(a) Integrate the following with respect to $x$. (i) $$\mathrm{e}^{5x-2}$$ (ii) $$\frac{1}{4-3x} \quad \text{where } x < \frac{4}{3}$$ (b) Show that $$ \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sec^2 \left(\frac{1}{2}x\right) \,\mathrm{d}x = 2\left(1 - \frac{\sqrt{3}}{3}\right) $$

(a)(i) (1/5)e^(5x-2) + c · (a)(ii) -(1/3)ln(4 - 3x) + c · (b) 2(1 - √3/3), as required

4037/23 Oct/Nov 2025 — Additional Mathematics Solved Past Paper

Q15MMedium-EasyFactors of polynomials

It is given that $\mathrm{p}(x) = ax^3 - 7x^2 - bx + 9$ , where $a$ and $b$ are constants.
$x - 3$ is a factor of $\mathrm{p}(x)$ .
When $\mathrm{p}(x)$ is divided by $x + 2$ the remainder is $-35$ .

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

DifficultyMedium-Easy

Worked solution

Approach

Since $x - 3$ is a factor, the factor theorem gives $\mathrm{p}(3) = 0$ . Since division by $x + 2$ leaves remainder $-35$ , the remainder theorem gives $\mathrm{p}(-2) = -35$ . These two equations are linear in $a$ and $b$ ; solve them simultaneously.

Working

By the factor theorem:

\mathrm{p}(3) = a(3)^3 - 7(3)^2 - b(3) + 9 = 0

27a - 63 - 3b + 9 = 0 \quad\Rightarrow\quad 27a - 3b = 54

By the remainder theorem:

\mathrm{p}(-2) = a(-2)^3 - 7(-2)^2 - b(-2) + 9 = -35

-8a - 28 + 2b + 9 = -35 \quad\Rightarrow\quad -8a + 2b = -16

Eliminate $b$ : multiply the second equation by $\frac{3}{2}$ :

-12a + 3b = -24

Add to $27a - 3b = 54$ :

15a = 30 \quad\Rightarrow\quad a = 2

Substitute into $27a - 3b = 54$ :

54 - 3b = 54 \quad\Rightarrow\quad b = 0

Answer

a = 2, \qquad b = 0

Final answer

a = 2, b = 0

Detailed explanation

Walkthrough

The polynomial is $\mathrm{p}(x) = ax^3 - 7x^2 - bx + 9$ with two unknown constants. Each piece of information about factors and remainders converts into an equation.

First, the factor theorem: if $x - 3$ is a factor of $\mathrm{p}(x)$ , then substituting $x = 3$ must give zero. Substituting gives $27a - 63 - 3b + 9 = 0$ , which simplifies to $27a - 3b = 54$ . This is the first linear equation in $a$ and $b$ .

Second, the remainder theorem: when a polynomial is divided by $x + 2$ , the remainder equals the value of the polynomial at $x = -2$ (the root of the divisor). So $\mathrm{p}(-2) = -35$ , giving $-8a - 28 + 2b + 9 = -35$ , i.e. $-8a + 2b = -16$ . This is the second equation.

Now solve simultaneously. Multiplying the second equation by $\frac{3}{2}$ makes the coefficient of $b$ equal to $+3$ , matching the first equation's $-3b$ , so adding eliminates $b$ : $15a = 30$ , hence $a = 2$ . Back-substituting into $27a - 3b = 54$ gives $54 - 3b = 54$ , so $b = 0$ .

Key Takeaways

The factor theorem: $x - k$ is a factor of $\mathrm{p}(x)$ exactly when $\mathrm{p}(k) = 0$ .
The remainder theorem: dividing $\mathrm{p}(x)$ by $x - k$ leaves remainder $\mathrm{p}(k)$ — note that for divisor $x + 2$ you substitute $x = -2$ , not $+2$ .
Two independent conditions on two unknowns produce a pair of simultaneous linear equations solvable by elimination or substitution.

Common Mistakes

Substituting $x = +2$ instead of $x = -2$ for the divisor $x + 2$ — the remainder theorem uses the root of the divisor, so the sign flips.
Sign slips when expanding $(-2)^3 = -8$ and $(-2)^2 = 4$ , or with the term $-bx$ becoming $+2b$ at $x = -2$ .
Arithmetic errors in simplifying the constants ( $-63 + 9 = -54$ ; $-28 + 9 = -19$ , so $-8a + 2b = -16$ ).
The mark scheme awards "nfww" (no follow-through from wrong working): correct-looking answers reached via incorrect equations score nothing, so every substitution line must be shown correctly.

Things to Be Careful About

Both equations must be formed before solving; the M1 for elimination is only awarded when both equations are genuinely linear in $a$ and $b$ .
Show the substituted expressions (e.g. $27a - 63 - 3b + 9 = 0$ ) explicitly — these carry the B1 marks, not just the simplified forms.
Check the final answer: with $a = 2$ , $b = 0$ , $\mathrm{p}(x) = 2x^3 - 7x^2 + 9$ , and indeed $\mathrm{p}(3) = 54 - 63 + 9 = 0$ and $\mathrm{p}(-2) = -16 - 28 + 9 = -35$ . A quick verification guards against sign errors.

Techniques used

apply the factor theorem to form a linear equation in a and bapply the remainder theorem to form a second linear equationsolve simultaneous linear equations by elimination

Similar questions

Q2MediumFunctionsLogarithmic and exponential functions

(a)

2M

(i)

Fig. 1

The diagram shows the graph of $y = \mathrm{f}(x)$ .

On the same diagram sketch the graph of $y = \mathrm{f}^{-1}(x)$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

The graph of $y = \mathrm{f}^{-1}(x)$ is the mirror image of the graph of $y = \mathrm{f}(x)$ in the line $y = x$ . Every point $(a, b)$ on the original curve becomes the point $(b, a)$ on the inverse curve.

Working

The given curve $y = \mathrm{f}(x)$ starts at $(0, 0)$ and rises smoothly through the first quadrant, flattening as it goes (it passes near $(4, 4)$ and reaches about $y = 5$ at $x = 8$ ).

Reflecting each point in the line $y = x$ :

(0, 0) \mapsto (0, 0), \qquad (4, 4) \mapsto (4, 4)

so the inverse curve also starts at the origin and passes through $(4, 4)$ .

The inverse has the reflected shape: it rises steeply at first (near the $y$ -axis it climbs quickly, since the original was flat there) and then becomes steeper again for large $x$ , because the original was steep near $x = 0$ . Its domain is the range of $\mathrm{f}$ and its range is the domain of $\mathrm{f}$ , so it occupies the same first-quadrant region.

Answer

Draw the mirror image of the given curve in the line $y = x$ : a curve starting at $(0, 0)$ , passing through approximately $(4, 4)$ where it crosses the original curve, rising steeply close to the $y$ -axis and then bending upward more sharply than the original for large $x$ , defined over the same first-quadrant region.

Final answer

Inverse curve drawn as the reflection of y = f(x) in the line y = x, starting at (0, 0), crossing f(x) again at about (4, 4)

Detailed explanation

Walkthrough

An inverse function undoes the original function: if $\mathrm{f}$ sends an input $a$ to the output $b$ , then $\mathrm{f}^{-1}$ sends $b$ back to $a$ . On a graph this means every point $(a, b)$ on $y = \mathrm{f}(x)$ reappears as $(b, a)$ on $y = \mathrm{f}^{-1}(x)$ — the coordinates are swapped. Swapping coordinates is exactly what reflecting in the line $y = x$ does, so the whole inverse curve is obtained by folding the diagram along that diagonal line.

Here the original curve starts at $(0, 0)$ , which lies on the mirror line itself, so the inverse must also start at $(0, 0)$ . The curves cross wherever a point satisfies $y = x$ , i.e. on the line $y = x$ ; from the sketch the second crossing is at roughly $(4, 4)$ , which the mark scheme expects to be shown. The shape matters too: the original is steep near $x = 0$ and flat for large $x$ , so the inverse is flat near the origin's neighbourhood along the other axis and steep for large $x$ — the mark scheme pays for the correct shape over the correct domain and range, not just any curve through two points.

Key Takeaways

The graph of $\mathrm{f}^{-1}$ is always the reflection of the graph of $\mathrm{f}$ in the line $y = x$ .
Points on the line $y = x$ are fixed by the reflection, so $\mathrm{f}$ and $\mathrm{f}^{-1}$ intersect only on $y = x$ .
The domain of $\mathrm{f}^{-1}$ equals the range of $\mathrm{f}$ , and vice versa.

Common Mistakes

Reflecting in the $y$ -axis or the $x$ -axis instead of the line $y = x$ .
Drawing the inverse with the same shape as the original rather than the mirrored shape (steep/flat behaviour swapped).
Omitting the second intersection point near $(4, 4)$ , which the mark scheme explicitly looks for.
Extending the inverse into regions where the original function was never defined.

Things to Be Careful About

This is a B1 drawing mark: the scheme requires the correct shape over the correct domain and range, so a rough freehand curve through two points is not enough. Make sure the reflected curve visibly bends the opposite way to the original between the origin and $(4, 4)$ .

Techniques used

reflect a curve in the line y = x to obtain the inverse graphswap the domain and range of the original curveidentify the intersection points of a function and its inverse

(ii)

Describe the relationship between the graph of $\mathrm{f}(x)$ and the graph of $\mathrm{f}^{-1}(x)$ .

1M Similar

(b)

A function $\mathrm{g}$ is defined by $\mathrm{g}(x) = \mathrm{e}^{\sqrt{x-2}}$ for $x \geq 2$ .

6M

(i)

Find an expression for $\mathrm{g}^{-1}(x)$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Write $y = \mathrm{g}(x)$ , take natural logs of both sides to undo the exponential, make $x$ the subject, then swap $x$ and $y$ to obtain $\mathrm{g}^{-1}(x)$ .

Working

Let

y = \mathrm{e}^{\sqrt{x-2}}

Take natural logarithms of both sides:

\ln y = \sqrt{x - 2}

Square both sides:

(\ln y)^2 = x - 2

Make $x$ the subject:

x = (\ln y)^2 + 2

Swap the variables to obtain the inverse function:

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

Answer

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

Final answer

(ln x)^2 + 2

Detailed explanation

Walkthrough

To find an inverse function we start from $y = \mathrm{g}(x)$ and solve for $x$ in terms of $y$ . The obstacle is the exponential $\mathrm{e}^{\sqrt{x-2}}$ : the operation that undoes an exponential is the natural logarithm, since $\ln(\mathrm{e}^{u}) = u$ . Taking $\ln$ of both sides strips away the exponential and leaves $\ln y = \sqrt{x-2}$ . Squaring removes the square root, giving $x = (\ln y)^2 + 2$ . Finally, because the inverse function takes inputs named $x$ , we swap the letters: $\mathrm{g}^{-1}(x) = (\ln x)^2 + 2$ . The swap can be done at the start ( $\ln x = \sqrt{y-2}$ ) or at the end — the mark scheme accepts either, awarding M2 for the complete method including the change of subject and the swap of variables at some point.

Key Takeaways

To find $\mathrm{f}^{-1}$ : write $y = \mathrm{f}(x)$ , rearrange to get $x$ alone, then interchange $x$ and $y$ .
$\ln$ is the inverse operation of exponentiating base $\mathrm{e}$ , so it is the tool for undoing $\mathrm{e}^{(\cdot)}$ .
Undo operations happen in reverse order: the outermost operation (the exponential) is removed first, then the square root, then the subtraction of 2.

Common Mistakes

Forgetting to swap variables at some stage, leaving the answer as $x = (\ln y)^2 + 2$ .
Writing $\ln y^2$ or $2\ln y$ instead of $(\ln y)^2$ — squaring the logarithm is not the same as squaring its argument.
Taking log base 10 instead of natural log, giving $(\lg x)^2 + 2$ .
Mishandling the order: trying to remove the ' $-2$ ' before dealing with the exponential.

Things to Be Careful About

The mark scheme awards M1 specifically for seeing $\ln y = \sqrt{x-2}$ (or $\ln x = \sqrt{y-2}$ ), so that line must appear explicitly. The final A1 is for the exact form $(\ln x)^2 + 2$ ; keep the square on the whole logarithm and the $+2$ outside it.

Techniques used

rearrange an exponential equation by taking natural logarithmsmake x the subjectswap variables to obtain the inverse function

(ii)

Write down the range of $\mathrm{g}^{-1}$ .

1M Similar

(iii)

A function $\mathrm{h}$ is defined by $\mathrm{h}(x) = \frac{1}{x^2} + 2$ for $x > 0$ .

Find an expression for $\mathrm{gh}(x)$ in its simplest form.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute $\mathrm{h}(x)$ into $\mathrm{g}$ : replace $x$ in $\mathrm{g}(x) = \mathrm{e}^{\sqrt{x-2}}$ by $\frac{1}{x^2} + 2$ , then simplify.

Working

\mathrm{gh}(x) = \mathrm{g}\left(\frac{1}{x^2} + 2\right) = \mathrm{e}^{\sqrt{\left(\frac{1}{x^2} + 2\right) - 2}}

Simplify inside the square root:

\sqrt{\frac{1}{x^2}} = \frac{1}{x} \qquad (x > 0)

Therefore

\mathrm{gh}(x) = \mathrm{e}^{\frac{1}{x}}

Answer

\mathrm{gh}(x) = \mathrm{e}^{\frac{1}{x}}

Final answer

e^(1/x)

Detailed explanation

Walkthrough

The composite $\mathrm{gh}(x)$ means apply $\mathrm{h}$ first, then apply $\mathrm{g}$ to the result: $\mathrm{g}(\mathrm{h}(x))$ . Substituting $\mathrm{h}(x) = \frac{1}{x^2} + 2$ into $\mathrm{g}(x) = \mathrm{e}^{\sqrt{x-2}}$ gives $\mathrm{e}^{\sqrt{\frac{1}{x^2} + 2 - 2}}$ . The $+2$ and $-2$ cancel, leaving $\sqrt{\frac{1}{x^2}} = \frac{1}{x}$ (valid because $x > 0$ , so no modulus sign is needed). Hence $\mathrm{gh}(x) = \mathrm{e}^{1/x}$ . The mark scheme's B1 is for reaching $\mathrm{e}^{\frac{1}{x^2} + 2 - 2}$ , showing the substitution has been made correctly before simplifying.

Key Takeaways

$\mathrm{gh}(x) = \mathrm{g}(\mathrm{h}(x))$ : the function written next to $x$ acts first.
Simplify inside radicals before evaluating; $\sqrt{1/x^2} = |1/x|$ , which equals $1/x$ only when $x > 0$ .

Common Mistakes

Computing $\mathrm{hg}(x)$ instead of $\mathrm{gh}(x)$ — composition order matters.
Leaving the answer as $\mathrm{e}^{\sqrt{1/x^2}}$ without simplifying, losing the final B1.
Writing $\sqrt{1/x^2} = 1/x^2$ or forgetting the cancellation of $+2 - 2$ .

Things to Be Careful About

Both marks are accuracy marks (B2 for the fully simplified $\mathrm{e}^{1/x}$ ); the intermediate form $\mathrm{e}^{\frac{1}{x^2}+2-2}$ scores B1 only, so carry the simplification through. The condition $x > 0$ from $\mathrm{h}$ 's domain is what makes $\sqrt{1/x^2} = 1/x$ legitimate without absolute values.

Techniques used

evaluate a composite function by substituting h(x) into gsimplify using e raised to a sum of exponents

Q3MediumQuadratic functionsEquations, inequalities and graphs

(a)

4M

(i)

Write $x^2 - x - 6$ in the form $(x + a)^2 + b$ where $a$ and $b$ are constants.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Complete the square on $x^2 - x - 6$ : take half the coefficient of $x$ , square it, and adjust by the constant.

Working

Half of $-1$ is $-\frac{1}{2}$ , so

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

Therefore

x^2 - x - 6 = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} - 6 = \left(x - \frac{1}{2}\right)^2 - \frac{25}{4}

Answer

\left(x - \frac{1}{2}\right)^2 - \frac{25}{4}

Final answer

(x - 1/2)^2 - 25/4

Detailed explanation

Walkthrough

Completing the square rewrites a quadratic so the squared term is isolated. For $x^2 - x$ , the coefficient of $x$ is $-1$ ; half of it is $-\frac{1}{2}$ , giving $(x - \frac{1}{2})^2$ . But $(x - \frac{1}{2})^2 = x^2 - x + \frac{1}{4}$ , which is a quarter too big, so we must subtract $\frac{1}{4}$ . Then the original constant $-6$ is carried along: $-\frac{1}{4} - 6 = -\frac{25}{4}$ .

Key Takeaways

Completing the square: take half the coefficient of $x$ , square it, subtract the same amount to compensate.
The result is the vertex form, which immediately reveals the turning point.

Common Mistakes

Forgetting to subtract the squared half-coefficient, giving $(x - \frac{1}{2})^2 - 6$ instead of $(x - \frac{1}{2})^2 - \frac{25}{4}$ .
Sign slips: the bracket is $(x - \frac{1}{2})$ , not $(x + \frac{1}{2})$ , since the coefficient of $x$ is negative.
Writing $b = -6\frac{1}{4}$ is fine ("oe"), but leaving $b$ as $+\frac{25}{4}$ loses the mark.

Things to Be Careful About

The mark scheme accepts equivalent forms ("oe") and ignores later mis-simplification ("isw"), but the completed-square form must be exact — fractions, not decimals.
Partial credit (B1) is given for either the correct bracket alone or the correct constant alone, so show both clearly.

Techniques used

complete the square of a quadraticexpress the quadratic in vertex form

(ii)

Hence write down the coordinates of the stationary point on the curve $y = x^2 - x - 6$ .

2M Similar

(b)

On the axes, draw the graph of $y = |x^2 - x - 6|$ for $-4 \leq x \leq 4$ .

Fig. 2

3M Similar

DifficultyMedium

Worked solution

Approach

The curve $y = |x^2 - x - 6|$ is the parabola $y = x^2 - x - 6$ with every negative part reflected in the $x$ -axis. Find where the parabola is negative, its intercepts, and its vertex, then sketch.

Working

The parabola $y = x^2 - x - 6 = (x - 3)(x + 2)$ crosses the $x$ -axis at $x = -2$ and $x = 3$ , and is negative between these roots.

Key points:

$y$ -intercept: $y = |0 - 0 - 6| = 6$ , so $(0, 6)$ .
Cusps on the $x$ -axis at $(-2, 0)$ and $(3, 0)$ , where the reflection happens.
Vertex of the parabola from part (a)(ii): $\left(\frac{1}{2}, -\frac{25}{4}\right)$ , which reflects to a local maximum at $\left(\frac{1}{2}, \frac{25}{4}\right) \approx (0.5, 6.25)$ .
Endpoints: at $x = -4$ , $y = |16 + 4 - 6| = 14$ ; at $x = 4$ , $y = |16 - 4 - 6| = 6$ .

The sketch is a W-shaped curve: falling from $(-4, 14)$ to the cusp at $(-2, 0)$ , rising to the maximum $(0.5, 6.25)$ through $(0, 6)$ , falling to the cusp at $(3, 0)$ , then rising to $(4, 6)$ .

Answer

A W-shaped modulus curve with cusps at $(-2, 0)$ and $(3, 0)$ , $y$ -intercept $(0, 6)$ , local maximum $\left(\frac{1}{2}, \frac{25}{4}\right)$ , passing through $(-4, 14)$ and $(4, 6)$ .

Final answer

W-shaped modulus curve with cusps at (-2, 0) and (3, 0), y-intercept (0, 6), local maximum (0.5, 6.25), through (-4, 14) and (4, 6)

Detailed explanation

Walkthrough

The modulus $|f(x)|$ leaves $f(x)$ unchanged where $f(x) \geq 0$ and reflects it where $f(x) < 0$ . Since $x^2 - x - 6 = (x-3)(x+2)$ is negative only between $x = -2$ and $x = 3$ , that middle piece of the parabola is flipped above the axis, creating sharp cusps at $(-2, 0)$ and $(3, 0)$ . The parabola's minimum $-\frac{25}{4}$ at $x = \frac{1}{2}$ becomes a local maximum $+\frac{25}{4}$ there, and the $y$ -intercept $-6$ becomes $+6$ . Evaluating at the interval ends gives $(-4, 14)$ and $(4, 6)$ to anchor the outer arms.

Key Takeaways

$|f(x)|$ reflects only the below-axis parts of the graph of $f$ .
Reflection points occur at the roots of $f$ , producing cusps there.
A reflected minimum becomes a local maximum of the modulus curve.

Common Mistakes

Drawing smooth minima at $(-2, 0)$ and $(3, 0)$ instead of sharp cusps — the mark scheme explicitly penalises this (B1 only for "correct shape with no cusps").
Reflecting the whole curve or getting intercepts wrong: the scheme requires intercepts at $(-2, 0)$ , $(3, 0)$ and $(0, 6)$ for full credit.
Plotting the unreflected parabola dipping below the axis.

Things to Be Careful About

The mark scheme awards B3 only for the fully correct graph; B2 requires correct shape with cusps AND correct intercepts AND the maximum in the first quadrant. Make the cusps visibly sharp and label the intercepts.
Keep the curve within $-4 \leq x \leq 4$ as instructed.

Techniques used

reflect the negative part of a parabola in the x-axisidentify cusps at the rootsplot key intercepts and sketch the modulus curve

(c)

Use your graph to solve the inequality $|x^2 - x - 6| < 4$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Draw the line $y = 4$ on the graph from part (b) and read off the $x$ -values where the modulus curve lies below it.

Working

Solve $|x^2 - x - 6| = 4$ to locate the crossings. For the outer (positive) branches:

x^2 - x - 6 = 4 \implies x^2 - x - 10 = 0 \implies x = \frac{1 \pm \sqrt{41}}{2}

giving $x \approx -2.7$ and $x \approx 3.7$ . For the middle (reflected) branch:

x^2 - x - 6 = -4 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0

giving $x = -1$ and $x = 2$ .

The curve is below $y = 4$ between $-2.7$ and $-1$ , and between $2$ and $3.7$ .

Answer

-2.7 < x < -1 \quad \text{or} \quad 2 < x < 3.7

Final answer

-2.7 < x < -1 or 2 < x < 3.7

Detailed explanation

Walkthrough

The inequality $|x^2 - x - 6| < 4$ asks where the modulus curve sits below the horizontal line $y = 4$ . Drawing that line on the graph from part (b), it cuts the curve four times: twice on the left outer arm and reflected hump, twice on the right. The crossings on the reflected middle branch come from $x^2 - x - 6 = -4$ , which factorises neatly as $(x-2)(x+1) = 0$ , giving $x = -1$ and $x = 2$ . The outer crossings come from $x^2 - x - 6 = 4$ , i.e. $x^2 - x - 10 = 0$ , whose roots $\frac{1 \pm \sqrt{41}}{2}$ are approximately $-2.7$ and $3.7$ . The curve is below the line in the two intervals between consecutive crossings: $-2.7 < x < -1$ and $2 < x < 3.7$ .

Key Takeaways

A modulus inequality is solved graphically by intersecting with a horizontal line and reading intervals where the curve is below (or above) it.
$|f(x)| = k$ splits into $f(x) = k$ and $f(x) = -k$ .

Common Mistakes

Giving only one of the two intervals — the mark scheme awards one mark per interval (STRICT FTB1 for either one).
Reading the outer crossing values inaccurately from the graph; they are $-2.7$ and $3.7$ to 2 significant figures.
Including the endpoints with $\leq$ : the inequality is strict, so the crossing values themselves are excluded.

Things to Be Careful About

The marks are STRICT follow-through from the candidate's own graph, and depend on part (b) earning at least B1 with correct intercepts and the curve crossing $y = 4$ four times — an inaccurate graph in (b) costs marks here too.
State the answer as a pair of inequalities on $x$ , not as $y$ -values.

Techniques used

draw the horizontal line y = 4 on the graphread off intersection intervalsexpress the solution as a union of x-intervals

Q4MediumCalculusTrigonometry

(a)

Integrate the following with respect to $x$ .

4M

(i)

$\mathrm{e}^{5x-2}$

2M Similar

(ii)

$\frac{1}{4-3x} \quad \text{where } x < \frac{4}{3}$

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The integral of $\frac{1}{ax+b}$ is $\frac{1}{a}\ln|ax+b|$ . Here $a = -3$ , so divide by $-3$ ; since $x < \frac{4}{3}$ , the quantity $4-3x$ is positive and $\ln(4-3x)$ may be written without modulus.

Working

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

Answer

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

Final answer

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

Detailed explanation

Walkthrough

We use the standard result $\int \frac{1}{ax+b}\,\mathrm{d}x = \frac{1}{a}\ln|ax+b| + c$ . Differentiating $\ln(4-3x)$ gives $\frac{-3}{4-3x}$ , so to obtain just $\frac{1}{4-3x}$ we must multiply by $-\frac{1}{3}$ . The condition $x < \frac{4}{3}$ guarantees $4 - 3x > 0$ , so the logarithm is defined and no modulus sign is needed.

Key Takeaways

$\int \frac{1}{ax+b}\,\mathrm{d}x = \frac{1}{a}\ln|ax+b| + c$ , and the sign of $a$ matters.
A given domain condition can justify dropping the modulus.

Common Mistakes

Writing $+\frac{1}{3}\ln(4-3x)$ and missing the minus sign from the coefficient $-3$ .
Writing $-\frac{1}{3}\ln 4 - 3x$ (misreading the argument of the logarithm).
Omitting the constant of integration.

Things to Be Careful About

The mark scheme gives B1 for $-\frac{1}{3}\ln 4 - 3x$ or $k\ln(4-3x)$ with $k$ non-zero, so the sign and the factor $-\frac{1}{3}$ are both essential; the final answer must be $-\frac{1}{3}\ln(4-3x)$ .

Techniques used

integrate the reciprocal of a linear functionhandle the sign when the linear factor has a negative coefficient

(b)

Show that

\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sec^2 \left(\frac{1}{2}x\right) \,\mathrm{d}x = 2\left(1 - \frac{\sqrt{3}}{3}\right)

3M Similar

DifficultyMedium

Worked solution

Approach

Integrate $\sec^2\left(\frac{1}{2}x\right)$ : since $\frac{\mathrm{d}}{\mathrm{d}x}\tan\left(\frac{1}{2}x\right) = \frac{1}{2}\sec^2\left(\frac{1}{2}x\right)$ , the antiderivative is $2\tan\left(\frac{1}{2}x\right)$ . Then evaluate at the limits using exact values $\tan\frac{\pi}{4} = 1$ and $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ .

Working

\int \sec^2\left(\frac{1}{2}x\right)\,\mathrm{d}x = 2\tan\left(\frac{1}{2}x\right)

Evaluate from $\frac{\pi}{3}$ to $\frac{\pi}{2}$ :

\left[2\tan\left(\frac{1}{2}x\right)\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}} = 2\tan\frac{\pi}{4} - 2\tan\frac{\pi}{6}

Using $\tan\frac{\pi}{4} = 1$ and $\tan\frac{\pi}{6} = \frac{\sqrt{3}}{3}$ :

2(1) - 2\left(\frac{\sqrt{3}}{3}\right) = 2\left(1 - \frac{\sqrt{3}}{3}\right)

This is exactly the required result.

Answer

\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sec^2\left(\frac{1}{2}x\right)\,\mathrm{d}x = 2\left(1 - \frac{\sqrt{3}}{3}\right)

Final answer

2(1 - √3/3), as required

Detailed explanation

Walkthrough

This is a "show that" question, so we must derive the printed answer from scratch. First find the antiderivative: we know $\int \sec^2 u\,\mathrm{d}u = \tan u$ , but the argument here is $\frac{1}{2}x$ , so the chain rule in reverse means we divide by the inner derivative $\frac{1}{2}$ , giving $2\tan\left(\frac{1}{2}x\right)$ . Next we apply the limits: at $x = \frac{\pi}{2}$ the argument is $\frac{\pi}{4}$ , where $\tan = 1$ ; at $x = \frac{\pi}{3}$ the argument is $\frac{\pi}{6}$ , where $\tan = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ . Subtracting lower from upper gives $2(1) - 2\cdot\frac{\sqrt{3}}{3}$ , which factors as $2\left(1 - \frac{\sqrt{3}}{3}\right)$ — exactly the printed target, so the result is established.

Key Takeaways

$\int \sec^2(kx)\,\mathrm{d}x = \frac{1}{k}\tan(kx) + c$ .
Exact values: $\tan\frac{\pi}{4} = 1$ , $\tan\frac{\pi}{6} = \frac{\sqrt{3}}{3}$ .
Definite integrals are evaluated as antiderivative at upper limit minus antiderivative at lower limit.

Common Mistakes

Writing $\tan\left(\frac{1}{2}x\right)$ without the factor $2$ (forgetting to divide by the inner derivative $\frac{1}{2}$ ).
Subtracting the limits the wrong way round, giving $2\left(\frac{\sqrt{3}}{3} - 1\right)$ .
Leaving $\tan\frac{\pi}{6}$ as $\frac{1}{\sqrt{3}}$ without converting to the printed form $\frac{\sqrt{3}}{3}$ , so the completion to the given answer is not fully justified.
Working in degrees instead of radians.

Things to Be Careful About

The mark scheme awards M1 for $k\tan\left(\frac{1}{2}x\right)$ with $k$ non-zero and M2 for the correct $2\tan\left(\frac{1}{2}x\right)$ , then A1 for a fully justified completion to $2\left(1 - \frac{\sqrt{3}}{3}\right)$ — every line between the antiderivative and the printed answer must be shown, including the exact values of the tangents. Since this is an AG (answer given) part, the derivation must run forward to exactly the printed form.

Techniques used

integrate sec squared of a multiple angleevaluate the antiderivative at exact limitsuse exact values of tan at special anglesderive forward to the given printed answer

Q5MediumPermutations and combinations

Six different digits are chosen from the nine digits $1, 2, 3, 4, 5, 6, 7, 8, 9$ .
These digits are used to form a 6-digit number.
Find how many 6-digit numbers can be formed in the following cases.

(a)

There are no restrictions.

1M Similar

(b)

The number is greater than $700\,000$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

A number greater than $700\,000$ must start with 7, 8 or 9. For each choice of first digit, arrange 5 of the remaining 8 digits.

Working

First digit: 3 choices ( $7$ , $8$ or $9$ ). Remaining 5 places filled from the other 8 digits:

3 \times {}^{8}\mathrm{P}_{5} = 3 \times (8 \times 7 \times 6 \times 5 \times 4) = 3 \times 6720 = 20\,160

(Equivalently, since exactly one third of all numbers start with 7, 8 or 9: $\tfrac{1}{3} \times {}^{9}\mathrm{P}_{6}$ .)

Answer

20\,160

Final answer

20160

Detailed explanation

Walkthrough

For the number to exceed $700\,000$ , its first digit must be 7, 8 or 9 — three choices. Once the first digit is fixed, the remaining five positions are filled with an ordered arrangement of 5 digits chosen from the 8 that are left:

{}^{8}\mathrm{P}_{5} = 8 \times 7 \times 6 \times 5 \times 4 = 6720

Multiplying by the three choices of leading digit gives $3 \times 6720 = 20\,160$ . A neat check: by symmetry, exactly one third of all unrestricted numbers start with 7, 8 or 9, giving $\tfrac{1}{3} \times 60\,480 = 20\,160$ .

Key Takeaways

Inequality conditions on large numbers usually constrain the leading digit(s); fix those first, then permute the rest.
Symmetry arguments (each digit equally likely to lead) provide quick alternative counts.

Common Mistakes

Counting only numbers starting with 7, forgetting 8 and 9 also work.
Using ${}^{8}\mathrm{P}_{5}$ without multiplying by 3.
Reusing the leading digit among the remaining five (must be 5 digits from the remaining 8).

Things to Be Careful About

The M1 requires seeing the structure $3 \times {}^{8}\mathrm{P}_{5}$ (or equivalent); show this line explicitly before evaluating.

Techniques used

split cases by the leading digitcount remaining positions with permutations

(c)

The number is greater than $750\,000$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Numbers greater than $750\,000$ either start with 8 or 9 (any arrangement), or start with 7 followed by a second digit of 5, 6, 8 or 9. Alternatively, take the answer to (b) and remove those starting $71$ , $72$ , $73$ or $74$ .

Working

Case 1: starts with 8 or 9:

2 \times {}^{8}\mathrm{P}_{5} = 2 \times 6720 = 13\,440

Case 2: starts with 7, then second digit is 5, 6, 8 or 9 (4 choices), then 4 remaining digits arranged in the last 4 places:

4 \times {}^{7}\mathrm{P}_{4} = 4 \times (7 \times 6 \times 5 \times 4) = 4 \times 840 = 3360

Total:

13\,440 + 3360 = 16\,800

(Alternative using part (b): $20\,160 - 3360 = 16\,800$ , where $3360$ counts numbers starting 7 then 1, 2, 3 or 4.)

Answer

16\,800

Final answer

16800

Detailed explanation

Walkthrough

The condition 'greater than $750\,000$ ' now constrains the first two digits, so we split into disjoint cases.

Case 1: if the number starts with 8 or 9, it automatically exceeds $750\,000$ . That gives $2 \times {}^{8}\mathrm{P}_{5} = 13\,440$ numbers.

Case 2: if it starts with 7, the second digit must be 5, 6, 8 or 9 — four choices (note 7 itself is used, so it cannot be the second digit). After fixing these two digits, the last four places hold an arrangement of 4 digits from the remaining 7: $4 \times {}^{7}\mathrm{P}_{4} = 4 \times 840 = 3360$ .

Adding the disjoint cases: $13\,440 + 3360 = 16\,800$ .

A slicker route reuses part (b): from the $20\,160$ numbers above $700\,000$ , remove those starting 7 followed by 1, 2, 3 or 4, which count $4 \times {}^{7}\mathrm{P}_{4} = 3360$ , leaving $16\,800$ .

Key Takeaways

When a threshold fixes more than one leading digit, split into cases where the constraint is automatic versus where further digits matter.
'Hence'-style reuse: earlier answers can be adjusted by subtracting a counted sub-case.
Cases must be disjoint so their counts can simply be added.

Common Mistakes

Allowing the second digit after 7 to be 7 again (digit already used), or forgetting 8 and 9 as valid second digits.
Subtracting the wrong sub-case in the alternative method (must be 7 followed by 1–4, not 5–9).
Double-counting: including numbers starting 75, 76, 78, 79 within the 'starts with 8 or 9' case.

Things to Be Careful About

The mark scheme awards separate M1s for the case starting 7 then 5/6/8/9 ( $3360$ ) and the case starting 8 or 9 ( $13\,440$ ) — show both counts explicitly. If using the subtraction route, the value $3360$ must appear.

Techniques used

split cases by leading digitssubtract a counted sub-case from a previous resultcount remaining positions with permutations

Q69MMediumSimultaneous equationsStraight-line graphs

The line $y = 3x + 4$ meets the curve $y = 2x^2 + 8x + 1$ at two points $A$ and $B$ .

Find the equation of the perpendicular bisector of $AB$ , giving your answer in the form $ax + by + c = 0$ , where $a$ , $b$ and $c$ are integers.

DifficultyMedium

Worked solution

Approach

Find $A$ and $B$ by equating the line and the curve. Then find the midpoint of $AB$ and the gradient perpendicular to $AB$ , and write the equation of the perpendicular bisector through that midpoint.

Working

Set the equations equal:

3x + 4 = 2x^2 + 8x + 1

2x^2 + 5x - 3 = 0

Factorise:

(2x - 1)(x + 3) = 0 \quad\Rightarrow\quad x = \frac{1}{2} \text{ or } x = -3

On the line $y = 3x + 4$ :

x = \frac{1}{2}:\; y = \frac{3}{2} + 4 = \frac{11}{2}, \qquad x = -3:\; y = -9 + 4 = -5

So $A\left(\frac{1}{2}, \frac{11}{2}\right)$ and $B(-3, -5)$ .

Midpoint of $AB$ :

\left(\frac{\frac{1}{2} + (-3)}{2}, \frac{\frac{11}{2} + (-5)}{2}\right) = \left(-\frac{5}{4}, \frac{1}{4}\right)

Gradient of $AB$ is $3$ , so the perpendicular gradient is $-\frac{1}{3}$ .

Equation of the perpendicular bisector:

y - \frac{1}{4} = -\frac{1}{3}\left(x + \frac{5}{4}\right)

Multiply through by $12$ :

12y - 3 = -4x - 5

4x + 12y + 2 = 0

Divide by $2$ :

2x + 6y + 1 = 0

Answer

2x + 6y + 1 = 0

Final answer

2x + 6y + 1 = 0

Detailed explanation

Walkthrough

The points where a line meets a curve are found by setting their $y$ -values equal — this eliminates one variable and gives a single equation in $x$ . Here $3x + 4 = 2x^2 + 8x + 1$ rearranges to $2x^2 + 5x - 3 = 0$ , which factorises as $(2x - 1)(x + 3) = 0$ , giving $x = \frac{1}{2}$ or $x = -3$ . Substituting each back into the simpler equation (the line) gives the full coordinates $A\left(\frac{1}{2}, \frac{11}{2}\right)$ and $B(-3, -5)$ .

The perpendicular bisector must pass through the midpoint of $AB$ : averaging the coordinates gives $\left(-\frac{5}{4}, \frac{1}{4}\right)$ . The gradient of $AB$ is $3$ (it is the given line), so the perpendicular gradient is the negative reciprocal, $-\frac{1}{3}$ .

Using point–slope form with the midpoint and the perpendicular gradient:

y - \frac{1}{4} = -\frac{1}{3}\left(x + \frac{5}{4}\right)

Clearing fractions by multiplying by $12$ gives $12y - 3 = -4x - 5$ , i.e. $4x + 12y + 2 = 0$ , which simplifies to $2x + 6y + 1 = 0$ in the required form with integer coefficients.

Key Takeaways

Intersections of a line and curve are found by equating the expressions and solving the resulting polynomial.
A perpendicular bisector needs exactly two ingredients: the midpoint of the segment and the negative reciprocal of its gradient.
Point–slope form plus clearing denominators converts any line into $ax + by + c = 0$ with integers.

Common Mistakes

Substituting the $x$ -values back into the curve instead of the line — both work, but arithmetic slips are more likely in the curve; either is accepted if correct.
Forgetting to take the negative reciprocal, giving a parallel line rather than a perpendicular one.
Using one endpoint instead of the midpoint when forming the equation.
Leaving fractional coefficients ( $x + 3y + \frac{1}{2} = 0$ ) — the question demands integers, so clear the fractions at the end.
Sign errors when rearranging into $ax + by + c = 0$ form.

Things to Be Careful About

The final answer must be in the exact form $ax + by + c = 0$ with $a$ , $b$ , $c$ integers — equivalent forms like $y = -\frac{1}{3}x - \frac{1}{6}$ score only partial credit per the mark scheme's "A1 for correct equation in a different form" note.
Follow-through applies to the midpoint and perpendicular gradient steps, but only from correctly found intersection points.
Keep fractions exact throughout; decimals like $(0.5, 5.5)$ are acceptable but make the final integer conversion harder.

Techniques used

eliminate one unknown to form a quadraticfactorise the quadratic to find the intersection pointsfind the midpoint of the two pointsuse the negative reciprocal of the gradient for the perpendicular bisectorrearrange the line equation into integer-coefficient form

Similar questions

Q76MMediumTrigonometry

Solve the equation

\sec^2 3x + \tan 3x - 3 = 0

for $0° \leq x \leq 120°$ .

DifficultyMedium

Worked solution

Approach

Use the identity $\sec^2 A = 1 + \tan^2 A$ to rewrite the whole equation in terms of $\tan 3x$ only, solve the resulting quadratic, then find every angle in the interval.

Working

Using $\sec^2 3x = 1 + \tan^2 3x$ :

(1 + \tan^2 3x) + \tan 3x - 3 = 0

\tan^2 3x + \tan 3x - 2 = 0

Factorising:

(\tan 3x + 2)(\tan 3x - 1) = 0

So $\tan 3x = -2$ or $\tan 3x = 1$ .

For $0° \leq x \leq 120°$ , we need $0° \leq 3x \leq 360°$ . Since $\tan$ has period $180°$ , each value gives two angles in this range.

For $\tan 3x = 1$ : principal value $45°$ , and $45° + 180° = 225°$ :

3x = 45°,\ 225° \quad\Rightarrow\quad x = 15°,\ 75°

For $\tan 3x = -2$ : reference angle $\arctan 2 = 63.435°$ , so angles in the second and third quadrants:

3x = 180° - 63.435° = 116.565°, \qquad 3x = 360° - 63.435° = 296.565°

x = 38.855°,\ 98.855°

Answer

x = 15°,\ 38.9°,\ 75°,\ 98.9°

Final answer

x = 15, 38.9, 75, 98.9 (degrees)

Detailed explanation

Walkthrough

The equation mixes $\sec^2 3x$ and $\tan 3x$ . The identity $\sec^2 A = 1 + \tan^2 A$ converts the secant term into tangent terms, giving a quadratic in $\tan 3x$ : $\tan^2 3x + \tan 3x - 2 = 0$ . This factorises as $(\tan 3x + 2)(\tan 3x - 1) = 0$ , so $\tan 3x = -2$ or $\tan 3x = 1$ .

Because the argument is $3x$ , the interval $0° \leq x \leq 120°$ becomes $0° \leq 3x \leq 360°$ , which is one full turn — so each tangent value yields two angles (tangent repeats every $180°$ ).

For $\tan 3x = 1$ : the acute solution is $45°$ , and adding $180°$ gives $225°$ . Dividing by 3 gives $x = 15°$ and $x = 75°$ .

For $\tan 3x = -2$ : the reference angle is $\arctan 2 \approx 63.435°$ . Tangent is negative in the second and third quadrants, so $3x = 180° - 63.435° = 116.565°$ and $3x = 360° - 63.435° = 296.565°$ . Dividing by 3 gives $x = 38.855°$ and $x = 98.855°$ , i.e. $38.9°$ and $98.9°$ to 1 decimal place.

All four values lie inside $0° \leq x \leq 120°$ , and there are no extras.

Key Takeaways

The identity $\sec^2 A = 1 + \tan^2 A$ turns mixed sec/tan equations into quadratics in $\tan A$ .
When the unknown appears as $kx$ , multiply the given interval by $k$ before listing angles, then divide the angles by $k$ at the end.
Tangent has period $180°$ , so each value of $\tan 3x$ produces two solutions over a full $360°$ range of $3x$ .
For negative tangent values, use the second and third quadrants via the reference angle.

Common Mistakes

Forgetting to scale the interval: solving only for $0° \leq 3x \leq 120°$ loses half the answers.
Taking only the principal value of each arctan and missing the second angle from the $180°$ periodicity.
Using the wrong quadrants for $\tan 3x = -2$ (tangent is positive in the first and third quadrants, negative in the second and third).
Dividing by 3 before finding both angles, or rounding $3x$ before dividing (rounding too early changes the final answer).
Sign errors when factorising or expanding $(\tan 3x + 2)(\tan 3x - 1)$ .
Including extra angles outside $0° \leq x \leq 120°$ — the mark scheme penalises extras in range.

Things to Be Careful About

The mark scheme awards A2 for all four correct angles with no extras in range, and A1 for any two correct angles ignoring extras — so completeness matters, but so does not inventing extra roots.
Angles are required to 1 decimal place ( $38.9°$ , $98.9°$ ); keep full accuracy on $3x$ until the division by 3 is done.
Check that every answer satisfies $0° \leq x \leq 120°$ : here $98.9°$ is valid but anything beyond $120°$ must be rejected.

Techniques used

apply the identity sec^2 A = 1 + tan^2 Areduce the equation to a quadratic in tan 3xfactorise the quadraticsolve for all angles in the required interval using the period of tan

Similar questions

Q8Medium-HardCircular measureTrigonometry

In this question the units are metres.

Fig. 3

The diagram shows a circle, centre $O$ and radius 2.
The chord $AB$ has length $2\sqrt{3}$ .
The point $Q$ lies on the circle such that $AQ = BQ$ .
The arc $APB$ is part of a circle, centre $Q$ .

(a)

Find the exact value of angle $AQB$ in radians.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $M$ be the midpoint of $AB$ . Since $OA = OB = 2$ , the line $OM$ bisects angle $AOB$ , and $AM = \sqrt{3}$ . Find half of angle $AOB$ from the right triangle $OMA$ , then double it.

Working

\sin\left(\frac{1}{2}\angle AOB\right) = \frac{AM}{OA} = \frac{\sqrt{3}}{2}

\frac{1}{2}\angle AOB = \frac{\pi}{3} \quad\Rightarrow\quad \angle AOB = \frac{2\pi}{3}

Since $Q$ lies on the circle with $AQ = BQ$ , the points $O$ and $Q$ both lie on the perpendicular bisector of $AB$ . In triangle $AOQ$ , $OA = OQ = 2$ , so it is isosceles and

\angle AQO = \angle QAO = \frac{1}{2}\left(\pi - \frac{2\pi}{3}\right) = \frac{\pi}{6}

By symmetry about $OQ$ ,

\angle AQB = 2 \times \frac{\pi}{6} = \frac{\pi}{3}

Answer

\angle AQB = \frac{\pi}{3}

Final answer

pi/3 radians

Detailed explanation

Walkthrough

The chord $AB$ has length $2\sqrt{3}$ in a circle of radius $2$ . Dropping a perpendicular from the centre $O$ to the chord splits the chord into two equal halves of length $\sqrt{3}$ and splits the angle at $O$ into two equal parts. In the resulting right triangle, $\sin$ of the half-angle equals opposite over hypotenuse, i.e. $\sqrt{3}/2$ , which is the exact value of $\sin\frac{\pi}{3}$ , so the full angle $AOB$ is $\frac{2\pi}{3}$ .

To get angle $AQB$ , note that $Q$ is on the circle and equidistant from $A$ and $B$ , so $Q$ sits on the same perpendicular bisector as $O$ . Triangle $AOQ$ has two sides equal to the radius ( $OA = OQ = 2$ ), so its base angles are equal, and since the angle at $O$ in this triangle is $\frac{2\pi}{3}$ , each base angle is $\frac{\pi}{6}$ . The line $OQ$ is an axis of symmetry for the whole figure, so angle $AQB$ is twice $\frac{\pi}{6}$ , giving $\frac{\pi}{3}$ .

Key Takeaways

A perpendicular from the centre to a chord bisects both the chord and the subtended angle.
Special-angle values ( $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ ) let you read off angles exactly in radians.
Isosceles triangles formed by two radii have equal base angles.

Common Mistakes

Giving the half-angle $\frac{\pi}{3}$ as the final answer instead of doubling to $\frac{2\pi}{3}$ for $AOB$ , or confusing $AOB$ with the required $AQB$ .
Working in degrees when the question demands radians — the mark scheme only accepts the exact radian form (a decimal $1.047$ scores only SC1).
Assuming triangle $AQB$ is equilateral without justifying it via the symmetry/isosceles argument.

Things to Be Careful About

The answer must be the exact value $\frac{\pi}{3}$ ; the decimal $1.047$ earns only a special-case mark.
Keep track of which angle is asked for: $AQB$ , not $AOB$ .
All reasoning here uses exact surd values, so no calculator rounding should appear.

Techniques used

drop a perpendicular from the centre to the chorduse the sine ratio in a right-angled trianglerecognise the special-angle value

(b)

Hence find the area of the shaded region. Give your answer in terms of $\pi$ .

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

The shaded region is the segment of the circle centre $O$ cut off by chord $AB$ , minus the segment of the circle centre $Q$ cut off by the same chord. Each segment is (sector) $-$ (triangle), using $\frac{1}{2}r^2\theta$ and $\frac{1}{2}ab\sin C$ with the exact angles $\angle AOB = \frac{2\pi}{3}$ and $\angle AQB = \frac{\pi}{3}$ from part (a).

Working

The radius of the circle centre $Q$ is $QA = QB = 2\sqrt{3}$ (found from the right triangle with legs $\sqrt{3}$ and $3$ : $QA^2 = 3 + 9 = 12$ ).

Segment of circle centre $O$ :

\text{sector } AOB - \triangle AOB = \frac{1}{2}(2)^2\left(\frac{2\pi}{3}\right) - \frac{1}{2}(2)^2\sin\frac{2\pi}{3} = \frac{4\pi}{3} - \sqrt{3}

Segment of circle centre $Q$ :

\text{sector } AQB - \triangle AQB = \frac{1}{2}(2\sqrt{3})^2\left(\frac{\pi}{3}\right) - \frac{1}{2}(2\sqrt{3})^2\sin\frac{\pi}{3} = 2\pi - 3\sqrt{3}

Shaded area:

\left(\frac{4\pi}{3} - \sqrt{3}\right) - \left(2\pi - 3\sqrt{3}\right) = \frac{4\pi}{3} - 2\pi + 2\sqrt{3} = 2\sqrt{3} - \frac{2\pi}{3}

Answer

2\sqrt{3} - \frac{2\pi}{3}

Final answer

2√3 − 2π/3

Detailed explanation

Walkthrough

The shaded sliver sits between the arc of the original circle (centre $O$ ) and the arc $APB$ of the circle centred at $Q$ , both spanning the same chord $AB$ . So its area is simply the $O$ -circle segment above the chord minus the $Q$ -circle segment above the chord.

Each segment is found as sector minus triangle. For the $O$ -circle, the radius is $2$ and the angle is $\frac{2\pi}{3}$ carried forward from part (a): the sector is $\frac{1}{2}\cdot 4 \cdot \frac{2\pi}{3} = \frac{4\pi}{3}$ and the triangle is $\frac{1}{2}\cdot 4 \cdot \sin\frac{2\pi}{3} = \sqrt{3}$ , giving segment $\frac{4\pi}{3} - \sqrt{3}$ .

For the $Q$ -circle we need its radius $QA$ . The midpoint $M$ of $AB$ satisfies $AM = \sqrt{3}$ and $OM = 2\cos\frac{\pi}{3} = 1$ , so $MQ = MO + OQ = 1 + 2 = 3$ , and $QA^2 = 3 + 9 = 12$ , i.e. $QA = 2\sqrt{3}$ . With angle $AQB = \frac{\pi}{3}$ from part (a), the sector is $\frac{1}{2}\cdot 12 \cdot \frac{\pi}{3} = 2\pi$ and the triangle is $\frac{1}{2}\cdot 12 \cdot \sin\frac{\pi}{3} = 3\sqrt{3}$ , giving segment $2\pi - 3\sqrt{3}$ .

Subtracting and collecting terms gives $2\sqrt{3} - \frac{2\pi}{3}$ , exactly as required in terms of $\pi$ .

Key Takeaways

Area of a sector: $\frac{1}{2}r^2\theta$ with $\theta$ in radians; area of a triangle: $\frac{1}{2}ab\sin C$ .
A circular segment = sector $-$ triangle.
Overlapping figures sharing a common chord can be handled by subtracting one segment from another.
"Hence" means the angles from part (a) must be reused, not recomputed.

Common Mistakes

Using degree-mode values or leaving $\theta$ in degrees inside $\frac{1}{2}r^2\theta$ .
Subtracting the segments the wrong way round, obtaining a negative area.
Taking the radius of the $Q$ -circle as $2$ instead of $2\sqrt{3}$ .
Forgetting the triangle term and using whole sectors instead of segments.
Giving a decimal answer — the question explicitly requires the answer in terms of $\pi$ , and the mark scheme flags the result "nfww" (nothing for wrong working).

Things to Be Careful About

The final answer must be the exact form $2\sqrt{3} - \frac{2\pi}{3}$ (equivalently $\frac{6\sqrt{3}-2\pi}{3}$ ); decimals score nothing here.
Every mark-scheme route requires all four areas (two sectors, two triangles) computed correctly, so show each one explicitly.
Keep $\pi$ and surds separate until the final collection of terms to avoid sign slips.
This part depends on part (a): use $\angle AOB = \frac{2\pi}{3}$ and $\angle AQB = \frac{\pi}{3}$ directly.

Techniques used

compute sector areas using half r squared thetacompute triangle areas using half ab sin Cdecompose the shaded region as a difference of circular segmentssimplify to an exact form in terms of pi and surds

Q9MediumStraight-line graphsLogarithmic and exponential functions

Two variables, $x$ and $y$ , are related by an equation of the form $y = Ax^b$ , where $A$ and $b$ are constants.
The following pairs of values of $x$ and $y$ are given.

$x$	0.61	4.48	12.18	33.1
$y$	1.65	4.47	7.39	12.17

(a)

On the axes below, use these values to draw the straight-line graph of $\ln y$ against $\ln x$ .

Fig. 4

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Taking natural logs of $y = Ax^b$ gives $\ln y = \ln A + b\ln x$ , so plotting $\ln y$ against $\ln x$ gives a straight line. Compute $\ln x$ and $\ln y$ for each pair, plot the points, and draw a single ruled line of best fit.

Working

\begin{array}{c|cccc} \ln x & -0.5 & 1.5 & 2.5 & 3.5 \\ \hline \ln y & 0.5 & 1.5 & 2.0 & 2.5 \end{array}

For example, $\ln 0.61 = -0.494 \approx -0.5$ , $\ln 1.65 = 0.501 \approx 0.5$ , $\ln 4.48 = 1.499 \approx 1.5$ , $\ln 4.47 = 1.497 \approx 1.5$ , $\ln 12.18 = 2.500$ , $\ln 7.39 = 2.000$ , $\ln 33.1 = 3.500$ , $\ln 12.17 = 2.499 \approx 2.5$ .

Plot the four points $(-0.5, 0.5)$ , $(1.5, 1.5)$ , $(2.5, 2.0)$ , $(3.5, 2.5)$ on the grid and draw a single straight ruled line of best fit through them.

Answer

The four points are plotted at $(-0.5, 0.5)$ , $(1.5, 1.5)$ , $(2.5, 2.0)$ and $(3.5, 2.5)$ , with a single ruled straight line of best fit drawn through them.

Final answer

Points plotted at (-0.5, 0.5), (1.5, 1.5), (2.5, 2.0) and (3.5, 2.5) with a single ruled straight line of best fit

Detailed explanation

Walkthrough

The equation $y = Ax^b$ is a power law, which is curved on ordinary axes. Taking natural logarithms of both sides gives $\ln y = \ln A + b\ln x$ — a straight-line form in the variables $\ln y$ and $\ln x$ . So for each data pair we compute the natural log of the $x$ value and the natural log of the $y$ value, plot these transformed points on the given grid, and draw one straight line of best fit. The points come out at $(-0.5, 0.5)$ , $(1.5, 1.5)$ , $(2.5, 2.0)$ and $(3.5, 2.5)$ , which lie very close to a straight line, confirming the model.

Key Takeaways

A power law $y = Ax^b$ linearises under logarithms: plotting $\ln y$ against $\ln x$ gives a straight line whose gradient is $b$ and whose intercept on the $\ln y$ -axis is $\ln A$ .

Common Mistakes

Using $\log_{10}$ instead of $\ln$ (the axes are labelled $\ln x$ and $\ln y$ ); plotting the raw $x$ and $y$ values instead of their logarithms; joining the points dot-to-dot instead of drawing one straight line of best fit; misreading the grid scales.

Things to Be Careful About

The mark scheme gives B2 for all four points correctly plotted (soi) with a single ruled line of best fit, and B1 for at least three correctly plotted points. The line must be ruled and straight — a freehand curve scores nothing. Values may be read to the nearest half-unit as shown.

Techniques used

take natural logarithms of each data pairplot the transformed points on the given griddraw a ruled line of best fit

(b)

Use your graph to find the values of $A$ and $b$ .

4M Similar

DifficultyMedium

Worked solution

Approach

From $\ln y = b\ln x + \ln A$ , the gradient of the line of best fit is $b$ and the intercept on the $\ln y$ -axis is $\ln A$ . Read two points off the line, find the gradient, then find the intercept and exponentiate.

Working

Using the end points of the line, $(-0.5, 0.5)$ and $(3.5, 2.5)$ :

b = \frac{2.5 - 0.5}{3.5 - (-0.5)} = \frac{2}{4} = 0.5

For the intercept, substitute one point into $\ln y = b\ln x + \ln A$ :

0.5 = 0.5 \times (-0.5) + \ln A \quad\Rightarrow\quad \ln A = 0.5 + 0.25 = 0.75

A = \mathrm{e}^{0.75} = 2.1170\ldots \approx 2.12

Answer

b = 0.5, \qquad A = \mathrm{e}^{0.75} \approx 2.12

Final answer

b = 0.5, A = 2.12

Detailed explanation

Walkthrough

Since $\ln y = b\ln x + \ln A$ is a straight line in $\ln x$ and $\ln y$ , the gradient equals $b$ . Using the two end points of the line of best fit, $(-0.5, 0.5)$ and $(3.5, 2.5)$ , the gradient is $(2.5 - 0.5)/(3.5 + 0.5) = 2/4 = 0.5$ , so $b = 0.5$ . The intercept $\ln A$ is found by substituting a point on the line: $0.5 = 0.5 \times (-0.5) + \ln A$ gives $\ln A = 0.75$ , so $A = \mathrm{e}^{0.75} \approx 2.12$ . The original relationship is therefore $y = 2.12\,x^{0.5}$ .

Key Takeaways

For a log-log plot of $y = Ax^b$ : gradient $= b$ , vertical intercept $= \ln A$ , and $A = \mathrm{e}^{\text{intercept}}$ . Always undo the logarithm at the end — $\ln A = 0.75$ is not itself the value of $A$ .

Common Mistakes

Reporting $\ln A = 0.75$ as the value of $A$ instead of exponentiating; using the gradient of the raw data instead of the transformed points; reading the intercept off a mis-scaled axis; using $10^{0.75}$ instead of $\mathrm{e}^{0.75}$ (the logs are natural); rounding $A$ too early.

Things to Be Careful About

The mark scheme accepts $b = 0.50$ (awrt) via the gradient $\frac{2.5-0.5}{3.5+0.5}$ or by eliminating $c$ from two simultaneous equations, and accepts $A = \mathrm{e}^{0.7\text{ to }0.8}$ , i.e. $2.01$ to $2.23$ , with later mis-simplification ignored (isw). Answers must come from the graph — values obtained directly from the exponential equation only earn the SC1 fallback. Give $A$ to 3 significant figures: $2.12$ .

Techniques used

read the gradient of the log-log line as bread the intercept as ln Aexponentiate to recover A

Q10MediumCalculusEquations, inequalities and graphs

In this question the units are metres and seconds.

A particle moves along a straight line through a point $A$ .
Its displacement, $s$ , from $A$ at time $t$ is given by

s = \frac{10t + 100}{\sqrt{2t^2 + 100}}

The diagram shows the displacement–time graph for the first 30 seconds of the motion.

Fig. 5

(a)

Find the value of $t$ when $s$ is a maximum.

6M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $s = \dfrac{10t + 100}{\sqrt{2t^2 + 100}}$ using the quotient rule, set $\dfrac{\mathrm{d}s}{\mathrm{d}t} = 0$ and solve for $t$ .

Working

First differentiate the inner square root by the chain rule:

\frac{\mathrm{d}}{\mathrm{d}t}\left(\sqrt{2t^2 + 100}\right) = \frac{1}{2}(2t^2 + 100)^{-\frac{1}{2}} \times 4t

By the quotient rule:

\frac{\mathrm{d}s}{\mathrm{d}t} = \frac{(\sqrt{2t^2 + 100})(10) - (10t + 100)\left(\frac{1}{2}(2t^2 + 100)^{-\frac{1}{2}} \times 4t\right)}{\left(\sqrt{2t^2 + 100}\right)^2}

At the maximum, $\dfrac{\mathrm{d}s}{\mathrm{d}t} = 0$ , so the numerator is zero:

10(2t^2 + 100) - 2t(10t + 100) = 0

20t^2 + 1000 - 20t^2 - 200t = 0 \implies 1000 - 200t = 0

t = 5

Answer

t = 5

Final answer

t = 5

Detailed explanation

Walkthrough

The displacement is a quotient, so the quotient rule is the natural tool. Before applying it we need the derivative of the denominator: writing $\sqrt{2t^2+100}$ as $(2t^2+100)^{1/2}$ and using the chain rule gives $\frac{1}{2}(2t^2+100)^{-1/2}\times 4t$ . Substituting this into the quotient rule gives the full expression for $\frac{\mathrm{d}s}{\mathrm{d}t}$ . A maximum of $s$ occurs where the gradient is zero, so we set the numerator equal to zero (the denominator is never zero). Multiplying through by $(2t^2+100)$ clears the fractional powers and leaves the linear equation $1000 - 200t = 0$ , giving $t = 5$ seconds.

Key Takeaways

The quotient rule combined with the chain rule handles quotients involving square roots.
Setting a numerator to zero is valid when the denominator cannot vanish.
Stationary points of a displacement–time graph correspond to zero velocity.

Common Mistakes

Forgetting the factor $4t$ from the chain rule when differentiating $(2t^2+100)^{-1/2}$ .
Writing the quotient rule with the terms in the wrong order or omitting the squared denominator.
Squaring incorrectly when clearing the root: the correct step is multiplying by $(2t^2+100)$ , since $(\sqrt{2t^2+100})^2 = 2t^2+100$ .
The answer must come from correct working ('nfww') — an unsupported $t=5$ scores nothing.

Things to Be Careful About

Show every intermediate line: the scheme awards B2 for the chain-rule derivative, M1 for the quotient/product structure, A1 for the correct derivative, M1 for equating to zero, and A1 for $t=5$ supported by e.g. $10(2t^2+100)-2t(10t+100)=0$ .
'isw' applies after the correct derivative, so later mis-simplification is ignored there, but the final value $t=5$ is cao.

Techniques used

differentiate a square root by the chain ruleapply the quotient rule to form the derivative of sset the derivative equal to zero and solve for t

(b)

The particle passes through its starting point again at time $t = T$ .

5M

(i)

Find the total distance travelled by the particle during the first $T$ seconds of its motion.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

From part (a), $s$ is a maximum at $t = 5$ . Evaluate $s$ at $t = 0$ and $t = 5$ ; the particle moves out to the maximum and returns, so the total distance is twice the gain in displacement.

Working

At $t = 0$ :

s = \frac{100}{\sqrt{100}} = 10

At $t = 5$ :

s = \frac{50 + 100}{\sqrt{50 + 100}} = \frac{150}{\sqrt{150}} = \sqrt{150} = 5\sqrt{6} \approx 12.247

Distance travelled out:

\sqrt{150} - 10

Total distance (out and back):

2\left(\sqrt{150} - 10\right) \approx 4.49

Answer

2\left(\sqrt{150} - 10\right) \approx 4.49 \text{ m}

Final answer

4.49 m (exactly 2(√150 − 10))

Detailed explanation

Walkthrough

The particle starts at displacement $s = 10$ (putting $t=0$ into the formula gives $100/\sqrt{100} = 10$ ). It moves away from that starting point until the maximum at $t = 5$ , where $s = 150/\sqrt{150} = \sqrt{150} = 5\sqrt{6} \approx 12.25$ . The distance covered on the way out is the change in displacement, $\sqrt{150}-10$ . Since it then returns to its starting point, the same distance is covered on the way back, so the total distance is twice the outward distance: $2(\sqrt{150}-10) \approx 4.49$ m.

Key Takeaways

Total distance is the integral of speed, which here reduces to twice the outward excursion because the motion reverses at the maximum.
Evaluating $s$ at key times ( $t=0$ and the turning time) is enough; no integration is needed.

Common Mistakes

Giving just $√{150}-10 \approx 2.25$ , forgetting the return journey.
Confusing displacement change with total distance when the particle reverses direction.
Arithmetic slips such as taking $\sqrt{150}$ as $75$ or mis-evaluating $s$ at $t=0$ as $0$ instead of $10$ .

Things to Be Careful About

The B1 requires both $t=0, s=10$ and $t=5, s=√{150}$ (or $5√{6}$ or $12.24[74]$ ) to be seen.
The final answer is B2: exact form $2(√{150}-10)$ , or the decimal $4.49$ (or $4.494$ to $4.495$ ); 'isw' allows later mis-simplification to be ignored.

Techniques used

evaluate the displacement at the start and at the maximum pointdouble the outward displacement to obtain total distance

(ii)

Use algebra to find $T$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The starting displacement is $s = 10$ (from part (b)(i)). Set the formula equal to $10$ and solve for $T$ algebraically.

Working

\frac{10T + 100}{\sqrt{2T^2 + 100}} = 10

Square both sides:

(10T + 100)^2 = 10^2(2T^2 + 100)

100T^2 + 2000T + 10000 = 200T^2 + 10000

2000T = 100T^2

Since $T > 0$ (and $T > 5$ ):

T = 20

Answer

T = 20

Final answer

T = 20

Detailed explanation

Walkthrough

From part (b)(i) the starting displacement is $s = 10$ . Passing through the starting point again means setting the displacement formula equal to $10$ : $\frac{10T+100}{\sqrt{2T^2+100}} = 10$ . Squaring both sides removes the surd and gives $(10T+100)^2 = 100(2T^2+100)$ . Expanding: $100T^2 + 2000T + 10000 = 200T^2 + 10000$ , so $2000T = 100T^2$ , i.e. $T(T-20)=0$ . The solution $T=0$ is the original start, so the second passage is at $T = 20$ seconds.

Key Takeaways

'Passes through its starting point again' translates into setting the displacement equal to its initial value.
Squaring both sides is a standard way to clear a square-root denominator.
Reject the trivial root corresponding to the initial instant.

Common Mistakes

Using the wrong starting value of $s$ (it is $10$ , not $0$ ).
Expanding $(10T+100)^2$ incorrectly, e.g. missing the middle term $2000T$ .
Keeping $T = 0$ as an answer instead of rejecting it as the initial time.

Things to Be Careful About

The M1 requires the equation with their value of $10$ (which must be positive and greater than their answer to part (a)); the second M1 is the squaring step, dependent on the first; the A1 is $T = 20$ exactly.

Techniques used

set the displacement formula equal to the starting displacementsquare both sides to remove the surdsolve the resulting linear equation for T

Q115MMediumCoordinate geometry of the circle

A circle has equation $x^2 + y^2 - 25 = 0$ .
A second circle has the same radius as the first circle, and the coordinates of its centre are both positive.
The two circles intersect at the points $A$ and $B$ .
The line $AB$ has length 6 and is parallel to the line $y = -x$ .

Find the equation of the second circle in the form $x^2 + y^2 + ax + by + c = 0$ , where $a$ , $b$ and $c$ are constants.

DifficultyMedium

Worked solution

Approach

The common chord $AB$ of two equal circles is their line of symmetry, so it lies midway between the two centres. From the chord length $6$ and radius $5$ , find how far $AB$ sits from each centre; this gives the distance between the centres, and since $AB$ is parallel to $y = -x$ the line joining the centres runs in the direction $y = x$ . Both coordinates of the new centre are positive, fixing its position.

Working

The first circle is

x^2 + y^2 - 25 = 0,

centre $(0, 0)$ and radius $r^2 = 25$ , so $r = 5$ .

For the common chord $AB$ : the half-chord is $3$ , so by Pythagoras the distance from each centre to the line $AB$ is

\sqrt{5^2 - 3^2} = \sqrt{16} = 4.

Hence the distance between the two centres is $4 + 4 = 8$ .

Since $AB$ is parallel to $y = -x$ , the line through the centres is perpendicular to it, i.e. in the direction $y = x$ . The second centre therefore has equal positive coordinates $(k, k)$ with

k^2 + k^2 = 8^2 \implies 2k^2 = 64 \implies k^2 = 32 \implies k = 4\sqrt{2}.

So the second centre is $(4\sqrt{2},\ 4\sqrt{2})$ .

With the same radius $5$ :

\left(x - 4\sqrt{2}\right)^2 + \left(y - 4\sqrt{2}\right)^2 = 25.

Expanding:

x^2 - 8\sqrt{2}x + 32 + y^2 - 8\sqrt{2}y + 32 = 25,

x^2 + y^2 - 8\sqrt{2}x - 8\sqrt{2}y + 39 = 0.

Answer

x^2 + y^2 - 8\sqrt{2}x - 8\sqrt{2}y + 39 = 0

Final answer

x^2 + y^2 - 8√2 x - 8√2 y + 39 = 0

Detailed explanation

Walkthrough

First read off what the given equation tells us: $x^2 + y^2 - 25 = 0$ is a circle centred at the origin with $r^2 = 25$ , so the radius is $5$ . This earns the B1 for the radius.

The key geometric fact is that when two circles of equal radius intersect, their common chord $AB$ is the perpendicular bisector of the segment joining the centres — the chord sits exactly halfway between them. So if we can find the distance from one centre to the chord, doubling it gives the distance between the centres.

To find that distance, drop a perpendicular from the origin to the chord. This splits the chord of length $6$ into two halves of length $3$ , forming right-angled triangles with hypotenuse $5$ (the radius). Pythagoras gives the perpendicular distance as $\sqrt{25 - 9} = 4$ . So the centres are $8$ apart — this is the alternative B1 route in the mark scheme.

Now place the second centre. A line parallel to $y = -x$ has gradient $-1$ , so any perpendicular to it has gradient $+1$ , i.e. lies along $y = x$ . The second centre must lie on the line through the origin in the direction $y = x$ , at distance $8$ , so it has the form $(k, k)$ with $k > 0$ (both coordinates are stated to be positive). Then $k^2 + k^2 = 64$ , giving $k = \sqrt{32} = 4\sqrt{2}$ .

Finally write the circle with centre $(4\sqrt{2}, 4\sqrt{2})$ and radius $5$ in standard form and expand: $(x - 4\sqrt{2})^2 + (y - 4\sqrt{2})^2 = 25$ becomes $x^2 + y^2 - 8\sqrt{2}x - 8\sqrt{2}y + 32 + 32 - 25 = 0$ , i.e. $x^2 + y^2 - 8\sqrt{2}x - 8\sqrt{2}y + 39 = 0$ .

Key Takeaways

For two equal circles, the common chord is the perpendicular bisector of the line joining the centres.
Half-chord, perpendicular distance from centre, and radius always satisfy Pythagoras' theorem: $d^2 + (c/2)^2 = r^2$ .
Perpendicular lines have gradients whose product is $-1$ : perpendicular to $y = -x$ means direction $y = x$ .
Converting between standard form $(x-a)^2 + (y-b)^2 = r^2$ and general form $x^2 + y^2 + ax + by + c = 0$ requires expanding carefully, keeping surds exact.

Common Mistakes

Using the full chord length $6$ instead of the half-chord $3$ in Pythagoras, giving a wrong centre distance ( $\sqrt{25-36}$ fails, or students fudge it).
Taking the distance between centres as $4$ instead of $8$ (forgetting the chord is midway between the centres).
Placing the centre in the wrong quadrant — the question states both coordinates are positive, which rules out $(-4\sqrt{2}, -4\sqrt{2})$ .
Writing $k = \sqrt{32}$ but leaving it unsimplified or decimalising it; the mark scheme demands the exact form $4\sqrt{2}$ and the final answer "Must be exact".
Sign errors when expanding $(x - 4\sqrt{2})^2$ : the linear term is $-8\sqrt{2}x$ , not $+8\sqrt{2}x$ .
Arithmetic slip in the constant: $32 + 32 - 25 = 39$ , not $64 - 25 = 39$ miscomputed as something else, or forgetting to subtract the $25$ .

Things to Be Careful About

The final answer must be exact — decimals such as $5.66$ may earn the intermediate B1 for the centre but the final equation must use $8\sqrt{2}$ exactly.
The mark scheme allows follow-through only for a centre of the form $(k,k)$ with non-zero $k$ ; a centre not on $y = x$ scores no M1 for the equation.
Show the substitution into the standard circle form before expanding so the method mark is visible.
Check the constant term: it comes from $2(4\sqrt{2})^2 - 25 = 64 - 25 = 39$ .

Techniques used

read the radius from the completed circle equationuse the perpendicular bisector property of a common chordapply Pythagoras to the half-chord to find the distance between centresplace the second centre using symmetry about the line y = xexpand the standard circle equation into general form

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