A circle has equation $(x - 3)^2 + (y - 4)^2 = 20$. (a) Write down the coordinates of the centre of the circle. (b) Write down the radius of the circle. (c) The line $AB$ is a diameter of this circle. The point $A$ has coordinates $(-1, 6)$. Find the coordinates of the point $B$.

(a) (3, 4) · (b) √20 (= 2√5) · (c) B = (7, 2)

The function $\mathrm{f}$ is such that $\mathrm{f}(x) = 3x^2 + 8x$ for $x \geq k$, where $k$ is a constant. It is given that $\mathrm{f}^{-1}$ exists. $k$ has the least possible value for which $\mathrm{f}^{-1}$ exists. (a) Find the value of $k$. (b) Find the range of $\mathrm{f}$. (c) Find an expression for $\mathrm{f}^{-1}$. (d) The function $\mathrm{g}$ is such that $\mathrm{g}(x) = 3^x$ for $x > -6$. (i) Write down the domain of $\mathrm{gf}$. (ii) Solve the equation…

(a) -4/3 · (b) f(x) >= -16/3 · (c) f^-1(x) = (-8 + sqrt(64 + 12x))/6 · (d)(i) x >= -4/3 · (d)(ii) x = 1/3

A curve has equation $$ y = \frac{4 + \mathrm{e}^{2x}}{2 + \mathrm{e}^x} $$ (a) Show that, when $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, $\mathrm{e}^{2x} + 4\mathrm{e}^x - 4 = 0$. (b) Hence find the $x$-coordinate of the stationary point on the curve. Give your answer in the form $\ln(a\sqrt{b} + c)$, where $a$, $b$ and $c$ are integers.

(a) e^(2x) + 4e^x - 4 = 0 (shown) · (b) x = ln(2√2 - 2)

(a) Write down the value of $\lg(10000)$. (b) Solve the equation $\log_2(\log_3 x) = 2$. (c) Solve the equation $\log_x 5 = \log_5 x^4$.

(a) 4 · (b) 81 · (c) x = √5 or x = 1/√5

4037/11 May/June 2026 — Additional Mathematics Solved Past Paper

Q13MMedium-EasyEquations, inequalities and graphs

Solve the equation $|5x + 4| = |3x - 2|$ .

DifficultyMedium-Easy

Worked solution

Approach

An equation of the form $|A| = |B|$ is satisfied when $A = B$ or $A = -B$ . Write both linear equations and solve each.

Working

Case 1:

5x + 4 = 3x - 2

2x = -6 \quad\Rightarrow\quad x = -3

Case 2:

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

8x = -2 \quad\Rightarrow\quad x = -\frac{1}{4}

Both values check: at $x = -3$ , $|5(-3)+4| = 11 = |3(-3)-2|$ ; at $x = -\frac{1}{4}$ , $|5(-\tfrac{1}{4})+4| = \tfrac{11}{4} = |3(-\tfrac{1}{4})-2|$ .

Answer

x = -3 \quad \text{or} \quad x = -\frac{1}{4}

Final answer

x = -3 or x = -1/4

Detailed explanation

Walkthrough

The equation $|5x + 4| = |3x - 2|$ says two expressions have equal absolute value. Two numbers have equal absolute value exactly when they are equal, or when they are opposites. So we split into two linear equations.

First case: $5x + 4 = 3x - 2$ . Subtracting $3x$ and $4$ gives $2x = -6$ , so $x = -3$ .

Second case: $5x + 4 = -(3x - 2) = -3x + 2$ . Adding $3x$ to both sides gives $8x + 4 = 2$ , so $8x = -2$ and $x = -\frac{1}{4}$ .

Both solutions can be verified by substitution, and neither is extraneous because each came directly from a valid case of the modulus definition.

Key Takeaways

$|A| = |B|$ splits into the two cases $A = B$ and $A = -B$ ; both must be solved.
Each case of a modulus equation is an ordinary linear (or quadratic) equation.
Squaring both sides, $(5x+4)^2 = (3x-2)^2$ , is an equivalent alternative route leading to the same quadratic $16x^2 + 52x + 12 = 0$ .

Common Mistakes

Solving only one case (usually $5x + 4 = 3x - 2$ ) and giving just $x = -3$ — the mark scheme requires both equations to be stated.
Mishandling the negative case, e.g. writing $5x + 4 = -3x - 2$ instead of $-3x + 2$ , which gives a wrong root.
Introducing extra solutions by careless squaring or by mis-expanding $(3x-2)^2$ (the scheme allows only one slip per bracket).
Sign slips when factorising the quadratic route or when substituting into the quadratic formula.

Things to Be Careful About

Both solutions must be given; the mark scheme awards one accuracy mark for each root.
The negative case is $-(3x-2) = -3x + 2$ — distribute the minus over both terms.
Answers are exact fractions here: $x = -\frac{1}{4}$ should be left as a fraction, not a decimal.
If using the squaring method, the resulting quadratic must be set equal to zero and solved correctly for the method mark.

Techniques used

split the modulus equation into two linear casessolve each linear equationcheck that no extra solutions are introduced

Similar questions

Q2Medium-EasyCoordinate geometry of the circleStraight-line graphs

A circle has equation $(x - 3)^2 + (y - 4)^2 = 20$ .

(a)

Write down the coordinates of the centre of the circle.

1M Similar

(b)

Write down the radius of the circle.

1M Similar

(c)

The line $AB$ is a diameter of this circle.
The point $A$ has coordinates $(-1, 6)$ .

Find the coordinates of the point $B$ .

2M Similar

Q34MMedium-EasyTrigonometry

Show that $(3 + 5\cot\theta)^2 + (3 - 5\cot\theta)^2 - 50\operatorname{cosec}^2\theta$ , where $0 < \theta < \frac{\pi}{2}$ , can be simplified to a constant.

DifficultyMedium-Easy

Worked solution

Approach

Expand both squares, observe that the $\cot\theta$ terms cancel, then use the identity $\cot^2\theta = \operatorname{cosec}^2\theta - 1$ to reduce the expression to a constant.

Working

(3 + 5\cot\theta)^2 + (3 - 5\cot\theta)^2 = (9 + 30\cot\theta + 25\cot^2\theta) + (9 - 30\cot\theta + 25\cot^2\theta)

The middle terms cancel:

= 18 + 50\cot^2\theta

So the whole expression becomes

18 + 50\cot^2\theta - 50\operatorname{cosec}^2\theta

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

18 + 50(\cot^2\theta - \operatorname{cosec}^2\theta) = 18 + 50(-1) = -32

Answer

-32

which is a constant, as required.

Final answer

-32

Detailed explanation

Walkthrough

We must show the given expression simplifies to a constant — that is, all dependence on $\theta$ disappears.

Step 1: Expand both squares. Each bracket is of the form $(a \pm b)^2 = a^2 \pm 2ab + b^2$ with $a = 3$ and $b = 5\cot\theta$ :

(3 + 5\cot\theta)^2 = 9 + 30\cot\theta + 25\cot^2\theta, \qquad (3 - 5\cot\theta)^2 = 9 - 30\cot\theta + 25\cot^2\theta

Step 2: Add and collect. The $+30\cot\theta$ and $-30\cot\theta$ terms cancel exactly — this is why the expression was built as a sum of conjugate squares. We are left with

18 + 50\cot^2\theta - 50\operatorname{cosec}^2\theta

Step 3: Apply the Pythagorean identity. From $1 + \cot^2\theta = \operatorname{cosec}^2\theta$ we get $\cot^2\theta - \operatorname{cosec}^2\theta = -1$ . Factoring out the 50:

18 + 50(\cot^2\theta - \operatorname{cosec}^2\theta) = 18 - 50 = -32

Every term containing $\theta$ has vanished, so the expression is the constant $-32$ , which is what we were asked to show.

Key Takeaways

Expanding $(a+b)^2 + (a-b)^2$ always kills the mixed terms, leaving $2(a^2 + b^2)$ .
The identity $\cot^2\theta - \operatorname{cosec}^2\theta = -1$ is a rearrangement of $1 + \cot^2\theta = \operatorname{cosec}^2\theta$ ; recognising matching coefficients ( $50$ of each) is the key observation.
"Show that ... simplifies to a constant" means every $\theta$ -term must be eliminated by identities, not just evaluated at a particular angle.

Common Mistakes

Sign slips when expanding $(3 - 5\cot\theta)^2$ : the middle term is $-30\cot\theta$ , not $+30\cot\theta$ . The mark scheme allows only one slip in expanding each bracket.
Using the wrong coefficient on $\cot^2\theta$ or $\operatorname{cosec}^2\theta$ after collecting — the scheme explicitly disallows this ("Do not allow if wrong coefficient ...").
Misquoting the identity as $\cot^2\theta - \operatorname{cosec}^2\theta = +1$ , giving $68$ instead of $-32$ .
Dropping the argument $\theta$ : writing $\cot^2$ instead of $\cot^2\theta$ throughout. The mark scheme withholds the final mark for "persistent omission of $\theta$ ".
The answer is "nfww" (no follow-through from wrong working): an unsupported $-32$ scores nothing.

Things to Be Careful About

This is a "show that" task: every algebraic line between the original expression and $-32$ must appear — expansion, collection, and the identity step.
Keep $\theta$ attached to every trig function in every line of working; omitting it persistently costs the final accuracy mark.
The final answer is exact ( $-32$ ); no rounding issues arise here.
The condition $0 < \theta < \frac{\pi}{2}$ ensures $\sin\theta \neq 0$ , so $\cot\theta$ and $\operatorname{cosec}\theta$ are defined; it does not otherwise affect the algebra.

Techniques used

expand both squared bracketscollect like termsapply the identity cot^2 theta - cosec^2 theta = -1

Similar questions

Q45MMedium-EasyCalculusLogarithmic and exponential functions

Given that

\int_{0}^{a} \frac{1}{3x + 2}\,\mathrm{d}x = \ln 4

find the value of the constant $a$ .

DifficultyMedium-Easy

Worked solution

Approach

Integrate $\frac{1}{3x+2}$ using the standard result $\int \frac{1}{ax+b}\,\mathrm{d}x = \frac{1}{a}\ln(ax+b)$ , substitute the limits, then combine the logarithms and equate with $\ln 4$ .

Working

\int \frac{1}{3x+2}\,\mathrm{d}x = \frac{1}{3}\ln(3x+2)

Substituting the limits $0$ and $a$ :

\int_{0}^{a} \frac{1}{3x+2}\,\mathrm{d}x = \frac{1}{3}\left[\ln(3a+2) - \ln 2\right]

Using the quotient law of logarithms:

\frac{1}{3}\ln\left(\frac{3a+2}{2}\right) = \ln 4

So:

\ln\left(\frac{3a+2}{2}\right) = 3\ln 4 = \ln 64

Therefore:

\frac{3a+2}{2} = 64 \quad\Rightarrow\quad 3a + 2 = 128 \quad\Rightarrow\quad a = 42

Answer

a = 42

Final answer

a = 42

Detailed explanation

Walkthrough

The integral is of the form $\int \frac{1}{ax+b}\,\mathrm{d}x$ , whose antiderivative is $\frac{1}{a}\ln(ax+b)$ . Here $a = 3$ , so dividing by 3 gives $\frac{1}{3}\ln(3x+2)$ — the factor $\frac{1}{3}$ appears because differentiating $\ln(3x+2)$ by the chain rule brings down a 3, which must be cancelled.

Next we apply the limits. Evaluating at the upper limit $x = a$ gives $\ln(3a+2)$ ; at the lower limit $x = 0$ gives $\ln 2$ . Subtracting gives $\frac{1}{3}[\ln(3a+2) - \ln 2]$ .

The difference of two logs becomes a single log via the quotient law: $\ln(3a+2) - \ln 2 = \ln\left(\frac{3a+2}{2}\right)$ , so the equation reads $\frac{1}{3}\ln\left(\frac{3a+2}{2}\right) = \ln 4$ .

Multiplying both sides by 3 gives $\ln\left(\frac{3a+2}{2}\right) = 3\ln 4$ . Using the power law, $3\ln 4 = \ln 4^3 = \ln 64$ . Since equal logarithms imply equal arguments, $\frac{3a+2}{2} = 64$ , so $3a + 2 = 128$ and $a = 42$ .

Key Takeaways

The standard integral $\int \frac{1}{ax+b}\,\mathrm{d}x = \frac{1}{a}\ln|ax+b| + c$ — never forget the coefficient $\frac{1}{a}$ .
Definite integrals are evaluated as (upper limit) minus (lower limit).
Log laws convert differences of logs into quotients and multiples of logs into powers, allowing an equation in logs to be reduced to a linear equation.
If $\ln A = \ln B$ then $A = B$ .

Common Mistakes

Omitting the factor $\frac{1}{3}$ when integrating — writing just $\ln(3x+2)$ loses the accuracy mark.
Forgetting to subtract the lower-limit value $\ln 2$ .
Misapplying the power law: $3\ln 4$ is $\ln 64$ , not $\ln 12$ or $\ln 4^3$ mis-evaluated.
Dropping brackets when substituting limits, leading to sign errors.
Solving $\frac{3a+2}{2} = 4$ instead of $= 64$ (forgetting to multiply through by 3 first).

Things to Be Careful About

Show the substituted expression $\frac{1}{3}[\ln(3a+2) - \ln 2]$ explicitly before simplifying — this step carries its own mark.
The final answer must come from correct working (the scheme's answer mark requires valid preceding steps).
Equivalent forms such as $(3a+2)^{\frac{1}{3}} = ...$ or exponentiating both sides are accepted ('oe'), but the value must be exactly $a = 42$ .

Techniques used

integrate the reciprocal of a linear function to a logarithmsubstitute limits into the antiderivativecombine logarithms using the quotient lawequate logarithmic expressions and solve for a

Similar questions

Q5MediumFactors of polynomialsQuadratic functions

The polynomial $\mathrm{p}$ is such that $\mathrm{p}(x) = ax^3 + 4x^2 - ax - 28$ , where $a$ is a positive constant.
It is given that $\mathrm{p}(a) = 0$ .

(a)

Find the value of $a$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Since $\mathrm{p}(a) = 0$ , substitute $x = a$ into $\mathrm{p}(x) = ax^3 + 4x^2 - ax - 28$ . The resulting equation contains $a^4$ , $a^2$ and a constant, so it is a quadratic in $a^2$ .

Working

\mathrm{p}(a) = a(a)^3 + 4(a)^2 - a(a) - 28 = 0

a^4 + 4a^2 - a^2 - 28 = 0

a^4 + 3a^2 - 28 = 0

Factorising as a quadratic in $a^2$ :

(a^2 + 7)(a^2 - 4) = 0

a^2 = -7 \quad \text{or} \quad a^2 = 4

Since $a$ is a positive constant, $a^2 = -7$ is rejected, and $a^2 = 4$ gives $a = 2$ (positive root only).

Answer

a = 2

Final answer

a = 2

Detailed explanation

Walkthrough

The condition $\mathrm{p}(a) = 0$ says that when $x = a$ is substituted into the polynomial, the result is zero. Substituting $x = a$ into $ax^3 + 4x^2 - ax - 28$ gives $a \cdot a^3 + 4a^2 - a \cdot a - 28 = a^4 + 4a^2 - a^2 - 28 = a^4 + 3a^2 - 28 = 0$ .

The trick is to notice this is a quadratic in $a^2$ : writing $u = a^2$ gives $u^2 + 3u - 28 = 0$ , which factorises as $(u + 7)(u - 4) = 0$ . So $a^2 = -7$ or $a^2 = 4$ .

Since $a$ is a real positive constant, $a^2$ cannot be negative, so $a^2 = -7$ is discarded. From $a^2 = 4$ , the positive value is $a = 2$ (the negative root $a = -2$ is rejected because $a$ is stated to be positive).

Key Takeaways

$\mathrm{p}(k) = 0$ means $(x - k)$ is a factor of $\mathrm{p}(x)$ — here it is used in reverse to pin down the unknown coefficient.
Equations containing $a^4$ and $a^2$ are quadratics in disguise: substitute $u = a^2$ (or factorise directly in $a^2$ ).
Always check stated conditions on the unknown (here $a > 0$ ) before choosing the root.

Common Mistakes

Forgetting to multiply the leading term by $a$ : writing $a^3$ instead of $a^4$ when substituting $x = a$ into $ax^3$ .
Simplifying $4a^2 - a^2$ incorrectly; it gives $3a^2$ , not $5a^2$ or $4a^2$ .
Accepting $a^2 = -7$ or giving $a = \pm 2$ — the question states $a$ is positive, so only $a = 2$ is allowed.
Sign slips when factorising $a^4 + 3a^2 - 28$ : the factors must be $(a^2 + 7)(a^2 - 4)$ , since $7 \times (-4) = -28$ and $7 - 4 = 3$ .

Things to Be Careful About

The mark scheme awards the substitution B1 even unsimplified, but the factorisation M1 requires a correct quadratic in $a^2$ (or an explicit substitution such as $u = a^2$ ).
The final A1 is for $a = 2$ only — giving $a = \pm 2$ loses the mark because the positivity of $a$ is part of the question.

Techniques used

apply the factor theorem condition p(a) = 0substitute x = a into the cubictreat a^2 as the unknown to form a quadratic in a^2factorise the quadratic in a^2 and reject the negative root

(b)

Using your value of $a$ , find $\mathrm{p}(x)$ as a product of a linear factor and a quadratic factor.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute $a = 2$ to get $\mathrm{p}(x) = 2x^3 + 4x^2 - 2x - 28$ . Since $\mathrm{p}(2) = 0$ , the factor $(x - 2)$ divides the cubic; find the quadratic factor by division (or by matching coefficients).

Working

\mathrm{p}(x) = 2x^3 + 4x^2 - 2x - 28 = 2\left(x^3 + 2x^2 - x - 14\right)

Divide $x^3 + 2x^2 - x - 14$ by $(x - 2)$ :

x^3 + 2x^2 - x - 14 = (x - 2)\left(x^2 + 4x + 7\right)

Check: expanding, $(x-2)(x^2 + 4x + 7) = x^3 + 4x^2 + 7x - 2x^2 - 8x - 14 = x^3 + 2x^2 - x - 14$ . ✓

Therefore

\mathrm{p}(x) = 2(x - 2)\left(x^2 + 4x + 7\right)

Equivalently $\mathrm{p}(x) = (x - 2)\left(2x^2 + 8x + 14\right)$ .

Answer

\mathrm{p}(x) = 2(x - 2)\left(x^2 + 4x + 7\right)

Final answer

p(x) = 2(x - 2)(x^2 + 4x + 7)

Detailed explanation

Walkthrough

With $a = 2$ the polynomial becomes $\mathrm{p}(x) = 2x^3 + 4x^2 - 2x - 28$ . Part (a) showed $\mathrm{p}(2) = 0$ , so by the factor theorem $(x - 2)$ is a factor. Factor out the 2 first to make the division easier: $\mathrm{p}(x) = 2(x^3 + 2x^2 - x - 14)$ .

Long division (or coefficient matching) of $x^3 + 2x^2 - x - 14$ by $(x - 2)$ : the quotient starts with $x^2$ (since $x^2 \cdot x = x^3$ ), subtracting $x^3 - 2x^2$ leaves $4x^2 - x - 14$ ; the next term $+4x$ subtracts $4x^2 - 8x$ leaving $7x - 14$ ; the final term $+7$ subtracts $7x - 14$ leaving zero remainder. So the quotient is $x^2 + 4x + 7$ .

A quick check: the product of the constants in the two factors must give $-28$ — here $(-2) \times 7 \times 2 = -28$ . ✓

Key Takeaways

The factor theorem turns a known root into a known linear factor.
Cubic ÷ linear by long division (or by matching coefficients) gives the quadratic factor.
A useful self-check: the product of all constant terms in the factorisation must equal the constant term of the polynomial.

Common Mistakes

Forgetting the overall factor of 2, writing $\mathrm{p}(x) = (x-2)(x^2+4x+7)$ , which expands to the wrong cubic.
Sign errors in the division, e.g. writing $(x + 2)$ as the factor instead of $(x - 2)$ .
Not checking that the constant terms multiply to $-28$ — the mark scheme explicitly requires $a \times c = -28$ in the factor form $(x - a)(ax^2 + bx + c)$ .

Things to Be Careful About

The mark scheme's M1FT requires the form $(x - a)(ax^2 + bx + c)$ with $a \times c = -28$ ; both $2(x-2)(x^2+4x+7)$ and $(x-2)(2x^2+8x+14)$ satisfy this.
Expand your answer as a check before moving on — a wrong quadratic factor will also sink part (c).

Techniques used

substitute the known value of a into the polynomialdivide the cubic by the known linear factorexpress the cubic as a product of linear and quadratic factors

(c)

Hence show that the equation $\mathrm{p}(x) = 0$ has exactly one root.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Using the factorisation from part (b), $\mathrm{p}(x) = 2(x - 2)(x^2 + 4x + 7)$ . The factor $(x-2)$ gives the root $x = 2$ . The equation has exactly one root if the quadratic factor $x^2 + 4x + 7$ has no real roots, which is decided by its discriminant.

Working

The discriminant of $x^2 + 4x + 7$ is

b^2 - 4ac = 4^2 - 4(1)(7) = 16 - 28 = -12

Since the discriminant is negative, the quadratic $x^2 + 4x + 7 = 0$ has no real roots.

Therefore the only real solution of $\mathrm{p}(x) = 0$ is $x = 2$ , so the equation has exactly one root.

Answer

The discriminant of the quadratic factor is $-12 < 0$ , so $x^2 + 4x + 7 = 0$ has no real roots; hence $\mathrm{p}(x) = 0$ has exactly one root, $x = 2$ .

Final answer

Discriminant of x^2 + 4x + 7 is -12, which is negative, so the quadratic has no real roots and p(x) = 0 has exactly one root, x = 2

Detailed explanation

Walkthrough

This part says "Hence", so it must build on the factorisation $\mathrm{p}(x) = 2(x - 2)(x^2 + 4x + 7)$ from part (b). A product equals zero when one of its factors equals zero. The linear factor gives $x = 2$ — one real root. The quadratic factor $x^2 + 4x + 7$ may give further real roots, so we test its discriminant:

b^2 - 4ac = 4^2 - 4(1)(7) = 16 - 28 = -12

A negative discriminant means the quadratic never crosses the $x$ -axis: it has no real roots. So no further real solutions exist, and $x = 2$ is the only root of $\mathrm{p}(x) = 0$ .

Key Takeaways

A cubic with one known linear factor has exactly one real root precisely when the remaining quadratic has a negative discriminant.
The discriminant $b^2 - 4ac$ classifies the roots: positive → two distinct real roots, zero → one repeated root, negative → no real roots.
"Hence" questions expect the previous result to be used, not a fresh solve from scratch.

Common Mistakes

Computing $16 - 28$ as $+12$ — a sign slip that reverses the conclusion entirely.
Stating the discriminant is negative without evaluating it; the mark scheme requires the discriminant to be evaluated correctly for the A1.
Forgetting to state the logical link: negative discriminant → no real roots from the quadratic → only $x = 2$ remains.
Trying to factorise $x^2 + 4x + 7$ over the reals and concluding wrongly, or dividing the equation by a factor and losing information.

Things to Be Careful About

The M1 allows either the discriminant of $2x^2 + 8x + 14$ (giving $64 - 112 = -48$ ) or of $x^2 + 4x + 7$ (giving $-12$ ) — both are acceptable, but whichever you use must be evaluated correctly.
The A1 needs the explicit statement that the discriminant is negative and therefore there are no (real) roots — the conclusion, not just the number, earns the mark.

Techniques used

use the factorisation from part (b)compute the discriminant of the quadratic factorargue that a negative discriminant gives no real roots

Q6MediumFunctionsQuadratic functionsLogarithmic and exponential functions

The function $\mathrm{f}$ is such that $\mathrm{f}(x) = 3x^2 + 8x$ for $x \geq k$ , where $k$ is a constant.
It is given that $\mathrm{f}^{-1}$ exists.
$k$ has the least possible value for which $\mathrm{f}^{-1}$ exists.

(a)

Find the value of $k$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

For $\mathrm{f}^{-1}$ to exist, $\mathrm{f}$ must be one-to-one. A parabola becomes one-to-one when its domain starts at the $x$ -coordinate of its vertex, so find that stationary point.

Working

Differentiate:

\frac{\mathrm{d}y}{\mathrm{d}x} = 6x + 8

Set equal to zero for the stationary point:

6x + 8 = 0 \quad \Rightarrow \quad x = -\frac{8}{6} = -\frac{4}{3}

The least value of $k$ is this $x$ -coordinate.

Answer

k = -\frac{4}{3}

Final answer

-4/3

Detailed explanation

Walkthrough

A function has an inverse only if it is one-to-one: each output must come from exactly one input. The parabola $y = 3x^2 + 8x$ fails this over a wide domain because outputs on either side of the vertex repeat. If we cut the domain so it begins exactly at the vertex, every output is produced by just one input, so the smallest permissible $k$ is the $x$ -coordinate of the vertex.

To find the vertex we differentiate: $\frac{\mathrm{d}y}{\mathrm{d}x} = 6x + 8$ , and setting this to zero gives $x = -\frac{4}{3}$ . (Equivalently, completing the square gives $3\left(x+\frac{4}{3}\right)^2 - \frac{16}{3}$ , or factorising as $x(3x+8)$ shows symmetry about $x = -\frac{4}{3}$ .)

Key Takeaways

An inverse function exists only when the function is one-to-one.
For a quadratic, restricting the domain to one side of the vertex makes it one-to-one; the least such restriction starts at the vertex itself.
The vertex can be found by differentiation, completing the square, or using symmetry of the factors.

Common Mistakes

Giving the $y$ -coordinate ( $-\frac{16}{3}$ ) instead of the $x$ -coordinate of the vertex — the mark scheme awards the M1 for $x = -\frac{4}{3}$ seen.
Sign errors in solving $6x + 8 = 0$ .
Confusing 'least possible value' with an arbitrary larger value of $k$ .

Things to Be Careful About

The answer must be exact: $k = -\frac{4}{3}$ , not a rounded decimal.
The method mark requires a complete route to the stationary point (derivative set to zero, completed square, or symmetry argument), not just the answer.

Techniques used

differentiate the quadraticset the derivative to zero to locate the vertexrestrict the domain at the turning point to make the function one-to-one

(b)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

On the domain $x \geq k$ , the minimum of $\mathrm{f}$ occurs at $x = k$ . Substitute $k = -\frac{4}{3}$ into $\mathrm{f}(x) = 3x^2 + 8x$ .

Working

\mathrm{f}\left(-\frac{4}{3}\right) = 3\left(-\frac{4}{3}\right)^2 + 8\left(-\frac{4}{3}\right)

= 3 \times \frac{16}{9} - \frac{32}{3} = \frac{16}{3} - \frac{32}{3} = -\frac{16}{3}

Since the parabola opens upwards, all values are greater than or equal to this.

Answer

\mathrm{f}(x) \geq -\frac{16}{3}

Final answer

f(x) >= -16/3

Detailed explanation

Walkthrough

With the domain starting at the vertex $k = -\frac{4}{3}$ , the smallest output of the upward-opening parabola is the value there. Substituting: $3\left(-\frac{4}{3}\right)^2 = \frac{16}{3}$ and $8\left(-\frac{4}{3}\right) = -\frac{32}{3}$ , giving $\frac{16}{3} - \frac{32}{3} = -\frac{16}{3}$ . So the range is $\mathrm{f}(x) \geq -\frac{16}{3}$ .

Key Takeaways

The range of a restricted quadratic is found from the value at the boundary of the domain.
Correct notation matters: write $\mathrm{f}(x) \geq \ldots$ or $y \geq \ldots$ , not just the number.

Common Mistakes

Stating the number without the inequality or without correct notation (the mark scheme insists on $\mathrm{f}$ or $y$ with $\geq$ ).
Arithmetic slips combining thirds: $\frac{16}{3} - \frac{32}{3} = -\frac{16}{3}$ , not $+\frac{16}{3}$ .
Using the wrong sign of $k$ when substituting.

Things to Be Careful About

The mark scheme requires the substitution of your own $k$ to be shown explicitly for the M1.
Leave the answer as the exact fraction $-\frac{16}{3}$ .

Techniques used

evaluate the function at the restricted domain endpointstate the range using correct inequality notation

(c)

Find an expression for $\mathrm{f}^{-1}$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Write $x = 3y^2 + 8y$ (swapping the roles of $x$ and $y$ ), then solve this quadratic for $y$ using the quadratic formula. Since the domain of $\mathrm{f}$ is $x \geq -\frac{4}{3}$ , which lies to the right of the vertex, we need the positive square-root branch.

Working

Swap variables:

x = 3y^2 + 8y \quad \Rightarrow \quad 3y^2 + 8y - x = 0

Apply the quadratic formula with $a = 3$ , $b = 8$ , $c = -x$ :

y = \frac{-8 \pm \sqrt{64 - 4(3)(-x)}}{2(3)} = \frac{-8 \pm \sqrt{64 + 12x}}{6}

Since $x \geq -\frac{4}{3}$ corresponds to inputs on the right-hand branch of the parabola, take the $+$ sign:

\mathrm{f}^{-1}(x) = \frac{-8 + \sqrt{64 + 12x}}{6}

Answer

\mathrm{f}^{-1}(x) = \frac{-8 + \sqrt{64 + 12x}}{6}

Final answer

f^-1(x) = (-8 + sqrt(64 + 12x))/6

Detailed explanation

Walkthrough

To invert a function written as $y = \mathrm{f}(x)$ , interchange input and output: put $x = 3y^2 + 8y$ . This is a quadratic in $y$ , so rearrange to $3y^2 + 8y - x = 0$ and apply the quadratic formula with coefficients $a = 3$ , $b = 8$ , $c = -x$ . The discriminant is $64 + 12x$ , giving $y = \frac{-8 \pm \sqrt{64+12x}}{6}$ .

Two branches exist, but only one is the inverse: since the domain of $\mathrm{f}$ was restricted to $x \geq -\frac{4}{3}$ (the right-hand branch of the parabola), the inverse must return values $\geq -\frac{4}{3}$ . At $x = 0$ , the $+$ branch gives $0$ while the $-$ branch gives $-\frac{8}{3}$ , so the $+$ branch is correct.

Key Takeaways

To find an inverse algebraically, swap $x$ and $y$ and make $y$ the subject.
When the inverse comes from a quadratic, choose the branch matching the original function's restricted domain.

Common Mistakes

Keeping both the $+$ and $-$ branches — the mark scheme gives B2 only for the single correct branch, B1 for leaving $\pm$ .
Sign errors inside the discriminant: $c = -x$ contributes $+12x$ , not $-12x$ .
Forgetting to swap variables first and solving $3x^2 + 8x = y$ for $x$ instead.

Things to Be Careful About

The unsimplified form $\frac{-8+\sqrt{64+12x}}{6}$ is fully accepted ('Allow unsimplified'), as is $\frac{-4+\sqrt{16+3x}}{3}$ .
The choice of branch must be justified by the domain $x \geq -\frac{4}{3}$ .

Techniques used

swap variables to form the inverse relationsolve a quadratic for y using the quadratic formulaselect the root consistent with the domain

(d)

The function $\mathrm{g}$ is such that $\mathrm{g}(x) = 3^x$ for $x > -6$ .

5M

(i)

Write down the domain of $\mathrm{gf}$ .

1M Similar

(ii)

Solve the equation $\mathrm{gf}(x) = 27$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Form the composite $\mathrm{gf}(x) = 3^{\mathrm{f}(x)} = 3^{3x^2+8x}$ and set it equal to $27 = 3^3$ . Equate exponents, solve the quadratic, then reject any root outside the domain $x \geq -\frac{4}{3}$ .

Working

\mathrm{gf}(x) = 3^{3x^2 + 8x} = 27 = 3^3

Equating exponents:

3x^2 + 8x = 3 \quad \Rightarrow \quad 3x^2 + 8x - 3 = 0

Factorise:

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

Since the domain requires $x \geq -\frac{4}{3}$ , the root $x = -3$ is rejected.

Answer

x = \frac{1}{3}

Final answer

x = 1/3

Detailed explanation

Walkthrough

First build the composite: $\mathrm{gf}(x) = \mathrm{g}(\mathrm{f}(x)) = 3^{3x^2+8x}$ . Setting this equal to $27$ and writing $27 = 3^3$ lets us equate exponents, since $3^a = 3^b$ forces $a = b$ . That turns the exponential equation into the quadratic $3x^2 + 8x = 3$ , i.e. $3x^2 + 8x - 3 = 0$ .

Factorising gives $(3x-1)(x+3) = 0$ , so $x = \frac{1}{3}$ or $x = -3$ . But the domain of $\mathrm{gf}$ (part d(i)) is $x \geq -\frac{4}{3}$ , and $-3 < -\frac{4}{3}$ , so $x = -3$ is not a valid solution. Only $x = \frac{1}{3}$ remains.

Key Takeaways

A composite $\mathrm{gf}(x)$ means substitute $\mathrm{f}(x)$ into $\mathrm{g}$ .
Equal powers of the same base allow equating exponents.
Solutions must be checked against the domain of the composite; extraneous roots are discarded.

Common Mistakes

Computing $\mathrm{fg}(x)$ instead of $\mathrm{gf}(x)$ — order matters.
Sign errors moving the $3$ : the equation is $3x^2 + 8x - 3 = 0$ , leading to $(3x-1)(x+3)$ , not $(3x+1)(x-3)$ .
Failing to reject $x = -3$ , which lies outside the domain — the mark scheme awards the final A1 for $x = \frac{1}{3}$ only.
Solving $3^{3x^2+8x} = 27$ by taking logs unnecessarily instead of writing $27 = 3^3$ .

Things to Be Careful About

The final A1 depends on the earlier marks being earned correctly (dependent marking).
State why $x = -3$ is rejected: it violates $x \geq -\frac{4}{3}$ .

Techniques used

form the composite gf explicitlyequate exponents of equal powers of 3factorise the resulting quadraticreject roots outside the domain

Q7MediumCalculusQuadratic functionsLogarithmic and exponential functions

A curve has equation

y = \frac{4 + \mathrm{e}^{2x}}{2 + \mathrm{e}^x}

(a)

Show that, when $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ , $\mathrm{e}^{2x} + 4\mathrm{e}^x - 4 = 0$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Differentiate the quotient using the quotient rule, with the chain rule giving $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{2x}\right) = 2\mathrm{e}^{2x}$ . Set the derivative equal to zero, which means the numerator must be zero, then expand and simplify.

Working

With $u = 4 + \mathrm{e}^{2x}$ and $v = 2 + \mathrm{e}^x$ :

\frac{\mathrm{d}u}{\mathrm{d}x} = 2\mathrm{e}^{2x}, \qquad \frac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{e}^x

Quotient rule:

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(2+\mathrm{e}^x)\,2\mathrm{e}^{2x} - (4+\mathrm{e}^{2x})\,\mathrm{e}^x}{(2+\mathrm{e}^x)^2}

When $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ , the numerator must be zero:

2\mathrm{e}^{2x}(2+\mathrm{e}^x) - \mathrm{e}^x(4+\mathrm{e}^{2x}) = 0

Expand:

4\mathrm{e}^{2x} + 2\mathrm{e}^{3x} - 4\mathrm{e}^x - \mathrm{e}^{3x} = 0

\mathrm{e}^{3x} + 4\mathrm{e}^{2x} - 4\mathrm{e}^x = 0

Since $\mathrm{e}^x > 0$ for all $x$ , divide through by $\mathrm{e}^x$ :

\mathrm{e}^{2x} + 4\mathrm{e}^x - 4 = 0

as required.

Answer

\mathrm{e}^{2x} + 4\mathrm{e}^x - 4 = 0 \quad \text{(shown)}

Final answer

e^(2x) + 4e^x - 4 = 0 (shown)

Detailed explanation

Walkthrough

The curve is a quotient of two expressions in $\mathrm{e}^x$ , so the quotient rule is the natural tool. Writing $u = 4 + \mathrm{e}^{2x}$ and $v = 2 + \mathrm{e}^x$ , we need $\frac{\mathrm{d}u}{\mathrm{d}x}$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ . Differentiating $\mathrm{e}^{2x}$ needs the chain rule: the derivative of $\mathrm{e}^{kx}$ is $k\,\mathrm{e}^{kx}$ , so $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{2x}\right) = 2\mathrm{e}^{2x}$ , while $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^x\right) = \mathrm{e}^x$ .

The quotient rule gives the derivative as a fraction. A fraction equals zero exactly when its numerator equals zero (the denominator $(2+\mathrm{e}^x)^2$ is never zero), so setting $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ reduces to solving $2\mathrm{e}^{2x}(2+\mathrm{e}^x) - \mathrm{e}^x(4+\mathrm{e}^{2x}) = 0$ .

Expanding each product: $2\mathrm{e}^{2x}(2+\mathrm{e}^x) = 4\mathrm{e}^{2x} + 2\mathrm{e}^{3x}$ and $\mathrm{e}^x(4+\mathrm{e}^{2x}) = 4\mathrm{e}^x + \mathrm{e}^{3x}$ . Subtracting gives $\mathrm{e}^{3x} + 4\mathrm{e}^{2x} - 4\mathrm{e}^x = 0$ .

Every term contains a factor $\mathrm{e}^x$ , and since $\mathrm{e}^x$ is never zero, dividing by it loses nothing. This leaves $\mathrm{e}^{2x} + 4\mathrm{e}^x - 4 = 0$ , which is the required result.

Key Takeaways

The quotient rule: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}$ , with the order of subtraction mattering.
The chain rule on exponentials: $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{kx}\right) = k\,\mathrm{e}^{kx}$ .
A fraction is zero when its numerator is zero, so the denominator can be ignored when solving $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ .
$\mathrm{e}^x > 0$ always, so dividing an equation by $\mathrm{e}^x$ is safe and loses no solutions.

Common Mistakes

Subtracting the quotient-rule terms the wrong way round: the scheme condones writing the terms in the wrong order but the subtraction itself must be present.
Forgetting the factor 2 in $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{2x}\right) = 2\mathrm{e}^{2x}$ (this is a specific B1 in the scheme).
Omitting the $v^2$ denominator or not making a convincing attempt at both derivatives (the M1 requires this).
Dividing by $\mathrm{e}^x$ without justification — the final A1 requires the division (or factorising) to be stated, since dividing by a variable expression normally loses solutions.
Sign slips when expanding $-\mathrm{e}^x(4+\mathrm{e}^{2x})$ .

Things to Be Careful About

This is an "AG" (Answer Given) part: every algebraic line between the derivative and the target equation must be shown — the scheme says sufficient correct detail is required.
The final A1 explicitly demands justification for dividing by $\mathrm{e}^x$ (e.g. "since $\mathrm{e}^x \neq 0$ ").
The denominator $(2+\mathrm{e}^x)^2$ is always positive, so it can never make the derivative zero — only the numerator matters.

Techniques used

differentiate e^(2x) using the chain ruleapply the quotient ruleset the numerator of the derivative to zeroexpand and simplify the numeratordivide through by e^x to obtain the given equation

(b)

Hence find the $x$ -coordinate of the stationary point on the curve.
Give your answer in the form $\ln(a\sqrt{b} + c)$ , where $a$ , $b$ and $c$ are integers.

4M Similar

DifficultyMedium

Worked solution

Approach

Solve the quadratic $\mathrm{e}^{2x} + 4\mathrm{e}^x - 4 = 0$ for $\mathrm{e}^x$ using the quadratic formula, keep only the positive root (since $\mathrm{e}^x > 0$ ), then take natural logarithms.

Working

Using the quadratic formula with $a = 1$ , $b = 4$ , $c = -4$ :

\mathrm{e}^x = \frac{-4 \pm \sqrt{16 - 4 \times 1 \times (-4)}}{2} = \frac{-4 \pm \sqrt{32}}{2} = \frac{-4 \pm 4\sqrt{2}}{2}

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

Since $\mathrm{e}^x > 0$ , the root $-2 - 2\sqrt{2} < 0$ is rejected, leaving:

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

Taking natural logarithms of both sides:

x = \ln\left(2\sqrt{2} - 2\right)

This is in the form $\ln(a\sqrt{b} + c)$ with $a = 2$ , $b = 2$ , $c = -2$ .

Answer

x = \ln\left(2\sqrt{2} - 2\right)

Final answer

x = ln(2√2 - 2)

Detailed explanation

Walkthrough

Part (a) produced a quadratic equation in $\mathrm{e}^x$ : treating $\mathrm{e}^x$ as the unknown, we have $\mathrm{e}^{2x} + 4\mathrm{e}^x - 4 = 0$ , i.e. $u^2 + 4u - 4 = 0$ with $u = \mathrm{e}^x$ . The quadratic formula with $a = 1$ , $b = 4$ , $c = -4$ gives $\mathrm{e}^x = \frac{-4 \pm \sqrt{16+16}}{2}$ . The discriminant is $32$ , and $\sqrt{32} = 4\sqrt{2}$ , so $\mathrm{e}^x = -2 \pm 2\sqrt{2}$ .

Now $\mathrm{e}^x$ is positive for every real $x$ , so the root $-2 - 2\sqrt{2}$ (about $-4.83$ ) is impossible and must be discarded. The only valid value is $\mathrm{e}^x = 2\sqrt{2} - 2$ (about $0.83$ , positive as required).

Taking natural logarithms of both sides undoes the exponential: $x = \ln\left(2\sqrt{2} - 2\right)$ , which is exactly the form $\ln(a\sqrt{b} + c)$ requested, with $a = 2$ , $b = 2$ , $c = -2$ .

Key Takeaways

Equations like $\mathrm{e}^{2x} + 4\mathrm{e}^x - 4 = 0$ are quadratics in disguise — substitute $u = \mathrm{e}^x$ mentally and solve for $u$ .
Always check roots against $\mathrm{e}^x > 0$ ; a negative value of $\mathrm{e}^x$ has no real solution.
Taking $\ln$ of both sides converts $\mathrm{e}^x = k$ into $x = \ln k$ .
Exact surd forms should be kept rather than decimalised when the question demands a form like $\ln(a\sqrt{b} + c)$ .

Common Mistakes

Keeping the negative root $-2 - 2\sqrt{2}$ and trying to take its logarithm — this is invalid since $\mathrm{e}^x$ cannot be negative.
Sign errors in the discriminant: $-4 \times 1 \times (-4) = +16$ , so the discriminant is $32$ , not $0$ or negative.
Simplifying $\sqrt{32}$ incorrectly (it is $4\sqrt{2}$ , not $2\sqrt{8}$ unsimplified or $16\sqrt{2}$ ).
Writing $x = \ln(-2 + 2\sqrt{2})$ with the terms in an order that hides the required $a\sqrt{b} + c$ structure — the accepted form is $\ln(2\sqrt{2} - 2)$ or equivalently $\ln(\sqrt{8} - 2)$ .
Giving a decimal answer instead of the exact logarithmic form the question demands.

Things to Be Careful About

The final two marks depend on taking logarithms (M1, dependent on the quadratic being solved correctly) and on the exact form $\ln(2\sqrt{2} - 2)$ — the scheme also accepts $\ln(\sqrt{8} - 2)$ .
The answer must be exact: no decimals.
The word "Hence" signals that the quadratic from part (a) must be used — do not re-derive the stationary point condition from scratch.

Techniques used

solve a quadratic in e^x using the quadratic formulareject the negative root since e^x must be positivetake natural logarithms to solve for xexpress the answer in the form ln(a√b + c)

Q8MediumLogarithmic and exponential functions

(a)

Write down the value of $\lg(10000)$ .

1M Similar

(b)

Solve the equation $\log_2(\log_3 x) = 2$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Peel off the outer logarithm by writing $\log_2(\log_3 x) = 2$ in index form, then do the same again for the inner logarithm.

Working

Rewrite the equation in index form with base $2$ :

\log_3 x = 2^{2} = 4

Now rewrite in index form with base $3$ :

x = 3^{4} = 81

Check: $\log_3 81 = 4$ and $\log_2 4 = 2$ . ✓

Answer

x = 81

Final answer

81

Detailed explanation

Walkthrough

The equation has a logarithm inside a logarithm, so we undo them from the outside in. The statement $\log_2(y) = 2$ means exactly the same thing as $y = 2^2$ , so taking $y = \log_3 x$ gives $\log_3 x = 4$ . Applying the same idea again, $\log_3 x = 4$ means $x = 3^4 = 81$ . A quick check confirms both logarithms evaluate correctly at $x = 81$ .

Key Takeaways

$\log_b y = k \iff y = b^k$ : converting between logarithmic and index form is the fundamental tool for solving log equations.
Nested logarithms are handled one layer at a time, starting from the outside.

Common Mistakes

Writing $x = 2^4$ or $x = 4^3$ instead of $x = 3^4$ — the inner logarithm has base $3$ , so the final power must have base $3$ .
Stopping after finding $\log_3 x = 4$ and giving $4$ as the answer instead of solving for $x$ .
Confusing the order of operations and trying to combine the two logarithms with log laws (they have different bases, so no law applies).

Things to Be Careful About

The mark scheme awards M1 for correctly unwrapping the outer log to get $\log_3 x = k$ , then A1 for $k = 4$ and A1 for $x = 81$ . Both steps of index conversion must be shown; jumping straight to $81$ risks losing method marks.

Techniques used

rewrite a logarithmic equation in index formapply the definition of a logarithm twiceevaluate the resulting power

(c)

Solve the equation $\log_x 5 = \log_5 x^4$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the power law on $\log_5 x^4$ and the reciprocal identity $\log_x 5 = \dfrac{1}{\log_5 x}$ to turn the equation into a quadratic in $\log_5 x$ .

Working

By the power law:

\log_5 x^{4} = 4\log_5 x

Also, since $\log_x 5 = \dfrac{1}{\log_5 x}$ , let $u = \log_5 x$ . Then $\log_x 5 = \dfrac{1}{u}$ , and the equation becomes

\frac{1}{u} = 4u

Multiplying through by $u$ :

1 = 4u^{2}

u^{2} = \frac{1}{4}

u = \pm\frac{1}{2}

Converting back with $u = \log_5 x$ , i.e. $x = 5^{u}$ :

x = 5^{\frac{1}{2}} = \sqrt{5}

or

x = 5^{-\frac{1}{2}} = \frac{1}{\sqrt{5}}

Both values are positive, so both are valid bases/arguments.

Answer

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

Final answer

x = √5 or x = 1/√5

Detailed explanation

Walkthrough

The equation $\log_x 5 = \log_5 x^4$ involves two different bases, $x$ and $5$ . Two facts bridge them. First, the power law lets us pull the exponent out: $\log_5 x^4 = 4\log_5 x$ . Second, a logarithm in one base is the reciprocal of the logarithm with the bases swapped: $\log_x 5 = \frac{1}{\log_5 x}$ . Substituting $u = \log_5 x$ turns the whole equation into $\frac{1}{u} = 4u$ , which rearranges to $4u^2 = 1$ . Solving gives $u = \pm\frac{1}{2}$ — both signs matter, because squaring loses the sign. Finally, each value of $u$ converts back to an $x$ via $x = 5^u$ , giving $\sqrt{5}$ and $\frac{1}{\sqrt{5}}$ . Both are positive, so both are legitimate solutions.

Key Takeaways

Power law: $\log_b(a^n) = n\log_b a$ .
Reciprocal identity: $\log_a b = \dfrac{1}{\log_b a}$ — this is how you handle equations mixing two bases.
Substituting a single letter for a logarithm reduces many log equations to ordinary quadratics.
After squaring, always recover both roots; then check each against domain restrictions ( $x > 0$ , $x \neq 1$ ).

Common Mistakes

Writing $\log_5 x^4 = (\log_5 x)^4$ instead of $4\log_5 x$ — the exponent applies to the argument, not to the logarithm.
Forgetting the negative root $u = -\frac{1}{2}$ and losing $x = \frac{1}{\sqrt{5}}$ .
Giving decimal approximations instead of the exact surds $\sqrt{5}$ and $\frac{1}{\sqrt{5}}$ required on this non-calculator paper.
Ignoring the domain: $x$ must be positive and $x \neq 1$ ; here both answers satisfy this, but the check should be made.
Mark scheme warning: "Do not isw rejected solutions" — if a candidate later discards a correct solution, marks are lost.

Things to Be Careful About

The mark scheme pays B1 for the reciprocal step ( $\frac{4}{\log_x 5} = \log_x 5$ or equivalent), B1 for reaching $(\log_x 5)^2 = 4$ or $(\log_5 x)^2 = \frac{1}{4}$ , and B1 each for the two exact answers $\sqrt{5}$ and $\frac{1}{\sqrt{5}}$ (oe). Every algebraic line must be visible. Answers must be left in exact surd form — decimals score nothing here.

Techniques used

express a power inside a logarithm as a multiple using log lawsuse the reciprocal relation between logs in swapped basesform and solve a quadratic-type equation in a logarithmconvert back from logarithm to index form

Q9MediumSeriesTrigonometry

(a)

A geometric progression has common ratio $r$ and first term $a$ , where $a \neq 0$ .
The 4th term is $-8$ times the 7th term.

4M

(i)

Find the value of $r$ .

2M Similar

(ii)

The sum of the first 5 terms of this progression is $\frac{99}{32}$ .

Find the value of $a$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the sum formula $S_n = \dfrac{a(1 - r^n)}{1 - r}$ with $n = 5$ and $r = -\dfrac{1}{2}$ from part (i), then solve for $a$ .

Working

\frac{99}{32} = a\,\frac{1 - \left(-\frac{1}{2}\right)^5}{1 - \left(-\frac{1}{2}\right)}

Evaluate the fraction:

\left(-\frac{1}{2}\right)^5 = -\frac{1}{32}, \qquad 1 - \left(-\frac{1}{32}\right) = \frac{33}{32}, \qquad 1 - \left(-\frac{1}{2}\right) = \frac{3}{2}

\frac{99}{32} = a \times \frac{33}{32} \times \frac{2}{3} = a \times \frac{66}{96} = a \times \frac{11}{16}

Solve for $a$ :

a = \frac{99}{32} \times \frac{16}{11} = \frac{99 \times 16}{32 \times 11} = \frac{9 \times 16}{32} = \frac{9}{2}

Answer

a = \frac{9}{2}

Final answer

a = 9/2

Detailed explanation

Walkthrough

The sum of the first $n$ terms of a geometric progression is $S_n = \dfrac{a(1 - r^n)}{1 - r}$ . Carrying forward $r = -\dfrac{1}{2}$ from part (i) and setting $n = 5$ , we substitute into $S_5 = \dfrac{99}{32}$ . Now $\left(-\frac{1}{2}\right)^5 = -\frac{1}{32}$ , so the numerator becomes $1 + \frac{1}{32} = \frac{33}{32}$ and the denominator is $1 + \frac{1}{2} = \frac{3}{2}$ . The fraction simplifies to $\frac{33}{32} \times \frac{2}{3} = \frac{11}{16}$ , so $\frac{99}{32} = \frac{11a}{16}$ , giving $a = \frac{99}{32} \times \frac{16}{11} = \frac{9}{2}$ .

Key Takeaways

The finite sum formula $S_n = \dfrac{a(1 - r^n)}{1 - r}$ works for any $r \neq 1$ , including negative $r$ .
Odd powers of a negative fraction stay negative: $\left(-\frac{1}{2}\right)^5 = -\frac{1}{32}$ .

Common Mistakes

Sign errors with $\left(-\frac{1}{2}\right)^5$ — it is negative, so $1 - r^5 = 1 + \frac{1}{32}$ .
Using $S_5 = \dfrac{a(r^5 - 1)}{r - 1}$ incorrectly or mixing the two formula forms.
Arithmetic slips when simplifying $\frac{33}{32} \div \frac{3}{2}$ .

Things to Be Careful About

The mark scheme allows follow-through on an incorrect $r$ from part (i), provided the substitution into the formula is correct.
The answer must be exact: $a = \frac{9}{2}$ (equivalent forms accepted, e.g. $4.5$ ).

Techniques used

apply the sum of the first n terms formula for a geometric progressionsubstitute the known common ratiosolve the resulting linear equation for the first term

(b)

The first three terms of a different geometric progression are $\cos\theta\sin\theta$ , $\cos\theta\sin^3\theta$ and $\cos\theta\sin^5\theta$ , for $0 < \theta < \frac{\pi}{2}$ .

Show that the sum to infinity of this progression is $\tan\theta$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Find the common ratio by dividing consecutive terms, check that $|r| < 1$ using $0 < \theta < \frac{\pi}{2}$ , then apply $S_\infty = \dfrac{a}{1 - r}$ and simplify with $\cos^2\theta + \sin^2\theta = 1$ .

Working

The common ratio is

r = \frac{\cos\theta\sin^3\theta}{\cos\theta\sin\theta} = \sin^2\theta

Since $0 < \theta < \frac{\pi}{2}$ , we have $0 < \sin\theta < 1$ , so $0 < \sin^2\theta < 1$ and the sum to infinity exists.

Applying $S_\infty = \dfrac{a}{1 - r}$ with $a = \cos\theta\sin\theta$ :

S_\infty = \frac{\cos\theta\sin\theta}{1 - \sin^2\theta}

Using $\cos^2\theta + \sin^2\theta = 1$ , so $1 - \sin^2\theta = \cos^2\theta$ :

S_\infty = \frac{\cos\theta\sin\theta}{\cos^2\theta}

Cancelling one factor of $\cos\theta$ :

S_\infty = \frac{\sin\theta}{\cos\theta} = \tan\theta

as required.

Answer

S_\infty = \tan\theta

Final answer

Sum to infinity = tan(theta), shown via r = sin^2(theta) and 1 - sin^2(theta) = cos^2(theta)

Detailed explanation

Walkthrough

First identify the common ratio: dividing the second term by the first gives $\dfrac{\cos\theta\sin^3\theta}{\cos\theta\sin\theta} = \sin^2\theta$ . Because $0 < \theta < \frac{\pi}{2}$ , $\sin\theta$ lies strictly between 0 and 1, so $|r| = \sin^2\theta < 1$ — the condition for a sum to infinity to exist. Then apply $S_\infty = \dfrac{a}{1-r}$ with $a = \cos\theta\sin\theta$ , giving $\dfrac{\cos\theta\sin\theta}{1 - \sin^2\theta}$ . The Pythagorean identity rewrites the denominator as $\cos^2\theta$ , and cancelling one factor of $\cos\theta$ leaves $\dfrac{\sin\theta}{\cos\theta} = \tan\theta$ , which is the printed target.

Key Takeaways

The common ratio of a GP whose terms share a factor is found by dividing consecutive terms.
A sum to infinity only exists when $|r| < 1$ — here the interval $0 < \theta < \frac{\pi}{2}$ guarantees this and should be stated.
$1 - \sin^2\theta = \cos^2\theta$ is the standard route to converting such fractions into $\tan\theta$ .

Common Mistakes

Omitting the justification that $|\sin^2\theta| < 1$ from the given range of $\theta$ .
Writing $1 - \sin^2\theta = \sin^2\theta$ or confusing it with $1 + \tan^2\theta = \sec^2\theta$ .
Dropping the $\theta$ throughout and writing $\tan$ with no argument — the mark scheme withholds the final mark for persistent omission of $\theta$ .
Cancelling $\cos\theta$ incorrectly, e.g. leaving $\dfrac{\sin\theta}{\cos^2\theta}$ .

Things to Be Careful About

This is a 'show that' (AG) part: derive forward to exactly $\tan\theta$ ; never start from $\tan\theta$ and work backwards.
Every intermediate line matters — the scheme awards a mark each for $r = \sin^2\theta$ , the substituted fraction, the $\cos^2\theta$ denominator, and the final $\tan\theta$ , each dependent on the previous one.
Keep the angle $\theta$ on every trigonometric function throughout the working.

Techniques used

identify the common ratio of a geometric progressionapply the sum to infinity formulaverify the convergence conditionuse the identity cos squared plus sin squared equals onesimplify a trigonometric fraction to tan

Q108MMediumCalculus

A curve has equation $y = \mathrm{f}(x)$ , where $\mathrm{f}''(x) = (2x + 10)^{-\frac{1}{2}} - 2x$ .

The curve has a stationary point at $\left(3, \frac{64}{3}\right)$ .

Find the equation of the curve.

DifficultyMedium

Worked solution

Approach

The second derivative is given, so integrate once to obtain $\mathrm{f}'(x)$ (with an unknown constant), use the stationary point condition $\mathrm{f}'(3) = 0$ to find that constant, then integrate again to get $\mathrm{f}(x)$ and use the point $\left(3, \frac{64}{3}\right)$ to find the final constant.

Working

Integrate $\mathrm{f}''(x) = (2x+10)^{-\frac{1}{2}} - 2x$ once:

\mathrm{f}'(x) = m(2x+10)^{\frac{1}{2}} - nx^2 + c

Since $\displaystyle\int (2x+10)^{-\frac{1}{2}}\,\mathrm{d}x = \frac{(2x+10)^{\frac{1}{2}}}{\frac{1}{2} \times 2} = (2x+10)^{\frac{1}{2}}$ :

\mathrm{f}'(x) = (2x+10)^{\frac{1}{2}} - x^2 + c

At the stationary point $x = 3$ , so $\mathrm{f}'(3) = 0$ :

(16)^{\frac{1}{2}} - 9 + c = 0 \implies 4 - 9 + c = 0 \implies c = 5

So

\mathrm{f}'(x) = (2x+10)^{\frac{1}{2}} - x^2 + 5

Integrate again:

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

Since $\displaystyle\int (2x+10)^{\frac{1}{2}}\,\mathrm{d}x = \frac{(2x+10)^{\frac{3}{2}}}{\frac{3}{2} \times 2} = \frac{1}{3}(2x+10)^{\frac{3}{2}}$ :

\mathrm{f}(x) = \frac{1}{3}(2x+10)^{\frac{3}{2}} - \frac{x^3}{3} + 5x + d

Substitute $x = 3$ , $y = \frac{64}{3}$ :

\frac{64}{3} = \frac{1}{3}(16)^{\frac{3}{2}} - \frac{27}{3} + 15 + d

\frac{64}{3} = \frac{64}{3} - 9 + 15 + d \implies d = -6

Answer

y = \frac{1}{3}(2x+10)^{\frac{3}{2}} - \frac{x^3}{3} + 5x - 6

Final answer

y = (1/3)(2x+10)^(3/2) - x^3/3 + 5x - 6

Detailed explanation

Walkthrough

We are told how the curve bends ( $\mathrm{f}''(x)$ ) and one fact about its slope ( $\mathrm{f}'(3)=0$ , because $(3, \frac{64}{3})$ is stationary). Each integration introduces an unknown constant, and each piece of given information pins down exactly one constant.

First integration: integrating $(2x+10)^{-\frac{1}{2}}$ needs the reverse chain rule — divide by the derivative of the bracket, which is $2$ . Combined with the factor $\frac{1}{2}$ from the power rule this gives coefficient $1$ , so $\mathrm{f}'(x) = (2x+10)^{\frac{1}{2}} - x^2 + c$ .

Finding $c$ : a stationary point means the gradient is zero there. Substituting $x=3$ : $\sqrt{16} - 9 + c = 0$ , i.e. $4 - 9 + c = 0$ , giving $c = 5$ .

Second integration: integrate each term again — the bracket term becomes $\frac{1}{3}(2x+10)^{\frac{3}{2}}$ (divide by $\frac{3}{2} \times 2 = 3$ ), $-x^2$ becomes $-\frac{x^3}{3}$ , and $5$ becomes $5x$ , plus a new constant $d$ .

Finding $d$ : the curve passes through $\left(3, \frac{64}{3}\right)$ , so substitute $x=3$ : $\frac{1}{3}(16)^{\frac{3}{2}} = \frac{64}{3}$ and $-\frac{27}{3} + 15 = 6$ , giving $\frac{64}{3} = \frac{64}{3} + 6 + d$ , so $d = -6$ .

Key Takeaways

Integrating twice recovers the original function from the second derivative, introducing one constant per integration.
A stationary point gives the condition $\mathrm{f}'(x) = 0$ at that $x$ -value.
A known point on the curve fixes the remaining constant in $\mathrm{f}(x)$ .
Reverse chain rule: $\displaystyle\int (ax+b)^n\,\mathrm{d}x = \frac{(ax+b)^{n+1}}{a(n+1)} + k$ .

Common Mistakes

Forgetting the constant of integration after either integration — without $c$ you cannot use the stationary point, and without $d$ you cannot match the given point.
Dividing by $n+1$ but forgetting the inner derivative $a = 2$ when integrating $(2x+10)^{-\frac{1}{2}}$ (this would give $2(2x+10)^{\frac{1}{2}}$ instead of $(2x+10)^{\frac{1}{2}}$ ).
Using $y = \frac{64}{3}$ in $\mathrm{f}'(x)$ instead of $\mathrm{f}'(3) = 0$ — the stationary point fixes the first derivative, not the second.
Sign slips: $-2x$ integrates to $-x^2$ , not $+x^2$ .
Giving an expression rather than an equation — the mark scheme requires $y = \ldots$ for the final A mark.

Things to Be Careful About

The answer must be stated as an equation ( $y = \ldots$ or $\mathrm{f}(x) = \ldots$ ); an unsimplified equivalent form is accepted.
Both integrations must be shown as correct attempts in the required form for the M marks; the constants must be found by substitution, not guessed.
Keep exact values throughout: $16^{\frac{3}{2}} = 64$ exactly, so no decimals are needed anywhere.

Techniques used

integrate a function of a linear expression using the reverse chain ruleintegrate a power of x term by termuse the stationary point condition f'(3) = 0 to find the constant of integrationsubstitute the given point on the curve to find the second constant

Similar questions

Q118MMediumVectors in two dimensions

A triangle $OAB$ is such that $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ .
The point $X$ lies on $OA$ such that $\overrightarrow{OA} = 3\overrightarrow{OX}$ .
The point $Y$ lies on $XB$ such that $\overrightarrow{XB} = 5\overrightarrow{XY}$ .
The point $Z$ lies on $OB$ such that $\overrightarrow{OB} = m\overrightarrow{OZ}$ , where $m$ is a scalar.
$\overrightarrow{OX} = n\overrightarrow{ZY}$ , where $n$ is a scalar.

Use a vector method to find the values of $m$ and $n$ .

DifficultyMedium

Worked solution

Approach

Express every relevant vector in terms of $\mathbf{a}$ and $\mathbf{b}$ , form $\overrightarrow{ZY}$ as a route from $Z$ back through $O$ , $X$ and $Y$ , then impose $\overrightarrow{OX} = n\overrightarrow{ZY}$ and equate coefficients of $\mathbf{a}$ and $\mathbf{b}$ .

Working

Since $\overrightarrow{OA} = 3\overrightarrow{OX}$ :

\overrightarrow{OX} = \frac{1}{3}\mathbf{a}

Then

\overrightarrow{XB} = \overrightarrow{OB} - \overrightarrow{OX} = \mathbf{b} - \frac{1}{3}\mathbf{a}

and since $\overrightarrow{XB} = 5\overrightarrow{XY}$ :

\overrightarrow{XY} = \frac{1}{5}\left(\mathbf{b} - \frac{1}{3}\mathbf{a}\right)

Since $\overrightarrow{OB} = m\overrightarrow{OZ}$ :

\overrightarrow{OZ} = \frac{1}{m}\mathbf{b}

Now build $\overrightarrow{ZY}$ along the path $Z \to O \to X \to Y$ :

\begin{aligned} \overrightarrow{ZY} &= -\overrightarrow{OZ} + \overrightarrow{OX} + \overrightarrow{XY} \\ &= -\frac{1}{m}\mathbf{b} + \frac{1}{3}\mathbf{a} + \frac{1}{5}\left(\mathbf{b} - \frac{1}{3}\mathbf{a}\right) \\ &= \left(\frac{1}{3} - \frac{1}{15}\right)\mathbf{a} + \left(\frac{1}{5} - \frac{1}{m}\right)\mathbf{b} \\ &= \frac{4}{15}\mathbf{a} + \left(\frac{1}{5} - \frac{1}{m}\right)\mathbf{b} \end{aligned}

Using $\overrightarrow{OX} = n\overrightarrow{ZY}$ :

\frac{1}{3}\mathbf{a} = n\left[\frac{4}{15}\mathbf{a} + \left(\frac{1}{5} - \frac{1}{m}\right)\mathbf{b}\right]

Equating coefficients of $\mathbf{a}$ :

\frac{1}{3} = \frac{4n}{15} \quad\Rightarrow\quad n = \frac{15}{12} = \frac{5}{4}

Equating coefficients of $\mathbf{b}$ :

0 = n\left(\frac{1}{5} - \frac{1}{m}\right) \quad\Rightarrow\quad \frac{1}{5} = \frac{1}{m} \quad\Rightarrow\quad m = 5

Answer

m = 5, \qquad n = \frac{5}{4}

Final answer

m = 5, n = 5/4

Detailed explanation

Walkthrough

The whole question rests on writing every point's position vector using only the two base vectors $\mathbf{a}$ and $\mathbf{b}$ .

First, $X$ is one third of the way along $OA$ , so $\overrightarrow{OX} = \frac{1}{3}\mathbf{a}$ . Then $\overrightarrow{XB}$ is found by going from $O$ to $B$ and back to $X$ : $\mathbf{b} - \frac{1}{3}\mathbf{a}$ . Since $Y$ divides $XB$ in the ratio $1:4$ ( $\overrightarrow{XB} = 5\overrightarrow{XY}$ ), we get $\overrightarrow{XY} = \frac{1}{5}(\mathbf{b} - \frac{1}{3}\mathbf{a})$ .

For $Z$ , the condition $\overrightarrow{OB} = m\overrightarrow{OZ}$ means $OZ$ is a fraction $\frac{1}{m}$ of $OB$ , so $\overrightarrow{OZ} = \frac{1}{m}\mathbf{b}$ — note it is $\frac{1}{m}$ , not $m$ , that multiplies $\mathbf{b}$ .

To reach $Y$ from $Z$ without knowing anything else about $Z$ 's position relative to $Y$ , travel backwards through $O$ : $\overrightarrow{ZY} = -\overrightarrow{OZ} + \overrightarrow{OX} + \overrightarrow{XY}$ . Collecting the $\mathbf{a}$ terms gives $\frac{1}{3} - \frac{1}{15} = \frac{4}{15}$ , so $\overrightarrow{ZY} = \frac{4}{15}\mathbf{a} + (\frac{1}{5} - \frac{1}{m})\mathbf{b}$ .

Finally, $\overrightarrow{OX} = n\overrightarrow{ZY}$ says that $\frac{1}{3}\mathbf{a}$ equals $n$ times this whole expression. Because $\mathbf{a}$ and $\mathbf{b}$ are independent directions (they are sides of a triangle, not parallel), the coefficients on each side must match separately. The $\mathbf{a}$ equation gives $n = \frac{5}{4}$ ; the $\mathbf{b}$ equation forces $\frac{1}{5} - \frac{1}{m} = 0$ , i.e. $m = 5$ .

Geometrically this says $Z$ is one fifth of the way along $OB$ , and $ZY$ is parallel to $OX$ with $ZY = \frac{4}{5}OX$ .

Key Takeaways

Any point dividing a segment in a given ratio can be written directly in terms of the base vectors.
A vector between two unknown points can always be built from a route through known points (here $Z \to O \to X \to Y$ ).
If $p\mathbf{a} + q\mathbf{b} = r\mathbf{a} + s\mathbf{b}$ with $\mathbf{a}, \mathbf{b}$ non-parallel, then $p = r$ and $q = s$ — equating coefficients is the standard way to extract scalars.
The statement $\overrightarrow{OB} = m\overrightarrow{OZ}$ gives $\overrightarrow{OZ} = \frac{1}{m}\mathbf{b}$ , not $m\mathbf{b}$ .

Common Mistakes

Writing $\overrightarrow{OZ} = m\mathbf{b}$ instead of $\frac{1}{m}\mathbf{b}$ — the mark scheme explicitly rejects leaving $m\overrightarrow{OZ} = \mathbf{b}$ unless it is recovered later.
Sign errors when reversing direction: $\overrightarrow{ZO} = -\overrightarrow{OZ}$ , not $+\overrightarrow{OZ}$ .
Forgetting to expand $\frac{1}{5}(\mathbf{b} - \frac{1}{3}\mathbf{a})$ correctly — the $\mathbf{a}$ coefficient is $-\frac{1}{15}$ , giving $\frac{1}{3} - \frac{1}{15} = \frac{4}{15}$ .
Trying to divide by $\mathbf{a}$ or $\mathbf{b}$ instead of equating coefficients — vectors cannot be cancelled like numbers.
Dropping the factor $n$ when substituting into $\overrightarrow{OX} = n\overrightarrow{ZY}$ .

Things to Be Careful About

Both equations must be solved: the coefficient of $\mathbf{b}$ being zero is what determines $m$ , and it is easy to overlook because there is no $\mathbf{b}$ term on the left-hand side.
Answers must come from correct working — the mark scheme awards accuracy marks only for values reached by a valid method.
Keep all fractions exact throughout; there is no need for decimals at any stage.
Check the final answer makes sense geometrically: with $m = 5$ , $Z$ sits at $\frac{1}{5}\mathbf{b}$ , and indeed $\overrightarrow{ZY} = \frac{4}{15}\mathbf{a}$ , which is exactly $\frac{4}{5}$ of $\overrightarrow{OX} = \frac{1}{3}\mathbf{a}$ , consistent with $n = \frac{5}{4}$ .

Techniques used

express each position vector in terms of the base vectors a and bbuild ZY by adding and subtracting known vectors along the path O to Y via Xuse the scalar multiple relation OX = n·ZYequate coefficients of independent vectors a and b to form simultaneous equations

Similar questions

Q123MMedium-EasyPermutations and combinations

Find the value of $n$ such that $^{n+3}\mathrm{C}_{6} = 12 \times {}^{n+2}\mathrm{C}_{5}$ .

DifficultyMedium-Easy

Worked solution

Approach

Write each combination in factorial form, cancel the factorials common to both sides, and solve the simple linear equation that remains.

Working

Using $^{m}\mathrm{C}_{r} = \dfrac{m!}{(m-r)!\,r!}$ :

\frac{(n+3)!}{(n-3)!\,6!} = 12 \times \frac{(n+2)!}{(n-3)!\,5!}

The factor $(n-3)!$ cancels from both sides:

\frac{(n+3)!}{6!} = \frac{12\,(n+2)!}{5!}

Since $(n+3)! = (n+3)(n+2)!$ , divide both sides by $(n+2)!$ :

\frac{n+3}{6!} = \frac{12}{5!}

With $6! = 720$ and $5! = 120$ :

\frac{n+3}{720} = \frac{12}{120} = \frac{1}{10}

n + 3 = 72

n = 69

Answer

n = 69

Final answer

69

Detailed explanation

Walkthrough

The key tool is the definition of a combination written with factorials: choosing $r$ objects from $m$ gives $^{m}\mathrm{C}_{r} = \frac{m!}{(m-r)!\,r!}$ . Applying this to both sides turns the equation into

\frac{(n+3)!}{(n-3)!\,6!} = 12 \times \frac{(n+2)!}{(n-3)!\,5!}

Notice that $(n-3)!$ appears on both sides — it comes from the fact that $^{n+3}\mathrm{C}_{6}$ has $6$ factors below the top and $^{n+2}\mathrm{C}_{5}$ has $5$ , so both leave exactly the same leftover factorial. Cancelling it removes all the messy part of the expression.

Next, since $(n+3)! = (n+3)(n+2)!$ , we can divide both sides by $(n+2)!$ . What survives is remarkably clean:

\frac{n+3}{6!} = \frac{12}{5!}

Now substitute the numerical factorials, $6! = 720$ and $5! = 120$ , so the right-hand side is $\frac{12}{120} = \frac{1}{10}$ . Multiplying up gives $n + 3 = 72$ , hence $n = 69$ .

This cancellation pattern is worth remembering: for consecutive top values, $^{m}\mathrm{C}_{r} = ^{m-1}\mathrm{C}_{r-1} \times \frac{m}{r}$ , which here reads $^{n+3}\mathrm{C}_{6} = ^{n+2}\mathrm{C}_{5} \times \frac{n+3}{6}$ .

Key Takeaways

Convert binomial coefficients to factorials before comparing them algebraically.
Factorials of nearby numbers differ by a single multiplicative factor: $(k+1)! = (k+1)\cdot k!$ , so large factorials cancel almost completely.
The identity $^{m}\mathrm{C}_{r} = \frac{m}{r}\,{}^{m-1}\mathrm{C}_{r-1}$ shortcuts this whole type of question.

Common Mistakes

Writing the bottom factorial incorrectly, e.g. using $(n+3-6)! = (n-3)!$ on one side but $(n+2-5)! = (n-3)!$ wrongly on the other — they are actually equal here, but candidates often write $(n-2)!$ by miscounting.
Forgetting to divide out $(n+2)!$ and trying to expand huge factorials instead.
Mixing up $6!$ and $5!$ values ( $720$ and $120$ ), or computing $6!$ as $6 \times 3 = 18$ style slips.
Sign or arithmetic errors when finally solving $n + 3 = 72$ .
The mark scheme flags "nfww" (no follow-through from wrong working): an answer of $69$ reached from incorrect simplification scores nothing, so every cancellation step must be correct.

Things to Be Careful About

Both B marks for the simplification are dependent on the previous correct step, so the factorial form must be set up correctly before any cancelling earns credit.
Keep the exact fraction route: $\frac{12}{120} = \frac{1}{10}$ then $n+3 = 72$ ; avoid decimal approximations.
Check the final value makes sense: substituting $n = 69$ back, $^{72}\mathrm{C}_{6}$ should equal $12 \times {}^{71}\mathrm{C}_{5}$ , consistent with the ratio $\frac{72}{6} = 12$ .

Techniques used

write binomial coefficients in factorial formcancel common factorial factorssolve the resulting linear equation

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