Mathematics 9709 / 15 — October/November 2025

DifficultyEasy

Worked solution

Approach

The equation of a circle with centre $(a,b)$ and radius $r$ is

(x-a)^2 + (y-b)^2 = r^2

Substitute the given centre and radius into this standard form.

Working

The centre is $(2,6)$ and the radius is $10$ , so

(x-2)^2 + (y-6)^2 = 10^2

(x-2)^2 + (y-6)^2 = 100

Expanding gives the equivalent form:

x^2 + y^2 - 4x - 12y - 60 = 0

Answer

(x-2)^2 + (y-6)^2 = 100

(Equivalently, $x^2 + y^2 - 4x - 12y - 60 = 0$ .)

Final answer

(x-2)^2 + (y-6)^2 = 100

Detailed explanation

Walkthrough

A circle is the set of all points that are exactly a distance $r$ from its centre $(a, b)$ . If a point $(x, y)$ is on the circle, the distance from $(x, y)$ to $(a, b)$ is $r$ , so the equation is

(x-a)^2 + (y-b)^2 = r^2.

Here we are told the centre is $(2, 6)$ and the radius is $10$ , so $a=2$ , $b=6$ and $r=10$ . Substituting these values gives

(x-2)^2 + (y-6)^2 = 10^2 = 100.

That is the standard equation, and the mark scheme accepts it in this form, or expanded as

x^2 + y^2 - 4x - 12y - 60 = 0.

Key Takeaways

The general equation of a circle requires the centre and the radius.
The right-hand side is $r^2$ , not $r$ .
The signs inside the brackets are opposite to the coordinates of the centre.
Expanding the completed-square form must produce matching coefficients for $x^2$ , $y^2$ and the constant.

Common Mistakes

Writing $10$ instead of $100$ on the right-hand side.
Using $(x+2)^2$ or $(y+6)^2$ instead of $(x-2)^2$ and $(y-6)^2$ .
Expanding incorrectly: for example, writing $(x-2)^2$ as $x^2 - 4$ instead of $x^2 - 4x + 4$ .
Omitting the $y$ -coordinates entirely.

Things to Be Careful About

The radius must be squared. When expanding, the constant is $100$ , so the equation becomes

x^2 - 4x + 4 + y^2 - 12y + 36 = 100,

which simplifies to

x^2 + y^2 - 4x - 12y - 60 = 0.

The mark scheme gives B1 for two correct components and B2 for all three components of the equation.

Techniques used

recall the general equation of a circlesubstitute the centre and radius into the standard form

(b)

The circle passes through the point $(8, k)$ .

Find the two possible values of $k$ .

DifficultyMedium-Easy

Worked solution

Approach

A point lies on the circle exactly when its coordinates satisfy the circle equation. Substitute $x=8$ and $y=k$ into the equation from part (a) and solve the resulting equation for $k$ .

Working

The circle equation is

(x-2)^2 + (y-6)^2 = 100.

Substitute $(8,k)$ :

(8-2)^2 + (k-6)^2 = 100.

Simplify:

6^2 + (k-6)^2 = 100

36 + (k-6)^2 = 100

Subtract $36$ :

(k-6)^2 = 64.

Take square roots:

k - 6 = \pm 8.

Therefore

k = 6 + 8 = 14 \quad \text{or} \quad k = 6 - 8 = -2.

(Expanding $(k-6)^2 = 64$ gives $k^2 - 12k - 28 = 0$ , i.e. $(k-14)(k+2)=0$ , leading to the same values.)

Answer

k = 14 \quad \text{or} \quad k = -2

Final answer

k = 14 or k = -2

Detailed explanation

Walkthrough

We know that $(8,k)$ lies on the circle, so its coordinates must satisfy the circle equation from part (a):

(x-2)^2 + (y-6)^2 = 100.

Substitute $x=8$ and $y=k$ :

(8-2)^2 + (k-6)^2 = 100.

The first term is $6^2 = 36$ , so:

36 + (k-6)^2 = 100.

Subtract $36$ from both sides:

(k-6)^2 = 64.

Taking square roots, remember that both positive and negative square roots are possible:

k-6 = \pm 8.

Therefore

k = 14 \quad \text{or} \quad k = -2.

Geometrically, the vertical line $x=8$ meets the circle in two points: one above the centre and one below it.

Key Takeaways

A point is on a curve only if its coordinates satisfy the curve's equation.
Substituting a known coordinate gives an equation in the unknown coordinate.
Solving $(k-6)^2 = 64$ requires the $\pm$ sign, producing two values.
The two answers correspond to the two intersections of the vertical line $x=8$ with the circle.

Common Mistakes

Forgetting to square the radius: using $100$ is correct, not $10$ .
Taking only the positive square root and missing $k = -2$ .
Rearranging incorrectly: $(k-6)^2 = 64$ , not $k-6 = \sqrt{64}$ with only one sign.
If using expansion, making a sign error in forming $k^2 - 12k - 28 = 0$ .

Things to Be Careful About

Since $(k-6)^2$ is a square, $k-6$ can be $8$ or $-8$ . This gives two different points on the same vertical line. If you write only the two answers without any working, the mark scheme awards only SC B2 for part (b), so show the substitution and the simplification to gain the method marks.

Techniques used

substitute a point into a circle equationisolate a squared term by rearrangingtake square roots to solve for a coordinateinterpret the two roots as the two positions of the point

Q2Medium-EasySeries

A geometric progression has first term $3 + 4\sqrt{2}$ and second term $5 - \sqrt{2}$ .

(a)

Find the common ratio of the geometric progression. Give your answer in the form $\sqrt{2} + p$ , where $p$ is an integer to be found.

DifficultyMedium-Easy

Worked solution

Approach

The common ratio of a geometric progression is the quotient of any term by the previous term: $r = \frac{u_2}{u_1}$ . The denominator contains a surd, so rationalise it by multiplying numerator and denominator by the conjugate $3 - 4\sqrt{2}$ , then simplify to the required form.

Working

Let the first term be $a = 3 + 4\sqrt{2}$ and the second term be $u_2 = 5 - \sqrt{2}$ .

r = \frac{5-\sqrt{2}}{3+4\sqrt{2}}

Rationalise the denominator:

r = \frac{(5-\sqrt{2})(3-4\sqrt{2})}{(3+4\sqrt{2})(3-4\sqrt{2})}

Expand the numerator:

(5-\sqrt{2})(3-4\sqrt{2}) = 15 - 20\sqrt{2} - 3\sqrt{2} + 8 = 23 - 23\sqrt{2}

Expand the denominator using the difference of two squares:

(3+4\sqrt{2})(3-4\sqrt{2}) = 9 - 32 = -23

Therefore

r = \frac{23 - 23\sqrt{2}}{-23} = \sqrt{2} - 1

So $p = -1$ .

Answer

r = \sqrt{2} - 1, \quad p = -1

Final answer

r = sqrt(2) - 1, p = -1

Detailed explanation

Walkthrough

We are given the first two terms of a geometric progression. In a geometric progression, each term is obtained from the previous one by multiplying by a fixed common ratio $r$ . Therefore the common ratio is found by dividing the second term by the first term: $r = \frac{5-\sqrt{2}}{3+4\sqrt{2}}$ .

The denominator is a surd, so the answer has not yet been written in a useful form. To simplify it, multiply both the numerator and denominator by the conjugate of the denominator, $3 - 4\sqrt{2}$ . This uses the identity $(a+b)(a-b)=a^2-b^2$ , so the denominator becomes $(3)^2 - (4\sqrt{2})^2 = 9 - 32 = -23$ .

The numerator expands term by term: $5 \times 3 = 15$ , $5 \times (-4\sqrt{2}) = -20\sqrt{2}$ , $(-\sqrt{2}) \times 3 = -3\sqrt{2}$ , and $(-\sqrt{2}) \times (-4\sqrt{2}) = +8$ . Adding these gives $23 - 23\sqrt{2}$ .

Since the denominator is $-23$ , the whole fraction simplifies to $-1 + \sqrt{2}$ , which is more neatly written as $\sqrt{2} - 1$ . Comparing this with $\sqrt{2} + p$ gives $p = -1$ .

Key Takeaways

The common ratio of a geometric progression is found by dividing any term by the previous term.
Rationalising a denominator with a surd means multiplying top and bottom by the conjugate.
The difference of two squares is the key to removing a surd from a denominator such as $3 + 4\sqrt{2}$ .

Common Mistakes

Multiplying only the numerator by the conjugate and not the denominator; this changes the value.
Sign errors when expanding $(-\sqrt{2})(-4\sqrt{2}) = 8$ , not $-8$ .
Forgetting that the denominator is negative, so the final simplification is $\sqrt{2} - 1$ , not $-\sqrt{2} + 1$ with the wrong sign.
Not showing the rationalisation process clearly; the mark scheme requires the process to be visible.

Things to Be Careful About

The answer must be exact and in the required form $\sqrt{2} + p$ , so ensure the final fraction has been fully simplified.
Check the sign carefully when simplifying $-\frac{23 - 23\sqrt{2}}{23}$ ; each term is divided by $-23$ .

Techniques used

divide a term by the previous term to find the common ratiorationalise a surd denominatorsimplify to the required form

(b)

Find the sum to infinity of the geometric progression.

DifficultyMedium-Easy

Worked solution

Approach

A geometric progression has a sum to infinity only when $|r| < 1$ . Its sum is $S_\infty = \frac{a}{1-r}$ . Here $a = 3 + 4\sqrt{2}$ and $r = \sqrt{2} - 1$ from part (a). Substitute these values and rationalise the result.

Working

Check that the series converges:

|\sqrt{2} - 1| = \sqrt{2} - 1 \approx 0.414 < 1

Substitute into the formula for the sum to infinity:

S_\infty = \frac{3+4\sqrt{2}}{1-(\sqrt{2}-1)} = \frac{3+4\sqrt{2}}{2-\sqrt{2}}

Rationalise by multiplying numerator and denominator by $2+\sqrt{2}$ :

S_\infty = \frac{(3+4\sqrt{2})(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}

Expand the numerator:

(3+4\sqrt{2})(2+\sqrt{2}) = 6 + 3\sqrt{2} + 8\sqrt{2} + 4(2) = 14 + 11\sqrt{2}

Expand the denominator:

(2-\sqrt{2})(2+\sqrt{2}) = 4 - 2 = 2

Therefore

S_\infty = \frac{14 + 11\sqrt{2}}{2} = 7 + \frac{11\sqrt{2}}{2}

Answer

S_\infty = 7 + \frac{11\sqrt{2}}{2}

Final answer

7 + 11√2/2

Detailed explanation

Walkthrough

For a geometric progression, the sum to infinity exists only if the common ratio $r$ satisfies $|r| < 1$ . Here $r = \sqrt{2} - 1$ , and since $\sqrt{2} \approx 1.414$ , $r \approx 0.414$ , which is less than $1$ . Therefore the series converges.

The sum to infinity formula is $S_\infty = \frac{a}{1-r}$ , where $a$ is the first term. Substituting $a = 3 + 4\sqrt{2}$ and $r = \sqrt{2} - 1$ gives

S_\infty = \frac{3+4\sqrt{2}}{1-(\sqrt{2}-1)}.

It is essential to keep the brackets around $r$ : $1-(\sqrt{2}-1) = 1 - \sqrt{2} + 1 = 2 - \sqrt{2}$ . If the brackets are missed, the expression becomes $1 - \sqrt{2} - 1 = -\sqrt{2}$ , which is wrong.

The resulting fraction has a surd in the denominator, so multiply numerator and denominator by the conjugate $2 + \sqrt{2}$ . The numerator becomes $14 + 11\sqrt{2}$ and the denominator becomes $2$ . Hence

S_\infty = \frac{14 + 11\sqrt{2}}{2},

which is equivalent to $7 + \frac{11\sqrt{2}}{2}$ .

Key Takeaways

A geometric progression has a finite sum to infinity only when the common ratio satisfies $|r| < 1$ .
The formula for the sum to infinity is $S_\infty = \frac{a}{1-r}$ .
Rationalising is also useful when applying the sum-to-infinity formula if the common ratio contains a surd.

Common Mistakes

Forgetting to check whether $|r| < 1$ before using the sum-to-infinity formula.
Omitting the brackets in $1-(\sqrt{2}-1)$ , which changes the denominator from $2 - \sqrt{2}$ to $-\sqrt{2}$ .
Making sign errors when rationalising $(2-\sqrt{2})(2+\sqrt{2}) = 4 - 2 = 2$ ; the result is not negative.
Not simplifying the final answer to the printed equivalent forms $\frac{14+11\sqrt{2}}{2}$ or $7 + \frac{11\sqrt{2}}{2}$ .

Things to Be Careful About

The mark scheme allows follow-through from part (a), but only if $r$ has been evaluated and satisfies $-1 < r < 1$ .
Do not condone missing brackets unless the substitution correctly uses them later in the calculation.
Equivalent answers are accepted, but it is safest to write the form with the rationalised denominator.

Techniques used

check the convergence conditionsubstitute into the sum-to-infinity formularationalise the resulting fraction

Q3Medium-EasyQuadratics

(a)

Express $4x^2 + 10x + 6$ in the form $a(x + b)^2 + c$ , where $a$ , $b$ and $c$ are rational constants to be determined.

DifficultyMedium-Easy

Worked solution

Approach

Complete the square on $4x^2+10x+6$ . Because the coefficient of $x^2$ is $4$ , factor $4$ out of the $x^2$ and $x$ terms before completing the square inside the brackets.

Working

4x^2+10x+6 = 4\left(x^2+\frac{5}{2}x\right)+6

Complete the square inside the bracket:

x^2+\frac{5}{2}x = \left(x+\frac{5}{4}\right)^2 - \left(\frac{5}{4}\right)^2 = \left(x+\frac{5}{4}\right)^2 - \frac{25}{16}

Therefore,

4\left(x^2+\frac{5}{2}x\right)+6 = 4\left(\left(x+\frac{5}{4}\right)^2-\frac{25}{16}\right)+6 = 4\left(x+\frac{5}{4}\right)^2-\frac{25}{4}+6 = 4\left(x+\frac{5}{4}\right)^2-\frac{1}{4}

Thus

a=4,\quad b=\frac{5}{4},\quad c=-\frac{1}{4}.

Answer

4x^2+10x+6 = 4\left(x+\frac{5}{4}\right)^2-\frac{1}{4}

with $a=4$ , $b=\frac{5}{4}$ and $c=-\frac{1}{4}$ .

Final answer

a = 4, b = 5/4, c = -1/4

Detailed explanation

Walkthrough

We need to rewrite the quadratic in completed-square form. Because the leading coefficient is $4$ , we cannot simply take half of $10$ ; instead, factor $4$ out of the $x^2$ and $x$ terms first. This leaves $x^2+\frac{5}{2}x$ inside the bracket. To complete the square, use

\left(x+\frac{5}{4}\right)^2 = x^2+\frac{5}{2}x+\frac{25}{16},

so

x^2+\frac{5}{2}x = \left(x+\frac{5}{4}\right)^2-\frac{25}{16}.

Multiplying by the outer $4$ gives $4\left(x+\frac{5}{4}\right)^2-\frac{25}{4}$ . Adding the original constant $6$ gives

-\frac{25}{4}+6 = -\frac{25}{4}+\frac{24}{4} = -\frac{1}{4}.

Thus the required constants are $a=4$ , $b=\frac{5}{4}$ and $c=-\frac{1}{4}$ .

Key Takeaways

Completing the square reorganises a quadratic into a perfect square plus a constant. When the coefficient of $x^2$ is not $1$ , factor it out before completing the square. The constants $a$ , $b$ and $c$ can then be read directly from the final form.

Common Mistakes

Forgetting to factor out the leading coefficient before halving the coefficient of $x$ , which gives the wrong value of $b$ .
Sign errors when subtracting the squared half-coefficient inside the bracket.
Arithmetic errors when combining $-\frac{25}{4}$ and $6$ .
Not simplifying to a single rational constant for $c$ .

Things to Be Careful About

The bracket must contain $x+\frac{5}{4}$ , not $x+\frac{5}{2}$ . Also, expanding the final answer is a quick way to catch sign mistakes. Decimals are allowed by the mark scheme, but exact fractions are safer.

Techniques used

factor out the leading coefficientcomplete the square inside the bracketsimplify the constant term

(b)

The curve with equation $y = 4x^2 + 10x + 6$ and the line $y = k$ have exactly one point of intersection.

Using your answer to part (a) or otherwise, state the value of the constant $k$ .

1M Similar

DifficultyEasy

Worked solution

Approach

The completed-square form shows the curve is a parabola with vertex at $x=-\frac{5}{4}$ and minimum value $-\frac{1}{4}$ . A horizontal line $y=k$ has exactly one intersection with the parabola precisely when it passes through the vertex. Therefore $k$ equals the minimum value.

Working

From part (a),

y = 4x^2+10x+6 = 4\left(x+\frac{5}{4}\right)^2-\frac{1}{4}.

Since

4\left(x+\frac{5}{4}\right)^2 \ge 0,

the minimum value of $y$ is $-\frac{1}{4}$ , occurring at $x=-\frac{5}{4}$ .

The horizontal line $y=k$ meets the curve exactly once when it touches the vertex, so

k = -\frac{1}{4}.

Answer

k=-\frac{1}{4}

Final answer

k = -1/4

Detailed explanation

Walkthrough

From part (a), the curve can be written as $y=4\left(x+\frac{5}{4}\right)^2-\frac{1}{4}$ . Because a square cannot be negative, the smallest possible value of $4\left(x+\frac{5}{4}\right)^2$ is $0$ , which occurs when $x=-\frac{5}{4}$ . Therefore the lowest point on the parabola has $y$ -coordinate $-\frac{1}{4}$ .

A horizontal line $y=k$ intersects the curve at points satisfying

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

i.e.

4\left(x+\frac{5}{4}\right)^2=k+\frac{1}{4}.

The left-hand side is zero only at the vertex; on either side of the vertex it is positive and gives two intersections. Hence there is exactly one intersection exactly when $k+\frac{1}{4}=0$ , so $k=-\frac{1}{4}$ .

Key Takeaways

The completed-square form immediately identifies the vertex of a parabola. For a positive leading coefficient, the constant term in $a(x+b)^2+c$ is the minimum value. A horizontal line meets a parabola exactly once only when it touches the vertex.

Common Mistakes

Stating $x=-\frac{1}{4}$ instead of $k=-\frac{1}{4}$ : the question asks for the $y$ -value of the line, not the $x$ -coordinate of the point.
Using an incorrect value of $c$ from part (a) without following through.
Thinking the line must have gradient $-\frac{1}{4}$ or involve $x=-\frac{5}{4}$ .

Things to Be Careful About

The phrase "exactly one point of intersection" means the line is a tangent at the vertex for this curve, so $k$ is the minimum value of $y$ . If part (a) was answered with a different completed-square form, $k$ should be the corresponding constant. The answer $-\frac{1}{4}$ may also be written as $-0.25$ .

Techniques used

read the vertex from completed-square forminterpret a single intersection as a horizontal tangent at the vertexstate the required line value

Q4MediumCircular MeasureTrigonometry

Fig. 4.1

The diagram shows the design for a company’s new logo. The sector of the circle, centre $O$ , has radius $r\text{ cm}$ . The acute angle $AOC = \frac{1}{3}\pi$ radians. The quadrilateral $OABC$ is a rhombus.

(a)

Find an expression for the perimeter of the design. Give your answer in terms of $\pi$ and $r$ .

DifficultyMedium-Easy

Worked solution

Approach

The perimeter of the design is the major arc $AC$ together with the two sides $AB$ and $BC$ of the rhombus. Find the reflex angle at $O$ , apply the arc length formula $s = r\theta$ , then add the two rhombus sides (each of length $r$ because all four sides of a rhombus are equal).

Working

The reflex angle at $O$ — the central angle subtended by the major arc $AC$ — is:

2\pi - \frac{\pi}{3} = \frac{5\pi}{3}

The length of the major arc $AC$ is:

\text{arc }AC = r \times \frac{5\pi}{3} = \frac{5\pi r}{3}

Since $OABC$ is a rhombus, all four sides are equal. The two radii $OA$ and $OC$ both have length $r$ , so:

AB = BC = r

The perimeter of the design is the major arc plus sides $AB$ and $BC$ :

\text{Perimeter} = \frac{5\pi r}{3} + r + r = \frac{5\pi r}{3} + 2r

Answer

\text{Perimeter} = \frac{5\pi r}{3} + 2r \text{ cm}

Final answer

5πr/3 + 2r cm

Detailed explanation

Walkthrough

The design's boundary has three parts: the major arc $AC$ , side $AB$ , and side $BC$ . The key recognition is that the major (long) arc is the one on the boundary, not the minor arc.

The minor sector at $O$ has central angle $\frac{\pi}{3}$ , so the major sector (and its arc) corresponds to the reflex angle:

2\pi - \frac{\pi}{3} = \frac{5\pi}{3}

The arc length formula $s = r\theta$ (with $\theta$ in radians) gives the major arc length:

\text{arc }AC = \frac{5\pi r}{3}

Because $OABC$ is a rhombus, all four sides are equal. The two radii $OA$ and $OC$ are both length $r$ , so the remaining sides $AB$ and $BC$ also have length $r$ . Adding the three pieces of the boundary:

\frac{5\pi r}{3} + r + r = \frac{5\pi r}{3} + 2r

Key Takeaways

The major arc corresponds to the reflex angle $2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ at the centre.
In a rhombus, all four sides are equal.
The arc length formula $s = r\theta$ requires the angle in radians.

Common Mistakes

Using the acute angle $\frac{\pi}{3}$ for the arc length instead of the reflex angle $\frac{5\pi}{3}$ — that would compute the minor arc.
Forgetting to add both rhombus sides $AB$ and $BC$ .
Treating $OA$ and $OC$ as part of the design's boundary — only the major arc (not the radii) is on the boundary.

Things to Be Careful About

The final answer must be exact (in terms of $\pi$ and $r$ ), not a decimal approximation.
Reflex angles are $2\pi$ minus the acute angle, since angles around a point sum to $2\pi$ .

Techniques used

compute reflex angleapply arc length formulause rhombus property (all sides equal)sum perimeter components

(b)

It is now given that the perimeter of the design is $200\text{ cm}$ .

Find the area of the design. Give your answer to 3 significant figures.

5M Similar

DifficultyMedium

Worked solution

Approach

Set the perimeter expression from part (a) equal to $200$ and solve for $r$ . Then the design's area is the sum of two pieces: the major sector (with central angle $\frac{5\pi}{3}$ ) and the entire rhombus $OABC$ . The rhombus area is found by splitting the rhombus along diagonal $AC$ into two congruent triangles, each with two sides of length $r$ and included angle $\frac{\pi}{3} = 60°$ .

Working

Setting the perimeter equal to $200$ :

r\left(\frac{5\pi}{3} + 2\right) = 200

Solving for $r$ :

r = \frac{200}{\frac{5\pi}{3} + 2} = \frac{200 \times 3}{5\pi + 6} = \frac{600}{5\pi + 6}

r \approx \frac{600}{21.708} \approx 27.64 \text{ cm}

Major sector area (central angle $\frac{5\pi}{3}$ ):

A_{\text{sector}} = \frac{1}{2} r^2 \times \frac{5\pi}{3} = \frac{1}{2}(27.64)^2 \times \frac{5\pi}{3}

A_{\text{sector}} \approx \frac{1}{2} \times 763.97 \times 5.236 \approx 2000 \text{ cm}^2

Rhombus area (split by diagonal $AC$ into triangles $OAC$ and $ABC$ , each with sides $r$ , $r$ and included angle $60°$ ):

A_{\text{rhombus}} = 2 \times \frac{1}{2} r^2 \sin 60° = r^2 \cdot \frac{\sqrt{3}}{2}

A_{\text{rhombus}} \approx 763.97 \times 0.866 \approx 661.6 \text{ cm}^2

Total design area:

A_{\text{total}} = A_{\text{sector}} + A_{\text{rhombus}} \approx 2000 + 661.6 = 2661.6 \text{ cm}^2

To $3$ significant figures:

A_{\text{total}} \approx 2660 \text{ cm}^2

Answer

\text{Area} \approx 2660 \text{ cm}^2 \text{ (3 s.f.)}

Final answer

2660 cm² (3 s.f.)

Detailed explanation

Walkthrough

With the perimeter expression from part (a), we first recover the radius $r$ :

r\left(\frac{5\pi}{3} + 2\right) = 200 \quad\Rightarrow\quad r = \frac{600}{5\pi + 6} \approx 27.64 \text{ cm}

The design consists of two pieces: the major sector of the circle (with central angle $\frac{5\pi}{3}$ ) and the entire rhombus $OABC$ . The two pieces share the chord $AC$ as a common edge, but the chord is interior to the design (the boundary consists only of the major arc and the two rhombus sides $AB$ , $BC$ ). Adding the two areas gives the design's area.

Major sector area. Using $A = \frac{1}{2} r^2 \theta$ with $\theta = \frac{5\pi}{3}$ :

A_{\text{sector}} = \frac{1}{2}(27.64)^2 \times \frac{5\pi}{3} \approx 2000 \text{ cm}^2

Rhombus area. A rhombus can be split along a diagonal into two congruent triangles. The diagonal $AC$ produces triangles $OAC$ and $ABC$ , each with two sides of length $r$ and the included angle $\frac{\pi}{3} = 60°$ (the angle at $O$ in the rhombus and the equal opposite angle at $B$ ). Each triangle has area $\frac{1}{2}r^2 \sin 60°$ , so the rhombus area is:

A_{\text{rhombus}} = 2 \times \frac{1}{2}r^2 \sin 60° = r^2 \sin 60° = r^2 \cdot \frac{\sqrt{3}}{2} \approx 661.6 \text{ cm}^2

Total. Adding the two pieces:

A_{\text{total}} \approx 2000 + 661.6 = 2661.6 \text{ cm}^2 \approx 2660 \text{ cm}^2 \text{ (3 s.f.)}

Key Takeaways

The major sector area uses the reflex angle $\frac{5\pi}{3}$ in $A = \frac{1}{2}r^2 \theta$ .
A rhombus with side $r$ and interior angle $\theta$ has area $r^2 \sin\theta$ (or equivalently, two triangles each of area $\frac{1}{2}r^2 \sin\theta$ split along a diagonal).
$\sin 60° = \frac{\sqrt{3}}{2}$ is an exact value that should be used.
The design consists of the major sector plus the entire rhombus; the chord $AC$ lies inside the design.

Common Mistakes

Using $\frac{\pi}{3}$ instead of $\frac{5\pi}{3}$ for the central angle of the major sector.
Confusing $\sin 60° = \frac{\sqrt{3}}{2}$ with $\sin 30° = \frac{1}{2}$ .
Only adding the triangle $ABC$ (the half of the rhombus outside the circle) and ignoring the triangle $OAC$ that lies inside the major segment, giving $\approx 2330$ instead of $\approx 2660$ .
Forgetting to round the final answer to $3$ significant figures.

Things to Be Careful About

The reflex angle of the major sector is $\frac{5\pi}{3}$ , not $\frac{\pi}{3}$ .
The rhombus consists of two triangles: $OAC$ (inside the circle) and $ABC$ (outside the circle). Both must be included because the design's interior extends across the chord $AC$ — the part of the disk in the major segment is in the design, and the triangle $ABC$ sticking out of the circle is also in the design.
The final answer must be given to $3$ significant figures, as the question explicitly requires.

Techniques used

set perimeter equal to 200 and solve for rapply sector area formula with reflex anglesplit rhombus into two triangles along a diagonalapply (1/2)ab sin C with sin(60°) = √3/2sum the two areas and round to 3 s.f.

Q53MMedium-EasyQuadratics

Solve the equation

x^3 - 28 + \frac{27}{x^3} = 0

DifficultyMedium-Easy

Worked solution

Approach

Multiply both sides by $x^3$ to clear the fraction, then treat the result as a quadratic in $u = x^3$ . Factorise to find $u$ , then take cube roots to recover $x$ .

Working

Since $x \neq 0$ , multiply each term by $x^3$ :

x^6 - 28x^3 + 27 = 0

Let $u = x^3$ . Then:

u^2 - 28u + 27 = 0

Factorise:

(u - 1)(u - 27) = 0

So $u = 1$ or $u = 27$ . Replacing $u$ with $x^3$ :

x^3 = 1 \quad \text{or} \quad x^3 = 27

Taking cube roots:

x = 1 \quad \text{or} \quad x = 3

Answer

x = 1 \quad \text{or} \quad x = 3

Final answer

x = 1 or x = 3

Detailed explanation

Walkthrough

The equation contains the reciprocal term $\frac{27}{x^3}$ , which makes it awkward to solve directly. The first step is to multiply every term by $x^3$ ; this is valid because $x \neq 0$ (otherwise the fraction would be undefined). After multiplication the equation becomes the polynomial $x^6 - 28x^3 + 27 = 0$ .

Notice that this is really a quadratic in disguise: if we let $u = x^3$ , then $x^6 = (x^3)^2 = u^2$ . Substituting gives $u^2 - 28u + 27 = 0$ , a standard quadratic we can solve. Factorising: $(u - 1)(u - 27) = 0$ , so $u = 1$ and $u = 27$ .

Finally, undo the substitution. Since $u$ was $x^3$ , we solve $x^3 = 1$ and $x^3 = 27$ . The real cube root function is one-to-one, so each equation gives exactly one real value: $x = 1$ and $x = 3$ .

Key Takeaways

This question shows how to recognise an equation that is quadratic in a function of $x$ : after multiplying by $x^3$ , the expression $x^6 - 28x^3 + 27$ is a quadratic in $x^3$ . The substitution $u = x^3$ turns it into a familiar quadratic. It also reinforces that taking cube roots is a safe step for real numbers because every real number has exactly one real cube root.

Common Mistakes

Failing to multiply every term by $x^3$ , especially the constant term $28$ , giving an incorrect equation such as $x^6 - 28 + 27x^3 = 0$ .
Not recognising the quadratic-in- $x^3$ structure and trying to solve the original equation by inspection.
Omitting the factorisation step; the mark scheme requires a valid method for solving the quadratic (factorisation, formula, or completing the square).
Writing $x = \pm 1$ or $x = \pm 3$ ; for real equations with cube terms there are no negative roots because $x^3 = 1$ has only the real root $x = 1$ .

Things to Be Careful About

Since $x^3$ appears in the denominator, $x = 0$ is not in the domain; multiplication by $x^3$ is still valid.
When substituting $u = x^3$ , remember to convert back from $u$ to $x$ ; do not stop at $u = 1$ and $u = 27$ .
The mark scheme awards the final mark for the correct values $x = 1$ and $x = 3$ with valid working ( $WWW$ means 'with wrong working'), so show all steps clearly.

Techniques used

clear the fraction by multiplying through by the denominatorrecognise the resulting polynomial as quadratic in x^3solve a quadratic equation by factorisingtake cube roots to obtain final values

Approach

Use the identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$ to rewrite $\frac{1}{\tan\theta}$ , then multiply through by $\sin\theta$ . Apply $\sin^2\theta + \cos^2\theta = 1$ and rearrange into the required quadratic form.

Working

Start with

6\sin\theta + \frac{1}{\tan\theta} = \frac{4}{\sin\theta}.

Since $\frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$ , the equation becomes

6\sin\theta + \frac{\cos\theta}{\sin\theta} = \frac{4}{\sin\theta}.

Multiply through by $\sin\theta$ :

6\sin^2\theta + \cos\theta = 4.

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

6(1-\cos^2\theta) + \cos\theta = 4.

Expand and simplify:

6 - 6\cos^2\theta + \cos\theta = 4.

Rearrange:

6\cos^2\theta - \cos\theta - 2 = 0.

Answer

6\cos^2\theta - \cos\theta - 2 = 0

Final answer

6cos^2θ - cosθ - 2 = 0

Detailed explanation

Walkthrough

We are given the equation and asked to show it can be written in a quadratic form in $\cos\theta$ . The first step is to rewrite $\frac{1}{\tan\theta}$ as $\frac{\cos\theta}{\sin\theta}$ , using $\tan\theta = \frac{\sin\theta}{\cos\theta}$ . This puts the equation in terms of sine and cosine only.

Next, multiply every term by $\sin\theta$ to clear the denominators. This is valid because the original equation is only defined when $\sin\theta \neq 0$ .

Then use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to replace $\sin^2\theta$ with $1 - \cos^2\theta$ . This is the key step that changes the equation into one involving only $\cos\theta$ .

Finally, expand the bracket, simplify, and rearrange to obtain the required form.

Key Takeaways

The identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$ lets us rewrite tangent in terms of sine and cosine.
The identity $\sin^2\theta + \cos^2\theta = 1$ is used to convert between sine-squared and cosine-squared.
Multiplying through by a denominator is a standard way to clear fractions.

Common Mistakes

Writing $\frac{1}{\tan\theta}$ incorrectly as $\tan^{-1}\theta$ instead of $\frac{\cos\theta}{\sin\theta}$ .
Forgetting to multiply every term by $\sin\theta$ , leaving an incomplete equation.
Making a sign error when rearranging $6 - 6\cos^2\theta + \cos\theta = 4$ into the final quadratic form.

Things to Be Careful About

The final answer is given, so every step must be shown; an unsupported jump would not earn the marks.
The mark scheme requires the method: using the identity for $\tan\theta$ and multiplying by $\sin\theta$ , then using $\sin^2\theta + \cos^2\theta = 1$ , then simplifying.
Since $\sin\theta$ appears in denominators, the excluded values $\theta = 0^\circ, 180^\circ, 360^\circ$ are not solutions of the original equation.

Techniques used

rewrite tan in terms of sine and cosinemultiply through by sineapply the Pythagorean identityrearrange into a quadratic form

(b)

Hence, solve the equation

6\sin\theta + \frac{1}{\tan\theta} = \frac{4}{\sin\theta}

for $0^\circ \le \theta \le 360^\circ$ .

DifficultyMedium

Worked solution

Approach

Use the result from part (a), factorise the quadratic in $\cos\theta$ , solve for the two values of $\cos\theta$ , then find all angles in the interval $0^\circ \le \theta \le 360^\circ$ .

Working

From part (a),

6\cos^2\theta - \cos\theta - 2 = 0.

Factorise:

(3\cos\theta - 2)(2\cos\theta + 1) = 0.

So

3\cos\theta - 2 = 0 \quad \text{or} \quad 2\cos\theta + 1 = 0.

Hence

\cos\theta = \frac{2}{3} \quad \text{or} \quad \cos\theta = -\frac{1}{2}.

For $\cos\theta = \frac{2}{3}$ :

\theta = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2^\circ.

Cosine is positive in the first and fourth quadrants, so

\theta = 48.2^\circ \quad \text{or} \quad \theta = 360^\circ - 48.2^\circ = 311.8^\circ.

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

Since $\cos 60^\circ = \frac{1}{2}$ , the reference angle is $60^\circ$ . Cosine is negative in the second and third quadrants, so

\theta = 180^\circ - 60^\circ = 120^\circ \quad \text{or} \quad \theta = 180^\circ + 60^\circ = 240^\circ.

Answer

\theta = 48.2^\circ, \ 120^\circ, \ 240^\circ, \ 311.8^\circ

Final answer

θ = 48.2°, 120°, 240°, 311.8°

Detailed explanation

Walkthrough

After part (a), we have the quadratic $6\cos^2\theta - \cos\theta - 2 = 0$ . Factorise it into two linear factors. Setting each factor to zero gives the two possible values of $\cos\theta$ : $\frac{2}{3}$ and $-\frac{1}{2}$ .

For $\cos\theta = \frac{2}{3}$ , take the inverse cosine to find the principal angle $48.2^\circ$ . Since cosine is positive in the first and fourth quadrants, the other solution is $360^\circ - 48.2^\circ = 311.8^\circ$ .

For $\cos\theta = -\frac{1}{2}$ , use the exact value $\cos 60^\circ = \frac{1}{2}$ . Cosine is negative in the second and third quadrants, so the angles are $180^\circ - 60^\circ = 120^\circ$ and $180^\circ + 60^\circ = 240^\circ$ .

Finally, list all four angles in ascending order.

Key Takeaways

A quadratic equation in $\cos\theta$ can be factorised and solved like an ordinary quadratic.
The equation $\cos\theta = k$ has two solutions in $0^\circ \le \theta \le 360^\circ$ unless $k = \pm 1$ .
The sign of $\cos\theta$ determines which quadrants contain the solutions.
Use inverse cosine for non-special values and exact reference angles for special values.

Common Mistakes

Giving only the principal value $48.2^\circ$ and missing $311.8^\circ$ .
For $\cos\theta = -\frac{1}{2}$ , incorrectly giving $60^\circ$ or $300^\circ$ instead of $120^\circ$ and $240^\circ$ .
Mixing degrees and radians, or leaving the calculator in the wrong mode.
Not checking that all answers lie in the required interval $0^\circ \le \theta \le 360^\circ$ .

Things to Be Careful About

The original equation has $\sin\theta$ in the denominator, so $\theta = 0^\circ, 180^\circ, 360^\circ$ are excluded; none of the four solutions are these values.
The mark scheme awards B1 for any two correct angles and B1 for all four correct angles with no extra values in the range.
If working in radians, the equivalent answers are approximately $0.841$ , $\frac{2\pi}{3}$ , $\frac{4\pi}{3}$ , and $5.44$ .

Techniques used

factorise a quadratic in cos thetasolve for cos thetaapply inverse cosine to find a principal angleuse quadrant symmetry to find all angles in the interval

Q7MediumDifferentiation

Fig. 7.1

A manufacturer wishes to design an open cylindrical tank, as shown in the diagram. The tank will have a base but no top. The outside of the tank will have a fixed surface area of $600\pi\text{ cm}^2$ . The radius $r\text{ cm}$ and height $h\text{ cm}$ of the tank can vary.

(a)

Show that the volume, $V\text{ cm}^3$ , of the tank is given by

V = \frac{\pi r(600 - r^2)}{2}

DifficultyMedium-Easy

Worked solution

Approach

The outside of the open cylindrical tank consists of a circular base (area $\pi r^2$ ) and a curved side (area $2\pi r h$ ). The top is open, so it contributes nothing. Setting the total outside area equal to $600\pi$ gives an equation in $h$ and $r$ . Solve for $h$ , then substitute into $V = \pi r^2 h$ .

Working

Total outside surface area of the open tank:

2\pi r h + \pi r^2 = 600\pi

Solve for $h$ :

2\pi r h = 600\pi - \pi r^2

h = \frac{600\pi - \pi r^2}{2\pi r} = \frac{600 - r^2}{2r}

The volume of a cylinder is $V = \pi r^2 h$ . Substituting the expression for $h$ and cancelling one factor of $r$ :

V = \pi r^2 \times \frac{600 - r^2}{2r} = \frac{\pi r(600 - r^2)}{2}

Answer

V = \frac{\pi r(600 - r^2)}{2}

Final answer

V = πr(600 − r²)/2

Detailed explanation

Walkthrough

The outside of the tank is made of two surfaces: a curved side and a circular base. The top is open, so it contributes no area. The curved side has area $2\pi r h$ and the base has area $\pi r^2$ , so the total outside area is

2\pi r h + \pi r^2

The problem states this must equal $600\pi$ :

2\pi r h + \pi r^2 = 600\pi

We want $V$ as a function of $r$ alone, so we isolate $h$ . Move $\pi r^2$ to the right and divide by $2\pi r$ :

h = \frac{600\pi - \pi r^2}{2\pi r}

Every term on the right has a factor of $\pi$ , and these $\pi$ s cancel:

h = \frac{600 - r^2}{2r}

The volume of a cylinder is $V = \pi r^2 h$ . Substituting our expression for $h$ and cancelling a factor of $r$ :

V = \pi r^2 \cdot \frac{600 - r^2}{2r} = \frac{\pi r(600 - r^2)}{2}

This matches the required expression, completing the derivation.

Key Takeaways

An 'open' cylinder has only the curved side and one circular face, so the surface area is $2\pi r h + \pi r^2$ — not $2\pi r h + 2\pi r^2$ .
The cylinder volume formula $V = \pi r^2 h$ combined with a constraint lets you eliminate one variable.
'Show that' questions require every algebraic step, not just the final expression.

Common Mistakes

Including the top in the surface area calculation.
Sign errors when rearranging the constraint equation.
Forgetting to cancel the $r$ when substituting into $V = \pi r^2 h$ , leaving the result unsimplified.
Losing a $\pi$ factor when dividing.

Things to Be Careful About

The problem says explicitly that the tank has a base but no top.
The constraint $600\pi$ is in $\text{cm}^2$ , so $r$ and $h$ are in $\text{cm}$ and $V$ ends up in $\text{cm}^3$ .
Each factor of $\pi$ in the constraint and in the formulae must be tracked carefully.

Techniques used

apply surface area formula for an open cylinderrearrange the constraint to solve for h in terms of rsubstitute h into the cylinder volume formula

(b)

Find the exact value of $r$ which corresponds to the maximum value of $V$ .

DifficultyMedium

Worked solution

Approach

Expand $V$ as a polynomial in $r$ , differentiate using the power rule, set $\frac{dV}{dr} = 0$ to find the stationary point, and solve for $r$ . Take the positive root (since $r$ is a radius) and confirm it is a maximum.

Working

Expand the expression for $V$ :

V = \frac{\pi r(600 - r^2)}{2} = \frac{\pi}{2}(600r - r^3)

Differentiate with respect to $r$ using the power rule:

\frac{dV}{dr} = \frac{\pi}{2}(600 - 3r^2)

Set $\frac{dV}{dr} = 0$ . Since $\frac{\pi}{2} \neq 0$ :

600 - 3r^2 = 0

r^2 = \frac{600}{3} = 200

r = \sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}

(We take the positive root because $r > 0$ .)

To confirm this is a maximum, the second derivative is

\frac{d^2V}{dr^2} = \frac{\pi}{2}(-6r) = -3\pi r

For $r > 0$ , $\frac{d^2V}{dr^2} < 0$ , so $V$ is concave down here and the stationary point is a maximum.

Answer

r = 10\sqrt{2} \text{ cm}

Final answer

r = 10√2 cm

Detailed explanation

Walkthrough

The expression $V = \frac{\pi r(600 - r^2)}{2}$ is a cubic in $r$ (after expanding the product). To find its maximum, treat $V$ as a function of $r$ and apply the standard stationary-point procedure.

First, expand to make differentiation easier:

V = \frac{\pi}{2}(600r - r^3) = 300\pi r - \frac{\pi r^3}{2}

Differentiate term by term using the power rule $\frac{d}{dr}(r^n) = nr^{n-1}$ . The constant factors $\pi$ and $\frac{1}{2}$ pass through differentiation unchanged:

\frac{dV}{dr} = 300\pi - \frac{3\pi r^2}{2} = \frac{\pi}{2}(600 - 3r^2)

A stationary point occurs when $\frac{dV}{dr} = 0$ . The constant $\frac{\pi}{2}$ is non-zero, so we only need to solve $600 - 3r^2 = 0$ :

3r^2 = 600 \quad\Rightarrow\quad r^2 = 200 \quad\Rightarrow\quad r = \pm\sqrt{200}

Since $r$ is a radius, it must be positive. Hence $r = \sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}$ .

To check the nature of the stationary point, examine the second derivative:

\frac{d^2V}{dr^2} = -3\pi r

For $r > 0$ this is negative, so the curve of $V$ against $r$ is concave down here — confirming a maximum.

(Alternative method: a useful shortcut is to recognise that for an open cylinder with fixed surface area, the volume is maximised when the height equals the radius, $h = r$ . Substituting $h = r$ into the surface area constraint $2\pi r h + \pi r^2 = 600\pi$ gives $3\pi r^2 = 600\pi$ , so $r^2 = 200$ and $r = 10\sqrt{2}$ .)

Key Takeaways

The power rule is the key technique: $\frac{d}{dr}(r^n) = nr^{n-1}$ .
Constant factors (like $\frac{\pi}{2}$ ) pass through differentiation unchanged.
Physical constraints (here $r > 0$ ) determine which root to take.
The second derivative test confirms whether a stationary point is a maximum ( $-$ ) or minimum ( $+$ ).

Common Mistakes

Differentiating $-r^3$ as $+3r^2$ (forgetting the negative sign).
Dividing $600$ by $3$ incorrectly, e.g. giving $r^2 = 100$ or $r^2 = 450$ .
Stopping at $r^2 = 200$ and not taking the square root.
Including $-\sqrt{200}$ , which has no physical meaning for a radius.
Failing to confirm the stationary point is a maximum — it could in principle be a minimum.

Things to Be Careful About

The mark scheme accepts either $10\sqrt{2}$ or $\sqrt{200}$ as the exact value.
The derivative must be in the form $a - br^2$ (e.g. $\frac{\pi}{2}(600 - 3r^2)$ ), not in a form that hides the quadratic structure.
The alternative shortcut $h = r$ at the maximum is only valid for an open cylinder; for a closed cylinder the optimum is at $h = 2r$ .

Techniques used

differentiate a cubic polynomial using the power ruleset the derivative equal to zerosolve a quadratic equation for rselect the positive root for a physical radiusapply the second derivative test to confirm a maximum

(c)

Hence, find the maximum value of $V$ .

DifficultyMedium-Easy

Worked solution

Approach

Substitute $r = 10\sqrt{2}$ from part (b) into the expression $V = \frac{\pi r(600 - r^2)}{2}$ and simplify. The exact answer is $2000\pi\sqrt{2}$ ; a 3-sf approximation is $8890\,\text{cm}^3$ .

Working

V = \frac{\pi r(600 - r^2)}{2} = \frac{\pi}{2}\bigl(600 \cdot 10\sqrt{2} - (10\sqrt{2})^3\bigr)

Compute each piece:

600 \cdot 10\sqrt{2} = 6000\sqrt{2}

(10\sqrt{2})^3 = 10^3 \cdot (\sqrt{2})^3 = 1000 \cdot 2\sqrt{2} = 2000\sqrt{2}

Substitute back:

V = \frac{\pi}{2}(6000\sqrt{2} - 2000\sqrt{2}) = \frac{\pi}{2} \cdot 4000\sqrt{2} = 2000\pi\sqrt{2}

Numerically:

V = 2000 \cdot \pi \cdot \sqrt{2} \approx 2000 \cdot 3.14159 \cdot 1.41421 \approx 8885.8 \approx 8890\,\text{cm}^3 \text{ (3 sf)}

Answer

V = 2000\pi\sqrt{2} \approx 8890\,\text{cm}^3

Final answer

V = 2000π√2 ≈ 8890 cm³

Detailed explanation

Walkthrough

We have $V = \frac{\pi r(600 - r^2)}{2}$ from part (a), and from part (b) the value of $r$ that gives the maximum volume is $r = 10\sqrt{2}$ . Substituting:

V = \frac{\pi \cdot 10\sqrt{2} \cdot \bigl(600 - (10\sqrt{2})^2\bigr)}{2}

Compute the inner pieces in the right order. First, $(10\sqrt{2})^2 = 100 \cdot 2 = 200$ , so $600 - (10\sqrt{2})^2 = 600 - 200 = 400$ . Then:

V = \frac{\pi \cdot 10\sqrt{2} \cdot 400}{2} = \frac{4000\pi\sqrt{2}}{2} = 2000\pi\sqrt{2}

Equivalently, expanding $r(600 - r^2) = 600r - r^3$ first:

$600r = 600 \cdot 10\sqrt{2} = 6000\sqrt{2}$
$r^3 = (10\sqrt{2})^3 = 1000 \cdot 2\sqrt{2} = 2000\sqrt{2}$
$r(600 - r^2) = 6000\sqrt{2} - 2000\sqrt{2} = 4000\sqrt{2}$
$V = \frac{\pi \cdot 4000\sqrt{2}}{2} = 2000\pi\sqrt{2}$

Numerically, $2000\pi\sqrt{2} \approx 8885.77$ , which rounds to $8890$ (3 significant figures).

Key Takeaways

'Hence' in the question tells you to reuse the result of the previous part directly.
Simplify cubes of surds carefully: $(a\sqrt{b})^3 = a^3 \cdot b\sqrt{b}$ .
The exact form $2000\pi\sqrt{2}$ is preferred when the answer should be exact; a 3-sf approximation is appropriate for a numerical answer.

Common Mistakes

Confusing $(\sqrt{2})^3 = 2\sqrt{2}$ with $(\sqrt{2})^3 = 8$ .
Arithmetic slip in $6000 - 2000 = 4000$ or in the final division by 2.
Forgetting units: the answer is in $\text{cm}^3$ , not $\text{cm}^2$ or $\text{cm}$ .
Not converting the exact answer to a decimal when the mark scheme asks for a numerical value.

Things to Be Careful About

The mark scheme accepts either the exact form $2000\pi\sqrt{2}$ or the 3-sf form $8890$ ; give the form requested by the question.
'Hence' means the answer must follow from part (b); you should not re-derive $r$ from scratch here.
The cube $(10\sqrt{2})^3$ has three factors of $10$ and three factors of $\sqrt{2}$ : $10 \cdot 10 \cdot 10 = 1000$ and $\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2} = 2\sqrt{2}$ .

Techniques used

substitute the optimal r into the volume expressionsimplify a cube of a surdcombine like radical termscompute a numerical approximation

Q8MediumIntegrationQuadratics

Fig. 8.1

The diagram shows the curve with equation $y = \sqrt{6x + 5}$ . The shaded region is bounded by the curve, the $x$ -axis and the lines $x = a$ and $x = 2a$ , where $a$ is a positive constant.

The shaded region is rotated through $360^\circ$ about the $x$ -axis to form a solid. The solid has volume, $V$ , such that $V \ge 46\pi$ .

(a)

Show that $9a^2 + 5a - 46 \ge 0$ .

DifficultyMedium

Worked solution

Approach

The volume of a solid formed by rotating a region about the x-axis is given by $V = \pi \int y^2 \, dx$ . Since $y = \sqrt{6x + 5}$ , we have $y^2 = 6x + 5$ . We integrate from $x = a$ to $x = 2a$ , evaluate the definite integral, and use the condition $V \geq 46\pi$ to obtain the required inequality.

Working

The volume of revolution is:

V = \pi \int_{a}^{2a} y^2 \, dx

Substitute $y^2 = 6x + 5$ :

V = \pi \int_{a}^{2a} (6x + 5) \, dx

Integrate:

V = \pi \left[ 3x^2 + 5x \right]_{a}^{2a}

Substitute the upper limit $x = 2a$ :

3(2a)^2 + 5(2a) = 12a^2 + 10a

Substitute the lower limit $x = a$ :

3a^2 + 5a

Subtract lower from upper:

V = \pi \left( 12a^2 + 10a - 3a^2 - 5a \right) = \pi (9a^2 + 5a)

Given $V \geq 46\pi$ :

\pi(9a^2 + 5a) \geq 46\pi

Divide both sides by $\pi$ :

9a^2 + 5a \geq 46

9a^2 + 5a - 46 \geq 0

This is the required result.

Answer

$9a^2 + 5a - 46 \geq 0$ (shown as required)

Final answer

9a^2 + 5a - 46 >= 0

Detailed explanation

Walkthrough

The volume of a solid of revolution about the x-axis is $V = \pi \int y^2 \, dx$ . Here the curve is $y = \sqrt{6x + 5}$ , so squaring gives $y^2 = 6x + 5$ , which is a simple linear expression. The shaded region lies between $x = a$ and $x = 2a$ , so these are our limits of integration.

We compute the integral $\int (6x + 5) \, dx = 3x^2 + 5x$ , then evaluate at the upper limit $2a$ to get $12a^2 + 10a$ , and at the lower limit $a$ to get $3a^2 + 5a$ . The difference is $9a^2 + 5a$ . Multiplying by $\pi$ gives $V = \pi(9a^2 + 5a)$ . Setting this $\geq 46\pi$ and dividing through by $\pi$ yields $9a^2 + 5a - 46 \geq 0$ .

Key Takeaways

The volume of revolution about the x-axis uses $V = \pi \int y^2 \, dx$ , not $\pi \int y \, dx$ .
Squaring $\sqrt{6x + 5}$ removes the square root and gives a polynomial to integrate.
Always substitute both limits carefully and subtract (upper − lower).

Common Mistakes

Forgetting the $\pi$ factor in the volume formula.
Integrating $y$ instead of $y^2$ .
Sign errors when subtracting the lower limit evaluation from the upper limit evaluation.
Forgetting to divide by $\pi$ when simplifying $V \geq 46\pi$ .

Things to Be Careful About

The mark scheme awards B1 for the correct integral setup with $\pi$ and correct limits (CAO SOI). Ensure $\pi$ is present and limits are $a$ to $2a$ .
The final answer $9a^2 + 5a - 46 \geq 0$ is marked A1 AG (answer given), so the working leading to it must be complete and clear.
Do not skip the step of substituting both limits — this is where DM1 is awarded.

Techniques used

apply volume of revolution formulaevaluate definite integralsubstitute limits of integrationsimplify to obtain quadratic inequality

(b)

Find the range of possible values of $a$ .

DifficultyMedium-Easy

Worked solution

Approach

From part (a), we have the inequality $9a^2 + 5a - 46 \geq 0$ . Solve the corresponding quadratic equation to find the roots, determine which intervals satisfy the inequality (since the coefficient of $a^2$ is positive, the quadratic is $\geq 0$ outside the roots), and then apply the condition that $a$ is a positive constant ( $a > 0$ ) to select the valid range.

Working

Solve $9a^2 + 5a - 46 = 0$ .

Using the quadratic formula:

a = \frac{-5 \pm \sqrt{5^2 - 4(9)(-46)}}{2(9)}

a = \frac{-5 \pm \sqrt{25 + 1656}}{18}

a = \frac{-5 \pm \sqrt{1681}}{18}

a = \frac{-5 \pm 41}{18}

So:

a = \frac{-5 + 41}{18} = \frac{36}{18} = 2

a = \frac{-5 - 41}{18} = \frac{-46}{18} = -\frac{23}{9}

Since the coefficient of $a^2$ is $9 > 0$ , the parabola opens upwards, so $9a^2 + 5a - 46 \geq 0$ when:

a \leq -\frac{23}{9} \quad \text{or} \quad a \geq 2

The problem states that $a$ is a positive constant, so $a > 0$ . This eliminates the interval $a \leq -\frac{23}{9}$ .

Therefore:

a \geq 2

Answer

$a \geq 2$

Final answer

a >= 2

Detailed explanation

Walkthrough

From part (a), we need to solve $9a^2 + 5a - 46 \geq 0$ . First, find the roots of $9a^2 + 5a - 46 = 0$ using the quadratic formula. The discriminant is $25 + 1656 = 1681 = 41^2$ , giving roots $a = 2$ and $a = -23/9$ .

Since the leading coefficient is positive, the quadratic is non-negative outside the roots: $a \leq -23/9$ or $a \geq 2$ . The problem states $a$ is a positive constant, so $a > 0$ , which eliminates the negative interval. The final answer is $a \geq 2$ .

Key Takeaways

For a quadratic $Ax^2 + Bx + C \geq 0$ with $A > 0$ , the solution is $x \leq$ smaller root or $x \geq$ larger root.
Always apply any given domain restrictions (here $a > 0$ ) to the final answer.
The quadratic formula is reliable when factorisation is not obvious.

Common Mistakes

Writing $x \geq 2$ instead of $a \geq 2$ — the mark scheme explicitly says "Don't allow $x \geq 2$ ".
Forgetting to apply the condition $a > 0$ and giving both intervals.
Sign errors in the quadratic formula, particularly with $-4ac$ when $c$ is negative.

Things to Be Careful About

The final answer must be in terms of $a$ , not $x$ .
The mark scheme awards B1 WWW (without warning) for the final answer, so the reasoning must be clear.
Condone absence of $-23/9$ in the working, but the final answer must correctly restrict to $a \geq 2$ .

Techniques used

solve quadratic equation by factorisation or formuladetermine sign of quadratic outside rootsapply domain restriction a > 0

Q9MediumSeries

In the expansion of $(p + qx)^4$ , the coefficient of $x$ is equal to the coefficient of $x^2$ . The constants $p$ and $q$ are both positive.

(a)

Find the ratio $p : q$ . Give your answer in its simplest form.

DifficultyMedium-Easy

Worked solution

Approach

Write the expansion of $(p+qx)^4$ using the binomial theorem, identify the coefficients of $x$ and $x^2$ , equate them, and simplify the resulting ratio.

Working

The binomial expansion is

(p+qx)^4 = p^4 + 4p^3qx + 6p^2q^2x^2 + 4pq^3x^3 + q^4x^4

The coefficient of $x$ is $4p^3q$ and the coefficient of $x^2$ is $6p^2q^2$ . Equating them gives

4p^3q = 6p^2q^2

Since $p$ and $q$ are positive, $p^2q \neq 0$ , so divide both sides by $2p^2q$ :

2p = 3q

Thus

\frac{p}{q} = \frac{3}{2}

so the ratio is

p : q = 3 : 2

Answer

p : q = 3 : 2

Final answer

p : q = 3 : 2

Detailed explanation

Walkthrough

The binomial expansion of $(a+b)^4$ is

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

Here $a = p$ and $b = qx$ . The term containing $x$ comes from $4a^3b = 4p^3(qx)$ , so its coefficient is $4p^3q$ . The term containing $x^2$ comes from $6a^2b^2 = 6p^2(qx)^2$ , so its coefficient is $6p^2q^2$ . Because the coefficient of $x$ is equal to the coefficient of $x^2$ , we set these expressions equal:

4p^3q = 6p^2q^2

The constants $p$ and $q$ are positive, so $p^2q$ is not zero and we may divide by $2p^2q$ safely. This gives $2p = 3q$ , or equivalently $p/q = 3/2$ . In simplest ratio form this is $p : q = 3 : 2$ .

Key Takeaways

This question tests binomial coefficients and how to interpret the coefficient of a particular power of $x$ . Once the coefficients are written correctly, equating them gives a simple linear relation between $p$ and $q$ . The positivity of the constants justifies cancellation without worrying about zero division.

Common Mistakes

Forgetting the binomial coefficients $4$ and $6$ , and writing incorrect terms such as $p^3qx$ and $p^2q^2x^2$ .
Failing to square the $q$ when forming the $x^2$ term: the term is $6p^2q^2x^2$ , not $6p^2q x^2$ .
Simplifying to $p/q = 3/2$ but then writing the ratio in the wrong order, e.g. $q : p = 3 : 2$ .
Not simplifying the ratio, e.g. leaving $6 : 4$ instead of $3 : 2$ .

Things to Be Careful About

The question asks for $p : q$ , so the final ratio must have $p$ first and $q$ second. Because $p$ and $q$ are positive, dividing by $p^2q$ is valid. Make sure the final ratio is in simplest form and contains no powers of $x$ .

Techniques used

expand a binomial using binomial coefficientsequate coefficients of like powerscancel common factorsexpress a linear relation as a simplified ratio

(b)

It is given that the coefficient of $x^3$ is 486.

Find the values of $p$ and $q$ .

DifficultyMedium

Worked solution

Approach

Find the coefficient of $x^3$ in the binomial expansion, set it equal to 486, substitute the relation $p = \frac{3q}{2}$ from part (a), and solve for the positive values of $p$ and $q$ .

Working

The term in $x^3$ in $(p+qx)^4$ is

\binom{4}{3}p(qx)^3 = 4pq^3x^3

so its coefficient is $4pq^3$ . Given the coefficient of $x^3$ is 486,

4pq^3 = 486

From part (a), $2p = 3q$ , so

p = \frac{3q}{2}

Substitute into the coefficient equation:

4\left(\frac{3q}{2}\right)q^3 = 486

Simplify:

6q^4 = 486

q^4 = 81

Since $q > 0$ , take the positive fourth root:

q = 3

Then

p = \frac{3(3)}{2} = \frac{9}{2}

Answer

p = \frac{9}{2}, \quad q = 3

Final answer

p = 9/2, q = 3

Detailed explanation

Walkthrough

From part (a), the binomial expansion already gives the relation $2p = 3q$ . We now use the information about the $x^3$ term. In the expansion, the term in $x^3$ is

\binom{4}{3}p(qx)^3 = 4pq^3x^3

so its coefficient is $4pq^3$ . The condition says this coefficient is 486:

4pq^3 = 486

To solve for $p$ and $q$ , use $p = \frac{3q}{2}$ from part (a). Substituting gives

4\left(\frac{3q}{2}\right)q^3 = 486

This simplifies to $6q^4 = 486$ , so $q^4 = 81$ . Since $q$ is positive, $q = 3$ , not $-3$ . Finally, substituting into $p = \frac{3q}{2}$ gives $p = \frac{9}{2}$ .

Key Takeaways

This part combines the result of part (a) with another binomial coefficient. It also shows the importance of the positivity condition: solving $q^4 = 81$ gives both $q = 3$ and $q = -3$ , but only the positive value is valid. The final step is substitution to find the other unknown.

Common Mistakes

Using the wrong coefficient for $x^3$ ; the correct coefficient is $4pq^3$ .
Forgetting to use part (a), or substituting the ratio backwards as $q = \frac{3p}{2}$ .
Taking $q = -3$ as well as $q = 3$ when the constants are stated to be positive.
Dividing incorrectly when simplifying $4\left(\frac{3q}{2}\right)q^3 = 486$ ; note that $4 \times \frac{3}{2} = 6$ .

Things to Be Careful About

Since $p$ and $q$ are both positive, reject the negative fourth root. The answer should be given as exact values: $p = \frac{9}{2}$ and $q = 3$ . Show the substitution step clearly because the mark scheme gives credit for substituting the part (a) relation. If answers are given without working, the mark scheme only awards a special case mark, so full working is needed.

Techniques used

identify the coefficient of a chosen termsubstitute a known linear relationsolve a quartic equationapply positivity to select the correct root

Q10MediumFunctions

The function $f$ is defined by

f(x) = 3 + \frac{7}{x - 2}

for $x > 2$ .

(a)

It is given that $f(a) = 4$ .

Find the value of $a$ .

(b)

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

DifficultyMedium

Worked solution

Approach

Write $y=f(x)$ and swap $x$ and $y$ , then rearrange to make $y$ the subject. The domain of the inverse is the range of $f$ . For $x>2$ , the fraction $\frac{7}{x-2}$ is positive and tends to $0$ as $x$ increases, so the range is all values greater than $3$ .

Working

Let

y = 3 + \frac{7}{x-2}

Swap $x$ and $y$ :

x = 3 + \frac{7}{y-2}

Subtract $3$ from both sides:

x - 3 = \frac{7}{y-2}

Take reciprocals:

y - 2 = \frac{7}{x-3}

Add $2$ :

y = 2 + \frac{7}{x-3}

Hence

f^{-1}(x) = 2 + \frac{7}{x-3}

The domain of $f^{-1}$ is the range of $f$ . Since $x>2$ , $x-2>0$ , so $\frac{7}{x-2}>0$ and $f(x)>3$ . As $x\to\infty$ , $\frac{7}{x-2}\to 0$ , so the range is $(3,\infty)$ . Therefore the domain of $f^{-1}$ is $x>3$ .

Answer

f^{-1}(x) = 2 + \frac{7}{x-3}, \quad x > 3

Final answer

f^{-1}(x) = 2 + 7/(x-3), domain x > 3

Detailed explanation

Walkthrough

To find the inverse, start by writing the function as $y=3+\frac{7}{x-2}$ . The standard method is to swap $x$ and $y$ , then rearrange to make $y$ the subject. This gives $y=2+\frac{7}{x-3}$ , which is the inverse function. The domain of the inverse is the range of the original function. For $x>2$ , the denominator $x-2$ is positive, so $\frac{7}{x-2}$ is positive. As $x$ approaches $2$ from above, the fraction becomes arbitrarily large; as $x$ increases without bound, the fraction tends to $0$ . Hence $f(x)$ takes all values greater than $3$ , so the inverse is defined for $x>3$ .

Key Takeaways

The inverse is found by swapping the variables and rearranging.
The domain of the inverse equals the range of the original function.
For a rational function of the form $a+\frac{b}{x-c}$ , the horizontal asymptote gives the boundary of the range.

Common Mistakes

Forgetting to swap $x$ and $y$ before rearranging.
Stopping at $y=3+\frac{7}{x-3}$ without adding $2$ .
Writing the domain of the inverse as $x>2$ instead of $x>3$ ; the domain of the inverse is the range of $f$ , not the domain of $f$ .
Not writing $f^{-1}(x)=$ or $f^{-1}:x\mapsto$ ; the mark scheme requires one of these forms.

Things to Be Careful About

The inverse may also be written as $\frac{2x+1}{x-3}$ , but $2+\frac{7}{x-3}$ is usually easier to relate to the range.
One algebraic slip can lose the final mark, so show each stage clearly.
The domain statement must involve $x$ ; an equivalent such as $x>3$ is acceptable.

Techniques used

swap variables to find the inverserearrange a rational expressiondetermine the range of the original function to state the inverse domain

(c)

The function $g$ is defined by

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

for $x > \frac{3}{2}$ .

Show that $fg(x) \equiv kx$ , where $k$ is a constant to be determined.

DifficultyMedium

Worked solution

Approach

Substitute the expression for $g(x)$ into $f(x)$ to form $fg(x)$ . Then simplify the compound fraction in the denominator to show that $fg(x)$ reduces to $kx$ .

Working

Since $f(x)=3+\frac{7}{x-2}$ and $g(x)=\frac{1+4x}{2x-3}$ ,

\begin{aligned} fg(x) &= 3 + \frac{7}{g(x)-2} \\ &= 3 + \frac{7}{\frac{1+4x}{2x-3} - 2} \end{aligned}

Simplify the denominator:

\frac{1+4x}{2x-3} - 2 = \frac{1+4x - 2(2x-3)}{2x-3} = \frac{1+4x-4x+6}{2x-3} = \frac{7}{2x-3}

Therefore

fg(x) = 3 + \frac{7}{\frac{7}{2x-3}} = 3 + (2x-3) = 2x

So $fg(x)\equiv 2x$ , and the constant is $k=2$ .

Answer

k = 2

Final answer

k = 2

Detailed explanation

Walkthrough

This part asks for the composition $fg(x)$ , which means evaluate $f$ at $g(x)$ . Start with $f(x)=3+\frac{7}{x-2}$ and replace $x$ with $g(x)=\frac{1+4x}{2x-3}$ . The denominator becomes $g(x)-2$ , a compound fraction. To simplify it, subtract $2$ by writing it with the common denominator $2x-3$ : $2=\frac{2(2x-3)}{2x-3}$ . The numerator becomes $1+4x-4x+6=7$ , so $g(x)-2=\frac{7}{2x-3}$ . Dividing by this fraction is equivalent to multiplying by its reciprocal, so $\frac{7}{\frac{7}{2x-3}}=2x-3$ . Adding $3$ gives $2x$ , so $fg(x)=2x$ and $k=2$ .

Key Takeaways

Composition $fg(x)$ means substitute $g(x)$ into $f(x)$ .
Simplifying compound fractions requires a common denominator.
The identity sign $\equiv$ means the expression is true for all $x$ in the domain.

Common Mistakes

Forgetting the $-2$ when subtracting $2$ from $\frac{1+4x}{2x-3}$ .
Inverting the fraction incorrectly: $\frac{7}{\frac{7}{2x-3}}=2x-3$ , not $\frac{1}{2x-3}$ .
Stating $k=2$ without showing the algebra; the mark scheme requires supporting working.

Things to Be Careful About

The domain of $g$ is $x>\frac{3}{2}$ , but the composition remains valid on that domain.
Keep the simplification of the compound fraction explicit to earn the method mark.
The final answer must identify $k=2$ after the algebra, not just state it.

Techniques used

substitute one function into anothersimplify a compound fractionidentify the constant multiplier

Q11Medium-HardDifferentiationCoordinate GeometryIntegrationFunctions

Fig. 11.1

The diagram shows the curve with equation $y = 4x^2 - x^3$ and the tangent to the curve at the point $P$ . The point $P$ has $x$ -coordinate 3.

(a)

Find the equation of the tangent to the curve at the point $P$ . Give your answer in the form $y = mx + c$ .

5M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate the curve equation to find the gradient function, evaluate it at $x = 3$ to get the gradient of the tangent, find the $y$ -coordinate of $P$ , and then use the point-gradient form to find the tangent equation.

Working

The curve is $y = 4x^2 - x^3$ .

Differentiate with respect to $x$ :

\frac{dy}{dx} = 8x - 3x^2

At $P$ , $x = 3$ :

\frac{dy}{dx} = 8(3) - 3(3)^2 = 24 - 27 = -3

Find the $y$ -coordinate of $P$ by substituting $x = 3$ into the curve equation:

y = 4(3)^2 - (3)^3 = 36 - 27 = 9

So $P$ is $(3, 9)$ .

The gradient of the tangent is $m = -3$ . Using the point-gradient form $y - y_1 = m(x - x_1)$ :

y - 9 = -3(x - 3)

y - 9 = -3x + 9

y = -3x + 18

Answer

y = -3x + 18

Final answer

y = -3x + 18

Detailed explanation

Walkthrough

First, we find the gradient function by differentiating the curve equation $y = 4x^2 - x^3$ with respect to $x$ , giving $\frac{dy}{dx} = 8x - 3x^2$ . Next, we evaluate this gradient at $x = 3$ to find the slope of the tangent at point $P$ , which is $-3$ . We then find the $y$ -coordinate of $P$ by substituting $x = 3$ into the original curve equation, yielding $y = 9$ . With the point $(3, 9)$ and gradient $-3$ , we use the point-gradient form $y - y_1 = m(x - x_1)$ to derive the tangent equation $y = -3x + 18$ .

Key Takeaways

To find the equation of a tangent to a curve at a given point, differentiate to find the gradient function, evaluate it at the given $x$ -coordinate, find the corresponding $y$ -coordinate on the curve, and apply the point-gradient formula.

Common Mistakes

Forgetting to find the $y$ -coordinate of the point by substituting into the curve equation.
Sign errors when expanding $-3(x - 3)$ .
Incorrect differentiation, such as differentiating $x^3$ as $3x$ instead of $3x^2$ .

Things to Be Careful About

Ensure the final equation is in the requested form $y = mx + c$ . Double-check arithmetic when evaluating the gradient and coordinates at $x = 3$ .

Techniques used

differentiate polynomial functionevaluate gradient at a given pointfind point coordinates on curveuse point-gradient form for tangent equation

(b)

The shaded region is bounded by the curve, the $x$ -axis and the tangent to the curve at $P$ .

Find the exact area of the shaded region.

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

The shaded region is bounded by the curve $y = 4x^2 - x^3$ , the $x$ -axis, and the tangent $y = -3x + 18$ . First, find the $x$ -intercepts of the curve and the tangent to determine the boundaries. The curve meets the $x$ -axis at $x = 4$ (since $4x^2 - x^3 = x^2(4-x) = 0$ ). The tangent meets the $x$ -axis at $x = 6$ (since $-3x + 18 = 0$ ). The shaded region can be found by calculating the area under the tangent from $x = 3$ to $x = 6$ and subtracting the area under the curve from $x = 3$ to $x = 4$ .

Working

Find the $x$ -intercept of the tangent $y = -3x + 18$ :

-3x + 18 = 0 \implies x = 6

Find the $x$ -intercept of the curve $y = 4x^2 - x^3$ :

x^2(4 - x) = 0 \implies x = 4 \text{ (for } x > 0\text{)}

The area under the tangent from $x = 3$ to $x = 6$ is a triangle with base $(6 - 3) = 3$ and height $y = 9$ at $x = 3$ :

\text{Area}_{\text{triangle}} = \frac{1}{2} \times 3 \times 9 = \frac{27}{2}

Alternatively, using integration:

\int_3^6 (-3x + 18) \, dx = \left[ -\frac{3x^2}{2} + 18x \right]_3^6 = \left( -54 + 108 \right) - \left( -\frac{27}{2} + 54 \right) = 54 - \frac{81}{2} = \frac{27}{2}

Now find the area under the curve from $x = 3$ to $x = 4$ :

\int_3^4 (4x^2 - x^3) \, dx = \left[ \frac{4x^3}{3} - \frac{x^4}{4} \right]_3^4

Substitute the limits:

= \left( \frac{4(64)}{3} - \frac{256}{4} \right) - \left( \frac{4(27)}{3} - \frac{81}{4} \right)

= \left( \frac{256}{3} - 64 \right) - \left( 36 - \frac{81}{4} \right)

= \frac{64}{3} - \frac{63}{4} = \frac{256 - 189}{12} = \frac{67}{12}

The shaded area is the area under the tangent minus the area under the curve:

\text{Shaded Area} = \frac{27}{2} - \frac{67}{12} = \frac{162}{12} - \frac{67}{12} = \frac{95}{12}

Answer

\frac{95}{12}

Final answer

95/12

Detailed explanation

Walkthrough

The shaded region is bounded above by the tangent and the curve, and below by the $x$ -axis. The tangent intersects the $x$ -axis at $x = 6$ , and the curve intersects the $x$ -axis at $x = 4$ . From $x = 3$ to $x = 6$ , the area under the tangent forms a triangle with area $\frac{27}{2}$ . From $x = 3$ to $x = 4$ , the area under the curve is found by integrating $4x^2 - x^3$ , giving $\frac{67}{12}$ . Subtracting the area under the curve from the area under the tangent yields the shaded area $\frac{95}{12}$ .

Key Takeaways

To find the area of a region bounded by a curve, a line, and the $x$ -axis, identify the $x$ -intercepts of all boundaries. Calculate the total area under the upper boundary and subtract the area under any lower boundaries within the region.

Common Mistakes

Using incorrect limits of integration.
Forgetting to subtract the area under the curve from the area under the tangent.
Arithmetic errors when evaluating fractions.

Things to Be Careful About

Ensure the limits of integration match the actual boundaries of the shaded region. The curve and tangent only bound the region between $x = 3$ and $x = 4$ , while the tangent extends to $x = 6$ . The final answer must be exact.

Techniques used

set up definite integral for area under curveevaluate definite integral with limitsfind x-intercept of tangent linecalculate area of trianglesubtract area under curve from area under tangent

(c)

The graph of $y = 4x^2 - x^3$ is transformed by a stretch of scale factor $\frac{1}{3}$ in the $x$ -direction. The point $Q$ is the image of $P$ under this transformation. The transformed shaded region is bounded by the transformed curve, the $x$ -axis and the tangent to the transformed curve at $Q$ .

3M

(i)

Find the equation of the transformed curve in the form $y = mx^2 + nx^3$ , where $m$ and $n$ are integers to be found.

1M Similar

(ii)

State the coordinates of $Q$ and the area of the transformed shaded region.