - \frac{\pi}{3} \leq \theta \leq 2\left(\frac{\pi}{2}\right) - \frac{\pi}{3}

-\frac{4\pi}{3} \leq \theta \leq \frac{2\pi}{3}

The angles in $\left[-\frac{4\pi}{3},\ \frac{2\pi}{3}\right]$ with $\sin\theta = \pm\frac{1}{2}$ are:

\theta = -\frac{7\pi}{6},\ -\frac{5\pi}{6},\ -\frac{\pi}{6},\ \frac{\pi}{6}

(Note $\theta = \frac{5\pi}{6}$ also has sine $\frac{1}{2}$ but lies outside the interval, since $\frac{5\pi}{6} > \frac{2\pi}{3}$ .)

Solving $\alpha = \frac{1}{2}\left(\theta + \frac{\pi}{3}\right)$ for each:

\begin{aligned} \theta = \frac{\pi}{6}:\quad & \alpha = \frac{1}{2}\left(\frac{\pi}{6} + \frac{\pi}{3}\right) = \frac{\pi}{4} \\ \theta = -\frac{\pi}{6}:\quad & \alpha = \frac{1}{2}\left(-\frac{\pi}{6} + \frac{\pi}{3}\right) = \frac{\pi}{12} \\ \theta = -\frac{5\pi}{6}:\quad & \alpha = \frac{1}{2}\left(-\frac{5\pi}{6} + \frac{\pi}{3}\right) = -\frac{\pi}{4} \\ \theta = -\frac{7\pi}{6}:\quad & \alpha = \frac{1}{2}\left(-\frac{7\pi}{6} + \frac{\pi}{3}\right) = -\frac{5\pi}{12} \end{aligned}

All four values lie in $\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right]$ .

Answer

\alpha = -\frac{5\pi}{12},\ -\frac{\pi}{4},\ \frac{\pi}{12},\ \frac{\pi}{4}

Final answer

alpha = -5pi/12, -pi/4, pi/12, pi/4

Detailed explanation

Walkthrough

The equation has a squared sine, so the first move is to take the square root. This gives two equations at once: $\sin\left(2\alpha - \frac{\pi}{3}\right) = \frac{1}{2}$ and $\sin\left(2\alpha - \frac{\pi}{3}\right) = -\frac{1}{2}$ . Forgetting the negative case is the most common way to lose solutions here.

Next, work out what values the inner angle $\theta = 2\alpha - \frac{\pi}{3}$ can actually take. Because $\alpha$ runs from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ , doubling gives $-\pi$ to $\pi$ , and subtracting $\frac{\pi}{3}$ shifts this to $-\frac{4\pi}{3} \leq \theta \leq \frac{2\pi}{3}$ . Doing this interval transformation first tells you exactly how far around the circle to hunt for solutions, so you neither miss any nor include extras.

Within that interval, the angles with sine $\frac{1}{2}$ are $-\frac{7\pi}{6}$ and $\frac{\pi}{6}$ (note $\frac{5\pi}{6}$ is just outside, since $\frac{5\pi}{6} > \frac{2\pi}{3}$ ), and the angles with sine $-\frac{1}{2}$ are $-\frac{5\pi}{6}$ and $-\frac{\pi}{6}$ . That gives four values of $\theta$ in total.

Finally, undo the substitution: $\alpha = \frac{1}{2}\left(\theta + \frac{\pi}{3}\right)$ . Applying this to each of the four $\theta$ values gives $\alpha = \frac{\pi}{4},\ \frac{\pi}{12},\ -\frac{\pi}{4},\ -\frac{5\pi}{12}$ , all of which lie inside the required range, so all four are kept.

Key Takeaways

Squared trig equations must be split with a $\pm$ square root before solving.
When the unknown sits inside a function of the angle (here $2\alpha - \frac{\pi}{3}$ ), transform the given interval for the outer variable into an interval for the inner angle first — this fixes exactly which solutions exist.
Solve for the inner angle completely, then back-substitute; do not try to solve for $\alpha$ directly.
Always check each candidate against the original range and discard extras (here $\theta = \frac{5\pi}{6}$ had to be rejected).

Common Mistakes

Omitting the $\pm$ when taking the square root, losing the two negative-sine solutions (the mark scheme condones a missing $\pm$ only on the first B1, but the solutions themselves are still needed for the A marks).
Solving $\sin\theta = \pm\frac{1}{2}$ only in the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and missing $-\frac{5\pi}{6}$ and $-\frac{7\pi}{6}$ .
Failing to transform the interval, then either including $\theta = \frac{5\pi}{6}$ (which gives $\alpha = \frac{7\pi}{12}$ , outside the range) or missing solutions near the lower end.
Dividing by 2 before adding $\frac{\pi}{3}$ — the order of operations must be add $\frac{\pi}{3}$ first, then halve.
Giving decimal answers instead of exact multiples of $\pi$ , when the question explicitly demands answers in terms of $\pi$ .

Things to Be Careful About

The question requires answers in terms of $\pi$ — exact fractions of $\pi$ , not decimals.
All four solutions must be found and no extras included: the scheme awards A3 for four correct solutions with no extras in range, A2 for three, A1 for two.
Check the boundary values carefully: $\frac{7\pi}{12}$ exceeds $\frac{\pi}{2}$ and must be rejected, while $-\frac{5\pi}{12}$ is legitimately inside the range even though it is easy to miss.
The mark scheme's first mark (B1) is for stating $\sin\left(2\alpha - \frac{\pi}{3}\right) = \pm\frac{1}{2}$ , so that line must be visible in the working; the M1 requires a correct attempt at one solution with the correct order of operations (add $\frac{\pi}{3}$ , then divide by 2).

Techniques used

take the square root to split into two sine equationstransform the interval for the inner anglefind all angles with sine equal to plus or minus one half in the transformed intervalback-substitute to solve for alphareject solutions outside the given range

Q9MediumSimultaneous equationsLogarithmic and exponential functionsQuadratic functions

(a)

Solve the following simultaneous equations.

\mathrm{e}^{x+y} \times \mathrm{e}^{3x-2y} = 1

x^2 y = 256

5M Similar

DifficultyMedium

Worked solution

Approach

Use index laws to combine the left-hand side of the first equation into a single power of $\mathrm{e}$ , then use $\mathrm{e}^0 = 1$ to obtain a linear equation in $x$ and $y$ . Substitute this into $x^2 y = 256$ to get a cubic in one variable.

Working

Combine the powers:

\mathrm{e}^{x+y} \times \mathrm{e}^{3x-2y} = \mathrm{e}^{(x+y)+(3x-2y)} = \mathrm{e}^{4x-y}

Since $\mathrm{e}^{4x-y} = 1$ and $\mathrm{e}^0 = 1$ :

4x - y = 0 \quad \Rightarrow \quad y = 4x

Substitute into $x^2 y = 256$ :

x^2(4x) = 256 \quad \Rightarrow \quad 4x^3 = 256 \quad \Rightarrow \quad x^3 = 64

x = 4

Then:

y = 4x = 16

Answer

x = 4, \quad y = 16

Final answer

x = 4, y = 16

Detailed explanation

Walkthrough

The first equation multiplies two powers of $\mathrm{e}$ . The index law $a^m \times a^n = a^{m+n}$ lets us add the exponents, giving $\mathrm{e}^{4x-y}$ . Since $\mathrm{e}$ raised to any nonzero power is never equal to $1$ , the exponent must be zero: $4x - y = 0$ , so $y = 4x$ . This converts the exponential equation into a linear relation between $x$ and $y$ .

Substituting $y = 4x$ into the second equation gives $x^2 \cdot 4x = 4x^3 = 256$ , so $x^3 = 64$ and $x = 4$ (the real cube root). Then $y = 4 \times 4 = 16$ .

Key Takeaways

Index laws turn products of exponentials into a single exponential.
$\mathrm{e}^t = 1$ forces $t = 0$ — this is the standard way to convert an exponential equation into a linear one.
A linear relation can be substituted into a nonlinear equation to reduce it to one variable.

Common Mistakes

Writing $y = \pm 16$ : taking both cube roots or sign slips when substituting back. The mark scheme explicitly awards A0 for $y = \pm 16$ — since $y = 4x$ and $x^3 = 64 > 0$ , only positive values work.
Forgetting that $\mathrm{e}^{4x-y} = 1$ means the exponent is $0$ , not $1$ .
Arithmetic errors combining $(x+y) + (3x-2y)$ .

Things to Be Careful About

Both answers must be given: $x = 4$ and $y = 16$ , each worth an accuracy mark.
The cubic has exactly one real root here ( $x = 4$ ); do not invent complex or negative solutions.
Show the substitution step $4x^3 = 256$ explicitly — it carries the method mark.

Techniques used

combine powers of e using index lawsrecognise that e to a power equals 1 only when the power is zerosubstitute the linear relation into the nonlinear equationsolve a simple cubic equation

(b)

Solve the equation $10\mathrm{e}^{(2x-1)} - 11 = 6\mathrm{e}^{(1-2x)}$ , giving your answer in exact form.

DifficultyMedium

Worked solution

Approach

Use $\mathrm{e}^{1-2x} = \dfrac{1}{\mathrm{e}^{2x-1}}$ to rewrite the equation as a quadratic in $\mathrm{e}^{2x-1}$ , solve it, discard the negative root, then take natural logs.

Working

Since $\mathrm{e}^{1-2x} = \dfrac{1}{\mathrm{e}^{2x-1}}$ , let $u = \mathrm{e}^{2x-1}$ . The equation becomes:

10u - 11 = \frac{6}{u}

Multiply through by $u$ :

10u^2 - 11u - 6 = 0

Factorise:

(5u + 2)(2u - 3) = 0

u = \frac{3}{2} \quad \text{or} \quad u = -\frac{2}{5}

Since $\mathrm{e}^{2x-1} > 0$ always, reject $u = -\frac{2}{5}$ . So:

\mathrm{e}^{2x-1} = \frac{3}{2}

Taking natural logarithms:

2x - 1 = \ln\frac{3}{2}

x = \frac{1}{2} + \frac{1}{2}\ln\frac{3}{2}

Answer

x = \frac{1}{2} + \frac{1}{2}\ln\frac{3}{2}

Final answer

x = 1/2 + (1/2)ln(3/2)

Detailed explanation

Walkthrough

The equation contains $\mathrm{e}^{2x-1}$ and $\mathrm{e}^{1-2x}$ — exponents that are negatives of each other. Using the reciprocal property $\mathrm{e}^{-t} = 1/\mathrm{e}^{t}$ , we write $\mathrm{e}^{1-2x} = 1/\mathrm{e}^{2x-1}$ . Substituting $u = \mathrm{e}^{2x-1}$ turns the equation into $10u - 11 = 6/u$ , which after multiplying by $u$ becomes the quadratic $10u^2 - 11u - 6 = 0$ .

Factorising gives $(5u+2)(2u-3) = 0$ , so $u = 3/2$ or $u = -2/5$ . But an exponential is always strictly positive, so $u = -2/5$ must be rejected. Taking $\ln$ of both sides of $\mathrm{e}^{2x-1} = 3/2$ gives $2x - 1 = \ln(3/2)$ , hence $x = \frac{1}{2} + \frac{1}{2}\ln\frac{3}{2}$ .

Key Takeaways

When an equation involves $\mathrm{e}^{kx+c}$ and $\mathrm{e}^{-(kx+c)}$ , substitute $u = \mathrm{e}^{kx+c}$ to obtain a quadratic.
Exponentials are always positive, so any negative root for $u$ is extraneous and must be discarded.
Undoing $\mathrm{e}^{f(x)} = a$ requires taking $\ln$ of both sides: $f(x) = \ln a$ .

Common Mistakes

Keeping the negative root $\mathrm{e}^{2x-1} = -\frac{2}{5}$ and 'solving' it — the mark scheme awards A0 if the negative root is not discounted.
Sign errors when forming the quadratic, e.g. writing $10u^2 + 11u - 6 = 0$ instead of $10u^2 - 11u - 6 = 0$ .
Forgetting to divide by 2 at the end, giving $x = 1 + \ln\frac{3}{2}$ .
Giving a decimal answer instead of the exact form required.

Things to Be Careful About

The answer must be exact: $x = \frac{1}{2} + \frac{1}{2}\ln\frac{3}{2}$ , equivalently $\frac{1 + \ln\frac{3}{2}}{2}$ or $\frac{1}{2}\ln\left(\frac{3}{2}\mathrm{e}\right)$ — all are accepted.
Explicitly state why the negative root is rejected ( $\mathrm{e}^{2x-1} > 0$ ), as this justifies discarding it.
Show each stage: the reciprocal step (B1), the quadratic formation (M1), solving it (M1), and the final log step (A1).

Techniques used

rewrite an exponential as a reciprocalform a hidden quadratic by substitutionsolve the quadratic and discard the negative roottake natural logarithms to isolate x

Q10MediumCalculusTrigonometry

In this question, all distances are in metres and time, $t$ , is in seconds.

A particle $P$ is at a fixed point $O$ at time $t = 0$ .
The velocity, $v$ , of $P$ is given by $v = 3\sin 2t$ for $t \geq 0$ .

(a)

Find the exact value of $t$ for which the velocity is zero for the first time after $P$ leaves $O$ .

(b)

Find an expression, in terms of $t$ , for the displacement of $P$ from $O$ at time $t$ .

DifficultyMedium-Easy

Worked solution

Approach

Displacement is the integral of velocity. Integrate $v = 3\sin 2t$ with respect to $t$ , then use $s = 0$ at $t = 0$ to find the constant.

Working

s = \int 3\sin 2t \,\mathrm{d}t = -\frac{3}{2}\cos 2t + c

Since $P$ is at $O$ at $t = 0$ , so $s = 0$ when $t = 0$ :

0 = -\frac{3}{2}\cos 0 + c = -\frac{3}{2} + c \implies c = \frac{3}{2}

Therefore:

s = \frac{3}{2} - \frac{3}{2}\cos 2t

Answer

s = \frac{3}{2} - \frac{3}{2}\cos 2t

Final answer

s = 3/2 - (3/2)cos 2t

Detailed explanation

Walkthrough

Displacement is found by integrating velocity: $s = \int v\,\mathrm{d}t$ . Integrating $3\sin 2t$ gives $-\frac{3}{2}\cos 2t$ — dividing by the coefficient 2 of $t$ inside the sine. The result carries an unknown constant $c$ , which we pin down using the condition that the particle starts at $O$ : substituting $t = 0$ and $s = 0$ gives $c = \frac{3}{2}$ .

Key Takeaways

$\displaystyle\int \sin kt\,\mathrm{d}t = -\frac{1}{k}\cos kt + c$ .
An indefinite integral always needs an extra condition (here $s = 0$ at $t = 0$ ) to determine the constant.

Common Mistakes

Forgetting the constant of integration entirely, leaving $s = -\frac{3}{2}\cos 2t$ .
Forgetting to divide by 2 (the inner coefficient), giving $s = -3\cos 2t + c$ .
Sign error: integrating sine gives minus cosine.

Things to Be Careful About

The mark scheme requires the final expression for $s$ (A1), not just the antiderivative before finding $c$ .
The M1 for the constant depends on actually using their $s$ with $s = 0$ at $t = 0$ — show the substitution.

Techniques used

integrate the velocity function term by termapply the initial condition to find the constant of integration

(c)

Find the distance travelled by $P$ for $0 \leq t \leq \pi$ .

DifficultyMedium

Worked solution

Approach

From parts (a) and (b): $P$ moves forward until $t = \frac{\pi}{2}$ (where $v = 0$ ), then returns. By symmetry the distance travelled on each leg is equal, so the total distance is twice the displacement gained over $0 \leq t \leq \frac{\pi}{2}$ .

Working

Using the expression for $s$ from part (b):

\text{distance} = 2 \times \left[\frac{3}{2} - \frac{3}{2}\cos 2t\right]_{0}^{\frac{\pi}{2}}

Evaluate the bracket:

\left[\frac{3}{2} - \frac{3}{2}\cos 2t\right]_{0}^{\frac{\pi}{2}} = \left(\frac{3}{2} - \frac{3}{2}\cos\pi\right) - \left(\frac{3}{2} - \frac{3}{2}\cos 0\right)

= \left(\frac{3}{2} + \frac{3}{2}\right) - \left(\frac{3}{2} - \frac{3}{2}\right) = 3 - 0 = 3

Doubling by symmetry:

\text{distance} = 2 \times 3 = 6

Answer

6

Final answer

6

Detailed explanation

Walkthrough

Distance travelled is not the same as displacement when the particle changes direction. Part (a) showed the velocity is zero at $t = \frac{\pi}{2}$ — that is the turning point. From part (b), the displacement grows from $0$ to $3$ metres by $t = \frac{\pi}{2}$ , then shrinks back to $0$ by $t = \pi$ (since $\cos 2\pi = 1$ ). So the particle goes out 3 m and comes back 3 m. Using the symmetry of the motion, the total distance is twice the outward leg: $2 \times 3 = 6$ metres.

Key Takeaways

Total distance requires splitting the motion at every point where $v = 0$ and adding the magnitudes of each leg.
Symmetry of $\sin 2t$ about its zeros lets you double one leg instead of computing both.
Displacement over $0 \leq t \leq \pi$ would be $0$ — very different from the distance $6$ .

Common Mistakes

Computing only the net displacement over $[0, \pi]$ , getting $0$ instead of $6$ .
Using limits $0$ to $\pi$ directly without accounting for the reversal of direction (the scheme condones writing those limits but the symmetry/doubling must be correct).
Getting limits the wrong way round when evaluating the square bracket.
Not using the $s$ from part (b) — the M1 depends on it unless the candidate restarts properly.

Things to Be Careful About

The answer must be exactly $6$ (metres); no decimals involved.
The second M1 is dependent on correct substitution of limits at least once AND correct use of symmetry — both must be visible.
Remember the unit: distances are in metres.

Techniques used

recognise the change of direction at t = pi/2use symmetry to double one leg of the journeyevaluate the displacement expression between limits

Q1110MMedium-HardCalculusStraight-line graphs

The tangent to the curve $y = (3x-1)^{\frac{1}{3}}$ at the point where $x = 3$ meets the coordinate axes at the points $A$ and $B$ . The point with coordinates $(a, a)$ lies on the perpendicular bisector of the line $AB$ . Find the exact value of $a$ .

DifficultyMedium-Hard

Worked solution

Approach

Find the point on the curve at $x = 3$ , differentiate using the chain rule to get the tangent's gradient, write the tangent equation, read off its intercepts $A$ and $B$ , then build the perpendicular bisector of $AB$ and impose $x = y = a$ on it.

Working

When $x = 3$ :

y = (3(3) - 1)^{\frac{1}{3}} = 8^{\frac{1}{3}} = 2

Differentiate by the chain rule:

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

At $x = 3$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = 8^{-\frac{2}{3}} = \frac{1}{8^{\frac{2}{3}}} = \frac{1}{4}

Tangent at $(3, 2)$ :

y - 2 = \frac{1}{4}(x - 3)

Intercepts: setting $y = 0$ gives $-2 = \frac{1}{4}(x - 3)$ , so $x - 3 = -8$ and $x = -5$ . Setting $x = 0$ gives $y = 2 - \frac{3}{4} = \frac{5}{4}$ . So $A(-5, 0)$ and $B\left(0, \frac{5}{4}\right)$ .

Midpoint of $AB$ :

\left(\frac{-5 + 0}{2}, \frac{0 + \frac{5}{4}}{2}\right) = \left(-\frac{5}{2}, \frac{5}{8}\right)

The tangent has gradient $\frac{1}{4}$ , so the perpendicular bisector has gradient $-4$ :

y - \frac{5}{8} = -4\left(x + \frac{5}{2}\right)

The point $(a, a)$ lies on this line, so substitute $x = y = a$ :

a - \frac{5}{8} = -4\left(a + \frac{5}{2}\right) = -4a - 10

5a = -10 + \frac{5}{8} = -\frac{75}{8}

a = -\frac{15}{8}

Answer

a = -\frac{15}{8}

Final answer

a = -15/8

Detailed explanation

Walkthrough

First we locate the point of tangency: substituting $x = 3$ into $y = (3x-1)^{\frac{1}{3}}$ gives $y = 8^{\frac{1}{3}} = 2$ , so the tangent touches the curve at $(3, 2)$ .

Next we need the gradient of the tangent. The curve is a function of a linear expression, so the chain rule applies: differentiate the outer power to get $\frac{1}{3}(3x-1)^{-\frac{2}{3}}$ , then multiply by the derivative of the inner function $3x - 1$ , which is $3$ . The factors $\frac{1}{3}$ and $3$ cancel, leaving $\frac{\mathrm{d}y}{\mathrm{d}x} = (3x-1)^{-\frac{2}{3}}$ . At $x = 3$ this is $8^{-\frac{2}{3}} = \frac{1}{4}$ , since $8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 4$ .

With the point $(3, 2)$ and gradient $\frac{1}{4}$ , the tangent is $y - 2 = \frac{1}{4}(x - 3)$ . To find where it cuts the axes, set $y = 0$ (giving $x = -5$ ) and $x = 0$ (giving $y = \frac{5}{4}$ ). These are the points $A$ and $B$ .

The perpendicular bisector of $AB$ passes through the midpoint $\left(-\frac{5}{2}, \frac{5}{8}\right)$ and is perpendicular to $AB$ , which is the tangent itself. Since the tangent's gradient is $\frac{1}{4}$ , the perpendicular gradient is the negative reciprocal, $-4$ . Its equation is $y - \frac{5}{8} = -4\left(x + \frac{5}{2}\right)$ .

Finally, the point $(a, a)$ has equal coordinates, so putting $x = a$ and $y = a$ into this equation gives $a - \frac{5}{8} = -4a - 10$ , hence $5a = -\frac{75}{8}$ and $a = -\frac{15}{8}$ .

Key Takeaways

The chain rule on $(3x-1)^{\frac{1}{3}}$ : outer derivative times inner derivative, with the $\frac{1}{3} \times 3$ cancellation.
Fractional indices are evaluated as root then power: $8^{\frac{2}{3}} = 4$ .
A tangent needs a point and a gradient; intercepts follow by setting $y = 0$ and $x = 0$ .
A perpendicular bisector combines the midpoint of the segment with the negative reciprocal gradient.
A point $(a, a)$ on a line means substituting the same value for both $x$ and $y$ .

Common Mistakes

Forgetting the chain rule factor and writing $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{3}(3x-1)^{-\frac{2}{3}}$ without multiplying by $3$ (the scheme allows $k(3x-1)^{-\frac{2}{3}}$ for the M1 but the A1 for $\frac{1}{4}$ needs the correct value).
Mis-evaluating $8^{-\frac{2}{3}}$ — it is $\frac{1}{4}$ , not $-\frac{1}{4}$ or $\frac{1}{16}$ .
Sign slips when finding the $x$ -intercept: $-2 = \frac{1}{4}(x-3)$ gives $x = -5$ , not $5$ .
Taking the perpendicular gradient as $-\frac{1}{4}$ instead of $-4$ (the negative reciprocal of $\frac{1}{4}$ ).
Using the midpoint of the wrong points, or forgetting that the bisector must pass through the midpoint, not through $A$ or $B$ .
Giving a decimal only; the question demands the exact value $-\frac{15}{8}$ (the scheme accepts $-1.875$ as equivalent, but the exact fraction is the required form).

Things to Be Careful About

The question says "exact value", so keep everything in fractions throughout; do not round intermediate values.
The mark scheme's later M marks depend on earlier ones (the midpoint mark depends on the tangent equation mark; the substitution of $(a,a)$ depends on the bisector equation), so each stage must be shown explicitly.
Check the final equation carefully: $a - \frac{5}{8} = -4a - 10$ requires adding $4a$ to both sides and $\frac{5}{8}$ to both sides — sign errors here are the most common way to lose the final A1.
The perpendicular bisector is perpendicular to $AB$ , and $AB$ is the tangent line — not the curve — so its gradient comes from $\frac{1}{4}$ , not from the curve at any other point.

Techniques used

differentiate a function of a linear function using the chain ruleevaluate the gradient at a given pointform the tangent equation from a point and a gradientfind the axis intercepts of a linefind the midpoint of a segmentuse the negative reciprocal gradient for a perpendicular linesubstitute the point (a, a) into the perpendicular bisector equation

Q12MediumCalculus

(a)

It is given that $y = \frac{\ln 3x}{x^2}$ for $x > 0$ .

Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ . Give your answer in the form $\frac{A + B\ln 3x}{x^3}$ , where $A$ and $B$ are integers.

DifficultyMedium-Easy

Worked solution

Approach

Use the quotient rule on $y = \dfrac{\ln 3x}{x^2}$ , noting that $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\ln 3x\right) = \dfrac{1}{x}$ by the chain rule.

Working

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^2 \cdot \frac{1}{x} - 2x \ln 3x}{x^4}

Simplify the numerator and cancel one factor of $x$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x - 2x \ln 3x}{x^4} = \frac{1 - 2 \ln 3x}{x^3}

Answer

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1 - 2\ln 3x}{x^3}

Final answer

(1 - 2 ln 3x) / x^3

Detailed explanation

Walkthrough

We differentiate $y = \frac{\ln 3x}{x^2}$ using the quotient rule, which states that for $y = \frac{u}{v}$ ,

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}

Here $u = \ln 3x$ and $v = x^2$ . By the chain rule, $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{3x} \times 3 = \frac{1}{x}$ , and $\frac{\mathrm{d}v}{\mathrm{d}x} = 2x$ . Substituting gives

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^2 \cdot \frac{1}{x} - 2x \ln 3x}{x^4}

The numerator simplifies to $x - 2x\ln 3x$ , and cancelling a factor of $x$ top and bottom leaves the required form with integers $A = 1$ and $B = -2$ .

Key Takeaways

The derivative of $\ln kx$ is always $\frac{1}{x}$ for any positive constant $k$ , because the chain rule brings down the constant which cancels.
The quotient rule structure must be memorised exactly; mixing up the order in the numerator is the classic error.
Answers should be simplified into the requested form so the constants are clearly visible.

Common Mistakes

Writing the derivative of $\ln 3x$ as $\frac{1}{3x}$ without multiplying by 3 — this loses the B1 mark for $\frac{1}{x}$ or $\frac{3}{3x}$ .
Reversing the order in the quotient rule numerator ( $u v' - v u'$ instead of $v u' - u v'$ ), giving the wrong sign throughout.
Failing to cancel the common factor of $x$ , leaving an unsimplified answer that does not match the required form $\frac{A + B\ln 3x}{x^3}$ .

Things to Be Careful About

The mark scheme awards B1 specifically for seeing $\frac{1}{x}$ or $\frac{3}{3x}$ as the log derivative, M1 for a correct quotient or product rule attempt, A1 for all non-log terms correct, and a final A1 for $\frac{1 - 2\ln 3x}{x^3}$ — every step must be shown.
The final answer must be in the exact form $\frac{1 - 2\ln 3x}{x^3}$ (or equivalent); an unsimplified expression may not earn the last accuracy mark.

Techniques used

differentiate the natural logarithm of a linear functionapply the quotient rulesimplify to the required form with integer constants

(b)

Hence find $\int \frac{\ln 3x}{x^3}\,\mathrm{d}x$ .