(a) On the axes, sketch the graph of $y = (2x - 5)(x + 3)(1 - x)$, stating the intercepts with the coordinate axes. (b) Hence (i) solve the inequality $(2x - 5)(x + 3)(1 - x) \leq 0$ (ii) on the axes below, sketch the graph of $y = (2x - 5)(x + 3)(1 - x) $.

(a) Cubic sketch with x-intercepts at x = -3, 1 and 2.5 and y-intercept (0, -15) · (b)(i) -3 = 2.5 · (b)(ii) Modulus curve entirely on or above the x-axis with cusps at x = -3, 1 and 2.5 and y-intercept 15

(a) Evaluate $$ \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cos \frac{x}{4} \,\mathrm{d}x $$ You must show all your working. (b) Find $$ \int \left(\frac{1}{4x - 3} + \frac{1}{x^3}\right) \,\mathrm{d}x $$

(a) 0.495 · (b) (1/4)ln(4x-3) - x^(-2)/2 + c

The polynomial $\mathrm{p}$ is such that $\mathrm{p}(x) = 6x^3 + x^2 - 12x + 5$. (a) Find the remainder when $\mathrm{p}(x)$ is divided by $x - 2$. (i) Show that $2x - 1$ is a factor of $\mathrm{p}(x)$. (ii) Hence write $\mathrm{p}(x)$ as a product of linear factors. (iii) Hence solve the equation $6\sin^3 \theta + \sin^2 \theta - 12\sin \theta + 5 = 0$ for $0^\circ \leq \theta \leq 90^\circ$.

(a) 33 · (b)(i) p(1/2) = 0, so 2x - 1 is a factor of p(x) · (b)(ii) (2x - 1)(3x + 5)(x - 1) · (b)(iii) theta = 30, 90

(a) Show that $\sin^3 x\left(\frac{\operatorname{cosec} x}{\cot x}\right)$ can be written as $\sin^2 x \tan x$. (b) Solve the equation $\cos^2 x \tan x - \frac{1}{2}\tan x = 0$ for $-\pi < x < \pi$.

(a) Shown: sin^3 x (cosec x / cot x) = sin^2 x tan x · (b) x = 0, +/-pi/4, +/-3pi/4

The functions $\mathrm{f}$ and $\mathrm{fg}$ are defined by $$ \mathrm{f}(x) = \mathrm{e}^{x^2 + 3} \quad \text{for } x \frac{3}{2} $$ (a) Explain why $\mathrm{f}^{-1}$ exists. (b) Find an expression for $\mathrm{f}^{-1}(x)$ and state the domain and range of $\mathrm{f}^{-1}$. (c) Hence find and simplify an expression for $\mathrm{g}(x)$.

(a) f is one-one on its domain x e^3, range f^-1(x) < 0 · (c) g(x) = -sqrt(2x - 3)

4037/22 May/June 2024 — Additional Mathematics Solved Past Paper

Q1Medium-EasyEquations, inequalities and graphs

(a)

On the axes, sketch the graph of $y = (2x - 5)(x + 3)(1 - x)$ , stating the intercepts with the coordinate axes.

Fig. 1

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The cubic is already factorised, so read off the roots from the factors, evaluate the $y$ -intercept at $x = 0$ , and use the leading coefficient to fix the end behaviour before sketching.

Working

Roots: set each factor to zero.

2x - 5 = 0 \Rightarrow x = 2.5, \qquad x + 3 = 0 \Rightarrow x = -3, \qquad 1 - x = 0 \Rightarrow x = 1

$y$ -intercept: substitute $x = 0$ .

y = (0 - 5)(0 + 3)(1 - 0) = (-5)(3)(1) = -15

End behaviour: expanding the leading terms gives $2x \cdot x \cdot (-x) = -2x^3$ , so the curve rises from below on the left and falls below the $x$ -axis on the right.

Answer

A cubic sketch crossing the $x$ -axis at $x = -3$ , $x = 1$ and $x = 2.5$ , with $y$ -intercept $(0, -15)$ , coming from below on the left and going to below on the right.

Final answer

Cubic sketch with x-intercepts at x = -3, 1 and 2.5 and y-intercept (0, -15)

Detailed explanation

Walkthrough

Because the cubic is given fully factorised, the roots come straight from setting each bracket to zero: $x = 2.5$ , $x = -3$ and $x = 1$ . The $y$ -intercept is the value when $x = 0$ : substituting gives $(-5)(3)(1) = -15$ . To get the overall shape, look at the leading term: multiplying the highest powers of each factor, $2x \cdot x \cdot (-x) = -2x^3$ , a negative cubic. So as $x \to -\infty$ , $y \to +\infty$ and as $x \to +\infty$ , $y \to -\infty$ : the curve enters from the top left and leaves at the bottom right, crossing the axis at $-3$ (downwards), turning, crossing at $1$ (upwards), reaching a maximum, then crossing at $2.5$ (downwards). Note that $x = 0$ lies between $-3$ and $1$ , consistent with the negative $y$ -intercept.

Key Takeaways

A factorised cubic gives its roots directly from the brackets.
The $y$ -intercept is found by substituting $x = 0$ .
The sign of the leading coefficient determines the end behaviour: negative cubic means top-left to bottom-right.

Common Mistakes

Reading $x = 5$ instead of $x = 2.5$ from the factor $2x - 5$ .
Drawing a positive cubic shape (bottom-left to top-right) by ignoring the factor $(1 - x)$ .
Forgetting to state the $y$ -intercept, which carries its own mark.
Not extending the curve above and below the $x$ -axis — the mark scheme requires the ends to leave the axis region.

Things to Be Careful About

The sketch must show a genuine cubic shape with two turning points, and all three roots plus the intercept $-15$ must be labelled.
The curve must extend above the $x$ -axis on the far left and below it on the far right.

Techniques used

find the roots of a factorised cubicevaluate the y-interceptdetermine the end behaviour of a cubic with negative leading coefficientsketch the cubic through its intercepts

(b)

Hence

3M

(i)

solve the inequality $(2x - 5)(x + 3)(1 - x) \leq 0$

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the sketch from part (a): $(2x - 5)(x + 3)(1 - x) \leq 0$ where the curve is on or below the $x$ -axis.

Working

From the sketch, the curve is below or on the $x$ -axis between the roots $-3$ and $1$ , and again from the root $2.5$ onwards to the right. Since the inequality is $\leq$ , the endpoints are included.

-3 \leq x \leq 1 \quad \text{or} \quad x \geq 2.5

Answer

-3 \leq x \leq 1 \;\text{ or }\; x \geq 2.5

Final answer

-3 <= x <= 1 or x >= 2.5

Detailed explanation

Walkthrough

The inequality asks where the cubic is negative or zero. On the sketch from part (a), the curve dips below the axis between $x = -3$ and $x = 1$ , and again for all $x$ beyond $2.5$ . Because the inequality uses $\leq$ (not strict $<$ ), the roots themselves — where the curve equals zero — must be included, so the intervals are written with $\leq$ and $\geq$ signs.

Key Takeaways

An inequality in a factorised polynomial is solved by reading where its graph lies on the required side of the $x$ -axis.
$\leq$ or $\geq$ means the boundary roots are included; $<$ or $>$ means they are excluded.

Common Mistakes

Using strict inequalities $-3 < x < 1$ , $x > 2.5$ — the mark scheme awards only a special-casing SC1 for this, not full marks.
Giving only one of the two intervals (earns B1 only).
Including the interval between $1$ and $2.5$ , where the curve is above the axis.

Things to Be Careful About

Both intervals are required for full marks, and the endpoints must be inclusive to match the $\leq$ sign. The answer is marked on the final statement, so a correct sketch with a wrong final line scores nothing.

Techniques used

read the solution of an inequality from the sketch of the cubicidentify where the curve is on or below the x-axisexpress the answer as a union of intervals with correct endpoints

(ii)

on the axes below, sketch the graph of $y = |(2x - 5)(x + 3)(1 - x)|$ .

Fig. 2

1M Similar

DifficultyEasy

Worked solution

Approach

The graph of $y = |\mathrm{f}(x)|$ is the graph of $y = \mathrm{f}(x)$ with every part below the $x$ -axis reflected above it; parts already above stay unchanged.

Working

From part (a), the cubic lies below the $x$ -axis for $x < -3$ and between $x = 2.5$ and beyond, and also in the dip between $-3$ and $1$ near the origin. Reflecting these portions in the $x$ -axis produces sharp cusps at $x = -3$ , $x = 1$ and $x = 2.5$ , and turns the $y$ -intercept $-15$ into $+15$ .

Answer

The modulus graph: the whole cubic lies on or above the $x$ -axis, with cusps at $x = -3$ , $x = 1$ and $x = 2.5$ , a positive $y$ -intercept of $15$ , and the same $x$ -intercepts as part (a).

Final answer

Modulus curve entirely on or above the x-axis with cusps at x = -3, 1 and 2.5 and y-intercept 15

Detailed explanation

Walkthrough

Taking a modulus makes every output non-negative. Graphically, any part of the curve sitting below the $x$ -axis is flipped (reflected) above the axis, while the parts already above are untouched. For this cubic, the left-hand branch (below the axis for $x < -3$ ), the dip containing the $y$ -intercept $-15$ , and the right-hand branch (below the axis for $x > 2.5$ ) are all reflected. Where the curve crosses the axis, the original and reflected pieces meet at a sharp point — a cusp — so cusps appear at $x = -3$ , $x = 1$ and $x = 2.5$ . The $y$ -intercept becomes $|{-15}| = 15$ , positive as the mark scheme requires.

Key Takeaways

$y = |\mathrm{f}(x)|$ reflects only the below-axis portions of $y = \mathrm{f}(x)$ .
Cusps form at the $x$ -intercepts where reflection meets the unchanged part.
$x$ -intercepts are unchanged; $y$ -values become their absolute values.

Common Mistakes

Reflecting the whole curve (as in $y = \mathrm{f}(|x|)$ ) instead of only the below-axis parts.
Leaving the $y$ -intercept negative — the mark scheme explicitly requires a positive $y$ -intercept.
Drawing smooth turning points instead of cusps at the intercepts.

Things to Be Careful About

The sketch must match the roots from part (a) (follow-through allowed on an equivalent cubic), show cusps at the intercepts, and never dip below the $x$ -axis.

Techniques used

reflect the negative parts of the curve in the x-axisform cusps at the x-intercepts where the reflection meetskeep the x-intercepts unchanged under the modulus

Q2Medium-EasyCalculusTrigonometryLogarithmic and exponential functions

(a)

Evaluate

\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cos \frac{x}{4} \,\mathrm{d}x

You must show all your working.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Integrate $\cos \frac{x}{4}$ by dividing by the coefficient of $x$ inside the cosine, giving $4\sin\frac{x}{4}$ , then evaluate between the limits $\frac{\pi}{3}$ and $\frac{\pi}{2}$ .

Working

\int \cos \frac{x}{4} \,\mathrm{d}x = \frac{\sin \frac{x}{4}}{\frac{1}{4}} = 4\sin \frac{x}{4}

So the definite integral is

\left[ 4\sin \frac{x}{4} \right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}

Substituting the limits:

4\sin \frac{\pi}{8} - 4\sin \frac{\pi}{12}

Evaluating:

4\sin \frac{\pi}{8} = 4 \times 0.382683\ldots = 1.530734\ldots

4\sin \frac{\pi}{12} = 4 \times 0.258819\ldots = 1.035276\ldots

1.530734\ldots - 1.035276\ldots = 0.495457\ldots

Answer

0.495 \;\text{(to 3 significant figures)}

Final answer

0.495

Detailed explanation

Walkthrough

The integral is of $\cos \frac{x}{4}$ , where the angle is $\frac{1}{4}x$ . When integrating $\cos(kx)$ , the result is $\frac{\sin(kx)}{k}$ — you divide by the coefficient of $x$ to undo the chain rule. Here $k = \frac{1}{4}$ , so dividing by $k$ is the same as multiplying by $4$ , giving $4\sin\frac{x}{4}$ . This is the step the mark scheme rewards with B2.

Next we write the definite integral in square-bracket form with the limits attached, then substitute the upper limit $x = \frac{\pi}{2}$ and subtract the value at the lower limit $x = \frac{\pi}{3}$ . This gives $4\sin\frac{\pi}{8} - 4\sin\frac{\pi}{12}$ , which is the M1 step.

Finally we evaluate each sine (angles are in radians, as the limits are multiples of $\pi$ ) and subtract: $1.530734\ldots - 1.035276\ldots = 0.495457\ldots$ , which rounds to $0.495$ to 3 significant figures.

Key Takeaways

Integrating $\cos(kx)$ gives $\frac{\sin(kx)}{k}$ ; the coefficient inside the trig function must be divided out.
Definite integrals are evaluated as (value at upper limit) − (value at lower limit).
Radian limits such as $\frac{\pi}{8}$ and $\frac{\pi}{12}$ must be evaluated in radian mode.

Common Mistakes

Forgetting to divide by $\frac{1}{4}$ , writing $\sin\frac{x}{4}$ instead of $4\sin\frac{x}{4}$ — this loses the B2 (only B1 is awarded for $k\sin\frac{x}{4}$ with $k \neq 4$ ).
Writing $-4\sin\frac{x}{4}$ (wrong sign on the integrated cosine).
Evaluating the sines in degree mode instead of radian mode.
Subtracting in the wrong order (lower limit minus upper limit), giving $-0.495$ .
Rounding too early or giving fewer than 3 significant figures.

Things to Be Careful About

The mark scheme requires all working to be shown: the integrated form, the substituted expression $4\sin\frac{\pi}{8} - 4\sin\frac{\pi}{12}$ , and the final value. Jumping straight to $0.495$ loses method marks.
The final A1 depends on all previous marks being earned (dep on previous marks), so the integration must be correct.
The answer is a decimal here (calculator paper); give it to at least 3 significant figures — the scheme accepts answers rounding to $0.495$ or $0.4954[57\ldots]$ to 4 or more sf.

Techniques used

integrate cos of a multiple of x by adjusting the coefficientapply the square-bracket evaluation formevaluate sine at exact radian limitsround the final value to 3 significant figures

(b)

Find

\int \left(\frac{1}{4x - 3} + \frac{1}{x^3}\right) \,\mathrm{d}x

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Integrate each term separately. The reciprocal of a linear function integrates to a logarithm (dividing by the coefficient of $x$ ), and $\frac{1}{x^3} = x^{-3}$ integrates by the power rule. Include the constant of integration.

Working

For the first term, since $\int \frac{1}{ax+b}\,\mathrm{d}x = \frac{1}{a}\ln|ax+b|$ :

\int \frac{1}{4x-3}\,\mathrm{d}x = \frac{1}{4}\ln(4x-3)

For the second term, rewrite and apply the power rule $\int x^n \,\mathrm{d}x = \frac{x^{n+1}}{n+1}$ :

\int \frac{1}{x^3}\,\mathrm{d}x = \int x^{-3}\,\mathrm{d}x = \frac{x^{-2}}{-2} = -\frac{x^{-2}}{2}

Combining, with the constant of integration:

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

Answer

\frac{1}{4}\ln(4x-3) - \frac{x^{-2}}{2} + c

Final answer

(1/4)ln(4x-3) - x^(-2)/2 + c

Detailed explanation

Walkthrough

The integrand has two terms, so we integrate each one separately.

First term: $\frac{1}{4x-3}$ is the reciprocal of a linear function. The standard result is that $\int \frac{1}{ax+b}\,\mathrm{d}x = \frac{1}{a}\ln|ax+b|$ : the logarithm comes from integrating a reciprocal, and we must divide by the coefficient $a = 4$ to compensate for the chain rule. This gives $\frac{1}{4}\ln(4x-3)$ . Writing $\frac{1}{4}\ln(x - 0.75)$ is an equivalent accepted form, since $\frac{1}{4}\ln(4x-3) = \frac{1}{4}\ln 4 + \frac{1}{4}\ln(x-0.75)$ and the constant absorbs $\frac{1}{4}\ln 4$ .

Second term: $\frac{1}{x^3}$ is rewritten as $x^{-3}$ so the power rule applies. Raising the power by one gives $x^{-2}$ , and dividing by the new power $-2$ gives $-\frac{x^{-2}}{2}$ . A common slip is dividing by $+2$ instead of $-2$ .

Finally, because this is an indefinite integral, the constant of integration $+c$ must be included.

Key Takeaways

$\int \frac{1}{ax+b}\,\mathrm{d}x = \frac{1}{a}\ln|ax+b|$ — the coefficient of $x$ must be divided out.
Negative powers integrate by the same power rule: $\int x^{-3}\,\mathrm{d}x = -\frac{1}{2}x^{-2}$ .
Indefinite integrals always need $+c$ .

Common Mistakes

Writing $\ln(4x-3)$ without the factor $\frac{1}{4}$ — the scheme awards only B1 for $k\ln(4x-3)$ with $k \neq \frac{1}{4}$ .
Writing $\frac{1}{4}\ln 4x - 3$ (the $-3$ outside the logarithm) — this also scores only B1.
Dividing by $+2$ instead of $-2$ in the power term, giving $+\frac{x^{-2}}{2}$ .
Writing $\frac{x^{-2}}{-2}$ but failing to simplify the double negative, or integrating $\frac{1}{x^3}$ as $\ln x^3$ .
Omitting the constant of integration $+c$ .

Things to Be Careful About

The mark scheme says "isw" (ignore subsequent working), so an arithmetic slip after a correct answer is not penalised, but the printed answer form $\frac{1}{4}\ln(4x-3) - \frac{x^{-2}}{2}(+c)$ is what earns full marks.
The modulus bars are not required by the scheme here, but the argument of the logarithm must be $4x-3$ (or $x - 0.75$ ), not $4x$ with $-3$ outside.
Both terms must be integrated correctly to score all 3 marks: B2 for the logarithm term and B1 for the power term.

Techniques used

integrate the reciprocal of a linear function to a logarithmdivide by the coefficient of x inside the logarithmrewrite x to the power minus 3 and integrate using the power ruleinclude the constant of integration

Q3MediumQuadratic functionsEquations, inequalities and graphs

(a)

Determine whether the equation

\frac{(4x + 1)(3x + 2)}{5x - 3} = x + 1

has two distinct real roots, two equal roots or no real roots.

4M Similar

DifficultyMedium

Worked solution

Approach

Clear the fraction by multiplying both sides by $5x - 3$ , expand both sides, collect everything onto one side to get a quadratic, then examine its discriminant.

Working

Multiply both sides by $5x - 3$ :

(4x + 1)(3x + 2) = (x + 1)(5x - 3)

Expand each side:

12x^2 + 11x + 2 = 5x^2 + 2x - 3

Collect all terms on the left:

7x^2 + 9x + 5 = 0

Evaluate the discriminant $b^2 - 4ac$ with $a = 7$ , $b = 9$ , $c = 5$ :

9^2 - 4(7)(5) = 81 - 140 = -59

Since $-59 < 0$ , the quadratic has no real roots.

Answer

The equation has no real roots, because the discriminant is $-59 < 0$ .

Final answer

No real roots (discriminant = -59 < 0)

Detailed explanation

Walkthrough

The equation has a fraction, so the first move is to multiply both sides by the denominator $5x - 3$ . This is valid as long as $5x - 3 \neq 0$ , i.e. $x \neq \frac{3}{5}$ ; any root we find must be checked against this, though here none will exist anyway.

Expanding the left side gives $(4x+1)(3x+2) = 12x^2 + 8x + 3x + 2 = 12x^2 + 11x + 2$ . Expanding the right side gives $(x+1)(5x-3) = 5x^2 - 3x + 5x - 3 = 5x^2 + 2x - 3$ .

Subtracting the right side from the left collects everything into one quadratic: $(12x^2 - 5x^2) + (11x - 2x) + (2 + 3) = 7x^2 + 9x + 5 = 0$ .

Now the nature of the roots is decided entirely by the discriminant $b^2 - 4ac$ : positive means two distinct real roots, zero means two equal roots, negative means no real roots. Here $b^2 - 4ac = 81 - 140 = -59$ , which is negative, so there are no real roots.

Key Takeaways

To classify the roots of an equation that is not obviously quadratic, first rearrange it into the form $ax^2 + bx + c = 0$ .
The sign of $b^2 - 4ac$ alone determines whether the roots are distinct real, equal real, or non-real — you never need to solve the equation.
Careful expansion of both products before collecting like terms prevents sign errors.

Common Mistakes

Expanding only one side or forgetting to multiply the whole right-hand side by $5x - 3$ .
Sign slips when moving terms across: the scheme accepts either $7x^2 + 9x + 5 = 0$ or $-7x^2 - 9x - 5 = 0$ , but mixing signs mid-collection gives a wrong quadratic.
Computing $b^2 - 4ac$ as $81 - 35$ (forgetting to multiply $4 \times 7 \times 5$ fully).
Concluding 'no real roots' without stating the discriminant value or its sign — the A1 requires the negative value shown.

Things to Be Careful About

The mark scheme awards B2 for the correct three-term quadratic (B1 for two terms correct), so show the full expansion before collecting.
The M1 is follow-through on your own three-term quadratic, but the final A1 needs the correct discriminant $-59$ together with the conclusion 'no real roots'.
State the conclusion explicitly in words ('no real roots') — the discriminant value alone does not answer the question asked.

Techniques used

expand both sides of the equationcollect terms into a three-term quadraticevaluate the discriminantinterpret the sign of the discriminant

(b)

Solve the equation

\frac{12}{\sqrt[3]{x}} - \sqrt[3]{x} = 4

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $y = \sqrt[3]{x}$ . Then the equation becomes a quadratic in $y$ , which factorises; solve for $y$ and cube each value to recover $x$ .

Working

Substitute $y = \sqrt[3]{x}$ :

\frac{12}{y} - y = 4

Multiply through by $y$ :

12 - y^2 = 4y

Rearrange into standard form:

y^2 + 4y - 12 = 0

Factorise:

(y - 2)(y + 6) = 0

So $\sqrt[3]{x} = 2$ or $\sqrt[3]{x} = -6$ . Cubing each:

x = 2^3 = 8 \qquad \text{or} \qquad x = (-6)^3 = -216

Answer

x = 8 \quad \text{or} \quad x = -216

Final answer

x = 8 or x = -216

Detailed explanation

Walkthrough

The equation contains $\frac{12}{\sqrt[3]{x}}$ and $\sqrt[3]{x}$ — the same quantity appearing once 'on top' and once 'below'. Substituting $y = \sqrt[3]{x}$ converts it into an ordinary equation in one variable: $\frac{12}{y} - y = 4$ .

Multiplying every term by $y$ clears the fraction: $12 - y^2 = 4y$ . Rearranging all terms to one side gives the quadratic $y^2 + 4y - 12 = 0$ , which factorises neatly since $6 \times (-2) = -12$ and $6 + (-2) = 4$ : $(y-2)(y+6) = 0$ .

So $y = 2$ or $y = -6$ , meaning $\sqrt[3]{x} = 2$ or $\sqrt[3]{x} = -6$ . Cubing reverses the cube root: $x = 8$ or $x = -216$ . Note that cubing preserves the sign, so the negative root gives a valid negative solution — unlike square roots, cube roots are defined for negative numbers.

Key Takeaways

When an equation involves a root and its reciprocal, substituting a single letter for the root reduces it to a quadratic.
Cube roots are defined for all real numbers, so negative values of $y$ give valid solutions.
Always finish by undoing the substitution: cube each root to return to $x$ .

Common Mistakes

Forgetting to rearrange into standard form before factorising, e.g. trying to factorise $12 - y^2 = 4y$ directly.
Sign errors when moving terms: the scheme expects $y^2 + 4y - 12 = 0$ (or equivalent), not $y^2 - 4y - 12 = 0$ .
Discarding $y = -6$ because it looks 'invalid' — cube roots of negatives are fine, so $x = -216$ must be kept.
Stopping at $\sqrt[3]{x} = 2$ and $\sqrt[3]{x} = -6$ without cubing; the scheme awards A1 for these but the final A marks need $x = 8$ and $x = -216$ .

Things to Be Careful About

The B1 is awarded for the correct substituted quadratic ( $y^2 + 4y - 12 = 0$ or equivalent), seen or implied.
The M1 follows through on your own three-term quadratic, but the A2 requires both final values $x = 8$ and $x = -216$ ; losing the negative root costs an accuracy mark.
Check by substitution if unsure: $\frac{12}{2} - 2 = 4$ ✓ and $\frac{12}{-6} - (-6) = -2 + 6 = 4$ ✓.

Techniques used

substitute a new variable for the cube rootrearrange into a quadratic equationfactorise the quadraticcube to recover x

Q4MediumFactors of polynomialsTrigonometry

The polynomial $\mathrm{p}$ is such that $\mathrm{p}(x) = 6x^3 + x^2 - 12x + 5$ .

(a)

Find the remainder when $\mathrm{p}(x)$ is divided by $x - 2$ .

1M Similar

(b)

6M

(i)

Show that $2x - 1$ is a factor of $\mathrm{p}(x)$ .

1M Similar

(ii)

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

3M Similar

DifficultyMedium

Worked solution

Approach

Divide $\mathrm{p}(x) = 6x^3 + x^2 - 12x + 5$ by the known factor $2x - 1$ to obtain a quadratic, then factorise that quadratic.

Working

Divide $6x^3 + x^2 - 12x + 5$ by $2x - 1$ :

$6x^3 \div 2x = 3x^2$ ; subtract $3x^2(2x - 1) = 6x^3 - 3x^2$ , leaving $4x^2 - 12x + 5$ .
$4x^2 \div 2x = 2x$ ; subtract $2x(2x - 1) = 4x^2 - 2x$ , leaving $-10x + 5$ .
$-10x \div 2x = -5$ ; subtract $-5(2x - 1) = -10x + 5$ , leaving remainder $0$ .

So the quadratic factor is

3x^2 + 2x - 5

Factorise: $3x^2 + 2x - 5 = (3x + 5)(x - 1)$ , since $3x \times x = 3x^2$ , $3x \times (-1) + 5 \times x = 2x$ , and $5 \times (-1) = -5$ .

Hence

\mathrm{p}(x) = (2x - 1)(3x + 5)(x - 1)

Answer

\mathrm{p}(x) = (2x - 1)(3x + 5)(x - 1)

Final answer

(2x - 1)(3x + 5)(x - 1)

Detailed explanation

Walkthrough

Since part (i) established that $2x - 1$ is a factor, we divide the cubic by it (by algebraic long division or by equating coefficients) to find what remains. Each step of the division brings down the next term: the quotient builds up as $3x^2$ , then $+2x$ , then $-5$ , giving $3x^2 + 2x - 5$ with zero remainder — as expected since $2x - 1$ is an exact factor. That quadratic factors by inspection into $(3x + 5)(x - 1)$ because the numbers multiply to $-15$ and add to $+2$ . Combining everything gives the full product of linear factors.

Key Takeaways

A known linear factor reduces a cubic to a quadratic by division.
Factorise the quadratic by finding two numbers with the right product and sum.
Always check the final expansion matches the original cubic.

Common Mistakes

Sign errors during the subtraction steps of the division.
Writing $(2x - 1)(3x^2 + 2x - 5)$ and stopping without fully factorising.
Mis-factorising the quadratic, e.g. getting the signs of $5$ and $-1$ swapped.

Things to Be Careful About

The mark scheme awards M2 for the correct quadratic factor $3x^2 + 2x - 5$ (M1 for any two terms correct) and A1 for the complete product of three linear factors.
If nothing else scores, SC2 exists for justifying $x - 1$ as a factor and writing down $(2x - 1)(3x + 5)(x - 1)$ with no incorrect work seen.
Equivalent forms (oe) are accepted, but the answer must come from correct working.

Techniques used

divide the cubic by the known linear factor to find the quadratic factorfactorise the resulting quadraticwrite the cubic as a product of three linear factors

(iii)

Hence solve the equation $6\sin^3 \theta + \sin^2 \theta - 12\sin \theta + 5 = 0$ for $0^\circ \leq \theta \leq 90^\circ$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The equation has the same form as $\mathrm{p}(x) = 0$ with $x = \sin\theta$ . Use the factorisation from part (ii):

(2\sin\theta - 1)(3\sin\theta + 5)(\sin\theta - 1) = 0

Then solve each factor for $0° \leq \theta \leq 90°$ .

Working

2\sin\theta - 1 = 0 \implies \sin\theta = \frac{1}{2} \implies \theta = 30°

\sin\theta - 1 = 0 \implies \sin\theta = 1 \implies \theta = 90°

3\sin\theta + 5 = 0 \implies \sin\theta = -\frac{5}{3}

which is impossible since $-1 \leq \sin\theta \leq 1$ , so this factor gives no solution.

Answer

\theta = 30°, \; 90°

Final answer

theta = 30, 90

Detailed explanation

Walkthrough

The equation is exactly $\mathrm{p}(\sin\theta) = 0$ , so the factorisation from part (ii) applies directly with $x$ replaced by $\sin\theta$ : each linear factor set to zero gives either $\sin\theta = \frac{1}{2}$ , $\sin\theta = 1$ , or $\sin\theta = -\frac{5}{3}$ . Within $0° \leq \theta \leq 90°$ , the first gives $\theta = 30°$ and the second gives $\theta = 90°$ ; the third is impossible because sine cannot be less than $-1$ . Both valid angles lie inside the interval, so both are answers.

Key Takeaways

'Hence' means reuse the factorisation from part (ii) rather than solving from scratch.
Every factor must be checked: some give no real solution in range.

Common Mistakes

Ignoring the factor $3\sin\theta + 5$ entirely instead of noting it yields no solution.
Missing $\theta = 90°$ from the factor $\sin\theta - 1$ .
Giving angles outside $0° \leq \theta \leq 90°$ (e.g. $150°$ ).

Things to Be Careful About

The mark scheme marks each angle nfww (not from wrong working): $\theta = 30$ and $\theta = 90$ each earn B1 only if they follow correctly.
B1 alternatives: $\sin\theta = 0.5$ or $\theta = 30$ ; or $\sin\theta = 1$ or $\theta = 90$ .
Angles are required in degrees here, matching the interval given.

Techniques used

recognise the trigonometric equation as the factorised polynomial in sin thetasolve each linear factor for theta within the given interval

Q56MMediumCalculusStraight-line graphs

A curve has equation $y = 5\mathrm{e}^{2x - 1} + \mathrm{e}$ . The tangent to the curve at the point where $x = 1$ cuts the $x$ -axis at the point $P$ .

Find the equation of the tangent in the form $y = mx + c$ , where $m$ and $c$ are exact values, and hence find the $x$ -coordinate of $P$ .

DifficultyMedium

Worked solution

Approach

Differentiate the curve to find the gradient function, evaluate the gradient and the $y$ -coordinate at $x = 1$ , then build the tangent equation from the point-gradient form. Finally set $y = 0$ to find where the tangent cuts the $x$ -axis.

Working

Differentiate term by term:

\frac{\mathrm{d}y}{\mathrm{d}x} = 5 \times 2\,\mathrm{e}^{2x-1} = 10\mathrm{e}^{2x-1}

At $x = 1$ :

\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=1} = 10\mathrm{e}^{2(1)-1} = 10\mathrm{e}

The $y$ -coordinate on the curve at $x = 1$ :

y = 5\mathrm{e}^{2(1)-1} + \mathrm{e} = 5\mathrm{e} + \mathrm{e} = 6\mathrm{e}

Tangent through $(1, 6\mathrm{e})$ with gradient $10\mathrm{e}$ :

y - 6\mathrm{e} = 10\mathrm{e}(x - 1)

y = 10\mathrm{e}x - 10\mathrm{e} + 6\mathrm{e} = 10\mathrm{e}x - 4\mathrm{e}

At $P$ the tangent cuts the $x$ -axis, so $y = 0$ :

10\mathrm{e}x - 4\mathrm{e} = 0 \implies x = \frac{4\mathrm{e}}{10\mathrm{e}} = \frac{2}{5} = 0.4

Answer

The tangent is $y = 10\mathrm{e}x - 4\mathrm{e}$ , and the $x$ -coordinate of $P$ is $0.4$ .

Final answer

y = 10ex - 4e; x-coordinate of P is 0.4

Detailed explanation

Walkthrough

The curve is $y = 5\mathrm{e}^{2x-1} + \mathrm{e}$ . To find the tangent at $x = 1$ we need two things at that point: the gradient and the coordinates of the point itself.

The gradient function comes from differentiating. The derivative of $\mathrm{e}^{2x-1}$ is $2\mathrm{e}^{2x-1}$ by the chain rule (the inner function $2x - 1$ differentiates to $2$ ), so the first term gives $5 \times 2\mathrm{e}^{2x-1} = 10\mathrm{e}^{2x-1}$ . The second term $\mathrm{e}$ is a constant, so its derivative is $0$ .

At $x = 1$ the exponent is $2(1) - 1 = 1$ , so the gradient is $10\mathrm{e}^{1} = 10\mathrm{e}$ , and the point on the curve is $y = 5\mathrm{e}^{1} + \mathrm{e} = 6\mathrm{e}$ .

Using the point-gradient form $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = (1, 6\mathrm{e})$ and $m = 10\mathrm{e}$ :

y - 6\mathrm{e} = 10\mathrm{e}(x - 1)

Expanding and collecting the constant terms gives $y = 10\mathrm{e}x - 10\mathrm{e} + 6\mathrm{e} = 10\mathrm{e}x - 4\mathrm{e}$ , which is the required $y = mx + c$ form with exact values $m = 10\mathrm{e}$ and $c = -4\mathrm{e}$ .

The point $P$ is where the tangent meets the $x$ -axis, so we substitute $y = 0$ and solve for $x$ . The factors of $\mathrm{e}$ cancel, giving $x = \frac{4}{10} = 0.4$ .

Key Takeaways

Differentiating $\mathrm{e}^{ax+b}$ gives $a\,\mathrm{e}^{ax+b}$ — the chain rule brings down the coefficient of $x$ .
A constant term like $\mathrm{e}$ differentiates to zero; do not treat $\mathrm{e}$ as a variable.
A tangent needs both the gradient (from the derivative) and the point (from the original curve) evaluated at the given $x$ -value.
The $x$ -intercept of a line is found by setting $y = 0$ ; here the $\mathrm{e}$ factors cancel, leaving a clean exact value.

Common Mistakes

Writing $\frac{\mathrm{d}y}{\mathrm{d}x} = 5\mathrm{e}^{2x-1}$ , forgetting the chain-rule factor of $2$ (the scheme gives M1 for any $k\mathrm{e}^{2x-1}$ with $k \neq 10$ , but the final accuracy marks need $10$ ).
Adding a spurious constant of integration-style term, e.g. $10\mathrm{e}^{2x-1} + c$ — the scheme explicitly penalises this (SCM1 only).
Using $y = 5\mathrm{e} + \mathrm{e}$ incorrectly or forgetting the $+\mathrm{e}$ term when finding the point on the curve.
Substituting into the derivative instead of the curve (or vice versa) when finding the gradient and the point.
Giving a decimal approximation of $\mathrm{e}$ in the tangent equation — the question demands exact values, so $m = 10\mathrm{e}$ and $c = -4\mathrm{e}$ must stay in terms of $\mathrm{e}$ .
Sign slips when expanding $10\mathrm{e}(x - 1)$ , giving $+10\mathrm{e}$ instead of $-10\mathrm{e}$ before collecting constants.

Things to Be Careful About

The question requires exact values of $m$ and $c$ : leave them as $10\mathrm{e}$ and $-4\mathrm{e}$ , not $27.2$ and $-10.9$ . The final A1 for the $x$ -coordinate is "isw", so later mis-simplification after $0.4$ is ignored, but the tangent equation itself must be exact.
The scheme's A1 for the gradient and $y$ -value is FT on your derivative only if the M1 (or SCM1) was earned — so the chain-rule factor of $2$ matters from the very first line.
The final A1 depends on the tangent equation being correct with exact values, so an earlier error propagates.
Show the substituted expressions ( $10\mathrm{e}^{2(1)-1}$ , $5\mathrm{e}^{1} + \mathrm{e}$ ) before the simplified values so each method mark is visible.

Techniques used

differentiate an exponential function using the chain ruleevaluate the derivative and the curve at the given pointform the tangent equation from a point and a gradientset y = 0 to find the x-intercept

Approach

Rewrite $\operatorname{cosec} x$ and $\cot x$ in terms of $\sin x$ and $\cos x$ , simplify the fraction, and cancel to reach the printed target.

Working

Using $\operatorname{cosec} x = \frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$ :

\sin^3 x\left(\frac{\operatorname{cosec} x}{\cot x}\right) = \sin^3 x\left(\frac{1}{\sin x} \div \frac{\cos x}{\sin x}\right)

Dividing by $\frac{\cos x}{\sin x}$ multiplies by its reciprocal:

= \sin^3 x\left(\frac{1}{\sin x} \times \frac{\sin x}{\cos x}\right) = \sin^3 x\left(\frac{1}{\cos x}\right)

= \frac{\sin^3 x}{\cos x} = \sin^2 x \times \frac{\sin x}{\cos x} = \sin^2 x \tan x

as required.

Answer

\sin^3 x\left(\frac{\operatorname{cosec} x}{\cot x}\right) = \sin^2 x \tan x

Final answer

Shown: sin^3 x (cosec x / cot x) = sin^2 x tan x

Detailed explanation

Walkthrough

The bracket contains two reciprocal trigonometric functions, so the first move is to express them using their definitions: $\operatorname{cosec} x = \frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$ . Dividing by $\cot x$ means multiplying by its reciprocal $\frac{\sin x}{\cos x}$ , which is exactly $\tan x$ . The $\sin x$ in the numerator of the reciprocal cancels the $\frac{1}{\sin x}$ from $\operatorname{cosec} x$ , leaving $\frac{1}{\cos x}$ . Multiplying by $\sin^3 x$ gives $\frac{\sin^3 x}{\cos x}$ , and splitting this as $\sin^2 x \times \frac{\sin x}{\cos x}$ recognises $\tan x$ , producing the target $\sin^2 x \tan x$ . Since the question says "show that", the working must run forward from the given expression to the printed answer, with every cancellation line visible.

Key Takeaways

The definitions $\operatorname{cosec} x = \frac{1}{\sin x}$ , $\cot x = \frac{\cos x}{\sin x} = \frac{1}{\tan x}$ convert any compound trig fraction into sines and cosines.
Dividing by a fraction means multiplying by its reciprocal.
"Show that" proofs must be derived forward, showing all intermediate algebra.

Common Mistakes

Cancelling $\sin^3 x$ with the $\sin x$ inside $\cot x$ incorrectly, e.g. writing $\frac{\operatorname{cosec} x}{\cot x} = \frac{1}{\cos x}$ without showing the intermediate step — the scheme requires a convincing correct statement.
Treating $\frac{\operatorname{cosec} x}{\cot x}$ as $\operatorname{cosec} x \times \cot x$ instead of dividing.
Omitting intermediate lines: the mark scheme demands "convincing correct detail", so jumping straight to the answer loses the method mark.

Things to Be Careful About

This is an "Answer Given" (AG) part: the target is printed, so the marks are for the derivation, not the final line. Every algebraic step between the given expression and $\sin^2 x \tan x$ must appear.
Keep trigonometric function names upright and use $\operatorname{cosec}$ , not $\csc$ , matching 4037 notation.

Techniques used

rewrite cosec and cot in terms of sin and cossimplify the compound fractioncancel common factors to reach the target form

(b)

Solve the equation $\cos^2 x \tan x - \frac{1}{2}\tan x = 0$ for $-\pi < x < \pi$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Factorise out the common factor $\tan x$ , set each factor to zero, and solve each resulting equation for $-\pi < x < \pi$ .

Working

Factorise:

\cos^2 x \tan x - \frac{1}{2}\tan x = \tan x\left(\cos^2 x - \frac{1}{2}\right) = 0

First factor:

\tan x = 0 \quad \Rightarrow \quad x = 0

Second factor:

\cos^2 x = \frac{1}{2} \quad \Rightarrow \quad \cos x = \pm\sqrt{\frac{1}{2}} = \pm\frac{1}{\sqrt{2}}

For $\cos x = \frac{1}{\sqrt{2}}$ :

x = \frac{\pi}{4} \quad \text{or} \quad x = -\frac{\pi}{4}

For $\cos x = -\frac{1}{\sqrt{2}}$ :

x = \frac{3\pi}{4} \quad \text{or} \quad x = -\frac{3\pi}{4}

All values lie in $-\pi < x < \pi$ .

Answer

x = 0, \quad \pm\frac{\pi}{4}, \quad \pm\frac{3\pi}{4}

Final answer

x = 0, +/-pi/4, +/-3pi/4

Detailed explanation

Walkthrough

Both terms of $\cos^2 x \tan x - \frac{1}{2}\tan x$ contain the factor $\tan x$ , so the equation factorises as $\tan x\left(\cos^2 x - \frac{1}{2}\right) = 0$ . A product is zero when either factor is zero. The first factor gives $\tan x = 0$ , whose only solution in $-\pi < x < \pi$ is $x = 0$ . The second factor gives $\cos^2 x = \frac{1}{2}$ , so $\cos x = \pm\frac{1}{\sqrt{2}}$ . The positive value occurs at $x = \pm\frac{\pi}{4}$ and the negative value at $x = \pm\frac{3\pi}{4}$ (equivalently $\pm\frac{5\pi}{4}$ , but $\frac{5\pi}{4}$ lies outside the interval, and its reflection $-\frac{3\pi}{4}$ is the in-range solution). Collecting all five values gives the complete solution set. Crucially, the equation must be factorised, not divided by $\tan x$ : dividing would destroy the $x = 0$ solution.

Key Takeaways

Never divide a trigonometric equation by a trig function — factorise instead, so no solutions are lost.
From $\cos^2 x = k$ you get $\cos x = \pm\sqrt{k}$ , doubling the number of angle families.
In a symmetric interval $-\pi < x < \pi$ , solutions come in $\pm$ pairs plus possibly $0$ .

Common Mistakes

Dividing both sides by $\tan x$ — the mark scheme explicitly awards M0 for this, and it loses the solution $x = 0$ .
Dividing by $\sin x$ after rewriting in terms of $\sin x$ — also M0 for the same reason.
Finding only $\cos x = +\frac{1}{\sqrt{2}}$ and missing the $\pm$ , losing $\pm\frac{3\pi}{4}$ .
Giving $x = \pm\frac{5\pi}{4}$ instead of $\pm\frac{3\pi}{4}$ , or including values outside $-\pi < x < \pi$ — the A2 requires no extras in range.
Forgetting $x = 0$ from the $\tan x$ factor.
Writing $\cos x = \frac{1}{2}$ instead of $\frac{1}{\sqrt{2}}$ (forgetting the square root).

Things to Be Careful About

The mark scheme notes the A2 is "nfww" — no marks from wrong working — and requires no extra solutions in the interval; one error costs one of the two accuracy marks.
The interval is in radians, so answers must be given in radians ( $\pm\frac{\pi}{4} \approx \pm 0.785$ , $\pm\frac{3\pi}{4} \approx \pm 2.36$ ); exact multiples of $\pi$ are the cleanest form.
Check every solution lies strictly inside $-\pi < x < \pi$ — endpoints are excluded.

Techniques used

factorise the trigonometric equation instead of dividingsolve each factor separatelyfind all solutions in the symmetric interval about zero

Q7Medium-EasyPermutations and combinations

Find the number of different ways the 9 letters of the word POLYMATHS can be arranged when

(a)

the O and A are not next to each other

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

All 9 letters of POLYMATHS are distinct, so there are $9!$ arrangements in total. Count the arrangements where O and A ARE together (treat OA as one block, so $2! \times 8!$ ), then subtract from the total.

Working

9! = 362\,880

Arrangements with O and A together (as a block, in either order):

2! \times 8! = 2 \times 40\,320 = 80\,640

Subtracting:

9! - 2! \times 8! = 362\,880 - 80\,640 = 282\,240

Answer

282\,240

Final answer

282240

Detailed explanation

Walkthrough

Since all 9 letters are different, the total number of arrangements is $9! = 362\,880$ . It is easier to count the unwanted arrangements — those with O and A next to each other — and subtract. Glue O and A into a single block; the block plus the other 7 letters make 8 objects, arrangeable in $8!$ ways, and within the block O and A can swap, giving a factor $2!$ . So the unwanted count is $2! \times 8! = 80\,640$ , and the wanted count is $362\,880 - 80\,640 = 282\,240$ .

Key Takeaways

"Not next to each other" counting is almost always done by the complement: total minus together.
Treating adjacent letters as a single block reduces the number of objects to arrange.

Common Mistakes

Forgetting the internal $2!$ for the order of O and A inside the block.
Using $7!$ instead of $8!$ (forgetting the block counts as one object).
Subtracting the wrong way round.

Things to Be Careful About

The answer must come from correct working (nfww): the method $9! - 2! \times 8!$ must be visible.
All letters are distinct here — no division for repeats is needed.

Techniques used

count total unrestricted arrangementscount arrangements with the two letters together as a blocksubtract the complement from the total

(b)

the letters MATHS are together in this order.

2M Similar

Q8MediumStraight-line graphsLogarithmic and exponential functions

An experiment was carried out and values of $y$ for certain values of $x$ were recorded. The table shows the values recorded.

$x$	15	30	45	60	75
$y$	10	13	22	35	50

The relationship between $y$ and $x$ is modelled by $y = A\mathrm{e}^{kx}$ , where $A$ and $k$ are constants.

(a)

Draw a straight line graph for $\ln y$ against $x$ .

Fig. 3

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Taking natural logs of $y = A\mathrm{e}^{kx}$ gives $\ln y = \ln A + kx$ , so plotting $\ln y$ against $x$ should give a straight line. Compute each $\ln y$ value, plot the points on the given grid, and draw a single ruled line of best fit.

Working

The transformed values are:

\begin{array}{c|ccccc} x & 15 & 30 & 45 & 60 & 75 \\\hline \ln y & 2.30 & 2.56 & 3.09 & 3.56 & 3.91 \end{array}

Each point $(x, \ln y)$ is plotted on the grid: $(15, 2.30)$ , $(30, 2.56)$ , $(45, 3.09)$ , $(60, 3.56)$ , $(75, 3.91)$ . A single straight line is then ruled through them as a line of best fit, passing close to all five points.

Answer

Five points plotted at $(15, 2.3)$ , $(30, 2.6)$ , $(45, 3.1)$ , $(60, 3.6)$ , $(75, 3.9)$ with a single ruled straight line of best fit drawn through them.

Final answer

Points plotted at (15, 2.3), (30, 2.6), (45, 3.1), (60, 3.6), (75, 3.9) with a single ruled straight line of best fit

Detailed explanation

Walkthrough

The model $y = A\mathrm{e}^{kx}$ is exponential, so a direct plot of $y$ against $x$ would be a curve. Taking natural logarithms converts it to linear form:

\ln y = \ln(A\mathrm{e}^{kx}) = \ln A + kx

so if we plot $\ln y$ (vertical axis) against $x$ (horizontal axis), the points should lie near a straight line with gradient $k$ and intercept $\ln A$ . Each recorded $y$ value is converted using natural logs: $\ln 10 = 2.30$ , $\ln 13 = 2.56$ , $\ln 22 = 3.09$ , $\ln 35 = 3.56$ , $\ln 50 = 3.91$ . These five points are plotted on the printed grid and a single straight line is ruled so that it passes as close as possible to all of them.

Key Takeaways

An exponential model $y = A\mathrm{e}^{kx}$ becomes the straight line $\ln y = kx + \ln A$ when $\ln y$ is plotted against $x$ .
The gradient of such a plot estimates $k$ and the vertical intercept estimates $\ln A$ .
A line of best fit should be a single ruled line passing close to all plotted points, not point-to-point segments.

Common Mistakes

Using base-10 logarithms ( $\lg$ ) instead of natural logarithms ( $\ln$ ); the model uses $\mathrm{e}$ , so natural logs are required.
Plotting $y$ instead of $\ln y$ on the vertical axis.
Drawing a jagged line joining the points instead of one straight line of best fit.
Misreading the small grid subdivisions when locating values like 2.56 or 3.09.

Things to Be Careful About

The mark scheme awards B1 for at least four correctly plotted points and B1 for a single ruled straight line of best fit — both features must be present.
Values read to 2 decimal places are sufficient; the scheme accepts e.g. 2.5 or 2.6 for $\ln 13$ and 3.5 or 3.6 for $\ln 35$ .
The line must be ruled (use a ruler) and must not be forced through the origin.

Techniques used

take natural logarithms of the y-valuesplot ln y against xdraw a ruled line of best fit

(b)

Find the equation of the line in part (a) and hence find the values of $A$ and $k$ . Give each value correct to 1 significant figure.

5M Similar

DifficultyMedium

Worked solution

Approach

Since $\ln y = kx + \ln A$ , the gradient of the line in part (a) is $k$ and the vertical intercept is $\ln A$ . Read both from the drawn line, write the equation, then find $A = \mathrm{e}^{\ln A}$ and round each value to 1 significant figure.

Working

From the line of best fit, take two well-separated points on the line, e.g. $(0, 1.8)$ and $(60, 3.6)$ :

k = \frac{3.6 - 1.8}{60 - 0} = \frac{1.8}{60} = 0.03

The vertical intercept is $\ln A = 1.8$ . So the equation of the line is:

\ln y = 0.03x + 1.8

Comparing with $\ln y = kx + \ln A$ :

k = 0.03, \qquad \ln A = 1.8 \implies A = \mathrm{e}^{1.8} = 6.05

Rounded to 1 significant figure: $A = 6$ and $k = 0.03$ .

Answer

\ln y = 0.03x + 1.8, \qquad A = 6, \qquad k = 0.03

Final answer

ln y = 0.03x + 1.8, A = 6, k = 0.03

Detailed explanation

Walkthrough

Taking logs of the model gave $\ln y = kx + \ln A$ , which matches the straight-line form $y = mx + c$ with gradient $m = k$ and intercept $c = \ln A$ . So everything needed comes straight off the graph. Choosing two widely separated points that lie on the ruled line, such as $(0, 1.8)$ and $(60, 3.6)$ , the gradient is the change in $\ln y$ divided by the change in $x$ : $1.8/60 = 0.03$ . The point where the line crosses the vertical axis gives $\ln A = 1.8$ . Hence the equation of the line is $\ln y = 0.03x + 1.8$ . To recover $A$ we undo the logarithm by exponentiating: $A = \mathrm{e}^{1.8} \approx 6.05$ , which rounds to $6$ to 1 significant figure, while $k = 0.03$ is already given to 1 significant figure.

Key Takeaways

In a log-linear plot of an exponential model, gradient = $k$ and vertical intercept = $\ln A$ .
Use two far-apart points on the line (not necessarily data points) to compute the gradient accurately.
Undo $\ln A$ with $A = \mathrm{e}^{\ln A}$ to return to the original exponential model.

Common Mistakes

Reading the intercept as $A$ itself instead of $\ln A$ (giving $A = 1.8$ ).
Computing the gradient from two adjacent data points, which amplifies plotting error.
Forgetting to round to 1 significant figure — the mark scheme caps the marks at 2 if either value is not rounded to 1 sf.
Writing the line equation with $A$ instead of $\ln A$ as the constant term.

Things to Be Careful About

The final answers must be quoted to 1 significant figure: $A = 6$ (or 7) and $k = 0.03$ (or 0.02); unrounded answers lose marks.
The values must come from the linear graph or linear equation — values found directly from the original data score only SC marks.
Follow-through applies: your $A$ and $k$ may follow your own readings of gradient and intercept within the accepted ranges.

Techniques used

read the gradient and intercept from the line of best fitwrite the equation of the line in ln y = kx + ln A formexponentiate the intercept to recover A

(c)

Find the value of $x$ for which $y = 17$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute $y = 17$ into the linear equation $\ln y = 0.03x + 1.8$ from part (b), evaluate $\ln 17$ , and solve for $x$ .

Working

Using the equation of the line from part (b):

\ln 17 = 0.03x + 1.8

2.83 = 0.03x + 1.8

0.03x = 2.83 - 1.8 = 1.03

x = \frac{1.03}{0.03} = 34.4

Answer

x = 34.4

Final answer

34.4

Detailed explanation

Walkthrough

Part (b) established that the data satisfy $\ln y = 0.03x + 1.8$ . To find where $y = 17$ , substitute into this equation rather than the original exponential form: taking $\ln 17 = 2.83$ gives $2.83 = 0.03x + 1.8$ . Subtracting the intercept leaves $0.03x = 1.03$ , and dividing by the gradient gives $x = 34.4$ . This sits comfortably inside the accepted range $33 \leqslant x \leqslant 37.5$ , confirming consistency with the graph readings.

Key Takeaways

Once a relationship has been linearised, further questions are answered most easily in the linear form.
Solving $\ln y = kx + \ln A$ for $x$ is just solving a linear equation: isolate the $kx$ term and divide.

Common Mistakes

Substituting $y = 17$ without taking its logarithm first (i.e. writing $17 = 0.03x + 1.8$ ).
Using rounded intermediate values too aggressively; keep $\ln 17 = 2.83[32\ldots]$ until the division.
Sign errors when moving the intercept across the equation.

Things to Be Careful About

The answer must be nfww (not from wrong working): the method must use the transformed equation correctly.
The accepted answer range is roughly $33 \leqslant x \leqslant 37.5$ because it depends on the candidate's own gradient and intercept; later mis-simplification is ignored (isw).
Give the answer to 3 significant figures: $x = 34.4$ .

Techniques used

take natural logarithms of both sidesrearrange the linear equation to solve for x

Q99MMediumCalculusStraight-line graphs

Fig. 4

The diagram shows part of the curve $y = 32x - 4x^2 - 48$ and the line $AB$ .
The curve and the line $AB$ meet the $x$ -axis at $A$ and meet again at the point $B(5, 12)$ .
The line $CD$ extended is parallel to the $y$ -axis and passes through the maximum point of the curve.
Find the area of the shaded region.

DifficultyMedium

Worked solution

Approach

Find where the curve meets the $x$ -axis at $A$ , locate the maximum point (giving the vertical line $CD$ ), find $D$ on the line $AB$ , then compute the shaded area as the integral of the curve from $A$ 's $x$ -coordinate to $x = 4$ minus the area of the triangle under $AB$ .

Working

Point $A$ : solve

32x - 4x^2 - 48 = 0 \quad\Rightarrow\quad x^2 - 8x + 12 = 0 \quad\Rightarrow\quad (x-2)(x-6)=0

so $x = 2$ or $x = 6$ . Since $B(5,12)$ lies to the right, $A(2,0)$ .

Maximum point: by symmetry about $x = \frac{2+6}{2} = 4$ (or differentiating $32 - 8x = 0$ ), so $CD$ is the line $x = 4$ .

Line $AB$ : gradient $m_{AB} = \frac{12-0}{5-2} = 4$ , so $y = 4(x-2)$ , i.e. $y = 4x - 8$ . At $x = 4$ : $D(4, 8)$ .

Area under the curve from $x=2$ to $x=4$ :

\int_{2}^{4}\left(32x - 4x^2 - 48\right)\mathrm{d}x = \left[16x^2 - \frac{4}{3}x^3 - 48x\right]_{2}^{4}

= \left(256 - \frac{256}{3} - 192\right) - \left(64 - \frac{32}{3} - 96\right)

= \left(64 - \frac{256}{3}\right) - \left(-32 - \frac{32}{3}\right) = 96 - \frac{224}{3} = \frac{64}{3}

Triangle under $AB$ from $x=2$ to $x=4$ :

\frac{1}{2} \times (4-2) \times 8 = 8

Shaded area:

\frac{64}{3} - 8 = \frac{40}{3}

Answer

\frac{40}{3}

Final answer

40/3 (13.3)

Detailed explanation

Walkthrough

The shaded region is bounded above by the parabola, below by the straight line $AB$ , on the left by their meeting point $A$ , and on the right by the vertical line through the curve's maximum. So its area equals the area under the curve minus the area under the line over the same interval.

First we need $A$ . Setting $y = 0$ gives $32x - 4x^2 - 48 = 0$ ; dividing by $-4$ gives $x^2 - 8x + 12 = 0$ , which factorises as $(x-2)(x-6) = 0$ . The diagram shows $A$ is the left-hand intercept, so $A(2,0)$ .

Next, the vertical line $CD$ passes through the maximum of the parabola. A parabola is symmetric, so the vertex sits midway between the roots: $x = \frac{2+6}{2} = 4$ . Equivalently, differentiating gives $\frac{\mathrm{d}y}{\mathrm{d}x} = 32 - 8x = 0$ , again $x = 4$ .

To find $D$ we need the equation of $AB$ . It passes through $A(2,0)$ and $B(5,12)$ , so its gradient is $\frac{12-0}{5-2} = 4$ , giving $y = 4x - 8$ . At $x = 4$ this gives $y = 8$ , so $D(4,8)$ .

Now integrate the curve from $x = 2$ to $x = 4$ : the antiderivative is $16x^2 - \frac{4}{3}x^3 - 48x$ , which evaluates to $\frac{64}{3}$ across these limits. The region under the line over the same interval is simply a right triangle with base $4 - 2 = 2$ and height $8$ , area $\frac{1}{2}(2)(8) = 8$ . Subtracting gives the shaded area $\frac{64}{3} - 8 = \frac{40}{3}$ .

Key Takeaways

Areas between a curve and a chord are found by integrating the curve and subtracting the area under the line (or integrating the difference directly).
The vertex of a parabola lies midway between its $x$ -intercepts — a quick alternative to differentiating.
When one boundary is a straight line, a triangle area formula can replace an integration.

Common Mistakes

Taking $x = 6$ instead of $x = 2$ for $A$ (the diagram shows $A$ left of the vertex).
Forgetting to subtract the triangle under $AB$ , giving just the area under the curve ( $\frac{64}{3}$ ).
Sign slips when evaluating the antiderivative at the lower limit $x = 2$ ; the bracketed evaluation must be done carefully.
Using the wrong height for the triangle (e.g. using $y = 12$ from $B$ rather than $y = 8$ at $D$ ).
The mark scheme notes "isw" for later mis-simplification but "nfww" for $13.3$ : a decimal answer scores nothing unless reached from correct working — keep the exact fraction $\frac{40}{3}$ .

Things to Be Careful About

Every mark-scheme step must be visible: solving the quadratic for $A$ , finding $x = 4$ , obtaining $D(4,8)$ or $y = 4x - 8$ , writing the integrated expression with limits shown, and the subtraction plan.
Give the exact answer $\frac{40}{3}$ ; decimals are accepted only if they follow correct working.
The final accuracy mark depends on all earlier marks being earned, so check each stage.

Techniques used

factorise the quadratic to find the x-interceptlocate the maximum point of a parabola by symmetry or differentiationfind the equation of the chord AB from two pointsintegrate the curve between limitssubtract the triangular area under the line

Approach

An inverse function exists only when the function is one-one (each output comes from exactly one input). Check whether $\mathrm{f}(x) = \mathrm{e}^{x^2+3}$ is one-one on its given domain.

Working

For $x < 0$ , $x^2$ is strictly decreasing as $x$ increases towards $0$ , so $x^2 + 3$ takes each value exactly once, and hence $\mathrm{e}^{x^2+3}$ also takes each value exactly once.

Therefore $\mathrm{f}$ is one-one on its domain $x < 0$ , so $\mathrm{f}^{-1}$ exists.

Answer

$\mathrm{f}$ is one-one for $x < 0$ (each value of $\mathrm{f}(x)$ arises from exactly one value of $x$ ), so its inverse exists.

Final answer

f is one-one on its domain x < 0, so f^-1 exists

Detailed explanation

Walkthrough

A function has an inverse function precisely when it is one-one: no horizontal line cuts its graph more than once. The formula $\mathrm{e}^{x^2+3}$ on its own would not be one-one over all real numbers, because $x = -2$ and $x = 2$ both give $\mathrm{e}^{7}$ . But the domain here is restricted to $x < 0$ , which keeps only the left-hand branch of the parabola inside the exponent. On that branch, as $x$ increases, $x^2$ decreases, so every output value occurs exactly once — the function is strictly decreasing and therefore one-one.

Key Takeaways

An inverse function exists if and only if the function is one-one.
A restricted domain can rescue an otherwise many-one formula by keeping just one branch.

Common Mistakes

Simply saying "because f is one-one" without connecting it to the given domain $x<0$ ; the mark scheme wants a valid explanation using $\mathrm{f}$ itself.
Ignoring the restriction and claiming $\mathrm{e}^{x^2+3}$ is one-one everywhere — it is not, since $x^2$ repeats values for $\pm x$ .

Things to Be Careful About

The explanation must reference the actual function and its restricted domain; a bare assertion scores nothing.

Techniques used

recognise that an inverse exists only for a one-one functionargue from the restricted domain

(b)

Find an expression for $\mathrm{f}^{-1}(x)$ and state the domain and range of $\mathrm{f}^{-1}$ .

5M Similar

DifficultyMedium

Worked solution

Approach

To find $\mathrm{f}^{-1}$ , write $y = \mathrm{e}^{x^2+3}$ , swap the variables, then solve for $y$ using logarithms. The sign of the square root is fixed by the fact that the range of $\mathrm{f}^{-1}$ equals the domain of $\mathrm{f}$ , namely $x < 0$ .

Working

Let

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

Swap the variables:

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

Take natural logarithms of both sides:

\ln x = y^2 + 3

y^2 = \ln x - 3

y = \pm\sqrt{\ln x - 3}

Since the range of $\mathrm{f}^{-1}$ must equal the domain of $\mathrm{f}$ , which is $x < 0$ , we take the negative square root:

\mathrm{f}^{-1}(x) = -\sqrt{\ln x - 3}

For this square root to be defined we need $\ln x - 3 \geq 0$ , i.e. $\ln x \geq 3$ , i.e. $x \geq \mathrm{e}^3$ . Since $\mathrm{f}$ never actually attains the value $\mathrm{e}^3$ (as $x^2 > 0$ for $x<0$ ), the domain is $x > \mathrm{e}^3$ .

Domain of $\mathrm{f}^{-1}$ : $x > \mathrm{e}^3$ .
Range of $\mathrm{f}^{-1}$ : $\mathrm{f}^{-1}(x) < 0$ .

Answer

\mathrm{f}^{-1}(x) = -\sqrt{\ln x - 3}, \qquad x > \mathrm{e}^3, \qquad \mathrm{f}^{-1}(x) < 0

Final answer

f^-1(x) = -sqrt(ln x - 3), domain x > e^3, range f^-1(x) < 0

Detailed explanation

Walkthrough

The standard method for finding an inverse is to write $y = \mathrm{f}(x)$ , interchange $x$ and $y$ , and then make $y$ the subject again. Swapping gives $x = \mathrm{e}^{y^2+3}$ . Because the variable sits in an exponent, the correct tool is the natural logarithm, which undoes $\mathrm{e}^{(\cdot)}$ : taking $\ln$ of both sides gives $\ln x = y^2 + 3$ . Rearranging, $y^2 = \ln x - 3$ , so $y = \pm\sqrt{\ln x - 3}$ .

Which sign? The inverse's outputs are the original function's inputs, and the domain of $\mathrm{f}$ was $x < 0$ . So $\mathrm{f}^{-1}(x)$ must be negative, forcing the minus sign: $\mathrm{f}^{-1}(x) = -\sqrt{\ln x - 3}$ .

The domain of the inverse is the range of the original. Since $x^2 + 3 > 3$ for $x < 0$ , we have $\mathrm{f}(x) > \mathrm{e}^3$ , so the inverse accepts inputs $x > \mathrm{e}^3$ — consistent with needing $\ln x - 3 > 0$ under the root. Its range is all negative numbers.

Key Takeaways

Inverse method: swap variables, rearrange, swap back into notation.
Logarithms invert exponentials; $\ln(\mathrm{e}^u) = u$ .
Domain of $\mathrm{f}^{-1}$ = range of $\mathrm{f}$ ; range of $\mathrm{f}^{-1}$ = domain of $\mathrm{f}$ .
The sign of a square root in an inverse is dictated by the original domain restriction.

Common Mistakes

Writing $+\sqrt{\ln x - 3}$ or $\pm\sqrt{\ln x - 3}$ — the A marks require the negative root specifically.
Giving the domain as $x \geq \mathrm{e}^3$ instead of $x > \mathrm{e}^3$ .
Stating the range as $\mathrm{f}^{-1}(x) > 0$ through sign confusion.
Forgetting to state the domain and range at all — they carry two separate B1 marks.
Sign or arithmetic slips in reaching $\sqrt{\ln x - 3}$ lose accuracy marks unless the order of operations is right.

Things to Be Careful About

The scheme allows follow-through after one sign or arithmetic error provided the order of operations is correct, but the final answer must be the negative root. Answers may equivalently be written as $-\sqrt{\ln(x/\mathrm{e}^3)}$ . Later mis-simplification is ignored (isw).

Techniques used

swap variables to set up the inversetake natural logarithms of both sidesselect the negative square root using the domainstate domain and range of the inverse from those of the original

(c)

Hence find and simplify an expression for $\mathrm{g}(x)$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $\mathrm{fg}(x) = \mathrm{e}^{2x}$ , applying $\mathrm{f}^{-1}$ to both sides gives $\mathrm{g}(x) = \mathrm{f}^{-1}\left(\mathrm{e}^{2x}\right)$ . Substitute into the expression found in part (b).

Working

Using the result of part (b):

\mathrm{g}(x) = \mathrm{f}^{-1}\left(\mathrm{e}^{2x}\right) = -\sqrt{\ln\left(\mathrm{e}^{2x}\right) - 3}

Since $\ln\left(\mathrm{e}^{2x}\right) = 2x$ :

\mathrm{g}(x) = -\sqrt{2x - 3}

This is consistent with the given condition $x > \frac{3}{2}$ , which ensures $2x - 3 > 0$ .

Answer

\mathrm{g}(x) = -\sqrt{2x - 3}

Final answer

g(x) = -sqrt(2x - 3)

Detailed explanation

Walkthrough

The composite $\mathrm{fg}$ means "apply $\mathrm{g}$ first, then $\mathrm{f}$ ": $\mathrm{fg}(x) = \mathrm{f}\big(\mathrm{g}(x)\big) = \mathrm{e}^{2x}$ . To undo the outer function $\mathrm{f}$ , apply its inverse to both sides: $\mathrm{g}(x) = \mathrm{f}^{-1}\left(\mathrm{e}^{2x}\right)$ . This is why the question says "Hence" — the answer must use the $\mathrm{f}^{-1}$ found in part (b).

Substituting $\mathrm{e}^{2x}$ in place of $x$ in $-\sqrt{\ln x - 3}$ gives $-\sqrt{\ln\left(\mathrm{e}^{2x}\right) - 3}$ . The key simplification is $\ln\left(\mathrm{e}^{2x}\right) = 2x$ , since $\ln$ and $\mathrm{e}^{(\cdot)}$ cancel. So $\mathrm{g}(x) = -\sqrt{2x - 3}$ .

As a check, feeding this back: $\mathrm{f}\left(-\sqrt{2x-3}\right) = \mathrm{e}^{(-\sqrt{2x-3})^2 + 3} = \mathrm{e}^{2x - 3 + 3} = \mathrm{e}^{2x}$ , as required. The condition $x > \frac{3}{2}$ guarantees the quantity under the root is positive.

Key Takeaways

$\mathrm{f}^{-1}$ undoes $\mathrm{f}$ : applying it to a composite isolates the inner function.
$\ln\left(\mathrm{e}^{u}\right) = u$ for any real $u$ .
Composite-function conditions (like $x > \frac{3}{2}$ ) often exist to keep expressions well-defined.

Common Mistakes

Trying to find $\mathrm{g}$ from scratch rather than using $\mathrm{f}^{-1}$ from part (b) — the word "Hence" signals the intended route.
Dropping the minus sign and writing $+\sqrt{2x-3}$ .
Mishandling $\ln\left(\mathrm{e}^{2x}\right)$ , e.g. writing $(\ln \mathrm{e})^{2x}$ or $2x\ln \mathrm{e}$ incorrectly simplified.

Things to Be Careful About

The M1 follows through on the candidate's own $\mathrm{f}^{-1}$ expression, but the A1 requires the exact form $-\sqrt{2x - 3}$ . If scored zero, a special case mark (SCB1) is available for obtaining $-\sqrt{2x-3}$ by solving $\big(\mathrm{g}(x)\big)^2 + 3 = 2x$ and arguing the root must be negative from the existence of the composite.

Techniques used

use fg composed with f-inverse to recover gcompose functions by substitutionsimplify ln of an exponential

Q118MMediumSeries

In the binomial expansion of $\left(2 + \frac{x}{2}\right)^n$ , the first three terms in increasing powers of $x$ are

b + abx + \frac{9}{8}abx^2

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

DifficultyMedium

Worked solution

Approach

Expand $\left(2 + \frac{x}{2}\right)^n$ using the general binomial term ${}^{n}\mathrm{C}_r\,2^{n-r}\left(\frac{x}{2}\right)^r$ , match the first three terms to $b + abx + \frac{9}{8}abx^2$ , and use the ratio of consecutive coefficients to eliminate $a$ and $b$ and find $n$ .

Working

The general term is

{}^{n}\mathrm{C}_r\,2^{n-r}\left(\frac{x}{2}\right)^r = {}^{n}\mathrm{C}_r\,2^{n-2r}x^r

So the first three terms are

2^n, \qquad n \times 2^{n-2}x, \qquad \frac{n(n-1)}{2} \times 2^{n-4}x^2

Matching with $b + abx + \frac{9}{8}abx^2$ :

b = 2^n

Taking the ratio of the coefficient of $x^2$ to that of $x$ eliminates both $a$ and $b$ :

\frac{\frac{n(n-1)}{2} \times 2^{n-4}}{n \times 2^{n-2}} = \frac{9}{8}

Simplifying the left-hand side:

\frac{(n-1)}{2} \times 2^{-2} = \frac{n-1}{8} = \frac{9}{8}

so $n - 1 = 9$ , giving

n = 10

Then

b = 2^{10} = 1024

and from the coefficient of $x$ ,

ab = n \times 2^{n-2} = 10 \times 2^8 = 2560

a = \frac{2560}{1024} = \frac{5}{2}

Answer

n = 10, \qquad a = \frac{5}{2}, \qquad b = 1024

Final answer

n = 10, a = 5/2, b = 1024

Detailed explanation

Walkthrough

The binomial expansion of $(p + q)^n$ has general term ${}^{n}\mathrm{C}_r\,p^{n-r}q^r$ . Here $p = 2$ and $q = \frac{x}{2}$ , so each term is ${}^{n}\mathrm{C}_r\,2^{n-r}\left(\frac{x}{2}\right)^r$ . Since $\left(\frac{x}{2}\right)^r = \frac{x^r}{2^r}$ , the power of 2 in each term is $2^{n-r} \times 2^{-r} = 2^{n-2r}$ , giving coefficients $2^n$ , $n\,2^{n-2}$ and $\frac{n(n-1)}{2}\,2^{n-4}$ for $x^0$ , $x$ and $x^2$ respectively.

Matching these against $b + abx + \frac{9}{8}abx^2$ : the constant term gives $b = 2^n$ directly. The other two equations involve all three unknowns, so instead of solving them simultaneously we take the ratio of the last two coefficients — this cancels both $a$ and $b$ at once, leaving an equation in $n$ alone. The ratio simplifies because $2^{n-4}/2^{n-2} = 2^{-2} = \frac{1}{4}$ and one factor of $n$ cancels, leaving $\frac{n-1}{8} = \frac{9}{8}$ , hence $n = 10$ .

With $n$ known, $b = 2^{10} = 1024$ . The middle coefficient then reads $ab = 10 \times 2^8 = 2560$ , so $a = \frac{2560}{1024} = \frac{5}{2}$ .

Key Takeaways

The general binomial term ${}^{n}\mathrm{C}_r\,a^{n-r}b^r$ lets you write down any term without expanding fully.
When several unknowns appear, ratios of consecutive coefficients often eliminate them cleanly — here one ratio removed both $a$ and $b$ .
Powers of 2 combine by index laws: $2^{n-r} \times 2^{-r} = 2^{n-2r}$ .

Common Mistakes

Writing the general term as ${}^{n}\mathrm{C}_r\,2^r\left(\frac{x}{2}\right)^{n-r}$ (powers swapped) or forgetting that $\left(\frac{x}{2}\right)^r$ contributes its own power of 2.
Trying to solve three simultaneous equations in $n$ , $a$ , $b$ directly instead of using the ratio trick — possible but far more error-prone.
Arithmetic slips in $2^{n-2}$ : with $n = 10$ it is $2^8 = 256$ , not $2^{10}$ or $512$ .
Cancelling $n$ incorrectly in the ratio, e.g. losing the factor $\frac{1}{2}$ from $\frac{n(n-1)}{2}$ .
Giving $ab = 2560$ as the value of $a$ instead of dividing by $b$ .

Things to Be Careful About

The answer must be exact: $a = \frac{5}{2}$ , not $2.5$ rounded, and $b = 1024$ exactly.
The ratio method depends on matching like powers of $x$ ; check that the constant term is matched with $x^0$ before writing any ratio.
Every mark-scheme step here is earned through working (M1/A1/B1 structure), so the expansion terms must be written out explicitly — a bare answer scores nothing (nfww).

Techniques used

write the general binomial termmatch the first three terms to the given formuse ratios of consecutive coefficients to eliminate unknownssolve for n and back-substitute for a and b

Additional Mathematics 4037/22 — May/June 2024

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