(a) Find the complete expansion of $\left(2x - \frac{3}{x}\right)^4$. (b) Hence determine the coefficient of $x^2$ in the expansion of $(x^2 + 5)\left(2x - \frac{3}{x}\right)^4$.

(a) 16x^4 - 96x^2 + 216 - 216x^-2 + 81x^-4 · (b) -264

(a) Show that $3\tan^2\theta + 5\sin^2\theta \equiv \frac{8\sin^2\theta - 5\sin^4\theta}{1 - \sin^2\theta}$. (b) Hence solve the equation $3\tan^2\theta + 5\sin^2\theta = 9$ for $0^\circ < \theta < 270^\circ$.

(a) 3tan^2θ + 5sin^2θ = (8sin^2θ - 5sin^4θ)/(1 - sin^2θ) · (b) 54.1°, 125.9°, 234.1°

A geometric progression is such that its second term is $-120$ and its sum to infinity is 160. (a) Find the common ratio. (b) The first nine terms of the progression are now removed. Find the sum to infinity of the remaining terms of the progression.

(a) r = -1/2 · (b) -5/16 (or -0.3125)

9709/12 Feb/Mar 2025 — Mathematics Solved Past Paper

Q14MMediumQuadratics

A curve has equation $y = 5 + 3x - 2x^2$ and a straight line has equation $y = kx + 13$ , where $k$ is a constant.

Find the set of values of $k$ for which the curve and the line do not meet.

DifficultyMedium

Worked solution

Approach

The curve and the line meet when their $y$ -values are equal. Setting the two equations equal and rearranging gives a quadratic equation in $x$ ; the number of intersection points is determined by the discriminant of this quadratic. For the curve and line not to meet, the quadratic must have no real roots, so its discriminant must be negative.

Working

Set the two expressions for $y$ equal:

kx + 13 = 5 + 3x - 2x^2

Rearrange to form a quadratic in $x$ :

2x^2 + (k-3)x + 8 = 0

Here $a = 2$ , $b = k - 3$ and $c = 8$ . For no intersection, the discriminant must be negative:

b^2 - 4ac < 0

(k-3)^2 - 4(2)(8) < 0

(k-3)^2 - 64 < 0

(k-3)^2 < 64

Taking square roots gives:

-8 < k - 3 < 8

Adding 3 throughout:

-5 < k < 11

Answer

-5 < k < 11

Final answer

-5 < k < 11

Detailed explanation

Walkthrough

We are given a curve and a family of straight lines $y = kx + 13$ with the same intercept but varying gradient. We need the values of $k$ for which the line does not intersect the curve.

First, eliminate $y$ by equating the two expressions. This is the standard way to find intersection points: any common point must satisfy both equations, so its $x$ -coordinate satisfies the resulting quadratic. Rearranging gives $2x^2 + (k-3)x + 8 = 0$ .

Next, count real solutions of this quadratic. A quadratic $ax^2 + bx + c = 0$ has real roots when $b^2 - 4ac \geq 0$ and no real roots when $b^2 - 4ac < 0$ . Since 'do not meet' means no common point, we need no real $x$ -solutions, so we require the discriminant to be negative.

Substitute $a = 2$ , $b = k-3$ , $c = 8$ into the discriminant:
$(k-3)^2 - 64 < 0$ , i.e. $(k-3)^2 < 64$ . This inequality means the distance of $k-3$ from 0 is less than 8, so $-8 < k-3 < 8$ . Adding 3 gives $-5 < k < 11$ .

Key Takeaways

Intersections of a line and a curve are found by solving their equations simultaneously; eliminating $y$ gives an equation in $x$ .
The discriminant of a quadratic determines how many real intersections occur: positive means two, zero means one (tangent), negative means none.
To translate 'do not meet' into algebra, use discriminant $< 0$ (not $\leq 0$ ).

Common Mistakes

Using $b^2 - 4ac > 0$ instead of $< 0$ . That would give the values for which the line meets the curve in two points.
Using $\leq 0$ : when the discriminant is zero, the line is tangent to the curve and they do meet (at exactly one point), so the endpoints must be excluded.
Forgetting to rearrange the quadratic into the form $ax^2 + bx + c = 0$ before identifying $a,b,c$ .
Sign errors when moving terms: $5 + 3x - 2x^2$ moved to the left gives $2x^2 + (k-3)x + 8 = 0$ .

Things to Be Careful About

The coefficient of $x$ is $k-3$ , not $k+3$ .
The constant term is $8$ , obtained from $13 - 5$ .
The final answer must be a strict inequality, $-5 < k < 11$ . Do not write 'or', and do not use $\le$ signs.
If the mark scheme says 'condone $\pm$ errors', it means small sign slips in the quadratic may still earn method marks, but the final interval must be correct.

Techniques used

eliminate y to form a quadratic equationapply the discriminant condition for no real rootssolve a quadratic inequalityexpress the solution as an interval

Approach

At a minimum point $M$ , the gradient is zero. Set $\frac{dy}{dx} = 0$ and solve for $x$ . Then substitute this $x$ -value back into the original equation to find the corresponding $y$ -coordinate.

Working

Set $\frac{dy}{dx} = 0$ :

4x + \frac{5}{x^2} = 0

Multiply by $x^2$ (noting $x \neq 0$ ):

4x^3 + 5 = 0

4x^3 = -5

x^3 = -1.25

x = \sqrt[3]{-1.25} \approx -1.07722

Correct to 3 significant figures, $x = -1.08$ .

Substitute $x = -1.07722$ into $y = 2x^2 - \frac{5}{x} + 3$ :

y = 2(-1.07722)^2 - \frac{5}{-1.07722} + 3

y = 2(1.16040) + 4.64156 + 3

y = 2.32080 + 4.64156 + 3 = 9.96236

Correct to 3 significant figures, $y = 9.96$ .

Coordinates of $M$ are $(-1.08, 9.96)$ .

Answer

(-1.08, 9.96)

Final answer

(-1.08, 9.96)

Detailed explanation

Walkthrough

Since $M$ is a minimum point, the gradient at $M$ is zero. We use the derivative found in part (a): $4x + \frac{5}{x^2} = 0$ . Multiplying the entire equation by $x^2$ clears the denominator, giving $4x^3 + 5 = 0$ . Solving for $x$ yields $x^3 = -1.25$ , so $x = \sqrt[3]{-1.25} \approx -1.07722$ . Rounding to 3 significant figures gives $x = -1.08$ . To find the $y$ -coordinate, we substitute the unrounded value $x = -1.07722$ into the original curve equation $y = 2x^2 - \frac{5}{x} + 3$ . This gives $y \approx 9.96236$ , which rounds to $9.96$ to 3 significant figures. Thus, $M$ is at $(-1.08, 9.96)$ .

Key Takeaways

Stationary points occur where the first derivative is zero.
When finding coordinates, always use unrounded intermediate values for subsequent calculations to maintain accuracy.
Rounding should only be applied at the final step as instructed by the question.

Common Mistakes

Rounding $x$ to $-1.08$ before substituting into the $y$ -equation, which leads to a slightly incorrect $y$ -value.
Forgetting that $x^3 = -1.25$ has a real negative cube root.
Algebraic errors when multiplying by $x^2$ to clear the fraction.

Things to Be Careful About

The mark scheme specifies AWRT (Any Correct Rounding Technique), so using 3-4 significant figures for intermediate steps is acceptable, but final answers must be exactly 3 significant figures.
Ensure the negative sign is preserved when taking the cube root of a negative number.

Techniques used

set derivative to zerosolve cubic equationsubstitute x back into curve equation

Q3Medium-EasySeries

(a)

Find the complete expansion of $\left(2x - \frac{3}{x}\right)^4$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the binomial theorem to expand $\left(2x - \frac{3}{x}\right)^4$ term by term, keeping the sign of the second term negative throughout.

Working

\left(2x - \frac{3}{x}\right)^4 = \binom{4}{0}(2x)^4 + \binom{4}{1}(2x)^3\left(-\frac{3}{x}\right) + \binom{4}{2}(2x)^2\left(-\frac{3}{x}\right)^2 + \binom{4}{3}(2x)\left(-\frac{3}{x}\right)^3 + \binom{4}{4}\left(-\frac{3}{x}\right)^4

Now simplify each term:

\begin{aligned} (2x)^4 &= 16x^4 \\ 4(2x)^3\left(-\frac{3}{x}\right) &= 4(8x^3)\left(-\frac{3}{x}\right) = -96x^2 \\ 6(2x)^2\left(-\frac{3}{x}\right)^2 &= 6(4x^2)\left(\frac{9}{x^2}\right) = 216 \\ 4(2x)\left(-\frac{3}{x}\right)^3 &= 4(2x)\left(-\frac{27}{x^3}\right) = -216x^{-2} \\ \left(-\frac{3}{x}\right)^4 &= \frac{81}{x^4} = 81x^{-4} \end{aligned}

Therefore the complete expansion is:

16x^4 - 96x^2 + 216 - 216x^{-2} + 81x^{-4}

Answer

16x^4 - 96x^2 + 216 - 216x^{-2} + 81x^{-4}

Final answer

16x^4 - 96x^2 + 216 - 216x^-2 + 81x^-4

Detailed explanation

Walkthrough

We need to expand a fourth power of a binomial. The binomial theorem for $(a+b)^4$ gives

(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

Here $a = 2x$ and $b = -\frac{3}{x}$ . The negative sign must be included every time the term appears, because the terms with odd powers of $b$ will be negative. Substitute each term into the formula and simplify separately. For example, the second term is $4(2x)^3\left(-\frac{3}{x}\right) = 4 \cdot 8x^3 \cdot \left(-\frac{3}{x}\right) = -96x^2$ . The middle term is $6(2x)^2\left(-\frac{3}{x}\right)^2 = 6 \cdot 4x^2 \cdot \frac{9}{x^2} = 216$ . The fourth term is negative because $\left(-\frac{3}{x}\right)^3$ is negative, and the final term is positive.

Key Takeaways

The binomial expansion requires applying the binomial coefficients $1, 4, 6, 4, 1$ and carefully combining the powers of $2x$ and $-\frac{3}{x}$ . The result here contains positive, zero, and negative powers of $x$ .

Common Mistakes

Forgetting the binomial coefficients.
Using $+\frac{3}{x}$ instead of $-\frac{3}{x}$ , which changes the signs of the odd-power terms.
Dropping the minus sign on terms with odd powers of $\frac{3}{x}$ .
Incorrectly simplifying powers such as $x^3 \cdot x^{-1} = x^2$ or $x \cdot x^{-3} = x^{-2}$ .

Things to Be Careful About

The signs in the final expansion are $+, -, +, -, +$ . In particular, $\left(-\frac{3}{x}\right)^2 = \frac{9}{x^2}$ , $\left(-\frac{3}{x}\right)^3 = -\frac{27}{x^3}$ , and $\left(-\frac{3}{x}\right)^4 = \frac{81}{x^4}$ . Negative powers of $x$ are acceptable in the final answer.

Techniques used

apply the binomial theoremsimplify powers of xhandle negative signs and negative exponents

(b)

Hence determine the coefficient of $x^2$ in the expansion of $(x^2 + 5)\left(2x - \frac{3}{x}\right)^4$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the expansion from part (a). When multiplying by $(x^2 + 5)$ , the coefficient of $x^2$ comes from $x^2$ times the constant term and $5$ times the $x^2$ term.

Working

From part (a),

\left(2x - \frac{3}{x}\right)^4 = 16x^4 - 96x^2 + 216 - 216x^{-2} + 81x^{-4}

The products that produce an $x^2$ term are:

x^2 \cdot 216 = 216x^2

and

5 \cdot (-96x^2) = -480x^2

Therefore the coefficient of $x^2$ is:

216 - 480 = -264

Answer

-264

Final answer

-264

Detailed explanation

Walkthrough

We already know the expansion of $\left(2x - \frac{3}{x}\right)^4$ . To find the coefficient of $x^2$ in $(x^2+5)$ times this expansion, look only at the combinations that give $x^2$ . Multiplying $x^2$ by the constant term $216$ gives $216x^2$ . Multiplying $5$ by the $x^2$ term $-96x^2$ gives $-480x^2$ . No other products give $x^2$ : for example, $x^2 \cdot 16x^4$ gives $x^6$ , and $5 \cdot 216$ gives a constant. Adding the two contributions gives $216 - 480 = -264$ .

Key Takeaways

When multiplying two polynomials, the coefficient of a chosen power is found by selecting all contributions whose exponents sum to that power. This is the standard method of coefficient extraction.

Common Mistakes

Forgetting to multiply the $-96x^2$ term by $5$ .
Using only $216$ or only $-216$ .
Confusing the constant term with the $x^2$ term from part (a).
Mixing up signs: the $x^2$ term is $-96x^2$ , not $96x^2$ .

Things to Be Careful About

The final answer may be written as $-264$ or $-264x^2$ ; both are accepted. Use the correct terms from part (a): the constant term is $216$ and the $x^2$ term is $-96x^2$ .

Techniques used

multiply a polynomial by a binomial expansionidentify terms contributing to a target powercombine coefficients

Q4Medium-EasyCircular MeasureTrigonometry

Fig. 4.1

The diagram shows a triangle $OAB$ where $OA = OB = 10\text{ cm}$ and angle $AOB = 0.8\text{ radians}$ . Points $C$ and $D$ on $OA$ and $OB$ respectively are such that the arc $CD$ is part of a circle with centre $O$ and radius $6\text{ cm}$ . The shaded region is bounded by the arc $CD$ and the line segments $CA$ , $AB$ and $BD$ .

(a)

Find the perimeter of the shaded region.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The perimeter of the shaded region consists of four parts: the straight segments $CA$ and $BD$ , the straight segment $AB$ , and the curved arc $CD$ . We are given $CA = 4$ cm and $BD = 4$ cm. We calculate the arc length $CD$ using the circular measure formula $s = r\theta$ , and find $AB$ using the cosine rule in $\triangle OAB$ .

Working

Arc length $CD$ :
Using the formula for arc length $s = r\theta$ with $r = 6$ cm and $\theta = 0.8$ radians:

\text{Arc } CD = 6 \times 0.8 = 4.8 \text{ cm}

Length $AB$ :
In $\triangle OAB$ , we have $OA = 10$ cm, $OB = 10$ cm, and $\angle AOB = 0.8$ radians. Applying the cosine rule:

AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot \cos(\angle AOB)

AB^2 = 10^2 + 10^2 - 2 \times 10 \times 10 \times \cos(0.8)

AB^2 = 200 - 200 \cos(0.8)

AB = \sqrt{200 - 200 \cos(0.8)} \approx 7.816 \text{ cm}

Total Perimeter:
The perimeter is the sum of $CA$ , $AB$ , $BD$ , and arc $CD$ :

\text{Perimeter} = CA + AB + BD + \text{Arc } CD

\text{Perimeter} = 4 + 7.816 + 4 + 4.8 = 20.616 \text{ cm}

Rounding to 3 significant figures:

\text{Perimeter} \approx 20.6 \text{ cm}

Answer

20.6 \text{ cm}

Final answer

20.6 cm

Detailed explanation

Walkthrough

To find the perimeter of the shaded region, we add the lengths of its four boundary components: $CA$ , $AB$ , $BD$ , and the arc $CD$ .

Arc $CD$ : The arc is part of a circle with radius $r = 6$ cm and subtends an angle $\theta = 0.8$ radians at the centre $O$ . The arc length formula $s = r\theta$ gives $6 \times 0.8 = 4.8$ cm.
Segments $CA$ and $BD$ : These are given directly in the diagram as $4$ cm each (since $OA = 10$ cm and $OC = 6$ cm, so $CA = 10 - 6 = 4$ cm; similarly $BD = 4$ cm).
Segment $AB$ : This is the base of the isosceles triangle $OAB$ . We know two sides ( $OA = OB = 10$ ) and the included angle ( $\angle AOB = 0.8$ rad). The cosine rule is the appropriate tool: $AB^2 = 10^2 + 10^2 - 2(10)(10)\cos(0.8)$ . Evaluating this gives $AB \approx 7.816$ cm.
Summing up: Adding $4 + 7.816 + 4 + 4.8$ yields $20.616$ , which rounds to $20.6$ cm to 3 significant figures.

Key Takeaways

The arc length formula $s = r\theta$ requires the angle to be in radians.
The cosine rule $c^2 = a^2 + b^2 - 2ab\cos C$ is used to find a side length when two sides and the included angle are known.
Perimeter problems involving mixed straight and curved boundaries require summing each distinct component separately.

Common Mistakes

Using degrees instead of radians in the arc length formula $s = r\theta$ .
Forgetting to include the arc $CD$ in the perimeter and only summing the straight lines.
Using the sine rule incorrectly or failing to set up the cosine rule properly (e.g., sign errors).

Things to Be Careful About

Always ensure the angle is in radians when using $s = r\theta$ or area of sector $= \frac{1}{2}r^2\theta$ .
The final answer should be given to an appropriate number of significant figures (3 s.f. is standard unless specified otherwise).
Ensure calculator is in radian mode when evaluating $\cos(0.8)$ .

Techniques used

calculate arc length using r * thetaapply cosine rule to find side lengthsum perimeter components

(b)

Find the area of the shaded region.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The shaded region is the area of triangle $OAB$ minus the area of the circular sector $OCD$ . We calculate each area separately and then subtract.

Working

Area of sector $OCD$ :
Using the formula for the area of a sector $A = \frac{1}{2}r^2\theta$ with $r = 6$ cm and $\theta = 0.8$ radians:

\text{Area of sector } OCD = \frac{1}{2} \times 6^2 \times 0.8 = \frac{1}{2} \times 36 \times 0.8 = 14.4 \text{ cm}^2

Area of triangle $OAB$ :
Using the formula for the area of a triangle $A = \frac{1}{2}ab\sin C$ with $a = OA = 10$ , $b = OB = 10$ , and $C = \angle AOB = 0.8$ radians:

\text{Area of } \triangle OAB = \frac{1}{2} \times 10 \times 10 \times \sin(0.8)

\text{Area of } \triangle OAB = 50 \times \sin(0.8) \approx 50 \times 0.71736 = 35.868 \text{ cm}^2

Area of shaded region:
Subtract the sector area from the triangle area:

\text{Area of shaded region} = \text{Area of } \triangle OAB - \text{Area of sector } OCD

\text{Area of shaded region} = 35.868 - 14.4 = 21.468 \text{ cm}^2

Rounding to 3 significant figures:

\text{Area of shaded region} \approx 21.5 \text{ cm}^2

Answer

21.5 \text{ cm}^2

Final answer

21.5 cm^2

Detailed explanation

Walkthrough

The shaded region is bounded by arc $CD$ , segment $CA$ , segment $AB$ , and segment $BD$ . This shape is exactly the triangle $OAB$ with the circular sector $OCD$ removed from it.

Area of sector $OCD$ : The sector has radius $r = 6$ cm and angle $\theta = 0.8$ radians. The formula $A = \frac{1}{2}r^2\theta$ gives $\frac{1}{2} \times 36 \times 0.8 = 14.4$ cm $^2$ .
Area of triangle $OAB$ : The triangle has two sides of length $10$ cm and an included angle of $0.8$ radians. The formula $A = \frac{1}{2}ab\sin C$ gives $\frac{1}{2} \times 10 \times 10 \times \sin(0.8) = 50\sin(0.8) \approx 35.868$ cm $^2$ .
Subtraction: The shaded area is $35.868 - 14.4 = 21.468$ cm $^2$ , which rounds to $21.5$ cm $^2$ to 3 significant figures.

Key Takeaways

The area of a sector formula $A = \frac{1}{2}r^2\theta$ requires the angle in radians.
The triangle area formula $A = \frac{1}{2}ab\sin C$ is useful when two sides and the included angle are known.
Complex shaded regions can often be found by adding or subtracting simple geometric shapes (triangle minus sector).

Common Mistakes

Using degrees instead of radians in $\frac{1}{2}r^2\theta$ or $\frac{1}{2}ab\sin C$ .
Calculating the area of the wrong sector (e.g., using radius $10$ instead of $6$ ).
Forgetting to subtract the sector area and instead adding it.

Things to Be Careful About

Always check that the calculator is in radian mode when evaluating $\sin(0.8)$ .
Use the exact radius for each shape: $r = 6$ for the sector, $r = 10$ for the triangle sides.
Round the final answer to 3 significant figures as is standard practice.

Techniques used

calculate area of circular sectorapply area of triangle formula 1/2 ab sin Csubtract sector area from triangle area

Q56MMediumSeries

An arithmetic progression has first term 5 and common difference 6.

For this progression, find the sum of all the terms that lie between 150 and 400.

DifficultyMedium

Worked solution

Approach

An arithmetic progression has first term $a=5$ and common difference $d=6$ . Write the $n$ th term, find the first term greater than 150 and the last term less than 400, then sum the resulting arithmetic series.

Working

The $n$ th term is

u_n = 5 + (n-1)6 = 6n - 1

First term greater than 150:

6n - 1 > 150 \implies 6n > 151 \implies n > \frac{151}{6} \approx 25.17

So the first included term is $n = 26$ :

u_{26} = 5 + 25(6) = 155

Last term less than 400:

6n - 1 < 400 \implies 6n < 401 \implies n < \frac{401}{6} \approx 66.83

So the last included term is $n = 66$ :

u_{66} = 5 + 65(6) = 395

The required terms are $u_{26}, u_{27}, \ldots, u_{66}$ , so the number of terms is

66 - 26 + 1 = 41

Their sum is

S = \frac{41}{2}(155 + 395) = \frac{41}{2}(550) = 41 \times 275 = 11275

Answer

11275

Final answer

11275

Detailed explanation

Walkthrough

The terms of this progression are formed by adding 6 to the preceding term, starting from 5. We need only the terms that are strictly greater than 150 and strictly less than 400. Because the progression is increasing, these terms form a consecutive block, so once we know the first and last term numbers in that block, we can sum them with the arithmetic series formula.

First, write the general term. For an arithmetic progression with first term $a$ and common difference $d$ , the $n$ th term is $u_n = a + (n-1)d$ . Here $a=5$ and $d=6$ , so

u_n = 5 + (n-1)6 = 6n - 1

This form is convenient because it lets us solve for $n$ when the term crosses 150 and 400.

To find the first term above 150, solve

6n - 1 > 150 \quad \Rightarrow \quad n > \frac{151}{6} \approx 25.17

The smallest integer $n$ satisfying this is $n=26$ . Hence the first included term is

u_{26} = 5 + 25 \times 6 = 155

To find the last term below 400, solve

6n - 1 < 400 \quad \Rightarrow \quad n < \frac{401}{6} \approx 66.83

The largest integer $n$ satisfying this is $n=66$ . The last included term is

u_{66} = 5 + 65 \times 6 = 395

Thus the required block is $u_{26}, u_{27}, \ldots, u_{66}$ . The number of terms is

66 - 26 + 1 = 41

Finally, use the sum formula for an arithmetic series, $S_n = \frac{n}{2}(a + l)$ , where $a$ is the first term and $l$ is the last term. Here $n=41$ , $a=155$ , $l=395$ :

S = \frac{41}{2}(155 + 395) = \frac{41}{2}(550) = 41 \times 275 = 11275

Equivalently, this is the same as $S_{66} - S_{25}$ : the sum of the first 66 terms minus the sum of the first 25 terms.

Key Takeaways

This question tests the formula for the $n$ th term of an arithmetic progression and the formula for the sum of an arithmetic series. It also tests how to convert a range condition such as "between 150 and 400" into strict inequalities on $n$ , then choose the correct integer indices. A consecutive block of an arithmetic progression can be summed either directly with its first and last terms, or by subtracting the sum of the terms before the block from the total sum.

Common Mistakes

Using $n=25$ or $n=67$ instead of $n=26$ and $n=66$ . Because "between" is exclusive, terms equal to 150 or 400 are not included.
Counting the terms incorrectly: from $n=26$ to $n=66$ inclusive there are $66-26+1=41$ terms, not $40$ .
Applying the sum formula with the original first term $5$ and $n=41$ ; that would give the sum of the first 41 terms of the whole progression, not the selected block from $n=26$ to $n=66$ .
Forgetting the boundary step; the mark scheme awards a method mark for attempting to solve $5+(n-1)6=150$ or $5+(n-1)6=400$ .

Things to Be Careful About

The word "between" is exclusive, so terms exactly equal to 150 or 400 are excluded. When solving $6n - 1 > 150$ , $n > 25.17$ , so the first included term is at index 26; when solving $6n - 1 < 400$ , $n < 66.83$ , so the last included term is at index 66. Use the first and last terms of the selected block, not the original first term, in the sum formula. If using the subtraction method, the correct expression is $S_{66} - S_{25}$ , not $S_{66} - S_{26}$ .

Techniques used

write the nth term formula of an arithmetic progressionapply inequalities to find term boundariescount terms in an inclusive rangeuse the arithmetic series sum formula

Approach

The circle lies entirely in the first quadrant ( $x > 0$ , $y > 0$ for all points on $C$ ). The least distance from any point on the circle to the $x$ -axis is the vertical distance from the lowest point of the circle to the $x$ -axis. Since the radius is $r$ , this equals the $y$ -coordinate of the centre minus $r$ . Similarly, the least distance to the $y$ -axis is the horizontal distance from the leftmost point of the circle to the $y$ -axis, which equals the $x$ -coordinate of the centre minus $r$ .

Working

The least distance from any point on $C$ to the $x$ -axis is 8 units:

y_{\text{centre}} - r = 8

y_{\text{centre}} = r + 8

The least distance from any point on $C$ to the $y$ -axis is 5 units:

x_{\text{centre}} - r = 5

x_{\text{centre}} = r + 5

Answer

(r + 5,\; r + 8)

Final answer

(r + 5, r + 8)

Detailed explanation

Walkthrough

The circle is drawn entirely in the first quadrant. The lowest point on the circle is directly below the centre, at a vertical distance of $r$ from it. If the least distance from the circle to the $x$ -axis is 8, then the centre must be $r + 8$ units above the $x$ -axis, giving $y = r + 8$ . Similarly, the leftmost point is $r$ units to the left of the centre, and if the least distance to the $y$ -axis is 5, the centre must be $r + 5$ units to the right of the $y$ -axis, giving $x = r + 5$ .

Key Takeaways

The least distance from a circle to an axis equals the centre coordinate minus the radius (when the circle is on the positive side of that axis).
Coordinates of a centre can be expressed in terms of the radius using geometric relationships.

Common Mistakes

Confusing the least distance with the greatest distance (which would be $r + \text{centre coordinate}$ ).
Forgetting that the circle is entirely in the first quadrant, which ensures the centre coordinates are both positive.

Things to Be Careful About

The question states $x > 0$ and $y > 0$ for all points on $C$ , confirming the circle does not cross either axis. This means the centre coordinates are strictly greater than $r$ .
The answer must be expressed in terms of $r$ as an ordered pair $(x, y)$ .

Techniques used

interpret least distance to axesdetermine centre coordinates in terms of radius

(b)

Given that the distance between the origin and the centre of the circle is 15 units, find the value of $r$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the distance formula between the origin $(0, 0)$ and the centre $(r + 5, r + 8)$ from part (a), set it equal to 15, and solve the resulting quadratic equation for $r$ .

Working

The distance from the origin to the centre is:

\sqrt{(r + 5)^2 + (r + 8)^2} = 15

Squaring both sides:

(r + 5)^2 + (r + 8)^2 = 225

Expanding:

r^2 + 10r + 25 + r^2 + 16r + 64 = 225

Combining like terms:

2r^2 + 26r + 89 = 225

2r^2 + 26r - 136 = 0

Dividing through by 2:

r^2 + 13r - 68 = 0

Factoring:

(r + 17)(r - 4) = 0

So $r = -17$ or $r = 4$ . Since the radius must be positive, $r = 4$ .

Answer

r = 4

Final answer

r = 4

Detailed explanation

Walkthrough

Part (a) established that the centre is at $(r + 5, r + 8)$ . The distance from the origin to this point is given as 15. Applying the distance formula $\sqrt{x^2 + y^2}$ gives $\sqrt{(r+5)^2 + (r+8)^2} = 15$ . Squaring both sides eliminates the square root and produces a quadratic in $r$ . Expanding and simplifying yields $r^2 + 13r - 68 = 0$ , which factors as $(r + 17)(r - 4) = 0$ . Since $r$ represents a radius, it must be positive, so $r = 4$ .

Key Takeaways

The distance from the origin to a point $(x, y)$ is $\sqrt{x^2 + y^2}$ .
Quadratic equations arising from geometry problems may have a negative root that must be rejected on physical grounds.

Common Mistakes

Forgetting to square the 15, writing $(r+5)^2 + (r+8)^2 = 15$ instead of $= 225$ .
Accepting $r = -17$ as a valid answer without checking the physical constraint $r > 0$ .
Algebra errors when expanding $(r+5)^2$ or $(r+8)^2$ .

Things to Be Careful About

Always reject negative values for physical quantities like radius.
The mark scheme awards follow-through marks based on the answer from part (a), so consistency is important.

Techniques used

apply distance formula from originexpand and simplify quadraticsolve quadratic by factorisation

(c)

The point on the circle furthest from the origin is denoted by $P$ .

Find the gradient of the tangent to the circle at $P$ .

2M Similar

DifficultyMedium

Worked solution

Approach

The point $P$ on the circle furthest from the origin lies on the line joining the origin to the centre of the circle, on the far side of the centre. The radius at $P$ is therefore along the line $OP$ . The tangent at $P$ is perpendicular to this radius, so its gradient is the negative reciprocal of the gradient of $OP$ .

Working

With $r = 4$ from part (b), the centre of the circle is at:

(r + 5,\; r + 8) = (9,\; 12)

The gradient of the line from the origin $(0, 0)$ to the centre $(9, 12)$ is:

m_{OP} = \frac{12 - 0}{9 - 0} = \frac{12}{9} = \frac{4}{3}

Since $P$ lies on the extension of $OP$ beyond the centre, the radius at $P$ has the same gradient $\frac{4}{3}$ .

The tangent at $P$ is perpendicular to the radius at $P$ , so its gradient is:

m_{\text{tangent}} = -\frac{1}{m_{OP}} = -\frac{1}{4/3} = -\frac{3}{4}

Answer

-\frac{3}{4}

Final answer

-3/4

Detailed explanation

Walkthrough

Once we know the centre is at $(9, 12)$ , we can find the gradient of the line from the origin to the centre: $m = 12/9 = 4/3$ . The point on the circle furthest from the origin lies along this line, on the opposite side of the centre from the origin. The radius at $P$ is therefore parallel to (in fact, collinear with) the line $OP$ , so it has gradient $4/3$ . A fundamental property of circles is that the tangent at any point is perpendicular to the radius at that point. The gradient of a line perpendicular to one with gradient $m$ is $-1/m$ . Therefore, the tangent gradient is $-1/(4/3) = -3/4$ .

Key Takeaways

The point on a circle furthest from (or nearest to) a given point lies on the line joining that point to the centre.
The tangent to a circle at any point is perpendicular to the radius at that point.
If two lines are perpendicular, the product of their gradients is $-1$ .

Common Mistakes

Trying to find the coordinates of $P$ explicitly and then computing the tangent gradient, which is unnecessarily complicated.
Forgetting the negative sign when computing the perpendicular gradient.
Using the gradient of the radius as the gradient of the tangent.

Things to Be Careful About

The mark scheme allows the gradient of the tangent to be written as $-\frac{r+5}{r+8}$ or equivalently $-\frac{(r \text{ from (b)}) + 8}{(r \text{ from (b)}) + 5}$ , showing that the perpendicular gradient relationship is the key idea.
Only positive values of $r$ are valid, so the gradient expression must use $r = 4$ .

Techniques used

find gradient of line from origin to centreuse perpendicular gradient property for tangent

Q7MediumTrigonometryQuadratics

(a)

Show that $3\tan^2\theta + 5\sin^2\theta \equiv \frac{8\sin^2\theta - 5\sin^4\theta}{1 - \sin^2\theta}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the identity $\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$ to rewrite the left-hand side, then replace $\cos^2\theta$ with $1-\sin^2\theta$ and combine over a common denominator.

Working

Starting with the left-hand side:

3\tan^2\theta + 5\sin^2\theta = \frac{3\sin^2\theta}{\cos^2\theta} + 5\sin^2\theta

Put both terms over the common denominator $\cos^2\theta$ :

\frac{3\sin^2\theta + 5\sin^2\theta\cos^2\theta}{\cos^2\theta}

Use $\cos^2\theta = 1-\sin^2\theta$ :

\frac{3\sin^2\theta + 5\sin^2\theta(1-\sin^2\theta)}{1-\sin^2\theta}

Expand the numerator and simplify:

\frac{3\sin^2\theta + 5\sin^2\theta - 5\sin^4\theta}{1-\sin^2\theta} = \frac{8\sin^2\theta - 5\sin^4\theta}{1-\sin^2\theta}

Hence the identity is shown.

Answer

3\tan^2\theta + 5\sin^2\theta \equiv \frac{8\sin^2\theta - 5\sin^4\theta}{1-\sin^2\theta}

Final answer

3tan^2θ + 5sin^2θ = (8sin^2θ - 5sin^4θ)/(1 - sin^2θ)

Detailed explanation

Walkthrough

We need to show the left-hand side equals the right-hand side. Start by replacing $\tan^2\theta$ with $\frac{\sin^2\theta}{\cos^2\theta}$ , using the standard identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$ .

Now the expression has two terms. To combine them, write $5\sin^2\theta$ with denominator $\cos^2\theta$ :

5\sin^2\theta = \frac{5\sin^2\theta\cos^2\theta}{\cos^2\theta}

Then use $\cos^2\theta = 1-\sin^2\theta$ to replace every $\cos^2\theta$ term. This gives a single fraction in terms of $\sin^2\theta$ only. Expanding the numerator and collecting like terms produces exactly the required right-hand side.

Key Takeaways

This question tests the two fundamental trigonometric identities:

\tan\theta = \frac{\sin\theta}{\cos\theta}, \qquad \sin^2\theta + \cos^2\theta = 1

It also tests algebraic manipulation of fractions, especially writing terms over a common denominator and simplifying.

Common Mistakes

Forgetting to put $5\sin^2\theta$ over the same denominator as the first term.
Using the Pythagorean identity incorrectly, such as writing $\cos^2\theta = 1+\sin^2\theta$ .
Losing the factor of $\sin^2\theta$ when expanding $5\sin^2\theta(1-\sin^2\theta)$ .
Stopping before the expression is fully simplified to the required form.

Things to Be Careful About

The identity is valid when $\cos\theta \neq 0$ , i.e. when $\theta \neq 90^\circ, 270^\circ$ , because $\tan\theta$ is undefined there. In the manipulation, the denominator becomes $1-\sin^2\theta$ , which is the same condition. Since this is an answer-given question, all necessary working must be shown.

Techniques used

rewrite tan^2 terms in terms of sin and cosapply the Pythagorean identity to replace cos^2combine terms over a common denominator

(b)

Hence solve the equation $3\tan^2\theta + 5\sin^2\theta = 9$ for $0^\circ < \theta < 270^\circ$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the identity from part (a) to replace $3\tan^2\theta + 5\sin^2\theta$ with $\frac{8\sin^2\theta - 5\sin^4\theta}{1-\sin^2\theta}$ . Set this equal to $9$ , clear the denominator, rearrange into a quadratic in $\sin^2\theta$ , solve it, and then find all angles in the required interval.

Working

From part (a):

\frac{8\sin^2\theta - 5\sin^4\theta}{1-\sin^2\theta} = 9

Multiply both sides by $1-\sin^2\theta$ :

8\sin^2\theta - 5\sin^4\theta = 9(1-\sin^2\theta)

Expand and rearrange:

8\sin^2\theta - 5\sin^4\theta = 9 - 9\sin^2\theta

5\sin^4\theta - 17\sin^2\theta + 9 = 0

Let $u = \sin^2\theta$ . Then:

5u^2 - 17u + 9 = 0

Using the quadratic formula:

u = \frac{17 \pm \sqrt{(-17)^2 - 4(5)(9)}}{2(5)} = \frac{17 \pm \sqrt{289 - 180}}{10} = \frac{17 \pm \sqrt{109}}{10}

So:

u \approx 2.744 \quad \text{or} \quad u \approx 0.6559

Since $\sin^2\theta \leq 1$ , discard $u \approx 2.744$ . Hence:

\sin^2\theta = 0.6559, \qquad \sin\theta = \pm 0.810

For $\sin\theta = +0.810$ :

\theta = \sin^{-1}(0.810) \approx 54.1^\circ

The second-quadrant angle is:

180^\circ - 54.1^\circ = 125.9^\circ

For $\sin\theta = -0.810$ , the third-quadrant angle is:

180^\circ + 54.1^\circ = 234.1^\circ

The fourth-quadrant angle $360^\circ - 54.1^\circ = 305.9^\circ$ is outside $0^\circ < \theta < 270^\circ$ .

Answer

\theta = 54.1^\circ, \ 125.9^\circ, \ 234.1^\circ

Final answer

54.1°, 125.9°, 234.1°

Detailed explanation

Walkthrough

The equation has the same left-hand side as the identity in part (a), so replace it with the fraction:

\frac{8\sin^2\theta - 5\sin^4\theta}{1-\sin^2\theta} = 9

Clearing the denominator gives an equation in powers of $\sin\theta$ . Because only even powers appear, it is actually a quadratic in $\sin^2\theta$ . Let $u = \sin^2\theta$ to make the quadratic structure clear:

5u^2 - 17u + 9 = 0

Solving by the quadratic formula gives two roots. One root is about $2.744$ , which is impossible because $\sin^2\theta$ cannot exceed $1$ . The other root is about $0.6559$ , so $\sin\theta = \pm 0.810$ .

For the positive value, the acute angle is $54.1^\circ$ . Since sine is positive in the first and second quadrants, the second solution is $180^\circ - 54.1^\circ = 125.9^\circ$ . For the negative value, sine is negative in the third and fourth quadrants. The interval $0^\circ < \theta < 270^\circ$ includes the third quadrant but not the fourth, so only $180^\circ + 54.1^\circ = 234.1^\circ$ is included.

Key Takeaways

The identity from part (a) converts an expression involving $\tan^2\theta$ and $\sin^2\theta$ into a single fraction.
Many trigonometric equations become quadratics in $\sin^2\theta$ or $\cos^2\theta$ ; substituting $u = \sin^2\theta$ makes them easier to solve.
Always reject roots that are impossible for the trigonometric function.
Use the quadrant rule to find all solutions in the required interval.

Common Mistakes

Not using the identity from part (a) and trying to solve the original equation directly.
Making a sign error when rearranging $8\sin^2\theta - 5\sin^4\theta = 9 - 9\sin^2\theta$ into $5\sin^4\theta - 17\sin^2\theta + 9 = 0$ .
Forgetting to reject $\sin^2\theta \approx 2.744$ .
Taking only the positive square root and missing the negative sine solutions.
Including $305.9^\circ$ , which is outside the required interval, or missing $125.9^\circ$ .

Things to Be Careful About

The interval is $0^\circ < \theta < 270^\circ$ , so $0^\circ$ and $270^\circ$ are not included. Also, $\tan\theta$ is undefined at $90^\circ$ and $270^\circ$ , so those values cannot be solutions. The mark scheme accepts answers rounded to the nearest degree, so $54.1^\circ, 125.9^\circ, 234.1^\circ$ are acceptable. If any additional values within the interval are included, the final answer mark is lost.

Techniques used

substitute the identity from part (a)rearrange into a quadratic in sin^2 thetasolve the quadratic and discard impossible rootstake square roots and find all angles in the interval

Q8MediumSeriesQuadratics

A geometric progression is such that its second term is $-120$ and its sum to infinity is 160.

(a)

Find the common ratio.

4M Similar

DifficultyMedium

Worked solution

Approach

Let the first term be $a$ and the common ratio be $r$ . The second term gives $ar = -120$ , and the sum to infinity gives $\frac{a}{1-r} = 160$ . Eliminate $a$ to obtain a quadratic in $r$ , solve it, then use the convergence condition $|r| < 1$ to choose the correct root.

Working

The second term is

ar = -120

The sum to infinity is

\frac{a}{1-r} = 160

From $ar = -120$ , $a = -\frac{120}{r}$ . Substitute into the sum to infinity:

\frac{-\frac{120}{r}}{1-r} = 160

Multiply by $r(1-r)$ :

-120 = 160r(1-r)

Expand and rearrange:

160r^2 - 160r - 120 = 0

Divide by 40:

4r^2 - 4r - 3 = 0

Factorise:

(2r-3)(2r+1) = 0

So

r = \frac{3}{2} \quad \text{or} \quad r = -\frac{1}{2}

Since the sum to infinity exists, $|r| < 1$ . Therefore $r = \frac{3}{2}$ is rejected, leaving

r = -\frac{1}{2}

Answer

r = -\frac{1}{2}

Final answer

r = -1/2

Detailed explanation

Walkthrough

We are told two facts about a geometric progression: the second term is $-120$ , and the sum to infinity is $160$ . A geometric progression with first term $a$ and common ratio $r$ has second term $ar$ , so this gives $ar = -120$ . Its sum to infinity is $\frac{a}{1-r}$ , valid only when $|r| < 1$ , so this gives $\frac{a}{1-r} = 160$ .

Since $ar = -120$ , we can write $a = -\frac{120}{r}$ and substitute into $\frac{a}{1-r} = 160$ . This gives an equation containing only $r$ . After multiplying by $r(1-r)$ and rearranging, we obtain $4r^2 - 4r - 3 = 0$ .

Factorising gives $r = \frac{3}{2}$ or $r = -\frac{1}{2}$ . The sum-to-infinity formula is valid only when $|r| < 1$ . Since $\frac{3}{2}$ has absolute value greater than 1, it cannot be the common ratio of a progression with a finite sum to infinity. Therefore $r = -\frac{1}{2}$ is the only valid value.

Key Takeaways

The $n$ th term of a geometric progression is $ar^{n-1}$ , so the second term is $ar$ .
The sum to infinity of a geometric progression is $\frac{a}{1-r}$ , valid only for $|r| < 1$ .
When a problem gives two conditions, forming two equations and eliminating a variable is often the key step.
Always check the convergence condition when choosing between possible values of $r$ .

Common Mistakes

Forgetting that the sum to infinity requires $|r| < 1$ , so accepting $r = \frac{3}{2}$ as well.
Making a sign error when substituting $a = -\frac{120}{r}$ , especially with the negative sign.
Rearranging the equation incorrectly, so the quadratic has the wrong signs.

Things to Be Careful About

The mark scheme requires the method of eliminating $a$ or $r$ to be shown.
The quadratic must be rearranged into a three-term quadratic before solving.
The final answer must be $r = -\frac{1}{2}$ only; the invalid root $r = \frac{3}{2}$ must be rejected using the convergence condition.

Techniques used

write equations for the second term and the sum to infinityeliminate the first termsolve the quadratic by factorisationapply the convergence condition to reject an invalid root

(b)

The first nine terms of the progression are now removed.

Find the sum to infinity of the remaining terms of the progression.

3M Similar

DifficultyMedium

Worked solution

Approach

Removing the first nine terms leaves the terms from the 10th onwards. The sum of these remaining terms is the original sum to infinity minus the sum of the first nine terms. First find the first term $a$ , then use the formula for the sum of the first $n$ terms of a geometric progression.

Working

From part (a), $r = -\frac{1}{2}$ . Since $ar = -120$ ,

a = \frac{-120}{r} = \frac{-120}{-\frac{1}{2}} = 240

The sum of the first nine terms is

S_9 = \frac{a(1-r^9)}{1-r}

Substitute $a = 240$ and $r = -\frac{1}{2}$ :

S_9 = \frac{240\left(1 - \left(-\frac{1}{2}\right)^9\right)}{1 - \left(-\frac{1}{2}\right)}

Evaluate the power:

\left(-\frac{1}{2}\right)^9 = -\frac{1}{512}

so

1 - \left(-\frac{1}{2}\right)^9 = 1 + \frac{1}{512} = \frac{513}{512}

and

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

Therefore

S_9 = \frac{240 \cdot \frac{513}{512}}{\frac{3}{2}} = 240 \cdot \frac{513}{512} \cdot \frac{2}{3} = \frac{2565}{16}

The sum to infinity of the remaining terms is

160 - \frac{2565}{16} = \frac{2560}{16} - \frac{2565}{16} = -\frac{5}{16}

Answer

-\frac{5}{16}

or

-0.3125

Final answer

-5/16 (or -0.3125)

Detailed explanation

Walkthrough

After $r = -\frac{1}{2}$ is known, the first term can be found from the second term: $ar = -120$ , so $a = \frac{-120}{-\frac{1}{2}} = 240$ .

Removing the first nine terms leaves the 10th term onwards. One way to find the sum of these remaining terms is to take the original sum to infinity, $160$ , and subtract the sum of the first nine terms. The sum of the first $n$ terms of a GP is $\frac{a(1-r^n)}{1-r}$ , so with $n = 9$ :

S_9 = \frac{240\left(1 - \left(-\frac{1}{2}\right)^9\right)}{1 - \left(-\frac{1}{2}\right)}

Since $\left(-\frac{1}{2}\right)^9 = -\frac{1}{512}$ , we have $1 - \left(-\frac{1}{2}\right)^9 = \frac{513}{512}$ , and $1 - \left(-\frac{1}{2}\right) = \frac{3}{2}$ . Thus $S_9 = \frac{2565}{16}$ . The remaining sum is $160 - \frac{2565}{16} = -\frac{5}{16}$ .

An equivalent check is to treat the 10th term as the new first term. The 10th term is $240\left(-\frac{1}{2}\right)^9 = -\frac{15}{32}$ , and the sum to infinity of the remaining terms is $\frac{-\frac{15}{32}}{1 - \left(-\frac{1}{2}\right)} = -\frac{5}{16}$ .

Key Takeaways

The sum of the first $n$ terms of a GP is $\frac{a(1-r^n)}{1-r}$ .
Removing the first nine terms of an infinite GP leaves a GP whose first term is the 10th term.
The sum of the remaining terms can be found as $S_\infty - S_9$ , or as $\frac{ar^9}{1-r}$ .

Common Mistakes

Using the original first term $a$ as the first term of the remaining progression instead of the 10th term.
Forgetting that $r^9$ is negative when $r = -\frac{1}{2}$ and $9$ is odd.
Subtracting in the wrong order, which can change the sign of the final answer.
Giving a rounded decimal like $-0.313$ without also showing $-\frac{5}{16}$ or $-0.3125$ .

Things to be Careful About

Use the follow-through value of $r$ from part (a) if it is different but still satisfies $-1 < r < 1$ and $r \neq 0$ .
If using the 10th term as the new first term, the first term is $ar^9$ , not $ar^{10}$ .
The denominator $1-r$ is $\frac{3}{2}$ , not $1+r$ , because $r$ is negative.
The final answer should be exact: $-\frac{5}{16}$ or $-0.3125$ .

Techniques used

find the first term from the common ratiocompute the sum of the first nine termssubtract the partial sum from the sum to infinity

Q9MediumDifferentiationIntegration

A curve is such that $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{6}{x^4} - \frac{5}{x^3}$ . It is given that the curve has a stationary point at $\left(\frac{1}{2}, 9\right)$ .

(a)

Use the expression for $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ to determine whether the stationary point is a maximum or a minimum point.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since the point is stationary, $\frac{dy}{dx}=0$ there. To decide whether it is a maximum or a minimum, substitute $x=\frac{1}{2}$ into the given $\frac{d^2y}{dx^2}$ and check its sign.

Working

At the stationary point, $x=\frac{1}{2}$ . Substitute into the second derivative:

\frac{d^2y}{dx^2} = \frac{6}{\left(\frac{1}{2}\right)^4} - \frac{5}{\left(\frac{1}{2}\right)^3}

Evaluate:

\frac{6}{\frac{1}{16}} - \frac{5}{\frac{1}{8}} = 96 - 40 = 56

Since $\frac{d^2y}{dx^2}=56>0$ , the curve is concave up at this point.

Answer

The stationary point is a minimum point.

Final answer

minimum point

Detailed explanation

Walkthrough

The question gives the second derivative and tells us there is a stationary point at $x=\frac{1}{2}$ . A stationary point is where $\frac{dy}{dx}=0$ . To classify it, we use the second derivative test: evaluate $\frac{d^2y}{dx^2}$ at the stationary point.

Substitute $x=\frac{1}{2}$ into the given expression. Since $\left(\frac{1}{2}\right)^4=\frac{1}{16}$ and $\left(\frac{1}{2}\right)^3=\frac{1}{8}$ , we get $\frac{6}{1/16}-\frac{5}{1/8}=96-40=56$ . Because this is positive, the gradient is increasing through the stationary point, so the point is a minimum.

Key Takeaways

The second derivative test is the standard way to classify a stationary point. If $\frac{d^2y}{dx^2}>0$ at the point, it is a minimum; if $\frac{d^2y}{dx^2}<0$ , it is a maximum.

Common Mistakes

Substituting into $\frac{dy}{dx}$ instead of $\frac{d^2y}{dx^2}$ .
Making arithmetic errors with fractional powers, such as writing $\left(\frac{1}{2}\right)^4=\frac{1}{8}$ .
Forgetting to give the conclusion (maximum or minimum).

Things to Be Careful About

The mark scheme requires evidence of the substitution and evaluation, not just a statement that $\frac{d^2y}{dx^2}>0$ . It awards a special B1 for a correct sign conclusion only if the numerical substitution is not shown, but full marks require $96-40=56$ or equivalent.

Techniques used

substitute x-coordinate into second derivativeevaluate sign of second derivativeclassify stationary point

(b)

Find the equation of the curve.

7M Similar

DifficultyMedium

Worked solution

Approach

Integrate the second derivative once to obtain $\frac{dy}{dx}$ . Use the fact that the point is stationary, so $\frac{dy}{dx}=0$ at $x=\frac{1}{2}$ , to find the first constant of integration. Then integrate again and use the given point $\left(\frac{1}{2},9\right)$ to find the second constant.

Working

Write the second derivative using negative powers:

\frac{d^2y}{dx^2} = 6x^{-4} - 5x^{-3}

Integrate term by term:

\frac{dy}{dx} = 6\frac{x^{-3}}{-3} - 5\frac{x^{-2}}{-2} + c_1 = -2x^{-3} + \frac{5}{2}x^{-2} + c_1

At the stationary point, $\frac{dy}{dx}=0$ when $x=\frac{1}{2}$ :

0 = -2\left(\frac{1}{2}\right)^{-3} + \frac{5}{2}\left(\frac{1}{2}\right)^{-2} + c_1

Evaluate:

0 = -2(8) + \frac{5}{2}(4) + c_1 = -16 + 10 + c_1

So $c_1=6$ . Hence:

\frac{dy}{dx} = -2x^{-3} + \frac{5}{2}x^{-2} + 6

Integrate again:

y = -2\frac{x^{-2}}{-2} + \frac{5}{2}\frac{x^{-1}}{-1} + 6x + c_2 = x^{-2} - \frac{5}{2}x^{-1} + 6x + c_2

Use $y=9$ when $x=\frac{1}{2}$ :

9 = \left(\frac{1}{2}\right)^{-2} - \frac{5}{2}\left(\frac{1}{2}\right)^{-1} + 6\left(\frac{1}{2}\right) + c_2

9 = 4 - 5 + 3 + c_2 = 2 + c_2

So $c_2=7$ .

Answer

y = x^{-2} - \frac{5}{2}x^{-1} + 6x + 7

or equivalently

y = \frac{1}{x^2} - \frac{5}{2x} + 6x + 7

Final answer

y = x^-2 - (5/2)x^-1 + 6x + 7

Detailed explanation

Walkthrough

We are given the second derivative and a point on the curve that is also a stationary point. To find the equation of the curve, we need to integrate twice.

First, rewrite $\frac{d^2y}{dx^2}$ as $6x^{-4}-5x^{-3}$ so each term can be integrated using the power rule. Integrating $x^n$ gives $\frac{x^{n+1}}{n+1}$ for $n\neq -1$ . Thus:

\int 6x^{-4}\,dx = 6\frac{x^{-3}}{-3} = -2x^{-3}, \quad \int -5x^{-3}\,dx = -5\frac{x^{-2}}{-2} = \frac{5}{2}x^{-2}

and we add a constant $c_1$ because integration is indefinite.

Since the point is stationary, $\frac{dy}{dx}=0$ at $x=\frac{1}{2}$ . Substituting this into the integrated expression gives $c_1=6$ . This is the key use of the stationary point information.

Then integrate $\frac{dy}{dx}$ again to obtain $y$ . The constant this time is $c_2$ . Finally, substitute the given coordinates $\left(\frac{1}{2},9\right)$ into the expression for $y$ to find $c_2=7$ , giving the final equation.

Key Takeaways

Integrating a second derivative twice recovers the original function, with a new constant of integration each time.
The stationary point condition $\frac{dy}{dx}=0$ can be used to determine the first constant of integration.
A known point on the curve determines the second constant.

Common Mistakes

Forgetting one of the constants of integration.
Making sign errors when dividing by the new negative exponent.
Substituting the point into $\frac{dy}{dx}$ instead of $y$ when finding $c_2$ .
Arithmetic errors with $\left(\frac{1}{2}\right)^{-3}=8$ and $\left(\frac{1}{2}\right)^{-2}=4$ .

Things to Be Careful About

The mark scheme allows equivalent forms, but the final equation should be simplified.
When integrating, keep the constants $c_1$ and $c_2$ distinct.
At $x=\frac{1}{2}$ , $x^{-1}=2$ , $x^{-2}=4$ , and $x^{-3}=8$ ; check these before substituting.
The final answer may be written as $y=x^{-2}-\frac{5}{2}x^{-1}+6x+7$ or $y=\frac{1}{x^2}-\frac{5}{2x}+6x+7$ .

Techniques used

integrate second derivative term by termuse stationary point condition to find first constantintegrate first derivative term by termsubstitute given point to find second constant

Q10Medium-HardIntegrationDifferentiationCoordinate Geometry

Fig. 10.1

The diagram shows the curve with equation

y = 4(3x + 4)^{\frac{1}{2}} - 2x - 6

for values of $x$ such that $0 \leqslant x \leqslant 7$ . The tangent to the curve at the point $P(7, 0)$ meets the $y$ -axis at the point $Q$ . Region $A$ is bounded by the curve and the two axes. Region $B$ is bounded by the curve, the line segment $PQ$ and the $y$ -axis.

(a)

Find the area of region $A$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Region A is bounded by the curve $y = 4(3x+4)^{\frac{1}{2}} - 2x - 6$ , the $x$ -axis, and the $y$ -axis. The curve meets the $x$ -axis at $P(7, 0)$ and the $y$ -axis at $x = 0$ . The area of region A is therefore given by the definite integral $\int_0^7 y \, dx$ .

Working

Set up the integral:

\int_0^7 \left(4(3x+4)^{\frac{1}{2}} - 2x - 6\right) dx

Find the antiderivative of each term. For $4(3x+4)^{\frac{1}{2}}$ , use the reverse chain rule:

\int 4(3x+4)^{\frac{1}{2}} \, dx = 4 \times \frac{1}{3} \times \frac{2}{3}(3x+4)^{\frac{3}{2}} = \frac{8}{9}(3x+4)^{\frac{3}{2}}

For the polynomial terms:

\int (-2x - 6) \, dx = -x^2 - 6x

So the antiderivative is:

F(x) = \frac{8}{9}(3x+4)^{\frac{3}{2}} - x^2 - 6x

Evaluate at the upper limit $x = 7$ :

F(7) = \frac{8}{9}(3 \times 7 + 4)^{\frac{3}{2}} - 7^2 - 6 \times 7

= \frac{8}{9}(25)^{\frac{3}{2}} - 49 - 42

= \frac{8}{9} \times 125 - 91

= \frac{1000}{9} - 91 = \frac{1000 - 819}{9} = \frac{181}{9}

Evaluate at the lower limit $x = 0$ :

F(0) = \frac{8}{9}(4)^{\frac{3}{2}} - 0 - 0 = \frac{8}{9} \times 8 = \frac{64}{9}

Compute the area:

A = F(7) - F(0) = \frac{181}{9} - \frac{64}{9} = \frac{117}{9} = 13

Answer

\text{Area of region } A = 13

Final answer

13

Detailed explanation

Walkthrough

Region A lies between the curve and both coordinate axes. Since the curve starts on the $y$ -axis at $x = 0$ and ends on the $x$ -axis at $P(7, 0)$ , the area is simply the definite integral of $y$ from $0$ to $7$ .

Step 1: Set up the integral. We integrate $y = 4(3x+4)^{\frac{1}{2}} - 2x - 6$ with respect to $x$ from $0$ to $7$ .

Step 2: Find the antiderivative. The term $4(3x+4)^{\frac{1}{2}}$ is a composite function. Using the reverse chain rule (or substitution $u = 3x+4$ ), we increase the power by 1 to get $\frac{3}{2}$ , divide by the new power, and divide by the derivative of the inner function ( $3$ ). This gives $\frac{8}{9}(3x+4)^{\frac{3}{2}}$ . The polynomial part integrates normally to $-x^2 - 6x$ .

Step 3: Evaluate at the limits. At $x = 7$ , we get $\frac{8}{9}(25)^{\frac{3}{2}} - 91$ . Since $25^{\frac{3}{2}} = (\sqrt{25})^3 = 5^3 = 125$ , this becomes $\frac{1000}{9} - 91 = \frac{181}{9}$ . At $x = 0$ , we get $\frac{8}{9}(4)^{\frac{3}{2}} = \frac{8}{9} \times 8 = \frac{64}{9}$ .

Step 4: Subtract. $\frac{181}{9} - \frac{64}{9} = \frac{117}{9} = 13$ .

Key Takeaways

When integrating $(ax+b)^n$ , the antiderivative is $\frac{(ax+b)^{n+1}}{a(n+1)}$ .
The area under a curve above the $x$ -axis between $x = a$ and $x = b$ is $\int_a^b y \, dx$ .
Careful arithmetic with fractions and powers is essential to avoid errors.

Common Mistakes

Forgetting to divide by the derivative of the inner function ( $3$ ) when integrating $4(3x+4)^{\frac{1}{2}}$ .
Incorrectly computing $25^{\frac{3}{2}}$ as $25 \times \frac{3}{2}$ instead of $(\sqrt{25})^3 = 125$ .
Sign errors when evaluating $-7^2 - 6 \times 7$ .
Forgetting to subtract the lower limit evaluation.

Things to Be Careful About

Ensure the curve is above the $x$ -axis throughout $[0, 7]$ so the integral directly gives the area (the diagram confirms this).
$25^{\frac{3}{2}}$ means $(25^{\frac{1}{2}})^3 = 5^3 = 125$ , not $25^{1.5}$ computed incorrectly.
The final answer $13$ is exact; no rounding is needed.

Techniques used

integrate composite power function using reverse chain ruleintegrate polynomial termsevaluate definite integral with upper and lower limitssimplify fractional arithmetic

(b)

Find the area of region $B$ .

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Region B is bounded by the curve, the line segment $PQ$ , and the $y$ -axis. From the diagram, region B plus region A together form the triangle $OPQ$ . Therefore, area of B = area of triangle $OPQ$ minus area of region A.

We need to:

Differentiate $y$ to find the gradient of the tangent at $P(7, 0)$ .
Write the equation of the tangent line and find where it meets the $y$ -axis (point $Q$ ).
Calculate the area of triangle $OPQ$ .
Subtract the area of region A (from part (a)) to get the area of region B.

Working

Step 1: Find $\frac{dy}{dx}$ .

\frac{dy}{dx} = 4 \times \frac{1}{2}(3x+4)^{-\frac{1}{2}} \times 3 - 2

= 6(3x+4)^{-\frac{1}{2}} - 2

Step 2: Evaluate the gradient at $x = 7$ .

\left.\frac{dy}{dx}\right|_{x=7} = 6(25)^{-\frac{1}{2}} - 2 = \frac{6}{5} - 2 = -\frac{4}{5}

Step 3: Equation of the tangent at $P(7, 0)$ .

Using $y - y_1 = m(x - x_1)$ :

y - 0 = -\frac{4}{5}(x - 7)

y = -\frac{4}{5}x + \frac{28}{5}

Step 4: Find point $Q$ (where the tangent meets the $y$ -axis).

Set $x = 0$ :

y = \frac{28}{5}

So $Q = \left(0, \frac{28}{5}\right)$ .

Step 5: Area of triangle $OPQ$ .

The triangle has base $OP = 7$ along the $x$ -axis and height $OQ = \frac{28}{5}$ along the $y$ -axis:

\text{Area of } \triangle OPQ = \frac{1}{2} \times 7 \times \frac{28}{5} = \frac{98}{5}

Step 6: Area of region B.

\text{Area of } B = \text{Area of } \triangle OPQ - \text{Area of } A = \frac{98}{5} - 13 = \frac{98 - 65}{5} = \frac{33}{5}

Answer

\text{Area of region } B = \frac{33}{5}

Final answer

33/5

Detailed explanation

Walkthrough

Region B is the area between the tangent line $PQ$ and the curve, bounded on the left by the $y$ -axis. Looking at the diagram, the triangle $OPQ$ (with vertices $O(0,0)$ , $P(7,0)$ , and $Q(0, 28/5)$ ) contains both region A and region B. Since region A is between the curve and the axes, and region B is between the curve and the tangent line, we have:

\text{Area of } \triangle OPQ = \text{Area of } A + \text{Area of } B

Therefore, $\text{Area of } B = \text{Area of } \triangle OPQ - \text{Area of } A$ .

Step 1: Differentiate. Using the chain rule on $4(3x+4)^{\frac{1}{2}}$ , we bring down the power $\frac{1}{2}$ , multiply by the outer coefficient $4$ , reduce the power to $-\frac{1}{2}$ , and multiply by the derivative of the inner function $(3x+4)' = 3$ . This gives $6(3x+4)^{-\frac{1}{2}}$ . The derivative of $-2x - 6$ is $-2$ .

Step 2: Gradient at $P$ . Substituting $x = 7$ gives $6(25)^{-1/2} - 2 = 6/5 - 2 = -4/5$ . This negative gradient makes sense as the curve is decreasing at $P$ .

Step 3: Tangent equation. Using point-slope form with $P(7, 0)$ and gradient $-4/5$ , we get $y = -\frac{4}{5}(x-7) = -\frac{4}{5}x + \frac{28}{5}$ .

Step 4: Point $Q$ . Setting $x = 0$ gives $y = 28/5$ , so $Q = (0, 28/5)$ .

Step 5: Triangle area. Triangle $OPQ$ is a right-angled triangle with legs $7$ and $28/5$ , so its area is $\frac{1}{2} \times 7 \times \frac{28}{5} = \frac{98}{5}$ .

Step 6: Region B. Subtracting the area from part (a): $\frac{98}{5} - 13 = \frac{33}{5}$ .

Key Takeaways

The gradient of a curve at a point is found by differentiating and substituting.
The equation of a tangent line can be written using point-slope form $y - y_1 = m(x - x_1)$ .
When a region is bounded by a curve and a tangent line, it can sometimes be found by subtracting the area under the curve from the area of a simple geometric shape.
Area relationships: $\text{Area}(B) = \text{Area}(\triangle OPQ) - \text{Area}(A)$ .

Common Mistakes

Forgetting the chain rule factor of $3$ when differentiating $4(3x+4)^{\frac{1}{2}}$ .
Computing $25^{-1/2}$ as $-5$ instead of $\frac{1}{5}$ .
Using the wrong point-slope form or making sign errors in the tangent equation.
Forgetting that region B = triangle area − region A, and instead trying to integrate the difference directly.
Arithmetic errors when computing $\frac{98}{5} - 13$ .

Things to Be Careful About

The gradient at $P$ is negative ( $-4/5$ ), which is consistent with the curve decreasing toward the $x$ -axis.
Point $Q$ is above the curve's $y$ -intercept, which is why region B exists between the tangent and the curve.
The area of region B must be positive; if your answer is negative, check your subtraction order.
Use exact fractions throughout to avoid rounding errors.

Techniques used

differentiate composite function using chain ruleevaluate gradient at a specific pointform equation of tangent linefind y-intercept of tangentsubtract area of region A from triangle area

Q11MediumFunctionsQuadratics

Functions $\mathrm{f}$ and $\mathrm{g}$ are defined for all real values of $x$ by

\mathrm{f}(x) = 4x^2 - c \quad \text{and} \quad \mathrm{g}(x) = 2x + k,

where $c$ and $k$ are positive constants. It is given that $\mathrm{g}^{-1}(3k + 1) = c$ .

(a)

Show that $\mathrm{gf}(x) = 8x^2 - k - 1$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Find the inverse of $\mathrm{g}$ , use the given condition $\mathrm{g}^{-1}(3k+1)=c$ to relate $c$ and $k$ , then form $\mathrm{gf}(x)$ and simplify.

Working

Since $\mathrm{g}(x)=2x+k$ , write $y=2x+k$ and solve for $x$ :

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

so

\mathrm{g}^{-1}(x)=\frac{1}{2}(x-k)

Using $\mathrm{g}^{-1}(3k+1)=c$ :

c=\frac{1}{2}(3k+1-k)=\frac{1}{2}(2k+1)

Hence

2c=2k+1

Now form the composite:

\mathrm{gf}(x)=\mathrm{g}(\mathrm{f}(x))=2(4x^2-c)+k=8x^2-2c+k

Substitute $2c=2k+1$ :

\mathrm{gf}(x)=8x^2-(2k+1)+k=8x^2-k-1

Answer

\mathrm{gf}(x)=8x^2-k-1

Final answer

gf(x) = 8x^2 - k - 1

Detailed explanation

Walkthrough

We are told the value of the inverse function $\mathrm{g}^{-1}$ at $3k+1$ is $c$ . To use this, first find the inverse of the linear function $\mathrm{g}(x)=2x+k$ by writing $y=2x+k$ and solving for $x$ ; this gives $\mathrm{g}^{-1}(x)=\frac{1}{2}(x-k)$ . Substituting $x=3k+1$ gives $c=\frac{1}{2}(2k+1)$ , so $2c=2k+1$ . This relation is the key link between the two constants.

Next, form the composite $\mathrm{gf}(x)$ , which means apply $\mathrm{f}$ first and then $\mathrm{g}$ : replace every $x$ in $\mathrm{g}(x)$ by $\mathrm{f}(x)=4x^2-c$ . This gives $2(4x^2-c)+k=8x^2-2c+k$ . Finally, use $2c=2k+1$ to replace $-2c$ by $-(2k+1)$ , which simplifies to $8x^2-k-1$ , exactly the required expression.

Key Takeaways

This question tests inverse functions, composite functions, and the importance of using a derived relationship between constants. It also shows that a 'show that' question needs every substitution and simplification written out.

Common Mistakes

Computing $\mathrm{fg}(x)$ instead of $\mathrm{gf}(x)$ ; the order matters.
Making an arithmetic slip when finding the inverse, especially forgetting to divide the constant $k$ by $2$ .
Not using the relation $2c=2k+1$ after deriving it, so the final expression cannot be simplified to the required form.
Omitting intermediate detail in a 'show that' question; the mark scheme requires all necessary detail.

Things to Be Careful About

$\mathrm{gf}(x)$ means apply $\mathrm{f}$ first, then $\mathrm{g}$ .
The inverse relation can also be written as $\mathrm{g}(c)=3k+1$ , which gives $2c+k=3k+1$ and hence $2c=2k+1$ .
Keep the constants $c$ and $k$ general until the final substitution; do not try to find their values in this part.

Techniques used

find the inverse of a linear functionuse the inverse condition to relate constantsform the composite functionsimplify using a derived relation

(b)

The curve with equation $y = 8x^2 - k - 1$ is transformed to the curve with equation $y = \mathrm{h}(x)$ by the following sequence of transformations.

\text{Translation of } \begin{pmatrix} 2 \\ 3 \end{pmatrix}

\text{Stretch in the } y\text{-direction by scale factor } k

\text{Reflection in the } x\text{-axis}

Find an expression for $\mathrm{h}(x)$ in terms of $x$ and $k$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Apply the three transformations in the order given: translation by the vector $(2,3)$ , stretch in the $y$ -direction by scale factor $k$ , then reflection in the $x$ -axis. Track how $x$ and $y$ change at each stage.

Working

Start with

y=8x^2-k-1

A translation by the vector $(2,3)$ moves the graph $2$ units right and $3$ units up. Replace $x$ by $x-2$ and add $3$ to $y$ :

y=8(x-2)^2-k-1+3=8(x-2)^2-k+2

A stretch in the $y$ -direction by scale factor $k$ multiplies the $y$ -value by $k$ :

y=k(8(x-2)^2-k+2)

A reflection in the $x$ -axis multiplies the $y$ -value by $-1$ :

h(x)=-k(8(x-2)^2-k+2)

This can be expanded as

h(x)=-8k(x-2)^2+k^2-2k

Answer

h(x)=-k(8(x-2)^2-k+2)=-8k(x-2)^2+k^2-2k

Final answer

h(x) = -k(8(x-2)^2 - k + 2) = -8k(x-2)^2 + k^2 - 2k

Detailed explanation

Walkthrough

We start from $y=8x^2-k-1$ and apply transformations in the exact order stated.

Translation by $(2,3)$ : a graph shifts right by $2$ and up by $3$ . In the equation, replace $x$ by $x-2$ (because the old $x$ equals new $x$ minus $2$ ) and add $3$ to the whole expression. This gives $y=8(x-2)^2-k+2$ .
Stretch in the $y$ -direction by scale factor $k$ : every $y$ -coordinate is multiplied by $k$ , so multiply the whole right-hand side by $k$ .
Reflection in the $x$ -axis: every $y$ -coordinate changes sign, so multiply the whole right-hand side by $-1$ .

The result is $h(x)=-k(8(x-2)^2-k+2)$ , which can be expanded to $-8k(x-2)^2+k^2-2k$ . The expanded form is useful for part (c).

Key Takeaways

This question tests the order of graph transformations. Translation inside the function affects $x$ ; translation outside affects $y$ . A vertical stretch multiplies $y$ -values, and a reflection in the $x$ -axis changes the sign of $y$ .

Common Mistakes

Applying the translation to $x$ as $x+2$ instead of $x-2$ .
Applying the stretch before the translation, which changes the effect of the translation.
Forgetting the reflection sign, or applying it before the stretch.
Not simplifying $-k-1+3$ correctly to $-k+2$ .

Things to Be Careful About

The order in the question is translation, then stretch, then reflection; this order is essential.
The stretch is in the $y$ -direction, so it multiplies the whole expression, not just the $x^2$ term.
The final expression can be left factored or expanded; both are acceptable, but the expanded form helps identify the maximum in part (c).

Techniques used

apply a translation to a functionapply a vertical stretchapply a reflection in the x-axis

(c)

The range of $\mathrm{h}$ is given by $\mathrm{h}(x) \leqslant 15$ .

Find the values of $c$ and $k$ .

3M Similar

DifficultyMedium

Worked solution

Approach

The expression for $\mathrm{h}(x)$ is a quadratic with negative leading coefficient, so its graph is a downward-opening parabola. Its maximum value occurs at the vertex, $x=2$ . Since the range is $\mathrm{h}(x)\leq 15$ , the maximum value must be exactly $15$ . Solve for $k$ , then use $2c=2k+1$ from part (a) to find $c$ .

Working

From part (b),

h(x)=-8k(x-2)^2+k^2-2k

Because $k>0$ , the coefficient $-8k$ is negative, so the maximum occurs at $x=2$ :

h(2)=k^2-2k

The range $h(x)\leq 15$ means the maximum is $15$ :

k^2-2k=15

Solve:

k^2-2k-15=0

(k-5)(k+3)=0

So $k=5$ or $k=-3$ . Since $k$ is positive, $k=5$ .

Using $2c=2k+1$ :

2c=2(5)+1=11

c=\frac{11}{2}

Answer

k=5,\quad c=\frac{11}{2}

Final answer

k = 5, c = 11/2

Detailed explanation

Walkthrough

The function $h(x)=-8k(x-2)^2+k^2-2k$ is a quadratic in $x$ . Its leading coefficient is $-8k$ , which is negative because $k>0$ , so the parabola opens downwards and has a single maximum at its vertex. The vertex is at $x=2$ because the squared term is $(x-2)^2$ . At $x=2$ , the squared term is $0$ , so the maximum value is $k^2-2k$ .

The statement that the range is $h(x)\leq 15$ tells us the largest output value is $15$ . Therefore the maximum $k^2-2k$ must equal $15$ . This gives $k^2-2k-15=0$ , which factorises as $(k-5)(k+3)=0$ . The solutions are $k=5$ and $k=-3$ , but $k$ is given as positive, so $k=5$ .

Finally, return to the relation from part (a), $2c=2k+1$ . Substituting $k=5$ gives $2c=11$ , so $c=\frac{11}{2}$ .

Key Takeaways

This part combines graph transformations with the range of a quadratic. A downward-opening quadratic has a maximum at its vertex, and the range is all values below or equal to that maximum. It also uses the earlier relation between $c$ and $k$ .

Common Mistakes

Taking $k=-3$ as a valid solution without checking the condition $k>0$ .
Treating the range condition as an inequality $k^2-2k\leq 15$ and stopping without realising the maximum must actually equal $15$ .
Forgetting to use $2c=2k+1$ to find $c$ after finding $k$ .
Misidentifying the vertex as $x=-2$ rather than $x=2$ .

Things to Be Careful About

The vertex is at $x=2$ , not $x=-2$ , because the expression is $(x-2)^2$ .
Since $k>0$ , the coefficient $-8k$ is definitely negative, so the parabola opens downwards.
The range condition $h(x)\leq 15$ means the maximum is exactly $15$ , so equality must be used.
Both constants are positive: $k=5$ and $c=\frac{11}{2}$ satisfy this.

Techniques used

identify the maximum of a downward-opening quadraticsolve a quadratic equation by factorisationsubstitute back to find the second constant

Mathematics 9709/12 — February/March 2025

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