The equation of a curve is $y = 2x^2 - \frac{1}{2x} + 3$. (a) Find the coordinates of the stationary point. (b) Determine the nature of the stationary point. (c) For positive values of $x$, determine whether the curve shows a function that is increasing, decreasing or neither. Give a reason for your answer.

(a) (-1/2, 9/2) · (b) minimum · (c) increasing

The diagram shows the curve with equation $y = \sqrt{2x^3 + 10}$. (a) Find the equation of the tangent to the curve at the point where $x = 3$. Give your answer in the form $ax + by + c = 0$ where $a$, $b$ and $c$ are integers. (b) The region shaded in the diagram is enclosed by the curve and the straight lines $x = 1$, $x = 3$ and $y = 0$. Find the volume of the solid obtained when the shaded region is rotated through $360^\circ$ about the $x$-axis.

(a) 27x - 8y - 17 = 0 · (b) 60π

The geometric progression $a_1, a_2, a_3, \dots$ has first term $2$ and common ratio $r$ where $r > 0$. It is given that $\frac{9}{2}a_5 + 7a_3 = 8$. (a) Find the value of $r$. (b) Find the sum of the first $20$ terms of the geometric progression. Give your answer correct to $4$ significant figures. (c) Find the sum to infinity of the progression $a_2, a_5, a_8, \dots$.

(a) r = 2/3 · (b) 5.998 · (c) 36/19

9709/13 May/June 2024 — Mathematics Solved Past Paper

Q14MMedium-EasySeries

Find the coefficient of $x^2$ in the expansion of

(2 - 5x)(1 + 3x)^{10}

DifficultyMedium-Easy

Worked solution

Approach

Expand $(1+3x)^{10}$ only as far as the $x^2$ term, then multiply by $(2-5x)$ and collect the coefficient of $x^2$ .

Working

The binomial expansion of $(1+3x)^{10}$ begins

(1+3x)^{10} = 1 + \binom{10}{1}(3x) + \binom{10}{2}(3x)^2 + \cdots

The $x$ term is

\binom{10}{1}(3x) = 10 \times 3x = 30x

The $x^2$ term is

\binom{10}{2}(3x)^2 = 45 \times 9x^2 = 405x^2

So

(1+3x)^{10} = 1 + 30x + 405x^2 + \cdots

Now multiply by $(2-5x)$ . The $x^2$ terms come from

2(405x^2) + (-5x)(30x) = 810x^2 - 150x^2

Therefore

\text{coefficient of } x^2 = 810 - 150 = 660

Answer

660

Final answer

660

Detailed explanation

Walkthrough

The expansion of $(1+3x)^{10}$ would normally contain many terms, but we only need the coefficient of $x^2$ in the final product. Since we are multiplying by the linear factor $(2-5x)$ , the $x^2$ term in the product can only come from two places:

the constant $2$ multiplied by the $x^2$ term of the expansion, and
the $-5x$ term multiplied by the $x$ term of the expansion.

So we only need the first three terms of the binomial expansion of $(1+3x)^{10}$ .

Using the binomial theorem, the $x$ term is

\binom{10}{1}(3x) = 10 \times 3x = 30x

and the $x^2$ term is

\binom{10}{2}(3x)^2 = 45 \times 9x^2 = 405x^2

Higher terms such as $x^3$ and above will not contribute to the coefficient of $x^2$ after multiplying by $(2-5x)$ , because the highest power needed in the product is $x^2$ .

Now multiply:

2 \times 405x^2 = 810x^2

and

(-5x) \times 30x = -150x^2

Adding these gives

810x^2 - 150x^2 = 660x^2

so the coefficient of $x^2$ is $660$ .

Key Takeaways

This question tests the ability to use a binomial expansion efficiently. You do not need to write out the whole expansion; you only need the terms that can contribute to the required power after multiplication. It also tests careful multiplication of polynomials and accurate collection of like terms.

Common Mistakes

Expanding the whole of $(1+3x)^{10}$ instead of stopping at the $x^2$ term. This wastes time and increases the chance of error.
Forgetting to multiply the $x^2$ term by $2$ and the $x$ term by $-5x$ .
Making a sign error when multiplying by $-5x$ : the term $(-5x)(30x)$ is $-150x^2$ , not $+150x^2$ .
Stating the final answer as $660x^2$ without clearly identifying the coefficient. The mark scheme accepts $660x^2$ if it is clearly identified as the coefficient, but it is safer to state the coefficient as $660$ .

Things to Be Careful About

Only the terms with $x$ and $x^2$ in the binomial expansion are needed; all higher powers can be ignored.
Keep track of signs throughout the multiplication.
The final coefficient is a number only, not the full term. The question asks for the coefficient of $x^2$ , so the answer is $660$ .
Show the intermediate binomial terms clearly, because the mark scheme awards separate marks for the $30x$ term, the $405x^2$ term, the multiplication step, and the final coefficient.

Techniques used

expand a binomial expansion to the required termmultiply the expansion by a linear factorcollect the coefficient of a specified power

Approach

To find the coordinates of point $A$ , set $y = 0$ and solve for $x$ , identifying the correct root from the diagram. To find the coordinates of point $B$ , determine where the cosine function reaches its minimum value of $-1$ .

Working

For point $A$ , the curve crosses the $x$ -axis, so $y = 0$ :

k\cos\left(x - \frac{\pi}{6}\right) = 0

Since $k$ is a positive constant, we can divide by $k$ :

\cos\left(x - \frac{\pi}{6}\right) = 0

The general solution for $\cos \theta = 0$ is $\theta = \frac{\pi}{2} + n\pi$ for integer $n$ . Thus:

x - \frac{\pi}{6} = \frac{\pi}{2} + n\pi

x = \frac{\pi}{2} + \frac{\pi}{6} + n\pi = \frac{2\pi}{3} + n\pi

From the diagram, the curve starts at a positive $y$ -intercept ( $x=0 \implies y = k\cos(-\pi/6) = k\sqrt{3}/2$ ), reaches a maximum at $x = \pi/6$ , and then descends. The first $x$ -intercept (where the curve is decreasing) is at $n=0$ , giving $x = 2\pi/3$ . Point $A$ is the next $x$ -intercept (where the curve is increasing), which corresponds to $n=1$ :

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

So the coordinates of $A$ are $\left(\frac{5\pi}{3}, 0\right)$ .

For point $B$ , it is a minimum point. The minimum value of $k\cos(x - \pi/6)$ is $-k$ , which occurs when $\cos(x - \pi/6) = -1$ :

x - \frac{\pi}{6} = \pi + 2n\pi

x = \pi + \frac{\pi}{6} + 2n\pi = \frac{7\pi}{6} + 2n\pi

The first minimum is at $n=0$ , giving $x = 7\pi/6$ . From the diagram, point $B$ is the second minimum point to the right of the origin, so we take $n=1$ :

x = \frac{7\pi}{6} + 2\pi = \frac{19\pi}{6}

The $y$ -coordinate at a minimum is $-k$ .

So the coordinates of $B$ are $\left(\frac{19\pi}{6}, -k\right)$ .

Answer

Point $A$ : $\left(\frac{5\pi}{3}, 0\right)$

Point $B$ : $\left(\frac{19\pi}{6}, -k\right)$

Final answer

A = (5π/3, 0), B = (19π/6, -k)

Detailed explanation

Walkthrough

First, we analyze the equation $y = k\cos(x - \pi/6)$ . The basic cosine curve $y = \cos x$ has $x$ -intercepts at $\pi/2 + n\pi$ and minima at $\pi + 2n\pi$ . The term $(x - \pi/6)$ represents a phase shift to the right by $\pi/6$ .

For point $A$ , we set $y = 0$ to find the $x$ -intercepts. Solving $\cos(x - \pi/6) = 0$ gives $x - \pi/6 = \pi/2 + n\pi$ , or $x = 2\pi/3 + n\pi$ . Looking at the diagram, the curve starts at a positive $y$ -value, goes up to a maximum at $x = \pi/6$ , then crosses the $x$ -axis at $x = 2\pi/3$ (decreasing). Point $A$ is the next crossing, where the curve is increasing. This corresponds to $n=1$ , giving $x = 5\pi/3$ . The $y$ -coordinate is $0$ .

For point $B$ , we need the minimum point. The cosine function reaches its minimum of $-1$ when the argument is $\pi + 2n\pi$ . Solving $x - \pi/6 = \pi + 2n\pi$ gives $x = 7\pi/6 + 2n\pi$ . The first minimum is at $x = 7\pi/6$ . The diagram shows $B$ is the second minimum, so we add one full period ( $2\pi$ ) to get $x = 7\pi/6 + 2\pi = 19\pi/6$ . The $y$ -coordinate is $k \times (-1) = -k$ .

Key Takeaways

Phase shifts in trigonometric functions translate the graph horizontally without changing its shape or amplitude.
Identifying specific points on a trigonometric graph requires matching the algebraic solutions to the visual features (increasing/decreasing, first/second occurrence).
The minimum value of $A\cos(Bx + C) + D$ is $D - |A|$ .

Common Mistakes

Selecting the wrong root for $x$ when solving $\cos \theta = 0$ . Always check the diagram to see if the curve is increasing or decreasing at the intercept.
Forgetting to add the full period $2\pi$ when looking for the second minimum point.
Assuming the first minimum shown in the diagram is at $n=0$ without verifying against the phase shift.

Things to Be Careful About

The diagram is crucial here. Point $A$ is not the first $x$ -intercept; it is the second one (where the curve is rising). Point $B$ is the second minimum, not the first.
Ensure angles are in radians as specified in the question.
The constant $k$ remains in the $y$ -coordinate for point $B$ ; do not attempt to find its numerical value as it is not given.

Techniques used

identify x-intercepts from cosine equationfind minimum point of cosine function

(b)

Find the exact value of $t$ that satisfies the equation

3\sin^{-1}(3t) + 2\cos^{-1}\left(\frac{1}{2}\sqrt{2}\right) = \pi

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Evaluate the exact value of $\cos^{-1}\left(\frac{1}{2}\sqrt{2}\right)$ , substitute it into the equation, and then isolate $\sin^{-1}(3t)$ to solve for $t$ .

Working

The given equation is:

3\sin^{-1}(3t) + 2\cos^{-1}\left(\frac{1}{2}\sqrt{2}\right) = \pi

First, evaluate $\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)$ . We know that $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ , and the range of $\cos^{-1}$ is $[0, \pi]$ , so:

\cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}

Substitute this back into the equation:

3\sin^{-1}(3t) + 2\left(\frac{\pi}{4}\right) = \pi

3\sin^{-1}(3t) + \frac{\pi}{2} = \pi

Subtract $\frac{\pi}{2}$ from both sides:

3\sin^{-1}(3t) = \pi - \frac{\pi}{2} = \frac{\pi}{2}

Divide by 3:

\sin^{-1}(3t) = \frac{\pi}{6}

Now, take the sine of both sides. Since $\sin^{-1}(3t) = \frac{\pi}{6}$ , we have:

3t = \sin\left(\frac{\pi}{6}\right)

We know that $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ , so:

3t = \frac{1}{2}

Divide by 3:

t = \frac{1}{6}

Answer

t = \frac{1}{6}

Final answer

t = 1/6

Detailed explanation

Walkthrough

The equation involves inverse trigonometric functions. The first step is to simplify the known constant term: $\cos^{-1}\left(\frac{1}{2}\sqrt{2}\right)$ . Recognizing that $\frac{1}{2}\sqrt{2} = \frac{\sqrt{2}}{2}$ and that $\cos(\pi/4) = \sqrt{2}/2$ , we find this equals $\pi/4$ .

Substituting $\pi/4$ into the equation gives $3\sin^{-1}(3t) + 2(\pi/4) = \pi$ , which simplifies to $3\sin^{-1}(3t) + \pi/2 = \pi$ .

Isolating the inverse sine term: $3\sin^{-1}(3t) = \pi/2$ , so $\sin^{-1}(3t) = \pi/6$ .

To solve for $t$ , we use the definition of the inverse sine function: if $\sin^{-1}(x) = \theta$ , then $x = \sin(\theta)$ . Thus, $3t = \sin(\pi/6)$ .

Since $\sin(\pi/6) = 1/2$ , we have $3t = 1/2$ , which gives $t = 1/6$ .

Key Takeaways

Memorize the exact values of inverse trigonometric functions for standard angles (e.g., $\cos^{-1}(\sqrt{2}/2) = \pi/4$ , $\sin^{-1}(1/2) = \pi/6$ ).
Inverse trigonometric equations can be solved by isolating the inverse function and then applying the corresponding direct trigonometric function to both sides.
Pay attention to the range of inverse functions; for $\cos^{-1}$ , the range is $[0, \pi]$ , and for $\sin^{-1}$ , it is $[-\pi/2, \pi/2]$ .

Common Mistakes

Incorrectly evaluating $\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)$ as $\pi/3$ or $\pi/6$ . Remember the special right triangles or unit circle values.
Forgetting to divide by the coefficient (3) when isolating $\sin^{-1}(3t)$ .
Algebraic errors when solving $3t = 1/2$ for $t$ .

Things to Be Careful About

Ensure consistency in units (radians vs. degrees). The problem uses $\pi$ , so work in radians.
The value $t = 1/6$ gives $3t = 1/2$ , which is within the domain of $\sin^{-1}$ (i.e., $[-1, 1]$ ), so the solution is valid.
Do not confuse $\cos^{-1}(\frac{1}{2}\sqrt{2})$ with $\cos^{-1}(\frac{1}{2})$ ; the $\sqrt{2}$ is crucial.

Techniques used

evaluate inverse cosine exact valueisolate inverse sine termconvert inverse sine to sine equationsolve for t

Q3MediumCircular MeasureTrigonometry

Fig. 3.1

The diagram shows a sector of a circle with centre $C$ . The radii $CA$ and $CB$ each have length $r\text{ cm}$ and the size of the reflex angle $ACB$ is $\theta$ radians. The sector, shaded in the diagram, has a perimeter of $65\text{ cm}$ and an area of $225\text{ cm}^2$ .

(a)

Find the values of $r$ and $\theta$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the standard formulas for the perimeter and area of a circular sector. The perimeter of the shaded region includes the two radii and the arc length corresponding to the reflex angle $\theta$ . Set up a system of two equations with two unknowns ( $r$ and $\theta$ ), eliminate one variable to form a quadratic equation, and solve. Finally, check which solutions are valid given that $\theta$ must be a reflex angle (i.e., $\pi < \theta < 2\pi$ ).

Working

The perimeter of the shaded sector is the sum of the two radii and the arc length:

2r + r\theta = 65

The area of the shaded sector is given by:

\frac{1}{2}r^2\theta = 225

From the perimeter equation, express $r\theta$ in terms of $r$ :

r\theta = 65 - 2r

Substitute this expression into the area equation:

\frac{1}{2}r(65 - 2r) = 225

Expand and rearrange into a standard quadratic form:

65r - 2r^2 = 450

2r^2 - 65r + 450 = 0

Factorise the quadratic equation:

(2r - 45)(r - 10) = 0

This gives two possible values for $r$ :

r = 22.5 \quad \text{or} \quad r = 10

Now find the corresponding values of $\theta$ and check if they are reflex angles (where $\pi \approx 3.1416 < \theta < 2\pi \approx 6.2832$ ):

If $r = 22.5$ , then $r\theta = 65 - 2(22.5) = 20$ , giving $\theta = \frac{20}{22.5} = \frac{8}{9} \approx 0.889$ radians. Since $\frac{8}{9} < \pi$ , this is not a reflex angle. Reject this solution.
If $r = 10$ , then $r\theta = 65 - 2(10) = 45$ , giving $\theta = \frac{45}{10} = 4.5$ radians. Since $\pi < 4.5 < 2\pi$ , this is a valid reflex angle.

Thus, the valid values are $r = 10$ and $\theta = 4.5$ .

Answer

r = 10, \quad \theta = 4.5

Final answer

r = 10, θ = 4.5

Detailed explanation

Walkthrough

First, recall the formulas for the perimeter and area of a circular sector. The perimeter of a sector is the sum of the two straight radii and the curved arc length. For a reflex angle $\theta$ , the arc length is $r\theta$ . This gives the equation $2r + r\theta = 65$ . The area of the sector is given by $\frac{1}{2}r^2\theta = 225$ .

Next, solve this system of equations. Isolate $r\theta$ from the perimeter equation to get $r\theta = 65 - 2r$ . Substitute this into the area equation to eliminate $\theta$ and form a quadratic in $r$ : $\frac{1}{2}r(65 - 2r) = 225$ , which simplifies to $2r^2 - 65r + 450 = 0$ . Factorising gives $(2r - 45)(r - 10) = 0$ , yielding $r = 22.5$ or $r = 10$ .

Finally, test both values of $r$ to find the corresponding $\theta$ . If $r = 22.5$ , $\theta = 8/9$ , which is less than $\pi$ and therefore not a reflex angle, so it is rejected. If $r = 10$ , $\theta = 4.5$ , which lies between $\pi$ and $2\pi$ , confirming it is a valid reflex angle. The correct values are $r = 10$ and $\theta = 4.5$ .

Key Takeaways

The perimeter of a sector includes the arc length plus two radii: $P = r\theta + 2r$ .
The area of a sector is $A = \frac{1}{2}r^2\theta$ .
Always check the validity of solutions against the geometric constraints of the problem (e.g., whether an angle is reflex, acute, or obtuse).

Common Mistakes

Forgetting to include the two radii in the perimeter calculation (using only $r\theta = 65$ ).
Failing to reject the solution $r = 22.5, \theta = 8/9$ because $\theta$ must be a reflex angle ( $> \pi$ ).
Algebraic errors when forming or factorising the quadratic equation.

Things to Be Careful About

Ensure $\theta$ is in radians when using the arc length and sector area formulae.
A reflex angle is strictly between $\pi$ and $2\pi$ radians ( $180^\circ$ and $360^\circ$ ). The value $\theta = 8/9$ is a minor angle, not a reflex angle, so it must be discarded.
The mark scheme awards a method mark for forming a 3-term quadratic or cubic, so show the substitution step clearly.

Techniques used

apply sector perimeter formulaapply sector area formulasubstitute to form quadratic equationreject extraneous solutions based on angle type

(b)

Find the area of triangle $ACB$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The area of triangle $ACB$ can be found using the formula $\frac{1}{2}ab\sin C$ , where $a$ and $b$ are the lengths of two sides and $C$ is the included angle. The sides $CA$ and $CB$ both have length $r = 10$ . The angle at $C$ inside the triangle is the non-reflex angle, which is $2\pi - \theta$ .

Working

The angle inside triangle $ACB$ at vertex $C$ is:

\alpha = 2\pi - \theta = 2\pi - 4.5 \approx 1.783 \text{ rad} \quad (\text{or } 102.2^\circ)

The area of triangle $ACB$ is:

\text{Area} = \frac{1}{2} \times CA \times CB \times \sin(\alpha)

Substitute $CA = CB = 10$ and $\alpha = 2\pi - 4.5$ :

\text{Area} = \frac{1}{2} \times 10 \times 10 \times \sin(2\pi - 4.5)

\text{Area} = 50 \times \sin(1.783...)

\text{Area} \approx 48.9 \text{ cm}^2

Answer

48.9 \text{ cm}^2

Final answer

48.9 cm^2

Detailed explanation

Walkthrough

To find the area of triangle $ACB$ , use the trigonometric formula for the area of a triangle: $\text{Area} = \frac{1}{2}ab\sin C$ . Here, the two known sides are the radii $CA$ and $CB$ , both equal to $r = 10$ cm. The included angle is the angle at $C$ inside the triangle. Since the reflex angle is $\theta = 4.5$ radians, the non-reflex angle (the one inside the triangle) is $2\pi - 4.5$ . Calculate this angle, take its sine, and multiply by $\frac{1}{2}(10)(10)$ to get the area.

Key Takeaways

The area of a triangle with two known sides and an included angle is $\frac{1}{2}ab\sin C$ .
When a reflex angle is given for a sector, the interior angle of the corresponding triangle is $2\pi - \theta$ .

Common Mistakes

Using the reflex angle $\theta = 4.5$ directly in the sine formula. Note that $\sin(4.5) \approx -0.978$ , which would give a negative area. Using $2\pi - \theta$ ensures a positive sine value since $2\pi - 4.5$ is in the second quadrant.
Forgetting to include units in the final answer.

Things to Be Careful About

Ensure your calculator is in radian mode when evaluating $\sin(2\pi - 4.5)$ , or convert $2\pi - 4.5$ to degrees ( $\approx 102.2^\circ$ ) first.
The mark scheme allows for answers to 3 significant figures (48.9) or greater accuracy.

Techniques used

identify the non-reflex angle at the centreapply the sine area formula for a triangle

Q4MediumTrigonometryQuadratics

(a)

Show that the equation $\cos\theta(7\tan\theta - 5\cos\theta) = 1$ can be written in the form $a\sin^2\theta + b\sin\theta + c = 0$ , where $a$ , $b$ and $c$ are integers to be found.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$ to remove the tangent, then use $\cos^2\theta = 1 - \sin^2\theta$ to obtain a quadratic in $\sin\theta$ .

Working

Start with

\cos\theta(7\tan\theta - 5\cos\theta) = 1

Using $\tan\theta = \frac{\sin\theta}{\cos\theta}$ :

\cos\theta\left(\frac{7\sin\theta}{\cos\theta} - 5\cos\theta\right) = 1

7\sin\theta - 5\cos^2\theta = 1

Now use $\cos^2\theta = 1 - \sin^2\theta$ :

7\sin\theta - 5(1 - \sin^2\theta) = 1

7\sin\theta - 5 + 5\sin^2\theta = 1

5\sin^2\theta + 7\sin\theta - 6 = 0

Answer

The equation is

5\sin^2\theta + 7\sin\theta - 6 = 0

so $a = 5$ , $b = 7$ , $c = -6$ .

Final answer

5 sin^2 θ + 7 sin θ - 6 = 0; a = 5, b = 7, c = -6

Detailed explanation

Walkthrough

The aim is to rewrite the left-hand side as a quadratic in $\sin\theta$ . Start by expanding the bracket using the identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$ . This is helpful because the factor $\cos\theta$ outside the bracket cancels with the $\cos\theta$ in the denominator of $\tan\theta$ , leaving $7\sin\theta$ . The second term becomes $-5\cos^2\theta$ . The equation is then $7\sin\theta - 5\cos^2\theta = 1$ .

To make this a quadratic in $\sin\theta$ , replace $\cos^2\theta$ using the Pythagorean identity $\cos^2\theta = 1 - \sin^2\theta$ . Substituting gives $7\sin\theta - 5(1 - \sin^2\theta) = 1$ . Expanding and rearranging gives $5\sin^2\theta + 7\sin\theta - 6 = 0$ , so $a=5$ , $b=7$ , $c=-6$ . This matches the required form.

Key Takeaways

This question tests the two fundamental trigonometric identities:

$\tan\theta = \frac{\sin\theta}{\cos\theta}$
$\cos^2\theta = 1 - \sin^2\theta$

It also shows how an expression involving different trig functions can be reduced to a single quadratic expression in one trig function.

Common Mistakes

Expanding $\cos\theta(7\tan\theta - 5\cos\theta)$ incorrectly, especially forgetting the negative sign on $-5\cos^2\theta$ .
Using $\cos^2\theta = 1 + \sin^2\theta$ or another incorrect identity.
Making a sign error when moving the constant $1$ to the left-hand side, which would give the wrong value of $c$ .
Not simplifying fully to the required form $a\sin^2\theta + b\sin\theta + c = 0$ .

Things to Be Careful About

The mark scheme requires the use of both identities to be shown. An answer that jumps straight to the final quadratic may not earn the method marks.

The equation could equivalently be written as $-5\sin^2\theta - 7\sin\theta + 6 = 0$ ; the mark scheme allows either sign. The integers are $a=5$ , $b=7$ , $c=-6$ (or all multiplied by $-1$ ).

Since the original equation contains $\tan\theta$ , values where $\cos\theta = 0$ are not in the domain, but this does not affect the algebraic rearrangement.

Techniques used

use the tangent identity to simplify the productapply the Pythagorean identityrearrange into a quadratic in sin theta

(b)

Hence solve the equation $\cos 2x(7\tan 2x - 5\cos 2x) = 1$ for $0^\circ < x < 180^\circ$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the result from part (a) with $\theta = 2x$ . Solve the quadratic in $\sin 2x$ , reject any impossible root, then solve $\sin 2x = \frac{3}{5}$ over the interval $0^\circ < 2x < 360^\circ$ and divide by 2.

Working

From part (a), with $\theta = 2x$ :

5\sin^2 2x + 7\sin 2x - 6 = 0

Let $s = \sin 2x$ . Then

5s^2 + 7s - 6 = 0

Factorise:

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

So

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

Since $s = \sin 2x$ must satisfy $-1 \leq s \leq 1$ , reject $s = -2$ . Hence

\sin 2x = \frac{3}{5}

For $0^\circ < x < 180^\circ$ , we have $0^\circ < 2x < 360^\circ$ . The solutions of $\sin y = \frac{3}{5}$ in this interval are

y = 36.87^\circ, \ 143.13^\circ

Therefore

2x = 36.87^\circ \quad \text{or} \quad 2x = 143.13^\circ

x = 18.4^\circ \quad \text{or} \quad x = 71.6^\circ

Both values lie in the required interval, so these are the only solutions.

Answer

x = 18.4^\circ,\ 71.6^\circ

Final answer

x = 18.4°, 71.6°

Detailed explanation

Walkthrough

Since part (a) proved the identity for any angle, replace $\theta$ by $2x$ to get $5\sin^2 2x + 7\sin 2x - 6 = 0$ . This is a quadratic in $\sin 2x$ .

Let $s = \sin 2x$ . Factorising $5s^2 + 7s - 6$ gives $(5s - 3)(s + 2) = 0$ , so $s = \frac{3}{5}$ or $s = -2$ . The second value is impossible because the sine of any angle is always between $-1$ and $1$ . Therefore $\sin 2x = \frac{3}{5}$ .

Now solve $\sin y = \frac{3}{5}$ where $y = 2x$ . Since $0^\circ < x < 180^\circ$ , doubling gives $0^\circ < 2x < 360^\circ$ . In this interval the sine function is positive in the first and second quadrants, so the two solutions are $y = 36.87^\circ$ and $y = 180^\circ - 36.87^\circ = 143.13^\circ$ . Dividing by $2$ gives $x = 18.4^\circ$ and $x = 71.6^\circ$ . Both lie in the required range, so these are the only solutions.

Key Takeaways

A trigonometric equation can often be reduced to a quadratic in one trig function.
The range of $\sin$ is $[-1,1]$ , so roots outside this range must be rejected.
When solving $\sin y = k$ in a given interval, use the symmetry of the sine graph: the second solution is $180^\circ - \text{principal value}$ .
If the variable is $2x$ , solve for $2x$ first and then divide by $2$ , remembering to double the interval.

Common Mistakes

Forgetting to reject $s = -2$ and trying to solve $\sin 2x = -2$ , which has no solution.
Only giving the principal-value solution $18.4^\circ$ and missing $71.6^\circ$ .
Solving for $2x$ and then forgetting to divide by $2$ , giving $36.9^\circ$ and $143.1^\circ$ as the final answers.
Using the interval $0^\circ < x < 180^\circ$ instead of $0^\circ < 2x < 360^\circ$ when finding sine solutions.

Things to Be Careful About

The question says 'Hence', so the solution should use the quadratic obtained in part (a). The mark scheme awards a method mark for attempting to solve a three-term quadratic and finding at least one value of $\sin 2x$ .

The final answers should be given in degrees. If working in radians, the corresponding values are $0.322$ and $1.25$ radians, but the question asks for $0^\circ < x < 180^\circ$ .

Because the interval for $2x$ is one full period from $0^\circ$ to $360^\circ$ (with endpoints excluded), there are exactly two solutions. Do not add extra solutions by continuing the sine pattern beyond this interval.

Techniques used

substitute 2x for thetafactorise the quadratic in sin 2xreject the root outside the sine rangesolve the sine equation in the doubled intervaldivide by 2 to obtain x

Q5Medium-EasyDifferentiation

The equation of a curve is $y = 2x^2 - \frac{1}{2x} + 3$ .

(a)

Find the coordinates of the stationary point.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate the curve term by term, set $\frac{\mathrm{d}y}{\mathrm{d}x}=0$ to locate stationary points, solve for $x$ , then substitute back to find $y$ .

Working

Given

y = 2x^2 - \frac{1}{2x} + 3 = 2x^2 - \frac12 x^{-1} + 3

Differentiate:

\frac{\mathrm{d}y}{\mathrm{d}x} = 4x + \frac12 x^{-2}

Set the derivative equal to zero:

4x + \frac{1}{2x^2} = 0

Multiply by $x^2$ :

4x^3 + \frac12 = 0

Solve:

x^3 = -\frac18

x = -\frac12

Substitute into $y$ :

y = 2\left(-\frac12\right)^2 - \frac{1}{2\left(-\frac12\right)} + 3

y = \frac12 + 1 + 3 = \frac92

Answer

\left(-\frac12, \frac92\right)

Final answer

(-1/2, 9/2)

Detailed explanation

Walkthrough

We need to find where the curve has a stationary point, i.e. where the gradient is zero. First rewrite the term $-\frac{1}{2x}$ as $-\frac12 x^{-1}$ so we can use the power rule. Differentiating gives $4x + \frac12 x^{-2}$ . Setting this equal to zero gives an equation involving $x$ and $\frac{1}{x^2}$ ; multiplying through by $x^2$ turns it into the simple cubic $4x^3 + \frac12 = 0$ . Solving gives $x = -\frac12$ . Substituting this back into the original equation gives $y = \frac92$ . Therefore the stationary point is $\left(-\frac12, \frac92\right)$ .

Key Takeaways

Stationary points occur where the derivative is zero.
Negative powers can be differentiated using the same power rule as positive powers.
Multiplying by $x^2$ removes the denominator and makes the equation easier to solve.

Common Mistakes

Forgetting that the derivative of the constant $3$ is $0$ .
Making a sign error when differentiating $-\frac12 x^{-1}$ ; the derivative is $+\frac12 x^{-2}$ .
Stopping after finding $x$ without substituting back to find $y$ .

Things to Be Careful About

The curve is not defined at $x=0$ , but the stationary point is at $x=-\frac12$ , so no issue arises.
The mark scheme gives only a special-case mark if the cubic is not solved visibly, so show the step $x^3 = -\frac18$ clearly.

Techniques used

differentiate a polynomial with negative powersset the derivative equal to zerosolve the resulting cubic equationsubstitute the x-coordinate to find the y-coordinate

(b)

Determine the nature of the stationary point.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate the first derivative to obtain the second derivative, evaluate it at the stationary point, and use its sign to determine the nature of the stationary point.

Working

From part (a),

\frac{\mathrm{d}y}{\mathrm{d}x} = 4x + \frac12 x^{-2}

Differentiate again:

\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4 - x^{-3}

At $x = -\frac12$ :

\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4 - \left(-\frac12\right)^{-3}

Since $\left(-\frac12\right)^{-3} = -8$ ,

\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4 - (-8) = 12

Because $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$ , the stationary point is a minimum.

Answer

Minimum

Final answer

minimum

Detailed explanation

Walkthrough

We already know the first derivative from part (a). To determine the nature of the stationary point, differentiate again to get the second derivative. Substituting $x = -\frac12$ gives $12$ , which is positive. A positive second derivative means the gradient is increasing at that point, so the curve is locally concave up and the stationary point is a minimum.

Key Takeaways

The second derivative tells us the nature of a stationary point: positive means minimum, negative means maximum.
Evaluating the second derivative at the stationary point is often quicker than testing nearby values.

Common Mistakes

Forgetting to differentiate the first derivative correctly; the derivative of $\frac12 x^{-2}$ is $-x^{-3}$ , not $+x^{-3}$ .
Confusing the sign rule: positive second derivative means minimum, not maximum.

Things to Be Careful About

The mark scheme allows a sign test on the first derivative as an alternative, but if using the second derivative, the substitution must be shown or implied by a correct inequality.
Ensure the final conclusion follows from correct working only.

Techniques used

differentiate the first derivativeevaluate the second derivative at the stationary pointinterpret the sign of the second derivative

(c)

For positive values of $x$ , determine whether the curve shows a function that is increasing, decreasing or neither. Give a reason for your answer.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

For $x>0$ , examine the sign of the first derivative. If it is positive for all positive $x$ , the function is increasing.

Working

The first derivative is

\frac{\mathrm{d}y}{\mathrm{d}x} = 4x + \frac{1}{2x^2}

For $x>0$ , both terms are positive:

4x > 0 \quad \text{and} \quad \frac{1}{2x^2} > 0

Therefore

\frac{\mathrm{d}y}{\mathrm{d}x} > 0 \quad \text{for all } x>0

So the curve is increasing for positive values of $x$ .

Answer

Increasing

Final answer

increasing

Detailed explanation

Walkthrough

For positive values of $x$ , we only need to know the sign of the gradient. The derivative is $4x + \frac{1}{2x^2}$ . When $x>0$ , the term $4x$ is positive and the term $\frac{1}{2x^2}$ is also positive because the denominator is positive. Hence the whole derivative is always positive for $x>0$ . A positive gradient means the function is increasing on that interval.

Key Takeaways

The sign of the derivative determines whether a function is increasing or decreasing.
A function is increasing on an interval if its derivative is positive throughout that interval.
It is enough to argue that each term of the derivative has the same sign, rather than substituting individual values.

Common Mistakes

Substituting a few values of $x$ is not sufficient; the mark scheme requires a clear reference to the derivative being always positive for $x>0$ .
Saying the function is neither increasing nor decreasing because it has a minimum for negative $x$ ; the question only asks about positive values of $x$ .

Things to Be Careful About

The derivative is undefined at $x=0$ , but the interval is strictly $x>0$ , so this does not affect the conclusion.
The conclusion must be based on the first derivative, not on the shape of the curve alone.

Techniques used

inspect the sign of the derivative for x > 0conclude monotonic behaviour from the sign of the gradient

Q6Medium-EasyIntegrationFunctions

A curve passes through the point $(\frac{4}{5}, -3)$ and is such that $\frac{dy}{dx} = \frac{-20}{(5x - 3)^2}$ .

(a)

Find the equation of the curve.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

We are given the derivative $\frac{dy}{dx}$ and a point on the curve. We integrate the derivative with respect to $x$ , then use the given point to determine the constant of integration.

Working

Write the derivative in power form:

\frac{dy}{dx} = -20(5x-3)^{-2}

Integrate with respect to $x$ using the rule $\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c$ :

y = \int -20(5x-3)^{-2}\,dx

y = -20 \cdot \frac{(5x-3)^{-1}}{5 \cdot (-1)} + c

y = 4(5x-3)^{-1} + c

Substitute $x = \frac{4}{5}$ and $y = -3$ :

-3 = 4\left(5 \cdot \frac{4}{5} - 3\right)^{-1} + c

-3 = 4(4 - 3)^{-1} + c

-3 = 4(1)^{-1} + c

-3 = 4 + c

c = -7

Therefore the equation of the curve is:

y = \frac{4}{5x-3} - 7

Final answer

y = 4/(5x-3) - 7

Detailed explanation

Walkthrough

We are given the derivative $\frac{dy}{dx} = \frac{-20}{(5x-3)^2}$ and a point $\left(\frac{4}{5}, -3\right)$ on the curve. To find the equation of the curve, we integrate the derivative. The integrand is of the form $k(ax+b)^n$ with $a = 5$ , $b = -3$ , $n = -2$ , and $k = -20$ . Using the integration rule $\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c$ :

\int -20(5x-3)^{-2}\,dx = -20 \cdot \frac{(5x-3)^{-1}}{5 \cdot (-1)} + c = 4(5x-3)^{-1} + c

Now we substitute the point $\left(\frac{4}{5}, -3\right)$ into $y = 4(5x-3)^{-1} + c$ :

-3 = 4\left(5 \cdot \frac{4}{5} - 3\right)^{-1} + c = 4(4-3)^{-1} + c = 4(1)^{-1} + c = 4 + c

So $c = -7$ , and the equation is $y = 4(5x-3)^{-1} - 7$ .

Key Takeaways

Integration is the reverse of differentiation: to recover $y$ from $\frac{dy}{dx}$ , integrate.
The rule $\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c$ applies to all $n \neq -1$ , including negative powers.
The constant of integration $c$ is determined by substituting a known point on the curve.

Common Mistakes

Forgetting the factor $a$ in the denominator when integrating $(ax+b)^n$ .
Substituting the point in the wrong order (e.g., $x = -3$ , $y = \frac{4}{5}$ ) — the mark scheme explicitly awards DM0 for this.
Forgetting to include the constant $c$ before substituting.

Things to Be Careful About

The coefficient $-20$ divided by $5 \cdot (-1) = -5$ gives $+4$ , not $-4$ .
The mark scheme condones the absence of $+c$ at the integration stage, but $c$ must appear before the substitution step.
The final answer should be written as $y = \frac{4}{5x-3} - 7$ or $y = 4(5x-3)^{-1} - 7$ .

Techniques used

integrate a power of a linear expressionsubstitute a point to find the constant of integration

(b)

The curve is transformed by a stretch in the $x$ -direction with scale factor $\frac{1}{2}$ followed by a translation of $\begin{pmatrix} 2 \\ 10 \end{pmatrix}$ .

Find the equation of the new curve.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Apply the two transformations in the order given. A stretch in the $x$ -direction with scale factor $\frac{1}{2}$ replaces $x$ by $2x$ . A translation by $\begin{pmatrix}2\\10\end{pmatrix}$ replaces $x$ by $x-2$ and $y$ by $y-10$ .

Working

Start from the equation found in part (a):

y = 4(5x-3)^{-1} - 7

Step 1: Stretch in the $x$ -direction, scale factor $\frac{1}{2}$ .

Replace $x$ by $2x$ :

y = 4(5(2x)-3)^{-1} - 7 = 4(10x-3)^{-1} - 7

Step 2: Translation by $\begin{pmatrix}2\\10\end{pmatrix}$ .

Replace $x$ by $x-2$ and $y$ by $y-10$ :

y - 10 = 4(10(x-2)-3)^{-1} - 7

Simplify inside the brackets:

y - 10 = 4(10x - 20 - 3)^{-1} - 7

y - 10 = 4(10x-23)^{-1} - 7

Add 10 to both sides:

y = 4(10x-23)^{-1} + 3

Answer

y = \frac{4}{10x-23} + 3

Final answer

y = 4/(10x-23) + 3

Detailed explanation

Walkthrough

We start from the equation $y = 4(5x-3)^{-1} - 7$ found in part (a).

The first transformation is a stretch in the $x$ -direction with scale factor $\frac{1}{2}$ . A stretch in the $x$ -direction with scale factor $k$ replaces $x$ by $\frac{x}{k}$ . With $k = \frac{1}{2}$ , we replace $x$ by $2x$ :

y = 4(5(2x)-3)^{-1} - 7 = 4(10x-3)^{-1} - 7

The second transformation is a translation by $\begin{pmatrix}2\\10\end{pmatrix}$ . A translation by $\begin{pmatrix}a\\b\end{pmatrix}$ replaces $x$ by $x-a$ and $y$ by $y-b$ . Here $a = 2$ and $b = 10$ , so we replace $x$ by $x-2$ and $y$ by $y-10$ :

y - 10 = 4(10(x-2)-3)^{-1} - 7

Simplify inside the bracket: $10(x-2) - 3 = 10x - 20 - 3 = 10x - 23$ .

y - 10 = 4(10x-23)^{-1} - 7

Add 10 to both sides:

y = 4(10x-23)^{-1} + 3

Key Takeaways

A stretch in the $x$ -direction with scale factor $k$ replaces $x$ by $\frac{x}{k}$ .
A translation by $\begin{pmatrix}a\\b\end{pmatrix}$ replaces $x$ by $x-a$ and $y$ by $y-b$ .
Transformations must be applied in the order given.

Common Mistakes

Applying the transformations in the wrong order — the mark scheme awards M1 for the stretch as the second transformation and M1 for the translation as the first.
Getting the stretch factor wrong: scale factor $\frac{1}{2}$ means replace $x$ by $2x$ , not by $\frac{1}{2}x$ .
Sign errors in the translation: replacing $x$ by $x+2$ instead of $x-2$ , or $y$ by $y+10$ instead of $y-10$ .

Things to be Careful About

The mark scheme says "Do not ignore sign errors" — the translation must use $x-2$ and $y-10$ .
The final answer must be simplified to $y = \frac{4}{10x-23} + 3$ .

Techniques used

apply a stretch in the x-directionapply a translationsimplify the transformed expression

Q7MediumSeries

The first term of an arithmetic progression is $1.5$ and the sum of the first ten terms is $127.5$ .

(a)

Find the common difference.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the formula for the sum of the first $n$ terms of an arithmetic progression, substitute the given values, and solve for the common difference $d$ .

Working

The sum of the first $n$ terms of an arithmetic progression with first term $a$ and common difference $d$ is:

S_n = \frac{n}{2}(2a + (n-1)d)

Here $a = 1.5$ and $S_{10} = 127.5$ , so:

127.5 = \frac{10}{2}(2(1.5) + (10-1)d)

127.5 = 5(3 + 9d)

127.5 = 15 + 45d

45d = 112.5

d = 2.5

Answer

d = 2.5

Final answer

d = 2.5

Detailed explanation

Walkthrough

We are told the first term is $1.5$ and the sum of the first ten terms is $127.5$ . The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}(2a + (n-1)d)$ . Substituting $a = 1.5$ , $n = 10$ and $S_{10} = 127.5$ gives the equation $127.5 = 5(3 + 9d)$ . This simplifies to $127.5 = 15 + 45d$ , so $45d = 112.5$ and $d = 2.5$ . The key idea is that the sum formula links $a$ , $d$ and $n$ , so with three known quantities we can solve for the unknown common difference.

Key Takeaways

The sum formula $S_n = \frac{n}{2}(2a + (n-1)d)$ is the fundamental tool for arithmetic progressions. Substituting the given values and solving the resulting linear equation gives the common difference.

Common Mistakes

Forgetting to halve the number of terms: writing $S_{10} = 10(2a + 9d)$ instead of $5(2a + 9d)$ .
Arithmetic slips when solving $15 + 45d = 127.5$ , e.g. subtracting incorrectly.

Things to Be Careful About

The bracket is $2a + (n-1)d$ , not $a + (n-1)d$ .
When $n = 10$ , the bracket contains $9d$ , not $10d$ .
Check the result: $5(3 + 9(2.5)) = 5(3 + 22.5) = 5(25.5) = 127.5$ . ✓

Techniques used

substitute known values into sum formulasolve linear equation for common difference

(b)

Find the sum of all the terms of the arithmetic progression whose values are between $25$ and $100$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use the nth term formula to find the term numbers that fall strictly between 25 and 100, then sum the arithmetic series formed by those terms.

Working

The nth term is:

a_n = a + (n-1)d = 1.5 + 2.5(n-1)

First term greater than 25:

1.5 + 2.5(n-1) > 25

2.5(n-1) > 23.5

n-1 > 9.4

n > 10.4

So the first term in the set is the 11th term:

a_{11} = 1.5 + 2.5(10) = 26.5

Last term less than 100:

1.5 + 2.5(n-1) < 100

2.5(n-1) < 98.5

n-1 < 39.4

n < 40.4

So the last term in the set is the 40th term:

a_{40} = 1.5 + 2.5(39) = 99

The terms from the 11th to the 40th inclusive form the required set, so there are $40 - 11 + 1 = 30$ terms.

Using the sum formula with the first and last terms of the set:

S = \frac{30}{2}(26.5 + 99) = 15(125.5) = 1882.5

Answer

1882.5

Final answer

1882.5

Detailed explanation

Walkthrough

We need the sum of all terms whose values lie strictly between 25 and 100. Using the nth term formula $a_n = 1.5 + 2.5(n-1)$ , we solve two inequalities.

For the lower bound: $1.5 + 2.5(n-1) > 25$ gives $n > 10.4$ , so the smallest integer term number is $n = 11$ , giving $a_{11} = 26.5$ .

For the upper bound: $1.5 + 2.5(n-1) < 100$ gives $n < 40.4$ , so the largest integer term number is $n = 40$ , giving $a_{40} = 99$ .

The set therefore runs from the 11th to the 40th term inclusive, which is $40 - 11 + 1 = 30$ terms. The sum of an arithmetic series with 30 terms and first and last terms $26.5$ and $99$ is $\frac{30}{2}(26.5 + 99) = 1882.5$ .

Key Takeaways

To find the sum of terms in a range, first identify the term numbers at the boundaries using the nth term inequality, then apply the sum formula with the correct number of terms.

Common Mistakes

Using $n = 10$ as the first term instead of $n = 11$ (the inequality $n > 10.4$ forces the first integer to be 11).
Using $n = 41$ as the last term instead of $n = 40$ (the inequality $n < 40.4$ forces the last integer to be 40).
Counting the number of terms incorrectly: it is $40 - 11 + 1 = 30$ , not 29.

What to Be Careful About

"Between 25 and 100" is strict: neither endpoint is included.
The number of terms is $40 - 11 + 1 = 30$ .
The sum can also be found as $S_{40} - S_{10}$ , which gives the same result: $S_{40} = 2010$ and $S_{10} = 127.5$ , so $2010 - 127.5 = 1882.5$ . ✓

Techniques used

set up inequalities from nth term formulafind first and last terms in rangesum arithmetic series with first and last terms

Q88MMedium-HardCoordinate Geometry

A circle with equation $x^2 + y^2 - 6x + 2y - 15 = 0$ meets the $y$ -axis at the points $A$ and $B$ . The tangents to the circle at $A$ and $B$ meet at the point $P$ .

Find the coordinates of $P$ .

DifficultyMedium-Hard

Worked solution

Approach

Find where the circle meets the $y$ -axis by setting $x = 0$ and solving the resulting quadratic. Complete the square to locate the centre $C$ . For each intersection point, use the fact that a tangent is perpendicular to the radius, so its gradient is the negative reciprocal of the gradient of the corresponding radius. Write the equations of the two tangents and solve them simultaneously to find $P$ .

Working

Set $x = 0$ in the circle equation:

y^2 + 2y - 15 = 0

Factorise:

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

So the circle meets the $y$ -axis at

A(0, -5), \quad B(0, 3)

Complete the square on the circle equation:

x^2 + y^2 - 6x + 2y - 15 = (x - 3)^2 + (y + 1)^2 - 25 = 0

Hence the centre is

C(3, -1)

and the radius is $5$ .

Gradient of $CA$ :

m_{CA} = \frac{-1 - (-5)}{3 - 0} = \frac{4}{3}

Gradient of $CB$ :

m_{CB} = \frac{-1 - 3}{3 - 0} = -\frac{4}{3}

Since each tangent is perpendicular to the radius at the point of contact, the tangent gradients are the negative reciprocals:

m_{\text{tangent at }A} = -\frac{3}{4}, \qquad m_{\text{tangent at }B} = \frac{3}{4}

Equation of tangent at $A(0,-5)$ :

y + 5 = -\frac{3}{4}x \quad \Rightarrow \quad y = -\frac{3}{4}x - 5

Equation of tangent at $B(0,3)$ :

y - 3 = \frac{3}{4}x \quad \Rightarrow \quad y = \frac{3}{4}x + 3

Solve the simultaneous equations:

-\frac{3}{4}x - 5 = \frac{3}{4}x + 3

-\frac{3}{2}x = 8

x = -\frac{16}{3}

Substitute into the tangent at $B$ :

y = \frac{3}{4}\left(-\frac{16}{3}\right) + 3 = -4 + 3 = -1

Answer

P\left(-\frac{16}{3}, -1\right)

Final answer

(-16/3, -1)

Detailed explanation

Walkthrough

We start by finding the two points where the circle cuts the $y$ -axis. On the $y$ -axis every point has $x = 0$ , so we substitute $x = 0$ into the circle equation. This gives a quadratic in $y$ : $y^2 + 2y - 15 = 0$ . Factorising gives $y = -5$ and $y = 3$ , so the two points are $A(0,-5)$ and $B(0,3)$ .

Next, we need the centre of the circle. Rearranging $x^2 + y^2 - 6x + 2y - 15 = 0$ by completing the square gives $(x-3)^2 + (y+1)^2 = 25$ , so the centre is $C(3,-1)$ and the radius is $5$ .

The key geometric fact is that a tangent to a circle is perpendicular to the radius drawn to the point of contact. Therefore, once we know the gradient of the radius $CA$ or $CB$ , the gradient of the tangent is the negative reciprocal. For $A$ , the gradient of $CA$ is $\frac{4}{3}$ , so the tangent at $A$ has gradient $-\frac{3}{4}$ . For $B$ , the gradient of $CB$ is $-\frac{4}{3}$ , so the tangent at $B$ has gradient $\frac{3}{4}$ .

Using point-slope form, we write the two tangent equations. Since both tangents pass through $P$ , their equations must both be true at $P$ . We therefore solve the two linear equations simultaneously. Equating the two expressions for $y$ gives $x = -\frac{16}{3}$ , and substituting back gives $y = -1$ . Thus $P$ is $\left(-\frac{16}{3}, -1\right)$ .

Key Takeaways

The question combines several coordinate-geometry skills: finding intercepts, completing the square to identify a circle's centre, using the perpendicular relationship between radius and tangent, and solving simultaneous linear equations. It also shows the value of drawing a clear mental picture of the circle and its tangents before algebra.

Common Mistakes

Substituting $y = 0$ instead of $x = 0$ when finding where the circle meets the $y$ -axis. The $y$ -axis is the line $x = 0$ .
Factorising $y^2 + 2y - 15$ incorrectly; the roots are $-5$ and $3$ , not $5$ and $-3$ .
Forgetting to complete the square correctly: the centre is $(3,-1)$ , not $(-3,1)$ .
Using the radius gradient as the tangent gradient instead of taking the negative reciprocal.
Making sign errors when forming the tangent equations, especially with the point $(0,-5)$ .
Not showing the method for solving the quadratic; the mark scheme only gives $B1$ as a special case for the correct $y$ -values with no working.

Things to Be Careful About

The tangent at $A$ has gradient $-\frac{3}{4}$ , while the tangent at $B$ has gradient $\frac{3}{4}$ ; swapping these will give the wrong equations.
When solving the simultaneous equations, be careful with fractions: $-\frac{3}{4}x - 5 = \frac{3}{4}x + 3$ leads to $-\frac{3}{2}x = 8$ .
The $y$ -coordinate of $P$ is the same as the $y$ -coordinate of the centre, $-1$ , because the two tangents are symmetric about the horizontal line through the centre. This can be used as a check.
The mark scheme allows alternative methods such as implicit differentiation or similar triangles, but the perpendicular-radius method is direct and reliable.

Techniques used

substitute x = 0 to find y-axis intersectionscomplete the square to find the centre of the circleuse the perpendicular radius-tangent relationship to find tangent gradientssolve simultaneous equations of the two tangents

Approach

Differentiate $y = (2x^3 + 10)^{1/2}$ using the chain rule to find $\frac{dy}{dx}$ , evaluate at $x = 3$ to get the gradient, find the corresponding $y$ -value, then use the point-gradient form to write the tangent equation in the form $ax + by + c = 0$ .

Working

The curve is $y = (2x^3 + 10)^{1/2}$ .

Differentiate using the chain rule:

\frac{dy}{dx} = \frac{1}{2}(2x^3 + 10)^{-1/2} \cdot 6x^2 = 3x^2(2x^3 + 10)^{-1/2}

At $x = 3$ , find the $y$ -coordinate:

y = \sqrt{2(3)^3 + 10} = \sqrt{54 + 10} = \sqrt{64} = 8

So the point of tangency is $(3, 8)$ .

Evaluate the gradient at $x = 3$ :

\frac{dy}{dx} = 3(3)^2(2(3)^3 + 10)^{-1/2} = 27 \cdot 64^{-1/2} = \frac{27}{8}

Use the point-gradient form of the tangent at $(3, 8)$ :

y - 8 = \frac{27}{8}(x - 3)

Multiply through by 8:

8y - 64 = 27x - 81

Rearrange to the form $ax + by + c = 0$ :

27x - 8y - 17 = 0

Answer

27x - 8y - 17 = 0

Final answer

27x - 8y - 17 = 0

Detailed explanation

Walkthrough

First, we rewrite $y = \sqrt{2x^3 + 10}$ as $y = (2x^3 + 10)^{1/2}$ so we can apply the chain rule. Differentiating gives $\frac{dy}{dx} = \frac{1}{2}(2x^3 + 10)^{-1/2} \cdot 6x^2 = 3x^2(2x^3 + 10)^{-1/2}$ . At $x = 3$ , we compute $y = \sqrt{64} = 8$ , so the point is $(3, 8)$ . Substituting $x = 3$ into the derivative gives the gradient $\frac{27}{8}$ . Using the point-gradient form $y - y_1 = m(x - x_1)$ , we get $y - 8 = \frac{27}{8}(x - 3)$ . Multiplying by 8 and rearranging yields $27x - 8y - 17 = 0$ , which is in the required integer form.

Key Takeaways

The chain rule is essential for differentiating composite functions like $\sqrt{f(x)}$ .
The tangent equation can be rearranged into the general form $ax + by + c = 0$ by clearing fractions and collecting all terms on one side.
Always verify that $a$ , $b$ , and $c$ are integers as required by the question.

Common Mistakes

Forgetting to multiply by the derivative of the inner function $2x^3 + 10$ when applying the chain rule.
Making arithmetic errors when evaluating $2(3)^3 + 10 = 64$ .
Not clearing the fraction when rearranging into the form $ax + by + c = 0$ , leaving non-integer coefficients.
Using the gradient of the normal instead of the tangent (this would lose the DM1 mark).

Things to Be Careful About

The question requires $a$ , $b$ , $c$ to be integers, so clear any fractions before rearranging.
The mark scheme awards DM1 for an attempt at the tangent equation with the correct numerical gradient; using the normal's gradient gives DM0.
The answer $27x - 8y - 17 = 0$ or any integer multiple such as $-27x + 8y + 17 = 0$ is accepted.

Techniques used

apply chain rule for differentiationevaluate derivative at given point to find gradientfind equation of tangent in general form

(b)

The region shaded in the diagram is enclosed by the curve and the straight lines $x = 1$ , $x = 3$ and $y = 0$ .

Find the volume of the solid obtained when the shaded region is rotated through $360^\circ$ about the $x$ -axis.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the volume of revolution formula $V = \pi \int y^2 \, dx$ about the $x$ -axis, with limits $x = 1$ to $x = 3$ . Since $y = \sqrt{2x^3 + 10}$ , we have $y^2 = 2x^3 + 10$ , which simplifies the integrand to a polynomial.

Working

The volume of the solid obtained by rotating the region about the $x$ -axis is:

V = \pi \int_1^3 y^2 \, dx

Substitute $y^2 = 2x^3 + 10$ :

V = \pi \int_1^3 (2x^3 + 10) \, dx

Integrate:

V = \pi \left[ \frac{2x^4}{4} + 10x \right]_1^3 = \pi \left[ \frac{x^4}{2} + 10x \right]_1^3

Evaluate at the upper limit $x = 3$ :

\frac{3^4}{2} + 10(3) = \frac{81}{2} + 30 = \frac{141}{2}

Evaluate at the lower limit $x = 1$ :

\frac{1^4}{2} + 10(1) = \frac{1}{2} + 10 = \frac{21}{2}

Subtract:

V = \pi \left( \frac{141}{2} - \frac{21}{2} \right) = \pi \cdot \frac{120}{2} = 60\pi

Answer

60\pi

Final answer

60π

Detailed explanation

Walkthrough

The shaded region is bounded by the curve $y = \sqrt{2x^3 + 10}$ , the $x$ -axis ( $y = 0$ ), and the vertical lines $x = 1$ and $x = 3$ . When this region is rotated $360°$ about the $x$ -axis, the volume is given by $V = \pi \int y^2 \, dx$ . Since $y^2 = 2x^3 + 10$ , the integrand becomes a simple polynomial. We integrate to get $\frac{x^4}{2} + 10x$ , then evaluate at the upper limit $x = 3$ and lower limit $x = 1$ , subtracting the results. The final answer is $60\pi$ .

Key Takeaways

The volume of revolution about the $x$ -axis uses $V = \pi \int y^2 \, dx$ .
Squaring $y = \sqrt{2x^3 + 10}$ eliminates the square root, giving a polynomial integrand.
Always apply the limits of integration correctly by subtracting the lower limit value from the upper limit value.

Common Mistakes

Forgetting the $\pi$ factor in the volume formula.
Attempting to integrate $\sqrt{2x^3 + 10}$ directly instead of using $y^2 = 2x^3 + 10$ .
Arithmetic errors when evaluating $\frac{81}{2} + 30$ or $\frac{1}{2} + 10$ .
Forgetting to subtract the lower limit evaluation from the upper limit evaluation.

Things to Be Careful About

The mark scheme does not allow an unsimplified form such as $\pi \int ((2x^3 + 10)^{1/2})^2 \, dx$ ; you must simplify $y^2$ to $2x^3 + 10$ before integrating.
Numerical answers in the range 188–189 are accepted as equivalent to $60\pi \approx 188.5$ .
The $\pi$ should appear only at the end of the final answer, not inside the integral.

Techniques used

set up volume of revolution integralintegrate polynomialapply limits of integration

Q10MediumSeriesQuadratics

The geometric progression $a_1, a_2, a_3, \dots$ has first term $2$ and common ratio $r$ where $r > 0$ . It is given that $\frac{9}{2}a_5 + 7a_3 = 8$ .

(a)

Find the value of $r$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Write the $n$ th term of the geometric progression, substitute the given terms into the condition, and solve the resulting quadratic in $r^2$ . Since $r > 0$ , discard the negative possibility.

Working

For a GP with first term $a = 2$ and common ratio $r$ ,

a_n = 2r^{n-1}

Hence

a_3 = 2r^2, \qquad a_5 = 2r^4

Substitute into $\frac{9}{2}a_5 + 7a_3 = 8$ :

\frac{9}{2}(2r^4) + 7(2r^2) = 8

9r^4 + 14r^2 = 8

9r^4 + 14r^2 - 8 = 0

Let $x = r^2$ . Then

9x^2 + 14x - 8 = 0

(9x - 4)(x + 2) = 0

So $x = \frac{4}{9}$ or $x = -2$ . Since $x = r^2 \geq 0$ and $r > 0$ ,

r = \frac{2}{3}

Answer

r = \frac{2}{3}

Final answer

r = 2/3

Detailed explanation

Walkthrough

We are told that the progression is geometric with first term $2$ and common ratio $r$ . The key is to express every term in the given condition using $r$ . The $n$ th term of a GP is $a_n = ar^{n-1}$ , so $a_3 = 2r^2$ and $a_5 = 2r^4$ . Substituting these into $\frac{9}{2}a_5 + 7a_3 = 8$ gives $9r^4 + 14r^2 = 8$ , or $9r^4 + 14r^2 - 8 = 0$ . This is not a quadratic in $r$ , but it is a quadratic in $r^2$ : writing $x = r^2$ turns it into $9x^2 + 14x - 8 = 0$ . Factorising gives $(9x - 4)(x + 2) = 0$ , so $x = \frac{4}{9}$ or $x = -2$ . Because $x = r^2$ cannot be negative, the only possible value is $x = \frac{4}{9}$ . Finally, since the question states $r > 0$ , we take the positive square root and obtain $r = \frac{2}{3}$ .

Key Takeaways

The $n$ th term of a geometric progression is $a_n = ar^{n-1}$ .
An equation such as $9r^4 + 14r^2 - 8 = 0$ can be solved by substituting $x = r^2$ .
Always use the given conditions such as $r > 0$ to select the correct root.

Common Mistakes

Substituting $a_3$ and $a_5$ incorrectly, for example forgetting the factor $2$ from the first term.
Treating $9r^4 + 14r^2 - 8 = 0$ as a quadratic in $r$ rather than in $r^2$ .
Keeping $r = -\frac{2}{3}$ as well as $r = \frac{2}{3}$ even though $r > 0$ .
Giving an unsupported answer: the mark scheme only allows a special-case mark for the final answer without working.

Things to Be Careful About

$r^2$ is never negative, so $x = -2$ must be discarded.
Check the factorisation: $(9x - 4)(x + 2) = 9x^2 + 18x - 4x - 8 = 9x^2 + 14x - 8$ .
Use the sign condition $r > 0$ before taking the square root.

Techniques used

write the nth term of a geometric progressionsubstitute terms into a given equationsolve a quadratic in r^2 by factorisationselect the positive root using r > 0

(b)

Find the sum of the first $20$ terms of the geometric progression. Give your answer correct to $4$ significant figures.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the formula for the sum of the first $n$ terms of a geometric progression with $a = 2$ , $r = \frac{2}{3}$ and $n = 20$ , then round to 4 significant figures.

Working

For a GP,

S_n = \frac{a(1 - r^n)}{1 - r}

Substitute $a = 2$ , $r = \frac{2}{3}$ , $n = 20$ :

S_{20} = \frac{2\left(1 - \left(\frac{2}{3}\right)^{20}\right)}{1 - \frac{2}{3}}

Simplify the denominator:

S_{20} = \frac{2\left(1 - \left(\frac{2}{3}\right)^{20}\right)}{\frac{1}{3}}

S_{20} = 6\left(1 - \left(\frac{2}{3}\right)^{20}\right)

Evaluating,

S_{20} = 5.9981956\ldots

Correct to 4 significant figures,

S_{20} = 5.998

Answer

5.998

Final answer

5.998

Detailed explanation

Walkthrough

Part (b) asks for the sum of the first 20 terms. The formula for the sum of the first $n$ terms of a GP is $S_n = \frac{a(1-r^n)}{1-r}$ . We substitute $a = 2$ , $r = \frac{2}{3}$ and $n = 20$ . The denominator is $1 - \frac{2}{3} = \frac{1}{3}$ , so dividing by $\frac{1}{3}$ is equivalent to multiplying by $3$ , giving $S_{20} = 6\left(1 - \left(\frac{2}{3}\right)^{20}\right)$ . Evaluating this gives approximately $5.9981956$ . The instruction asks for 4 significant figures, so the answer is $5.998$ ; the next digit is $1$ , so we do not round up.

Key Takeaways

The sum formula for a GP is $S_n = \frac{a(1-r^n)}{1-r}$ .
When $r$ is close to $1$ , the sum can be close to $6$ , but the exact value must still be evaluated.
Significant-figure rounding requires looking at the first digit after the required number of figures.

Common Mistakes

Using the formula for the $n$ th term instead of the sum formula.
Using $r^{n}$ in the numerator but forgetting that the denominator is $1-r$ .
Rounding to 4 decimal places instead of 4 significant figures.
Giving the unrounded value $5.9981956$ instead of the required $5.998$ .

Things to Be Careful About

Since $r = \frac{2}{3} < 1$ , the formula $\frac{a(1-r^n)}{1-r}$ is appropriate; the alternative form $\frac{a(r^n-1)}{r-1}$ gives the same result.
$\left(\frac{2}{3}\right)^{20}$ is small, so $S_{20}$ is slightly less than $6$ .
The mark scheme accepts answers within a range (AWRT), but $5.998$ is the exact 4-significant-figure value.

Techniques used

substitute values into the GP sum formulaevaluate a large power and simplifyround to a required number of significant figures

(c)

Find the sum to infinity of the progression $a_2, a_5, a_8, \dots$ .

3M Similar

DifficultyMedium

Worked solution

Approach

The sequence $a_2, a_5, a_8, \dots$ is itself a geometric progression. Find its first term and common ratio, then use the sum-to-infinity formula.

Working

For the original GP, $a_n = 2r^{n-1}$ with $r = \frac{2}{3}$ . The first term of the new progression is

a_2 = 2r = 2\left(\frac{2}{3}\right) = \frac{4}{3}

The common ratio of the new progression is

\frac{a_5}{a_2} = \frac{2r^4}{2r} = r^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27}

Since $\left|\frac{8}{27}\right| < 1$ , the sum to infinity exists:

S_\infty = \frac{a_2}{1 - r^3}

S_\infty = \frac{\frac{4}{3}}{1 - \frac{8}{27}}

S_\infty = \frac{\frac{4}{3}}{\frac{19}{27}}

S_\infty = \frac{4}{3} \times \frac{27}{19} = \frac{36}{19}

Answer

\frac{36}{19}

Final answer

36/19

Detailed explanation

Walkthrough

The progression $a_2, a_5, a_8, \dots$ is obtained by taking every third term of the original GP. Its first term is $a_2 = 2r = \frac{4}{3}$ . Its common ratio is the ratio of consecutive selected terms: $\frac{a_5}{a_2} = \frac{2r^4}{2r} = r^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$ . Since $\left|\frac{8}{27}\right| < 1$ , the sum to infinity exists. Using $S_\infty = \frac{a}{1-r}$ with $a = \frac{4}{3}$ and $r = \frac{8}{27}$ , we get $S_\infty = \frac{\frac{4}{3}}{1 - \frac{8}{27}} = \frac{\frac{4}{3}}{\frac{19}{27}} = \frac{4}{3} \times \frac{27}{19} = \frac{36}{19}$ .

Key Takeaways

Taking every third term of a GP produces another GP whose common ratio is $r^3$ .
The sum to infinity formula $S_\infty = \frac{a}{1-r}$ is valid only when $|r| < 1$ .
Compound fractions can be simplified by multiplying by the reciprocal of the denominator.

Common Mistakes

Using the original first term $2$ instead of the new first term $a_2 = \frac{4}{3}$ .
Using the original common ratio $\frac{2}{3}$ instead of the new common ratio $\frac{8}{27}$ .
Applying the sum-to-infinity formula without checking that $|r| < 1$ .
Arithmetic errors when dividing fractions, such as writing $\frac{4}{3} \div \frac{19}{27}$ as $\frac{4}{3} \times \frac{27}{19}$ incorrectly.

Things to Be Careful About

The common ratio of the selected progression is $r^3$ , not $r$ .
The sum to infinity exists because $\frac{8}{27}$ is less than $1$ in magnitude.
The answer $\frac{36}{19}$ can also be written as $1.894736\ldots$ ; the mark scheme accepts either exact or rounded forms.

Techniques used

identify the first term of the subsequencefind the common ratio as a power of rapply the sum-to-infinity formulasimplify a compound fraction

Q11Medium-HardFunctionsQuadratics

The function $f$ is defined by $f(x) = 10 + 6x - x^2$ for $x \in \mathbb{R}$ .

(a)

By completing the square, find the range of $f$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Complete the square on $10+6x-x^2$ to write it in the form $a-(x-3)^2$ . Since $(x-3)^2 \geq 0$ , the expression is at most $a$ , so the range is all real values up to and including that maximum.

Working

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

Complete the square inside the bracket:

x^2 - 6x = (x-3)^2 - 9

Therefore

\begin{aligned} f(x) &= 10 - \left((x-3)^2 - 9\right) \\ &= 19 - (x-3)^2 \end{aligned}

Since $(x-3)^2 \geq 0$ for all real $x$ ,

19 - (x-3)^2 \leq 19

So the range is

\mathrm{f}(x) \leq 19

or equivalently $(-\infty, 19]$ .

Answer

The range of $f$ is $(-\infty, 19]$ , i.e. $\mathrm{f}(x) \leq 19$ .

Final answer

(-∞, 19]

Detailed explanation

Walkthrough

We are asked to find the range of $f(x)=10+6x-x^2$ by completing the square. The range is the set of all possible output values $f(x)$ as $x$ ranges over the real numbers. Since the coefficient of $x^2$ is negative, the graph is a downward-opening parabola, so the function has a maximum value and the range is of the form $(-\infty, \text{maximum}]$ .

First rewrite the quadratic with the $x^2$ term factored out as a negative: $10+6x-x^2 = 10-(x^2-6x)$ . Then complete the square on $x^2-6x$ : $x^2-6x=(x-3)^2-9$ . Substituting back gives $f(x)=10-((x-3)^2-9)=19-(x-3)^2$ .

Now $(x-3)^2$ is never negative, so subtracting it from 19 can only make the value smaller than or equal to 19. The maximum value 19 occurs when $x=3$ . Therefore every output is at most 19, and every value below 19 is attainable, giving the range $(-\infty,19]$ .

Key Takeaways

Completing the square reveals the vertex of a quadratic: for $f(x)=a-(x-h)^2$ , the maximum is $a$ at $x=h$ .
For a downward-opening quadratic the range is $(-\infty, \text{maximum}]$ ; for an upward-opening quadratic it is $[\text{minimum}, \infty)$ .
The range must use $\leq$ (or $\geq$ ) at the endpoint, not $<$ , because the maximum (or minimum) value is actually attained.

Common Mistakes

Forgetting the negative sign when completing the square, e.g. writing $10+(x-3)^2-9$ instead of $10-((x-3)^2-9)$ .
Writing $f(x)<19$ instead of $f(x)\leq 19$ . The value 19 is achieved at $x=3$ , so it must be included.
Using calculus or a graph when the question explicitly requires completing the square; the mark scheme requires the completed-square method.

Things to Be Careful About

Complete the square correctly: $x^2-6x$ becomes $(x-3)^2-9$ , not $(x-3)^2+9$ .
The final range can be written as $\mathrm{f}(x) \leq 19$ , $y \leq 19$ , or $(-\infty,19]$ . It must contain the endpoint 19.
If the completed-square form is written as $19-(3-x)^2$ , it is equivalent because $(3-x)^2=(x-3)^2$ .

Techniques used

complete the squareidentify the maximum valuestate the range including the endpoint

(b)

The function $g$ is defined by $g(x) = 4x + k$ for $x \in \mathbb{R}$ where $k$ is a constant.

It is given that the graph of $y = g^{-1}f(x)$ meets the graph of $y = g(x)$ at a single point $P$ .

Determine the coordinates of $P$ .

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Find the inverse of the linear function $g$ , form the composite $g^{-1}f(x)$ , and set it equal to $g(x)$ . The condition that the two graphs meet at a single point means this equation has exactly one solution, so its quadratic has discriminant zero. Solve for $k$ , then solve for $x$ and substitute back to find $y$ .

Working

Find $g^{-1}$ :

g(x) = 4x + k \quad \Rightarrow \quad g^{-1}(x) = \frac{x-k}{4}

Form the composite:

g^{-1}f(x) = \frac{f(x)-k}{4} = \frac{10+6x-x^2-k}{4}

Set $g^{-1}f(x) = g(x)$ :

\frac{10+6x-x^2-k}{4} = 4x+k

Multiply by 4:

10+6x-x^2-k = 16x+4k

Rearrange into a quadratic in $x$ :

-x^2 -10x +10 - 5k = 0

or equivalently

x^2 + 10x - 10 + 5k = 0

For the graphs to meet at a single point, this quadratic must have exactly one solution, so its discriminant is zero:

10^2 - 4(1)(-10+5k) = 0

100 - 4(5k-10) = 0

100 - 20k + 40 = 0

140 - 20k = 0

k = 7

Substitute $k=7$ into the quadratic:

x^2 + 10x - 10 + 5(7) = 0

x^2 + 10x + 25 = 0

(x+5)^2 = 0

So

x = -5

Find the corresponding $y$ -coordinate using $y = g(x)$ :

y = 4(-5)+7 = -20+7 = -13

Therefore the single point of intersection is

P = (-5, -13)

Answer

The coordinates of $P$ are $(-5, -13)$ .

Final answer

(-5, -13)

Detailed explanation

Walkthrough

We need to find where the graph of $y=g^{-1}f(x)$ meets the graph of $y=g(x)$ exactly once. The word “single point” is the key: it tells us that the equation formed by equating the two expressions has exactly one solution.

First, find the inverse of $g(x)=4x+k$ . Writing $y=4x+k$ and solving for $x$ gives $x=(y-k)/4$ , so $g^{-1}(x)=(x-k)/4$ .

Next form $g^{-1}f(x)$ by replacing $x$ in $g^{-1}$ with $f(x)=10+6x-x^2$ :

g^{-1}f(x)=\frac{10+6x-x^2-k}{4}

Then set this equal to $g(x)=4x+k$ and multiply through by 4. Rearranging gives the quadratic $x^2+10x-10+5k=0$ (after multiplying by $-1$ ).

Because the two graphs meet at a single point, this quadratic must have a repeated root, so its discriminant must be zero:

10^2 - 4(1)(-10+5k)=0

This simplifies to $140-20k=0$ , so $k=7$ .

With $k=7$ , the quadratic becomes $x^2+10x+25=0$ , i.e. $(x+5)^2=0$ , so $x=-5$ . Finally, substitute $x=-5$ into $y=g(x)=4x+7$ to get $y=-13$ . Hence $P=(-5,-13)$ .

Key Takeaways

To find where two graphs meet, equate their expressions and solve.
“Meet at a single point” for a quadratic equation means the discriminant is zero: $b^2-4ac=0$ .
Finding an inverse of a linear function $y=mx+c$ is done by swapping and solving, giving $x=(y-c)/m$ .
A composite like $g^{-1}f(x)$ is formed by substituting $f(x)$ into $g^{-1}(x)$ .

Common Mistakes

Forgetting to include $k$ when forming the composite; the discriminant condition cannot be applied without $k$ .
Using the wrong sign when rearranging the quadratic. The equation can be written either as $-x^2-10x+10-5k=0$ or $x^2+10x-10+5k=0$ ; both give the same discriminant.
Stopping after finding $k=7$ ; the question asks for the coordinates of $P$ , so you must also solve for $x$ and substitute to find $y$ .
Substituting $x=-5$ into $f$ instead of $g$ ; the point lies on $y=g(x)$ (and also on $y=g^{-1}f(x)$ ), and the mark scheme uses $g(x)$ to find $y$ .

Things to Be Careful About

The discriminant must be applied to the quadratic in $x$ after both sides have been combined, not before rearranging.
Check the sign of the constant term: from $-x^2-10x+10-5k=0$ , multiplying by $-1$ gives $x^2+10x-10+5k=0$ , so $c=-10+5k$ .
When solving $(x+5)^2=0$ , there is only one root, $x=-5$ , consistent with the “single point” condition.
The final answer should be an ordered pair, not just the value of $k$ or $x$ .

Techniques used

find the inverse of a linear functionform a composite functionequate expressions for intersectionapply the zero discriminant conditionsolve a quadratic and substitute to find coordinates

Mathematics 9709/13 — May/June 2024

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