(a) Write $5\lg a - 4\lg b - 3$ as a single base 10 logarithm. (b) Solve the equation $\log_5(x+1) - \log_{(x+1)} 5 = 0$.

(a) lg(a^5 / (1000 b^4)) · (b) x = 4 or x = -4/5

A curve has equation $y = \frac{\mathrm{e}^{(x^2)}}{x-2}$ for $x < 2$. (a) Find the value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ when $x = 0$. (b) When $x = 0$, $y$ is increasing at the rate of 0.5 units per second. Find the corresponding rate of change of $x$.

(a) -1/4 · (b) -2 units per second

4037/12 Oct/Nov 2025 — Additional Mathematics Solved Past Paper

Q1Medium-EasyEquations, inequalities and graphs

(a)

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

Fig. 1

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The cubic is already factorised, so the $x$ -intercepts come straight from the factors, and the $y$ -intercept from substituting $x = 0$ . Since the leading term is $-2x^3$ (negative cubic), the curve falls from top-left to bottom-right with a minimum between $x = 1$ and $x = 2.5$ and a maximum between $x = 2.5$ and $x = 5$ .

Working

$x$ -intercepts: set each factor to zero.

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

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

y = (1-0)(0-5)(0-5) = (1)(-5)(-5) = 25

Shape: the leading term is $-2x^3$ , so $y \to +\infty$ as $x \to -\infty$ and $y \to -\infty$ as $x \to +\infty$ . The curve crosses at $x = 1$ , dips to a distinct minimum in the fourth quadrant between $x = 1$ and $x = 2.5$ , rises to cross at $x = 2.5$ , reaches a distinct maximum in the first quadrant between $x = 2.5$ and $x = 5$ , then falls through $x = 5$ into the fourth quadrant.

Answer

A negative cubic passing through $(1, 0)$ , $(2.5, 0)$ , $(5, 0)$ and $(0, 25)$ , with a minimum in the fourth quadrant between $x = 1$ and $x = 2.5$ and a maximum in the first quadrant between $x = 2.5$ and $x = 5$ .

Final answer

Negative cubic crossing the x-axis at x = 1, 2.5 and 5, y-intercept 25, with a minimum between x = 1 and 2.5 and a maximum between x = 2.5 and 5

Detailed explanation

Walkthrough

The cubic is given in factorised form, which makes the intercepts immediate. Each factor $(1-x)$ , $(x-5)$ and $(2x-5)$ gives one root: $x = 1$ , $x = 5$ and $x = 2.5$ respectively. These are the three places the curve crosses the $x$ -axis.

For the $y$ -intercept we put $x = 0$ : the product is $(1)(-5)(-5) = 25$ , since the two negatives cancel. So the curve starts high on the $y$ -axis at $(0, 25)$ .

To decide the overall shape, look at the leading term. Expanding just the leading parts, $(1-x)(x-5)(2x-5)$ has leading term $(-x)(x)(2x) = -2x^3$ . A negative cubic comes down from the top-left and goes off to the bottom-right. Between consecutive roots the sign of $y$ alternates: positive for $x < 1$ , negative for $1 < x < 2.5$ (a dip below the axis, the minimum in the fourth quadrant), positive for $2.5 < x < 5$ (a hump above the axis, the maximum in the first quadrant), and negative for $x > 5$ .

Key Takeaways

A factorised cubic hands you its $x$ -intercepts directly from the factors.
The sign of the leading coefficient determines the end behaviour: negative cubic falls left-to-right overall.
The $y$ -intercept is found by substituting $x = 0$ into the factorised form — no expansion needed.

Common Mistakes

Missing that the cubic is negative (leading term $-2x^3$ ) and sketching a positive cubic instead — this loses the shape mark.
Taking $2x - 5 = 0$ as $x = 5$ instead of $x = 2.5$ .
Sign slip at the $y$ -intercept: $(1)(-5)(-5) = 25$ , not $-25$ .
Adding extra turns at the ends of the sketch — the mark scheme penalises "turning over at end points, suggesting further stationary points"; a cubic has at most two turning points.
Omitting the stated intercept values; the marks for $1$ , $2.5$ , $5$ and $25$ are independent and must be marked or stated.

Things to Be Careful About

The mark scheme awards separate B1 marks for the shape (with a distinct minimum in the fourth quadrant and a distinct maximum in the first quadrant), for the three $x$ -intercepts, and for the $y$ -intercept $25$ — all three are needed for full marks.
The turning points must be distinct and correctly placed; a flat or wrong-signed sketch scores zero for shape.
No scale is required, but the intercept values must be marked on the axes or stated with no contradiction and no extra critical values.

Techniques used

find x-intercepts by setting each factor to zerofind the y-intercept by substituting x = 0sketch a negative cubic with two turning points

(b)

Hence solve the inequality $(1-x)(x-5)(2x-5) \leq 0$ .

2M Similar

DifficultyEasy

Worked solution

Approach

"Hence" means use the sketch from part (a). The inequality $(1-x)(x-5)(2x-5) \leq 0$ holds wherever the curve is on or below the $x$ -axis.

Working

From the sketch: the curve is below the $x$ -axis for $1 < x < 2.5$ and for $x > 5$ , and equals zero at $x = 1$ , $x = 2.5$ and $x = 5$ . Since the inequality is $\leq$ , the endpoints are included.

1 \leq x \leq 2.5 \quad \text{or} \quad x \geq 5

Answer

1 \leq x \leq 2.5 \quad \text{or} \quad x \geq 5

Final answer

1 <= x <= 2.5 or x >= 5

Detailed explanation

Walkthrough

The word "hence" tells us to use the graph from part (a) rather than testing intervals algebraically. The product $(1-x)(x-5)(2x-5)$ is exactly the $y$ -value of the sketched curve, so the inequality asks: where is the curve at or below the $x$ -axis?

From the sketch, the curve dips below the axis between $x = 1$ and $x = 2.5$ , and again after $x = 5$ . Because the inequality is $\leq 0$ (not $<$ ), the points where $y = 0$ — the intercepts themselves — are included, giving the closed intervals $1 \leq x \leq 2.5$ and $x \geq 5$ .

Key Takeaways

A "hence" instruction requires using the previous result — here, the sketch.
$\leq$ includes the roots; $<$ would exclude them.
The solution set of a cubic inequality is generally two disjoint intervals, both of which must be stated.

Common Mistakes

Omitting one of the two intervals — both $1 \leq x \leq 2.5$ and $x \geq 5$ carry a separate B1 mark.
Using strict inequalities and dropping the endpoints $x = 1$ , $2.5$ , $5$ , which the $\leq$ sign requires.
Including the region $x \leq 1$ where the curve is above the axis (positive $y$ ).
Solving algebraically from scratch instead of using the sketch, which wastes time and risks sign errors.

Things to Be Careful About

The mark scheme marks each interval separately as the final answer, so both must appear.
Endpoints must be included because the inequality is $\leq$ ; writing $1 < x < 2.5$ or $x > 5$ loses the mark.

Techniques used

read intervals where the curve is on or below the x-axis from the sketchexpress the solution as a union of intervals

Q2MediumFactors of polynomialsQuadratic functions

The polynomial $\mathrm{p}$ is such that $\mathrm{p}(x) = 2x^3 + ax^2 + 13x + b$ , where $a$ and $b$ are integers.
It is given that $x+2$ is a factor of $\mathrm{p}(x)$ .
When $\mathrm{p}(x)$ is divided by $x+1$ there is a remainder of 6.

(a)

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

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $x+2$ is a factor of $\mathrm{p}(x)$ , the factor theorem gives $\mathrm{p}(-2)=0$ . Since division by $x+1$ leaves remainder $6$ , the remainder theorem gives $\mathrm{p}(-1)=6$ . These two equations are then solved simultaneously for $a$ and $b$ .

Working

Using the factor theorem with $x = -2$ :

\mathrm{p}(-2) = 2(-2)^3 + a(-2)^2 + 13(-2) + b = -16 + 4a - 26 + b = 0

4a + b = 42

Using the remainder theorem with $x = -1$ :

\mathrm{p}(-1) = 2(-1)^3 + a(-1)^2 + 13(-1) + b = -2 + a - 13 + b = 6

a + b = 21

Subtracting:

(4a + b) - (a + b) = 42 - 21 \implies 3a = 21 \implies a = 7

Then:

b = 21 - 7 = 14

Answer

a = 7, \quad b = 14

Final answer

a = 7, b = 14

Detailed explanation

Walkthrough

The factor theorem says that if $(x+2)$ is a factor of $\mathrm{p}(x)$ , then substituting $x=-2$ must give zero. Substituting into $\mathrm{p}(x) = 2x^3 + ax^2 + 13x + b$ gives $-16 + 4a - 26 + b = 0$ , which simplifies to $4a + b = 42$ . The remainder theorem says that when $\mathrm{p}(x)$ is divided by $(x+1)$ , the remainder equals $\mathrm{p}(-1)$ ; here that remainder is given as $6$ , so $-2 + a - 13 + b = 6$ , i.e. $a + b = 21$ . Subtracting the second equation from the first eliminates $b$ , giving $3a = 21$ , so $a = 7$ , and back-substitution gives $b = 14$ .

Key Takeaways

The factor theorem: $(x - k)$ is a factor of $\mathrm{p}(x)$ exactly when $\mathrm{p}(k) = 0$ .
The remainder theorem: dividing by $(x-k)$ leaves remainder $\mathrm{p}(k)$ .
Two unknown coefficients require two independent conditions, each converted into a linear equation.

Common Mistakes

Sign slips when substituting negative values such as $(-2)^3 = -8$ or $13(-2) = -26$ .
Confusing the two theorems — using $\mathrm{p}(2) = 0$ instead of $\mathrm{p}(-2) = 0$ for the factor $x+2$ .
Forgetting that the remainder equation equals $6$ , not $0$ .
Arithmetic errors in solving the simultaneous equations.

Things to Be Careful About

The mark scheme awards B1 for each correctly formed equation, so both substitutions must be shown explicitly before solving.
Keep exact integer values throughout; no rounding is involved.

Techniques used

apply the factor theorem to form an equation from the factor x+2apply the remainder theorem to form an equation from division by x+1solve a pair of simultaneous linear equations

(b)

Show that the equation $\mathrm{p}(x) = 0$ has only one real root.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the known factor $x+2$ to write $\mathrm{p}(x) = (x+2)(2x^2 + 3x + 7)$ , then show the quadratic factor has a negative discriminant, so it has no real roots and the only real root comes from $x + 2 = 0$ .

Working

With $a = 7$ and $b = 14$ :

\mathrm{p}(x) = 2x^3 + 7x^2 + 13x + 14

Dividing by the known factor $x + 2$ (or comparing coefficients):

\mathrm{p}(x) = (x+2)(2x^2 + 3x + 7)

Check: $(x+2)(2x^2+3x+7) = 2x^3 + 3x^2 + 7x + 4x^2 + 6x + 14 = 2x^3 + 7x^2 + 13x + 14$ . ✓

For the quadratic factor $2x^2 + 3x + 7$ , the discriminant is:

b^2 - 4ac = 3^2 - 4(2)(7) = 9 - 56 = -47 < 0

Since the discriminant is negative, $2x^2 + 3x + 7 = 0$ has no real solutions. Therefore $\mathrm{p}(x) = 0$ reduces to $x + 2 = 0$ , giving the single real root $x = -2$ .

Answer

The discriminant of the quadratic factor is $-47 < 0$ , so it has no real roots; hence $\mathrm{p}(x) = 0$ has only one real root, $x = -2$ .

Final answer

(x+2)(2x^2+3x+7) = 0; discriminant = -47 < 0, so the only real root is x = -2

Detailed explanation

Walkthrough

From part (a) we know $a = 7$ and $b = 14$ , and we already know $x + 2$ is a factor. Dividing the cubic by $x+2$ (by long division or by matching coefficients) gives the factorisation $\mathrm{p}(x) = (x+2)(2x^2 + 3x + 7)$ . A product equals zero only when one of its factors equals zero. The linear factor gives the real root $x = -2$ . For the quadratic factor we compute its discriminant: $b^2 - 4ac = 9 - 56 = -47$ . Because this is negative, the quadratic never crosses the $x$ -axis and contributes no real roots. Hence the whole cubic has exactly one real root.

Key Takeaways

A cubic with one known linear factor can be fully factorised by division or coefficient matching.
The sign of the discriminant determines whether a quadratic has real roots: negative means none.
Counting real roots of a polynomial can be done factor by factor.

Common Mistakes

Sign errors in the discriminant formula, e.g. computing $b^2 + 4ac$ or mis-substituting $a = 2$ , $b = 3$ , $c = 7$ .
Factorising incorrectly and obtaining a quadratic with a non-negative discriminant.
Stating "the discriminant is negative" without evaluating it — the mark scheme requires the correct value.
Concluding without linking the negative discriminant back to the number of real roots of the original equation.

Things to Be Careful About

This part depends on the values of $a$ and $b$ found in part (a); follow-through applies to your own quadratic factor, but it must have a negative discriminant for the argument to work.
The conclusion must be stated explicitly: negative discriminant means no real roots from the quadratic, leaving only $x = -2$ .

Techniques used

factorise the cubic using the known linear factorcompute the discriminant of the quadratic factorargue from a negative discriminant that there is only one real root

Q3MediumSeries

The sum of the first two terms of a geometric progression is 9.
The sum to infinity of this geometric progression is 25.

(a)

Find the possible values of the common ratio of this geometric progression.

4M Similar

DifficultyMedium

Worked solution

Approach

Let the first term be $a$ and the common ratio be $r$ . Write the sum of the first two terms and the sum to infinity as two equations, then divide one by the other to eliminate $a$ and solve for $r$ .

Working

Sum of the first two terms:

a + ar = 9

Sum to infinity (valid since $|r| < 1$ ):

\frac{a}{1 - r} = 25

Factorising the first equation and dividing by the second:

\frac{a(1 + r)}{\dfrac{a}{1 - r}} = \frac{9}{25}

(1 + r)(1 - r) = \frac{9}{25}

1 - r^2 = \frac{9}{25}

r^2 = \frac{16}{25}

r = \pm \frac{4}{5}

Answer

r = \frac{4}{5} \quad \text{or} \quad r = -\frac{4}{5}

Final answer

r = 4/5 or r = -4/5

Detailed explanation

Walkthrough

A geometric progression with first term $a$ and common ratio $r$ has terms $a, ar, ar^2, \ldots$ . The sum of the first two terms is $a + ar = a(1 + r)$ , which the question tells us equals $9$ . The sum to infinity exists only when $|r| < 1$ , and equals $\frac{a}{1 - r}$ , which equals $25$ .

Rather than solving for $a$ first, the efficient route is to divide the two equations: the $a$ cancels, leaving an equation in $r$ alone. Dividing $a(1 + r) = 9$ by $\frac{a}{1 - r} = 25$ gives $(1 + r)(1 - r) = \frac{9}{25}$ , and the left side is the difference of two squares, $1 - r^2$ . Rearranging gives $r^2 = \frac{16}{25}$ , so $r = \pm \frac{4}{5}$ . Both roots are valid since both satisfy $|r| < 1$ , so both must be reported.

Key Takeaways

The sum to infinity formula $S_\infty = \frac{a}{1 - r}$ requires $|r| < 1$ .
Dividing two equations sharing an unknown is a quick way to eliminate it.
$(1 + r)(1 - r) = 1 - r^2$ is the difference of two squares.
A squared equation generally yields two solutions; both must be checked and given.

Common Mistakes

Forgetting the negative root and giving only $r = \frac{4}{5}$ (the mark scheme explicitly awards the final B1 for $\pm \frac{4}{5}$ ).
Using the wrong sum formula, e.g. $\frac{a}{r - 1}$ with the wrong sign, giving $r^2$ with the wrong value.
Not checking that both roots satisfy $|r| < 1$ (here they both do).
Cancelling $a$ incorrectly when dividing the fractions.

Things to Be Careful About

The mark scheme is dependent: the $r^2 = \frac{16}{25}$ mark requires both earlier equations to be correct, and the final mark requires all previous marks — so both starting equations must be right.
Both values of $r$ are required; giving only one loses the final mark.
Keep exact fractions: $\pm \frac{4}{5}$ or $\pm 0.8$ are both accepted, but the working should show the exact equation $r^2 = \frac{16}{25}$ .

Techniques used

write the sum of the first two terms of a GPwrite the sum to infinity formuladivide the equations to eliminate asolve for r and take both square roots

(b)

Find the first term of this geometric progression for each possible value of the common ratio.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute each value of $r$ from part (a) into the sum-to-infinity equation $\frac{a}{1 - r} = 25$ and solve for $a$ .

Working

For $r = \frac{4}{5}$ :

\frac{a}{1 - \frac{4}{5}} = 25 \implies \frac{a}{\frac{1}{5}} = 25 \implies 5a = 25 \implies a = 5

For $r = -\frac{4}{5}$ :

\frac{a}{1 + \frac{4}{5}} = 25 \implies \frac{a}{\frac{9}{5}} = 25 \implies \frac{5a}{9} = 25 \implies a = 45

Answer

a = 5 \quad \text{or} \quad a = 45

Final answer

a = 5 or a = 45

Detailed explanation

Walkthrough

Carrying forward the sum-to-infinity equation $\frac{a}{1 - r} = 25$ from part (a), each value of $r$ gives a different first term.

When $r = \frac{4}{5}$ , the denominator is $1 - \frac{4}{5} = \frac{1}{5}$ , so $\frac{a}{1/5} = 5a = 25$ , giving $a = 5$ .

When $r = -\frac{4}{5}$ , the denominator is $1 - \left(-\frac{4}{5}\right) = \frac{9}{5}$ , so $\frac{a}{9/5} = \frac{5a}{9} = 25$ , giving $a = 45$ .

A quick check: with $a = 5$ , $r = \frac{4}{5}$ , the first two terms sum to $5 + 4 = 9$ ✓. With $a = 45$ , $r = -\frac{4}{5}$ , they sum to $45 - 36 = 9$ ✓.

Key Takeaways

Each admissible value of the common ratio pairs with exactly one first term for fixed $S_\infty$ and fixed two-term sum.
Dividing by a fraction means multiplying by its reciprocal: $\frac{a}{1/5} = 5a$ .

Common Mistakes

Sign error in $1 - r$ when $r$ is negative: $1 - \left(-\frac{4}{5}\right) = \frac{9}{5}$ , not $\frac{1}{5}$ .
Pairing the wrong $a$ with the wrong $r$ ; the check $a + ar = 9$ catches this instantly.
Dividing by $\frac{9}{5}$ incorrectly instead of multiplying by $\frac{5}{9}$ .

Things to Be Careful About

The mark scheme awards one B1 for each correct value, so both $a = 5$ and $a = 45$ must be given, matched to the two values of $r$ from part (a).
Keep the arithmetic exact: $\frac{5a}{9} = 25$ gives $a = 45$ exactly, no decimals needed.

Techniques used

substitute each value of r into the sum-to-infinity equationsolve for the first term a

Q4MediumQuadratic functionsFunctions

(a)

Show that $2x^2 + 5x + 3$ can be written in the form $2(x+a)^2 + b$ , where $a$ and $b$ are constants to be found.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Factor out the leading coefficient $2$ from the quadratic and linear terms, complete the square inside the bracket, and simplify the constant term to write the expression in the form $2(x+a)^2 + b$ .

Working

\begin{aligned} 2x^2 + 5x + 3 &= 2\left(x^2 + \frac{5}{2}x\right) + 3 \\ &= 2\left[\left(x + \frac{5}{4}\right)^2 - \left(\frac{5}{4}\right)^2\right] + 3 \\ &= 2\left(x + \frac{5}{4}\right)^2 - 2\left(\frac{25}{16}\right) + 3 \\ &= 2\left(x + \frac{5}{4}\right)^2 - \frac{25}{8} + \frac{24}{8} \\ &= 2\left(x + \frac{5}{4}\right)^2 - \frac{1}{8} \end{aligned}

Answer

2\left(x + \frac{5}{4}\right)^2 - \frac{1}{8}

Final answer

2(x + 5/4)^2 - 1/8

Detailed explanation

Walkthrough

Factor out $2$ from the first two terms: $2x^2 + 5x + 3 = 2\left(x^2 + \frac{5}{2}x\right) + 3$ .
To complete the square for $x^2 + \frac{5}{2}x$ , halve the coefficient of $x$ , giving $\frac{5}{4}$ .
Write the bracket as $\left(x + \frac{5}{4}\right)^2 - \left(\frac{5}{4}\right)^2 = \left(x + \frac{5}{4}\right)^2 - \frac{25}{16}$ .
Distribute the factor of $2$ : $2\left(x + \frac{5}{4}\right)^2 - \frac{25}{8}$ .
Combine the constants: $-\frac{25}{8} + 3 = -\frac{25}{8} + \frac{24}{8} = -\frac{1}{8}$ .

This gives $a = \frac{5}{4}$ and $b = -\frac{1}{8}$ .

Key Takeaways

When completing the square for $Ax^2 + Bx + C$ , always factor out $A$ from the variable terms first: $A\left(x^2 + \frac{B}{A}x\right) + C$ .
Halve the middle term coefficient inside: $\left(x + \frac{B}{2A}\right)^2 - \left(\frac{B}{2A}\right)^2$ .

Common Mistakes

Forgetting to multiply the squared term $-\left(\frac{5}{4}\right)^2$ by $2$ when expanding the outer brackets.
Halving $\frac{5}{2}$ incorrectly (e.g. getting $\frac{5}{2}$ instead of $\frac{5}{4}$ ).

Things to Be Careful About

Ensure exact fractions are retained throughout since this is on a non-calculator paper.

Techniques used

factor out the coefficient of x^2complete the square inside the bracketexpand and simplify constant terms

(b)

Hence write down the coordinates of the stationary point on the curve $y = 2x^2 + 5x + 3$ .

2M Similar

(c)

A function $\mathrm{f}$ is such that $\mathrm{f}(x) = 2x^2 + 5x + 3$ , for $x \geq p$ , where $p$ is a constant.
It is given that $\mathrm{f}^{-1}$ exists.

6M

(i)

Write down the least possible value of $p$ .

1M Similar

(ii)

Using your value of $p$ , sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$ .
Label each graph.
State the intercepts of each of the graphs with the axes.

Fig. 2

5M Similar

DifficultyMedium

Worked solution

Approach

Find the intercepts of $y = \mathrm{f}(x)$ for $x \geq -\frac{5}{4}$ . Reflect the curve and its intercepts across the line $y = x$ to obtain the graph of $y = \mathrm{f}^{-1}(x)$ .

Working

For $y = \mathrm{f}(x) = 2x^2 + 5x + 3$ $y = f (x) = 2 x^{2} + 5 x + 3$ with $x \geq -\frac{5}{4}$ $x \geq - \frac{5}{4}$ :
- $y$ -intercept (set $x = 0$ ):

\mathrm{f}(0) = 3 \implies (0, 3)

$x$ -intercepts (set $y = 0$ ):

2x^2 + 5x + 3 = 0 \implies (2x + 3)(x + 1) = 0 \implies x = -\frac{3}{2} \text{ or } x = -1

Since the domain is $x \geq -\frac{5}{4} = -1.25$ , $x = -\frac{3}{2} = -1.5$ is outside the domain.
Thus, the only $x$ -intercept is $(-1, 0)$ .

For $y = \mathrm{f}^{-1}(x)$ $y = f^{- 1} (x)$ , the graph is the reflection of $y = \mathrm{f}(x)$ $y = f (x)$ in the line $y = x$ $y = x$ :
- $x$ -intercept: $(3, 0)$
- $y$ -intercept: $(0, -1)$

Answer

Intercepts for $y = \mathrm{f}(x)$ :

$x$ -intercept at $-1$ (or $(-1, 0)$ )
$y$ -intercept at $3$ (or $(0, 3)$ )

Intercepts for $y = \mathrm{f}^{-1}(x)$ :

$x$ -intercept at $3$ (or $(3, 0)$ )
$y$ -intercept at $-1$ (or $(0, -1)$ )

Final answer

y = f(x) has x-intercept -1 and y-intercept 3; y = f^-1(x) has x-intercept 3 and y-intercept -1

Detailed explanation

Walkthrough

Calculate the intercepts of $\mathrm{f}(x)$ $f (x)$ :
- When $x = 0$ , $y = 3$ , giving the $y$ -intercept $(0, 3)$ .
- When $y = 0$ , $2x^2 + 5x + 3 = 0 \implies (2x+3)(x+1) = 0$ , so $x = -\frac{3}{2}$ or $x = -1$ . Because the domain is restricted to $x \geq -\frac{5}{4}$ , the root $x = -\frac{3}{2} = -1.5$ is excluded, leaving only the $x$ -intercept $(-1, 0)$ .
The graph of $y = \mathrm{f}(x)$ starts at the vertex $\left(-\frac{5}{4}, -\frac{1}{8}\right)$ , passes through $(-1, 0)$ and $(0, 3)$ , and curves upwards in the first quadrant.
The inverse function $y = \mathrm{f}^{-1}(x)$ $y = f^{- 1} (x)$ is the reflection of $y = \mathrm{f}(x)$ $y = f (x)$ in the line $y = x$ $y = x$ . Swapping the coordinates of the intercepts gives:
- $x$ -intercept at $(3, 0)$
- $y$ -intercept at $(0, -1)$
- Starting point at $\left(-\frac{1}{8}, -\frac{5}{4}\right)$ .
Sketch both curves on the axes, labelling $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$ , showing their correct curvatures and clearly marking the intercepts.

Key Takeaways

The graph of $y = \mathrm{f}^{-1}(x)$ is always the reflection of $y = \mathrm{f}(x)$ across the line $y = x$ .
Reflecting across $y = x$ swaps the coordinates of every point, so an intercept $(0, c)$ becomes $(c, 0)$ and $(d, 0)$ becomes $(0, d)$ .
When the domain of $\mathrm{f}$ is restricted, make sure to check which intercepts lie inside the domain.

Common Mistakes

Including both roots $x = -1$ and $x = -1.5$ for $\mathrm{f}(x)$ even though $x = -1.5$ lies outside $x \geq -\frac{5}{4}$ .
Drawing a full parabola instead of only the restricted branch $x \geq -\frac{5}{4}$ .
Drawing $y = \mathrm{f}^{-1}(x)$ with incorrect curvature (not reflecting properly across $y = x$ ).

Things to Be Careful About

Clearly label each graph ( $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$ ).
Explicitly state the numerical values of the intercepts on the axes.

Techniques used

find axis intercepts of a functionfind axis intercepts of an inverse function by swapping coordinatessketch a restricted quadratic curve and its reflection in y = x

Q5MediumLogarithmic and exponential functions

(a)

Write $5\lg a - 4\lg b - 3$ as a single base 10 logarithm.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the power rule to absorb the coefficients, write the constant 3 as $\lg 1000$ , then combine using the subtraction rule.

Working

Apply the power rule to each term:

5\lg a - 4\lg b - 3 = \lg a^5 - \lg b^4 - \lg 1000

since $3 = \lg 10^3 = \lg 1000$ . Combine using the subtraction rule:

\lg a^5 - \lg b^4 - \lg 1000 = \lg\left(\frac{a^5}{b^4 \times 1000}\right)

Answer

\lg\left(\frac{a^5}{1000b^4}\right)

Final answer

lg(a^5 / (1000 b^4))

Detailed explanation

Walkthrough

The expression $5\lg a - 4\lg b - 3$ mixes multiples of logarithms with a plain number, so the goal is to make everything a logarithm first. The power rule $n\lg x = \lg x^n$ absorbs the coefficients 5 and 4, giving $\lg a^5 - \lg b^4$ . The constant 3 is not yet a logarithm, but since $\lg 10 = 1$ , we have $3 = \lg 10^3 = \lg 1000$ . Now all three terms are logarithms, and the subtraction rule $\lg x - \lg y = \lg\left(\frac{x}{y}\right)$ combines them in one step: subtracting $\lg b^4$ divides by $b^4$ , and subtracting $\lg 1000$ divides by 1000.

Key Takeaways

The power rule moves a coefficient inside the logarithm as an exponent.
Any integer $n$ can be written as $\lg 10^n$ , which is the key step for absorbing constants.
Subtraction of logarithms corresponds to division of the arguments.

Common Mistakes

Writing $\lg(a^5 - b^4 - 1000)$ : the laws combine products and quotients, never sums or differences inside one logarithm.
Forgetting to convert the constant 3 into $\lg 1000$ , leaving a stray term outside the logarithm.
Sign slips: $-4\lg b$ becomes $b^{-4}$ or division by $b^4$ , not $b^4$ in the numerator.

Things to Be Careful About

The answer must be a single base-10 logarithm; any equivalent form such as $\lg(10^{-3}a^5b^{-4})$ is accepted.
The mark scheme awards a B1 for recognising $3 = \lg 1000$ and a B1 for using the power rule, so both steps must be visible in the working.

Techniques used

apply the power rule to move coefficients into the logarithmsapply the subtraction rule to combine logarithmsrewrite the constant as a logarithm of a power of 10

(b)

Solve the equation $\log_5(x+1) - \log_{(x+1)} 5 = 0$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Change the base of $\log_{(x+1)} 5$ to base 5, giving a reciprocal. Setting the expression to zero produces a quadratic in $\log_5(x+1)$ , which yields two values of $x$ .

Working

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

\log_5(x+1) - \frac{1}{\log_5(x+1)} = 0

Multiplying through by $\log_5(x+1)$ :

\left(\log_5(x+1)\right)^2 = 1 \quad\Rightarrow\quad \log_5(x+1) = \pm 1

Case 1: $\log_5(x+1) = 1$ gives $x + 1 = 5$ , so:

x = 4

Case 2: $\log_5(x+1) = -1$ gives $x + 1 = 5^{-1} = \frac{1}{5}$ , so:

x = -\frac{4}{5}

Both values give $x + 1 > 0$ and $x + 1 \neq 1$ , so both are valid.

Answer

x = 4 \quad \text{or} \quad x = -\frac{4}{5}

Final answer

x = 4 or x = -4/5

Detailed explanation

Walkthrough

The two logarithms have different bases — base 5 and base $(x+1)$ — so they cannot be combined directly. The change-of-base rule converts the second logarithm into base 5: $\log_{(x+1)} 5 = \frac{\log_5 5}{\log_5(x+1)} = \frac{1}{\log_5(x+1)}$ . The equation is now $y - \frac{1}{y} = 0$ where $y = \log_5(x+1)$ . Multiplying by $y$ gives $y^2 = 1$ , so $y = \pm 1$ — the negative root must not be discarded. Converting back: $\log_5(x+1) = 1$ means $x+1 = 5$ , giving $x = 4$ ; $\log_5(x+1) = -1$ means $x+1 = 5^{-1} = \frac{1}{5}$ , giving $x = -\frac{4}{5}$ . Both are checked: each makes $x+1$ positive and not equal to 1, so both logarithms are defined.

Key Takeaways

Change of base turns a logarithm with a variable base into a reciprocal of a fixed-base logarithm.
An equation of the form $y - \frac{1}{y} = 0$ is a disguised quadratic, $y^2 = 1$ .
$\log_a b = -1$ means $b = a^{-1}$ , which is where the fractional answer comes from.

Common Mistakes

Rejecting or crossing out $x = -\frac{4}{5}$ : the mark scheme explicitly gives A0 if it is rejected — it is a valid solution.
Forgetting the $\pm$ when taking the square root of $y^2 = 1$ and finding only $x = 4$ .
Not changing the base at all and trying to subtract logarithms of different bases directly.
Failing to check that $x + 1 \neq 1$ (a base cannot be 1); here both roots pass, but the check should be made.

Things to Be Careful About

Both solutions must be given; the mark scheme awards one mark each and penalises rejecting the negative root.
The change-of-base step earns the B1, so it must be shown explicitly in the working.
The domain requirements $x + 1 > 0$ and $x + 1 \neq 1$ should be verified for each root found.

Techniques used

change the base of a logarithmform a quadratic in a logarithmic expressionsolve the quadratic and convert back to xcheck both solutions are valid in the original equation

Q65MMediumEquations, inequalities and graphs

Solve the equation $|2x^2 + x - 10| = 5$ .

DifficultyMedium

Worked solution

Approach

The equation $\left|2x^2 + x - 10\right| = 5$ splits into two cases: the expression inside the modulus equals $5$ , or it equals $-5$ . Solve each quadratic and keep every root.

Working

Case 1: $2x^2 + x - 10 = 5$

2x^2 + x - 15 = 0

Factorise:

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

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

Case 2: $2x^2 + x - 10 = -5$

2x^2 + x - 5 = 0

Using the quadratic formula with $a = 2$ , $b = 1$ , $c = -5$ :

x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-5)}}{2(2)} = \frac{-1 \pm \sqrt{41}}{4}

Answer

x = \frac{5}{2}, \quad x = -3, \quad x = \frac{-1 + \sqrt{41}}{4}, \quad x = \frac{-1 - \sqrt{41}}{4}

Final answer

x = 5/2, x = -3, x = (-1 + √41)/4, x = (-1 - √41)/4

Detailed explanation

Walkthrough

The modulus $\left|A\right|$ equals $5$ exactly when $A = 5$ or $A = -5$ , so the single equation becomes two ordinary quadratic equations.

For the first case, $2x^2 + x - 10 = 5$ rearranges to $2x^2 + x - 15 = 0$ . This factorises as $(2x - 5)(x + 3) = 0$ , giving $x = \frac{5}{2}$ and $x = -3$ . Both must be kept — the mark scheme explicitly penalises rejecting $-3$ just because it is negative; there is no reason to reject it since substituting back gives $|18 - 3 - 10| = |5| = 5$ .

For the second case, $2x^2 + x - 10 = -5$ rearranges to $2x^2 + x - 5 = 0$ . Its discriminant is $1 + 40 = 41$ , which is not a perfect square, so the quadratic formula gives the exact surd answers $x = \frac{-1 \pm \sqrt{41}}{4}$ .

All four values satisfy the original equation, so all four are solutions.

Key Takeaways

An equation of the form $|f(x)| = k$ (with $k > 0$ ) always splits into $f(x) = k$ and $f(x) = -k$ .
Every root from either branch is a genuine solution — never discard negative or irrational roots without checking them against the original equation.
When factorisation fails (non-square discriminant), use the quadratic formula and leave the answer in exact surd form.

Common Mistakes

Solving only one branch of the modulus and reporting just two roots instead of four.
Rejecting $x = -3$ because it is negative — the mark scheme awards A0 for this.
Writing the second case's answer with a negative denominator, e.g. $\frac{1 \mp \sqrt{41}}{-4}$ — the scheme requires the denominator to be positive.
Sign slips forming the rearranged quadratics: forgetting that moving $-10$ across changes its sign, or miscomputing $b^2 - 4ac$ as $1 - 40$ instead of $1 + 40$ .
Rounding the surd answers to decimals when an exact form is expected.

Things to Be Careful About

Both quadratics must be correctly written as three-term equations equal to zero before solving (this is what the M marks check).
The final answers should be given in exact form: $\frac{5}{2}$ , $-3$ and $\frac{-1 \pm \sqrt{41}}{4}$ .
Check each root against the original modulus equation if unsure — here all four genuinely work.

Techniques used

split the modulus equation into positive and negative casessolve each resulting quadratic equationkeep all valid roots including negative ones

Similar questions

Q7MediumStraight-line graphsCoordinate geometry of the circle

The point $A$ has coordinates $(-2, 4)$ .
The point $B$ has coordinates $(6, 10)$ .
The point $C$ has coordinates $(12, 2)$ .

(a)

Find the gradients of the lines $AB$ , $AC$ and $BC$ .

2M Similar

DifficultyEasy

Worked solution

Approach

Use the gradient formula $\dfrac{y_2 - y_1}{x_2 - x_1}$ for each pair of points.

Working

\text{Grad } AB = \frac{10 - 4}{6 - (-2)} = \frac{6}{8}

\text{Grad } AC = \frac{2 - 4}{12 - (-2)} = -\frac{2}{14}

\text{Grad } BC = \frac{2 - 10}{12 - 6} = -\frac{8}{6}

Answer

\text{Grad } AB = \frac{6}{8}, \quad \text{Grad } AC = -\frac{2}{14}, \quad \text{Grad } BC = -\frac{8}{6}

Final answer

Gradient AB = 6/8, gradient AC = -2/14, gradient BC = -8/6

Detailed explanation

Walkthrough

For each pair of points, the gradient is the change in $y$ divided by the change in $x$ . For $AB$ : from $A(-2,4)$ to $B(6,10)$ , the rise is $10 - 4 = 6$ and the run is $6 - (-2) = 8$ , giving $\frac{6}{8}$ . For $AC$ : the rise is $2 - 4 = -2$ and the run is $12 - (-2) = 14$ , giving $-\frac{2}{14}$ . For $BC$ : the rise is $2 - 10 = -8$ and the run is $12 - 6 = 6$ , giving $-\frac{8}{6}$ . The mark scheme accepts these unsimplified forms, so there is no need to cancel.

Key Takeaways

The gradient formula $\frac{y_2 - y_1}{x_2 - x_1}$ works for any ordered pair of points.
Subtract coordinates in a consistent order (second point minus first point).
Watch the double negative when a coordinate is negative, as with $6 - (-2)$ .

Common Mistakes

Subtracting coordinates in opposite orders for the numerator and denominator, flipping the sign.
Mishandling $6 - (-2)$ and writing $4$ instead of $8$ .
Simplifying unnecessarily and risking arithmetic slips — the unsimplified fractions earn full marks.

Things to Be Careful About

The mark scheme pays B1 for two correct unsimplified gradients and B1 for all three, so accuracy on all three pairs matters.
Keep the fractions exactly as formed; equivalent simplified forms are accepted (oe), but unsimplified is safest.

Techniques used

apply the gradient formula between two pointscompute three gradients from coordinates

(b)

Hence find the equation of the circle which passes through the points $A$ , $B$ and $C$ .

4M Similar

DifficultyMedium

Worked solution

Approach

From part (a), $\text{Grad } AB \times \text{Grad } BC = \frac{6}{8} \times \left(-\frac{8}{6}\right) = -1$ , so $AB \perp BC$ and angle $ABC = 90°$ . By the angle in a semicircle, $AC$ is a diameter of the circle. The centre is the midpoint of $AC$ , and the radius follows from the distance from the centre to $A$ .

Working

Midpoint of $AC$ :

\left(\frac{-2 + 12}{2}, \frac{4 + 2}{2}\right) = (5, 3)

Square of the radius, using centre $(5,3)$ and point $A(-2,4)$ :

r^2 = (5 - (-2))^2 + (3 - 4)^2 = 7^2 + (-1)^2 = 49 + 1 = 50

Equation of the circle:

(x - 5)^2 + (y - 3)^2 = 50

Answer

(x - 5)^2 + (y - 3)^2 = 50

Final answer

(x - 5)^2 + (y - 3)^2 = 50

Detailed explanation

Walkthrough

The word "Hence" tells us to use the gradients from part (a). Multiplying the gradients of $AB$ and $BC$ gives $\frac{6}{8} \times \left(-\frac{8}{6}\right) = -1$ , so the lines $AB$ and $BC$ are perpendicular: the angle at $B$ is a right angle. A standard circle theorem says the angle in a semicircle is a right angle, so the right angle at $B$ must stand on a diameter — meaning $AC$ is the diameter of the circle through $A$ , $B$ and $C$ .

The centre of a circle is the midpoint of any diameter, so the centre is the midpoint of $AC$ : averaging the coordinates of $A(-2,4)$ and $C(12,2)$ gives $\left(\frac{-2+12}{2}, \frac{4+2}{2}\right) = (5,3)$ .

The radius is the distance from the centre to any point on the circle, for instance $A$ . Using the squared-distance form directly (the equation needs $r^2$ , so no square root is needed): $r^2 = (5-(-2))^2 + (3-4)^2 = 49 + 1 = 50$ .

Substituting centre $(5,3)$ and $r^2 = 50$ into $(x-a)^2 + (y-b)^2 = r^2$ gives the equation.

Key Takeaways

If two gradients multiply to $-1$ , the lines are perpendicular.
A right angle inscribed in a circle subtends a diameter (angle in a semicircle theorem) — this is how three points determine the circle here.
The centre of the circle through three points with a right angle is the midpoint of the hypotenuse-side (the diameter).
The circle equation $(x-a)^2 + (y-b)^2 = r^2$ needs the centre and the square of the radius; working with $r^2$ avoids a square root entirely.

Common Mistakes

Not using the part (a) result and instead solving simultaneous equations for the centre — slower and error-prone.
Taking the wrong pair as the diameter (e.g. halving $AB$ or $BC$ ); the mark scheme only allows centres from the correct diameter, or $(2,7)$ / $(9,6)$ by a different (incorrect) route — the correct centre is $(5,3)$ .
Forgetting to square the coordinate differences, or subtracting in the wrong order and introducing sign errors (squared, so order does not matter if done consistently).
Writing $r = 50$ instead of $r^2 = 50$ in the equation.
Expanding the brackets instead of leaving the completed-square form — the standard form is expected.

Things to Be Careful About

The M1 requires using the centre together with one of $A$ , $B$ or $C$ to get $r^2$ (or halving a full side length); the centre must not be one of $A$ , $B$ , $C$ .
The A1 accepts $(5\sqrt{2})^2$ in place of $50$ for $r^2$ .
Leave the answer in the form $(x-5)^2 + (y-3)^2 = 50$ ; expanding to $x^2 + y^2 - 10x - 6y - 16 = 0$ is an equivalent form (oe) but the completed-square form is safest.
Since $r^2 = 50$ is exact, no decimal rounding is involved.

Techniques used

recognise perpendicular gradients so AC is a diameterfind the midpoint of the diameter to get the centreuse the distance formula to find the radius squaredwrite the circle equation in completed-square form

Q84MMedium-EasyLogarithmic and exponential functions

On the axes, sketch the graph of $y = 5\ln(4x+3)$ .
State the intercepts with the axes.
State the equation of any asymptote.

Fig. 3

DifficultyMedium-Easy

Worked solution

Approach

To sketch the graph of $y = 5\ln(4x+3)$ :

Find the vertical asymptote by setting the argument of the logarithm to zero: $4x + 3 = 0$ .
Find the $x$ -intercept by setting $y = 0$ .
Find the $y$ -intercept by setting $x = 0$ .
Sketch a strictly increasing logarithmic curve in the third, second, and first quadrants that approaches the vertical asymptote as $x \to -\frac{3}{4}^{+}$ and passes through the calculated intercepts.

Working

Vertical asymptote:

4x + 3 = 0 \implies x = -\frac{3}{4}

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

5\ln(4x+3) = 0

\ln(4x+3) = 0

4x + 3 = \mathrm{e}^0 = 1

4x = -2 \implies x = -\frac{1}{2} = -0.5

So the $x$ -intercept is $\left(-\frac{1}{2}, 0\right)$ or $x = -0.5$ .

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

y = 5\ln(4(0) + 3) = 5\ln 3

So the $y$ -intercept is $(0, 5\ln 3)$ or $y = 5\ln 3$ .

Answer

Asymptote: $x = -\frac{3}{4}$ (or $x = -0.75$ )
$x$ -intercept: $x = -\frac{1}{2}$ (or $x = -0.5$ )
$y$ -intercept: $y = 5\ln 3$
Graph: An increasing logarithmic curve starting from near the vertical line $x = -\frac{3}{4}$ in the third quadrant, crossing the $x$ -axis at $x = -0.5$ , crossing the $y$ -axis at $y = 5\ln 3$ , and extending into the first quadrant.

Final answer

Asymptote: x = -3/4; x-intercept: x = -0.5; y-intercept: y = 5 ln 3; Sketch: increasing curve through (-0.5, 0) and (0, 5 ln 3) with asymptote x = -0.75

Detailed explanation

Walkthrough

Vertical Asymptote:
The natural logarithm function $\ln u$ is only defined for $u > 0$ . As $u \to 0^{+}$ , $\ln u \to -\infty$ . Therefore, the vertical asymptote occurs where the argument is zero:
$4x + 3 = 0 \implies x = -\frac{3}{4}$
On the sketch, draw a dashed vertical line at $x = -\frac{3}{4}$ (in the second and third quadrants).
$x$ -intercept:
Setting $y = 0$ gives $5\ln(4x+3) = 0$ , so $\ln(4x+3) = 0$ . Using the exponential definition, $4x+3 = \mathrm{e}^0 = 1$ , which leads to $4x = -2$ , so $x = -0.5$ . The curve crosses the negative $x$ -axis to the right of the vertical asymptote ( $-0.5 > -0.75$ ).
$y$ -intercept:
Setting $x = 0$ gives $y = 5\ln(3)$ . Since $\ln 3 > 1$ , $5\ln 3 > 5 > 0$ , so the curve crosses the positive $y$ -axis.
Drawing the Curve:

Draw a smooth, concave-down, strictly increasing curve that approaches the vertical asymptote $x = -\frac{3}{4}$ downwards into the third quadrant, passes through $(-0.5, 0)$ on the $x$ -axis, passes through $(0, 5\ln 3)$ on the $y$ -axis, and continues growing at a decreasing rate into the first quadrant.

Key Takeaways

For any function of the form $y = a\ln(bx + c)$ , the vertical asymptote is always found at $bx + c = 0$ .
The domain is $x > -\frac{c}{b}$ when $b > 0$ .
Always leave values in exact logarithmic form (e.g., $5\ln 3$ ) when working on non-calculator papers.

Common Mistakes

Confusing the asymptote equation with a horizontal line (e.g. writing $y = -\frac{3}{4}$ instead of $x = -\frac{3}{4}$ ).
Forgetting that $\ln(1) = 0$ and setting $4x + 3 = 0$ to find the $x$ -intercept instead of the asymptote.
Drawing the curve crossing the asymptote or curling backwards.
Drawing the asymptote on the wrong side of the $y$ -axis.

Things to Be Careful About

Ensure the relative positions of the asymptote ( $x = -0.75$ ) and the $x$ -intercept ( $x = -0.5$ ) are clear and correct (the intercept must lie to the right of the asymptote).
State the equation of the asymptote explicitly as $x = -\frac{3}{4}$ and clearly mark or state both intercepts.

Techniques used

find the vertical asymptote of a logarithmic functioncalculate axis interceptssketch a logarithmic curve with asymptote and intercepts

Similar questions

Q9MediumCalculus

A curve $y = \mathrm{f}(x)$ is such that $\mathrm{f}''(x) = (2x+5)^{-\frac{3}{2}}$ .

The curve has gradient $\frac{2}{3}$ at the point $(2, 2)$ .

(a)

Find the coordinates of the stationary point on the curve.

11M Similar

DifficultyMedium

Worked solution

Approach

Integrate $\mathrm{f}''(x)$ once to get $\mathrm{f}'(x)$ , using the gradient condition at $x = 2$ to find the constant $c$ . Integrate again to get $\mathrm{f}(x)$ , using the point $(2, 2)$ to find the constant $d$ . Then solve $\mathrm{f}'(x) = 0$ for the stationary point and substitute into $\mathrm{f}(x)$ for its $y$ -coordinate.

Working

Integrating $\mathrm{f}''(x) = (2x+5)^{-\frac{3}{2}}$ with respect to $x$ :

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

Use the gradient condition $\mathrm{f}'(2) = \frac{2}{3}$ :

-(2(2)+5)^{-\frac{1}{2}} + c = \frac{2}{3} \implies -9^{-\frac{1}{2}} + c = \frac{2}{3} \implies -\frac{1}{3} + c = \frac{2}{3}

c = 1 \quad \text{so} \quad \mathrm{f}'(x) = -(2x+5)^{-\frac{1}{2}} + 1

Integrating again:

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

Use the point $(2, 2)$ :

2 - (2(2)+5)^{\frac{1}{2}} + d = 2 \implies 2 - 3 + d = 2 \implies d = 3

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

Stationary point where $\mathrm{f}'(x) = 0$ :

-(2x+5)^{-\frac{1}{2}} + 1 = 0 \implies (2x+5)^{-\frac{1}{2}} = 1 \implies 2x + 5 = 1 \implies x = -2

Then

y = \mathrm{f}(-2) = -2 - (2(-2)+5)^{\frac{1}{2}} + 3 = -2 - 1 + 3 = 0

Answer

(-2,\ 0)

Final answer

(-2, 0)

Detailed explanation

Walkthrough

We are given the second derivative $\mathrm{f}''(x) = (2x+5)^{-\frac{3}{2}}$ and two conditions: the gradient is $\frac{2}{3}$ at the point $(2, 2)$ . To recover the curve we integrate twice, and each integration introduces a constant that the conditions allow us to pin down.

First integration: the general rule for integrating $(ax+b)^n$ is to divide by $a(n+1)$ , raising the index by one. Here $n = -\frac{3}{2}$ , so the new index is $-\frac{1}{2}$ and we divide by $2 \times (-\frac{1}{2}) = -1$ , giving $-(2x+5)^{-\frac{1}{2}} + c$ .

To find $c$ , substitute $x = 2$ and set the derivative equal to $\frac{2}{3}$ : since $(2(2)+5)^{-\frac{1}{2}} = 9^{-\frac{1}{2}} = \frac{1}{3}$ , we get $c = 1$ . So $\mathrm{f}'(x) = -(2x+5)^{-\frac{1}{2}} + 1$ .

Second integration: integrating $-(2x+5)^{-\frac{1}{2}}$ raises the index to $\frac{1}{2}$ and divides by $2 \times \frac{1}{2} = 1$ , giving $-(2x+5)^{\frac{1}{2}}$ ; integrating the constant $1$ gives $x$ . So $\mathrm{f}(x) = x - (2x+5)^{\frac{1}{2}} + d$ .

To find $d$ , substitute the point $(2, 2)$ : $2 - 3 + d = 2$ , so $d = 3$ .

A stationary point occurs where the gradient is zero. Setting $\mathrm{f}'(x) = 0$ gives $(2x+5)^{-\frac{1}{2}} = 1$ , and raising both sides to the power $-2$ (or simply inverting) gives $2x + 5 = 1$ , so $x = -2$ . Substituting into $\mathrm{f}(x)$ : $y = -2 - 1 + 3 = 0$ .

Key Takeaways

Integrating a second derivative twice recovers the original function, with one constant of integration at each stage.
Each given condition (a gradient value, a point on the curve) determines exactly one constant.
Stationary points are found by solving $\mathrm{f}'(x) = 0$ ; the $y$ -coordinate comes from substituting into $\mathrm{f}(x)$ , not $\mathrm{f}'(x)$ .

Common Mistakes

Forgetting the chain-rule factor when integrating $(2x+5)^{-\frac{3}{2}}$ — you must divide by $2$ (the coefficient of $x$ ) as well as by the new index; the scheme's M1 is specifically for the correct $k(2x+5)^{-\frac{1}{2}}$ form.
Forgetting the constants of integration, or using the wrong condition for each one: the gradient condition fixes $c$ in $\mathrm{f}'(x)$ , the point fixes $d$ in $\mathrm{f}(x)$ .
Sign errors: $\mathrm{f}'(x) = -(2x+5)^{-\frac{1}{2}} + 1$ has a negative leading term; dropping the minus sign gives $c = -1$ and everything downstream fails.
Solving $(2x+5)^{-\frac{1}{2}} = 1$ incorrectly — the answer is $2x+5 = 1$ , not $2x+5 = 1^{-2}$ misapplied with a sign slip.
Substituting $x = -2$ into $\mathrm{f}'(x)$ instead of $\mathrm{f}(x)$ to get the $y$ -coordinate.

Things to Be Careful About

The mark scheme requires each integration step in the correct form ( $k(2x+5)^{-\frac{1}{2}}$ then $p(2x+5)^{\frac{1}{2}}$ ) with a visible attempt at substitution for each constant — show both substitutions in full.
The scheme's Dep M1 notes mean the constant-finding marks depend on the earlier integration being in the right form.
Note that $x = -2 > -\frac{5}{2}$ , so the point lies in the domain where $(2x+5)^{-\frac{3}{2}}$ is defined — the stationary point is valid.
Give the answer as a coordinate pair $(-2, 0)$ ; the $y$ -value alone is not the full answer.

Techniques used

integrate a function of a linear expression with a negative fractional indexdetermine the constant of integration from a given gradientintegrate again and determine the second constant from a given pointsolve the equation f'(x) = 0 to find the stationary pointevaluate the y-coordinate on the curve

(b)

Determine the nature of this stationary point.

1M Similar

DifficultyEasy

Worked solution

Approach

Substitute $x = -2$ into the given second derivative and interpret the sign.

Working

\mathrm{f}''(-2) = (2(-2)+5)^{-\frac{3}{2}} = 1^{-\frac{3}{2}} = 1 > 0

Since the second derivative is positive at $x = -2$ , the curve is concave upwards there, so the stationary point is a minimum.

Answer

The stationary point $(-2, 0)$ is a minimum point.

Final answer

Minimum point (since f''(-2) = 1 > 0)

Detailed explanation

Walkthrough

The second derivative $\mathrm{f}''(x) = (2x+5)^{-\frac{3}{2}}$ was given in the question stem, so no differentiation work is needed here. Substituting the stationary point's $x$ -value, $x = -2$ , gives $\mathrm{f}''(-2) = (1)^{-\frac{3}{2}} = 1$ . A positive second derivative means the gradient is increasing through the point, so the curve has a local minimum there. (Equivalently, the first-derivative test could be used with $x$ -values just greater than $-\frac{5}{2}$ , where $\mathrm{f}'$ changes from negative to positive.)

Key Takeaways

The second-derivative test: $\mathrm{f}'' > 0$ at a stationary point means a minimum, $\mathrm{f}'' < 0$ means a maximum.
When the second derivative is given in the question, evaluating it at the stationary point is the quickest nature test.

Common Mistakes

Stating "minimum" without showing the substitution — the mark scheme explicitly requires evidence of substitution (B1).
Using an incorrectly obtained $x = -2$ from part (a) scores B0 even if the test is otherwise right.
Misinterpreting the sign: positive second derivative means minimum (concave up), not maximum.

Things to Be Careful About

The mark scheme requires visible substitution of $x = -2$ into either the second-derivative test or the first-derivative test — the bare word "minimum" does not earn the mark.
The value $1^{-\frac{3}{2}} = 1$ must be evaluated correctly; a negative-base slip would be impossible here since $2(-2)+5 = 1 > 0$ .
This conclusion is consistent with part (a): the point $(-2, 0)$ is a minimum of the curve $y = x - \sqrt{2x+5} + 3$ .

Techniques used

evaluate the second derivative at the stationary pointapply the second-derivative test for a minimum

Q105MMedium-EasyTrigonometryCalculus

Show that

\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left( (\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 \right) \mathrm{d}\theta = k\pi

where $k$ is an integer to be found.

DifficultyMedium-Easy

Worked solution

Approach

Expand both squares; the cross terms $+2\sin\theta\cos\theta$ and $-2\sin\theta\cos\theta$ cancel, leaving $2(\sin^2\theta + \cos^2\theta) = 2$ . Then integrate the constant between the given limits.

Working

(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta

(\sin\theta - \cos\theta)^2 = \sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta

Adding, the cross terms cancel:

(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2\sin^2\theta + 2\cos^2\theta = 2(\sin^2\theta + \cos^2\theta) = 2

So the integral becomes

\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2\,\mathrm{d}\theta = \left[2\theta\right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}

= 2 \times \frac{3\pi}{2} - 2 \times \frac{\pi}{2} = 3\pi - \pi = 2\pi

Hence $k = 2$ , which is an integer as required.

Answer

k = 2

Final answer

k = 2

Detailed explanation

Walkthrough

The integrand looks complicated, but expanding each square reveals massive cancellation. The first square gives $\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$ and the second gives $\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta$ . When added, the cross terms cancel exactly, and the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ collapses everything to the constant $2$ . Integrating a constant is trivial: $\int 2\,\mathrm{d}\theta = 2\theta$ . Substituting the limits in radians gives $3\pi - \pi = 2\pi$ , so comparing with $k\pi$ shows $k = 2$ .

Key Takeaways

Expanding $(a+b)^2 + (a-b)^2$ always gives $2a^2 + 2b^2$ — the cross terms cancel.
The identity $\sin^2\theta + \cos^2\theta = 1$ can reduce a trigonometric integrand to a constant before any integration happens.
Definite integration of a constant: multiply by the difference of the limits.

Common Mistakes

Forgetting to double when adding the two expansions, or mishandling signs on the cross term $-2\sin\theta\cos\theta$ .
Not recognising that the result must be simplified using $\sin^2\theta + \cos^2\theta = 1$ — leaving the answer as an unsimplified trig expression loses the accuracy mark.
Arithmetic slips with limits such as writing $3\pi/2 - \pi/2 = \pi$ instead of evaluating $2\theta$ at each limit first ( $3\pi - \pi$ ).
Giving a decimal value for the integral instead of keeping it as an exact multiple of $\pi$ .

Things to Be Careful About

The mark scheme awards M1 for expanding both terms and attempting simplification via the correct identity or eliminating the $2\sin\theta\cos\theta$ terms, A1 for the antiderivative $2\theta$ , a dependent M1 for correct use of limits, and A1 for $2\pi$ — every one of these steps must appear in your working.
Work entirely in radians here since the limits are multiples of $\pi$ .
The final answer must be left in the form $k\pi$ with $k$ identified as the integer $2$ ; do not convert to a decimal.

Techniques used

expand both squared bracketscancel the cross terms using sin^2 + cos^2 = 1integrate a constantapply exact limits in radians

Similar questions

Q113MMediumPermutations and combinations

Solve the equation $(n-4)\,{}^{n+1}\mathrm{C}_{5} = {}^{n+2}\mathrm{C}_{7}$ .

DifficultyMedium

Worked solution

Approach

Write each binomial coefficient in factorial form, cancel the common factors $(n-5)!$ , simplify $\frac{7!}{5!}$ numerically, then solve the resulting linear equation.

Working

{}^{n+1}\mathrm{C}_{5} = \frac{(n+1)!}{(n-4)!,\,5!}, \qquad {}^{n+2}\mathrm{C}_{7} = \frac{(n+2)!}{(n-5)!,\,7!}

So the equation becomes

\frac{(n-4)(n+1)!}{(n-4)!\,5!} = \frac{(n+2)!}{(n-5)!\,7!}

Since $(n-4)(n-5)! = (n-4)!$ , the left side simplifies:

\frac{(n+1)!}{(n-5)!\,5!} = \frac{(n+2)!}{(n-5)!\,7!}

Multiplying both sides by $(n-5)!$ :

\frac{(n+1)!}{5!} = \frac{(n+2)!}{7!}

Rearranging:

\frac{7!}{5!} = \frac{(n+2)!}{(n+1)!}

The numerical side is $7 \times 6 = 42$ , and the algebraic side is $n + 2$ :

42 = n + 2

n = 40

Answer

n = 40

Final answer

n = 40

Detailed explanation

Walkthrough

The equation involves two binomial coefficients with unknown $n$ in their top entries. The only practical route is to expand each coefficient into its factorial definition.

First, ${}^{n+1}\mathrm{C}_{5} = \frac{(n+1)!}{(n-4)!\,5!}$ because $r!$ in the denominator uses $(n+1)-5 = n-4$ . Similarly ${}^{n+2}\mathrm{C}_{7} = \frac{(n+2)!}{(n-5)!\,7!}$ since $(n+2)-7 = n-5$ .

Substituting these gives the printed first line. The key cancellation follows from noticing that $(n-4)! = (n-4)(n-5)!$ , so multiplying the left side by $(n-4)$ cancels it entirely, leaving $(n+1)!$ over $(n-5)!\,5!$ on the left. Now both sides share the factor $(n-5)!$ , which can be removed by multiplying through.

What remains is $\frac{(n+1)!}{5!} = \frac{(n+2)!}{7!}$ . Cross-multiplying gives $\frac{7!}{5!} = \frac{(n+2)!}{(n+1)!}$ . The right side collapses to $n+2$ because consecutive factorials differ by one factor; the left side is $7 \times 6 = 42$ because $7! = 7 \times 6 \times 5!$ . Hence $n + 2 = 42$ , so $n = 40$ .

Key Takeaways

Every binomial coefficient can be written as a ratio of factorials, which is often the only way to compare two coefficients algebraically.
Factorials of nearby numbers telescope: $(k+1)! = (k+1)\cdot k!$ , so ratios like $\frac{(n+2)!}{(n+1)!}$ are just $n+2$ .
Ratios like $\frac{7!}{5!}$ reduce to a short product rather than evaluating large factorials.

Common Mistakes

Writing the bottom entry of the coefficient incorrectly: for ${}^{n+1}\mathrm{C}_{5}$ the denominator needs $(n-4)!$ , not $(n-5)!$ — the mark scheme's B1 marks specifically reward correct simplification of both the numerical factorials ( $7 \times 6$ or $42$ ) and the algebraic ones ( $n+2$ ).
Failing to spot that $(n-4)(n-5)! = (n-4)!$ , leaving an unsolvable-looking mess instead of cancelling.
Trying to evaluate $7!$ and $5!$ as huge numbers instead of forming the simple ratio $42$ .
Arithmetic slips such as writing $\frac{7!}{5!} = 7 \times 6 \times 5 = 210$ or concluding $n = 42$ without subtracting the $2$ .

Things to Be Careful About

Both B1 marks here are awarded for intermediate simplifications ( $42$ and $n+2$ ), so those steps must be visible in your working — jumping straight to $n = 40$ risks losing marks.
Keep the factorial definitions exact: ${}^{m}\mathrm{C}_{r} = \frac{m!}{(m-r)!\,r!}$ , and double-check $m - r$ for each coefficient separately.
The final answer must be $n = 40$ exactly (cao); check it is consistent, e.g. $36 \times {}^{41}\mathrm{C}_{5} = {}^{42}\mathrm{C}_{7}$ holds.

Techniques used

write binomial coefficients in factorial formcancel common factorial factorssimplify the numerical ratio of factorialssolve the resulting linear equation

Similar questions

Q12MediumCalculus

A curve has equation $y = \frac{\mathrm{e}^{(x^2)}}{x-2}$ for $x < 2$ .

(a)

Find the value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ when $x = 0$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $y = \dfrac{\mathrm{e}^{x^2}}{x-2}$ using the quotient rule, remembering that the derivative of $\mathrm{e}^{x^2}$ is $2x\,\mathrm{e}^{x^2}$ by the chain rule. Then substitute $x = 0$ .

Working

By the quotient rule:

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x-2)\cdot 2x\,\mathrm{e}^{x^2} - \mathrm{e}^{x^2}\cdot 1}{(x-2)^2}

At $x = 0$ , noting that $\mathrm{e}^{0} = 1$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(0-2)(2\times 0)\mathrm{e}^{0} - \mathrm{e}^{0}}{(0-2)^2} = \frac{0 - 1}{4}

\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{4}

Answer

-\frac{1}{4}

Final answer

-1/4

Detailed explanation

Walkthrough

The curve is a quotient: an exponential on top and a linear expression underneath. The right tool is the quotient rule, which needs the derivatives of both parts.

The top is $\mathrm{e}^{x^2}$ . This is a composite function — the exponential of $x^2$ — so the chain rule gives its derivative as $2x\,\mathrm{e}^{x^2}$ : differentiate the outer exponential (leaving it unchanged) and multiply by the derivative of the inner $x^2$ .

The bottom, $x - 2$ , has derivative $1$ .

Applying the quotient rule gives

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

Substituting $x = 0$ : the first term in the numerator vanishes because of the factor $2x = 0$ , leaving $-\mathrm{e}^{0} = -1$ ; the denominator is $(0-2)^2 = 4$ . So the gradient at $x = 0$ is $-\frac{1}{4}$ .

Key Takeaways

The quotient rule: $\left(\frac{u}{v}\right)' = \frac{vu' - uv'}{v^2}$ .
The chain rule for exponentials: $\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{\mathrm{f}(x)} = \mathrm{f}'(x)\,\mathrm{e}^{\mathrm{f}(x)}$ .
Evaluating a derivative at a point after differentiating in general form.

Common Mistakes

Forgetting the chain-rule factor $2x$ when differentiating $\mathrm{e}^{x^2}$ (the mark scheme awards B1 specifically for $2x\,\mathrm{e}^{x^2}$ ).
Reversing the order in the quotient rule numerator ( $uv' - vu'$ instead of $vu' - uv'$ ), giving a sign error.
Writing $(x-2)^2$ incorrectly as $x^2 - 4$ or similar when evaluating at $x = 0$ .
Dropping the minus sign: the answer must be negative, and the mark scheme says "Must be from correct working".

Things to Be Careful About

The mark scheme accepts either the single-fraction quotient-rule form or the product-rule form $2x\,\mathrm{e}^{x^2}(x-2)^{-1} - \mathrm{e}^{x^2}(x-2)^{-2}$ ; both earn full method credit.
The final A1 requires the value to come from correct working — a correct-looking $-\frac{1}{4}$ from wrong differentiation scores nothing.
Keep the exact fraction $-\frac{1}{4}$ ; do not round.

Techniques used

differentiate an exponential of a function using the chain ruleapply the quotient ruleevaluate the derivative at a given point

(b)

When $x = 0$ , $y$ is increasing at the rate of 0.5 units per second.

Find the corresponding rate of change of $x$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the chain rule for rates of change, $\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{\mathrm{d}x}{\mathrm{d}t}$ , with $\dfrac{\mathrm{d}y}{\mathrm{d}t} = 0.5$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{4}$ from part (a).

Working

\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}

Substituting the known values:

0.5 = -\frac{1}{4} \times \frac{\mathrm{d}x}{\mathrm{d}t}

Solving for $\dfrac{\mathrm{d}x}{\mathrm{d}t}$ :

\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{0.5}{-\frac{1}{4}} = -2

Answer

-2 \text{ units per second}

Final answer

-2 units per second

Detailed explanation

Walkthrough

This part connects the rate of change of $y$ with respect to time to the rate of change of $x$ with respect to time. The bridge between them is the chain rule written in rate form:

\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}.

We are told $y$ increases at $0.5$ units per second, so $\frac{\mathrm{d}y}{\mathrm{d}t} = 0.5$ . From part (a), the gradient at $x = 0$ is $\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{4}$ . Substituting and dividing through gives $\frac{\mathrm{d}x}{\mathrm{d}t} = 0.5 \div (-\frac{1}{4}) = -2$ .

The negative sign makes sense: since the curve is decreasing at this point ( $y$ falls as $x$ rises), $x$ must be decreasing over time while $y$ increases.

Key Takeaways

Connected rates of change via $\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}$ .
Interpreting the sign of a rate: a negative $\frac{\mathrm{d}x}{\mathrm{d}t}$ means $x$ is decreasing.

Common Mistakes

Inverting the relation and computing $\frac{\mathrm{d}y}{\mathrm{d}x} \div \frac{\mathrm{d}y}{\mathrm{d}t}$ instead of solving correctly.
Losing the negative sign and answering $+2$ .
Using a wrong value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ from part (a); the M1 allows follow-through on the candidate's own value, but the final A1 requires the correct $-\frac{1}{4}$ from correct working in part (a).

Things to Be Careful About

The mark scheme explicitly follows through on the candidate's part (a) value for the method mark, but the accuracy mark requires the correct derivative from correct working — so part (a) must be right for full marks here.
State the unit: the rate is in units per second, matching the wording of the question.

Techniques used

apply the chain rule linking rates of changesolve for the required rate

Q134MMediumTrigonometry

For $-1 \leq x \leq 1$ it is given that $x = 2 - 3\sin\theta$ and $y = 3\cot^2 \theta$ .

Find $y$ in terms of $x$ .

DifficultyMedium

Worked solution

Approach

Eliminate $\theta$ between the two parametric equations. Express $\sin\theta$ in terms of $x$ , then use trigonometric identities to write $y = 3\cot^2\theta$ entirely in terms of $x$ .

Working

From $x = 2 - 3\sin\theta$ :

\sin\theta = \frac{2 - x}{3}

Write $y$ using $\cot\theta = \frac{\cos\theta}{\sin\theta}$ :

y = 3\left(\frac{\cos^2\theta}{\sin^2\theta}\right)

Using $\cos^2\theta = 1 - \sin^2\theta$ and substituting $\sin\theta = \frac{2-x}{3}$ :

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

Dividing numerator and denominator by $\left(\frac{2-x}{3}\right)^2$ :

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

Answer

y = 3\left(\frac{9}{(2-x)^2} - 1\right)

Final answer

y = 3(9/(2-x)^2 - 1)

Detailed explanation

Walkthrough

The question gives $x$ and $y$ each in terms of a parameter $\theta$ , and asks for $y$ directly in terms of $x$ . The strategy is to eliminate $\theta$ : solve one equation for a trig function of $\theta$ , then rewrite the other equation so it only involves that same trig function.

First, rearrange $x = 2 - 3\sin\theta$ : subtract 2 and divide by $-3$ to get $\sin\theta = \frac{2-x}{3}$ .

Next, express $y = 3\cot^2\theta$ using $\cot\theta = \frac{\cos\theta}{\sin\theta}$ , so $y = 3\frac{\cos^2\theta}{\sin^2\theta}$ .

Now use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to replace $\cos^2\theta$ with $1 - \sin^2\theta$ , giving everything in terms of $\sin\theta$ alone:

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

Finally simplify the compound fraction: split the big fraction into $\frac{1}{\left(\frac{2-x}{3}\right)^2} - 1$ , and since dividing by $\frac{2-x}{3}$ is multiplying by $\frac{3}{2-x}$ , this becomes $\left(\frac{3}{2-x}\right)^2 = \frac{9}{(2-x)^2}$ . Hence

y = 3\left(\frac{9}{(2-x)^2} - 1\right).

Key Takeaways

To eliminate a parameter, isolate a trig function of the parameter from one equation and substitute into the other.
The identity $\sin^2\theta + \cos^2\theta = 1$ converts any mixture of $\sin\theta$ and $\cos\theta$ into a single variable.
Compound fractions should be simplified by splitting them or by multiplying top and bottom by a common factor — the final answer must not contain fractions within fractions.

Common Mistakes

Sign errors when rearranging $x = 2 - 3\sin\theta$ ; check that $\sin\theta = \frac{2-x}{3}$ , not $\frac{x-2}{3}$ .
Forgetting to square when substituting into $\cot^2\theta$ — the whole of $\frac{2-x}{3}$ must be squared.
Leaving the answer as a fraction within a fraction; the mark scheme explicitly requires no nested fractions.
Dropping the factor of 3 outside the bracket, or expanding incorrectly at the end.
The final mark is dependent on all earlier marks being correct, so an early slip loses everything downstream.

Things to Be Careful About

All four marks are B marks with dependencies: the last two are only awarded if the earlier steps are correct, so every line must be shown clearly.
The answer may be given in any equivalent form (oe), e.g. $y = \frac{27}{(2-x)^2} - 3$ , but must be fully simplified with no fractions inside fractions.
Keep $(2-x)^2$ intact rather than writing $(x-2)^2$ inconsistently — both are equal, but consistency avoids sign slips.

Techniques used

rearrange a parametric equation to express sin theta in terms of xrewrite cot squared as cos squared over sin squaredapply the identity sin squared plus cos squared equals onesimplify the compound fraction to a single expression

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