(a) Use completing the square to find the exact solutions of the equation $4x^2 - 4x - 1 = 0$. (b) Hence solve the equation $4\tan\theta = 4 + \frac{1}{\tan\theta}$ for $0^\circ < \theta < 180^\circ$.

(a) x = (1 ± √2)/2 · (b) θ = 50.4°, 168.3°

The equation of a curve is $y = x^3 + ax^2 + bx + 5$. The curve has a stationary point at $(1, 9)$. (a) Find the values of the constants $a$ and $b$. (b) Find the coordinates of the other stationary point. (c) A point $P$ is moving along part of the curve in such a way that the $y$-coordinate of $P$ is increasing at a constant rate of 6 units per second. Find the rate at which the $x$-coordinate of $P$ is increasing when $x = 5$.

(a) a = -6, b = 9 · (b) (3, 5) · (c) dx/dt = 1/4

9709/15 May/June 2025 — Mathematics Solved Past Paper

Q14MMedium-EasyIntegration

The equation of a curve is such that $\frac{\mathrm{d}y}{\mathrm{d}x} = 12(2x - 5)^2 + 8x$ . It is given that the curve passes through the point $(2, 4)$ .

Find an equation of the curve.

DifficultyMedium-Easy

Worked solution

Approach

Integrate each term with respect to $x$ . For the term $12(2x-5)^2$ , use the reverse of the chain rule: divide by the derivative of the inside function and by the new power. Then use the point $(2,4)$ to find the constant of integration.

Working

\begin{aligned} y &= \int \left(12(2x-5)^2 + 8x\right)\,dx \\ &= 12 \cdot \frac{(2x-5)^3}{2 \cdot 3} + 8 \cdot \frac{x^2}{2} + c \\ &= 2(2x-5)^3 + 4x^2 + c \end{aligned}

Use $(2,4)$ :

\begin{aligned} 4 &= 2(2(2)-5)^3 + 4(2)^2 + c \\ 4 &= 2(-1)^3 + 16 + c \\ 4 &= -2 + 16 + c \\ 4 &= 14 + c \\ c &= -10 \end{aligned}

Answer

y = 2(2x-5)^3 + 4x^2 - 10

Final answer

y = 2(2x-5)^3 + 4x^2 - 10 (or y = 16x^3 - 116x^2 + 300x - 260)

Detailed explanation

Walkthrough

We are given the gradient function $\frac{\mathrm{d}y}{\mathrm{d}x} = 12(2x-5)^2 + 8x$ and a point on the curve. To recover $y$ , we integrate both sides with respect to $x$ .

The first term is $12(2x-5)^2$ . This is a composite function: the outer function is squaring and the inner function is $2x-5$ . By the reverse chain rule,

\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c

Here $a = 2$ and $n = 2$ , so the integral of $(2x-5)^2$ is $\frac{(2x-5)^3}{2 \times 3}$ . Multiplying by the coefficient $12$ gives $2(2x-5)^3$ .

The second term, $8x$ , integrates to $4x^2$ because $\int x\,dx = \frac{x^2}{2}$ .

We must add an arbitrary constant $c$ because differentiation removes constants. Then we use the point $(2,4)$ : substitute $x = 2$ and $y = 4$ into the integrated expression. This gives $4 = 2(-1)^3 + 16 + c$ , so $4 = 14 + c$ and $c = -10$ .

Therefore the equation of the curve is $y = 2(2x-5)^3 + 4x^2 - 10$ .

Key Takeaways

Integration is the reverse of differentiation.
For a term of the form $(ax+b)^n$ , divide by $a(n+1)$ when integrating.
Always include the constant of integration $+c$ when finding an indefinite integral.
A point on the curve is used to determine the value of $c$ .

Common Mistakes

Forgetting to include $+c$ before substituting the point.
Forgetting to divide by the derivative of the inside function, $2$ , when integrating $(2x-5)^2$ .
Substituting the point into the derivative instead of the integrated expression.
Making a sign error when evaluating $2(2)-5 = -1$ and then cubing it.

Things to Be Careful About

The mark scheme awards marks for each correctly integrated term, so leaving terms unsimplified is acceptable.
When substituting $(2,4)$ , compute $2(2)-5 = -1$ carefully; $(-1)^3 = -1$ .
If you expand the final answer, check the expansion carefully: $2(2x-5)^3 + 4x^2 - 10 = 16x^3 - 116x^2 + 300x - 260$ .
The final answer must include an equals sign and the expression for $y$ .

Techniques used

integrate (ax + b)^n using the reverse chain ruleintegrate a polynomial termuse a given point to find the constant of integration

Approach

Find the coefficients of $x^2$ and $x$ in each expansion using the binomial theorem, add the contributions, use the given condition, then solve the resulting quadratic.

Working

For $(3+ax)^5$ , the coefficient of $x^2$ is:

\binom{5}{2}3^{3}a^2 = 10 \times 27 a^2 = 270a^2

and the coefficient of $x$ is:

\binom{5}{1}3^{4}a = 5 \times 81 a = 405a

For $(6-x)^4$ , the coefficient of $x^2$ is:

\binom{4}{2}6^{2}(-1)^2 = 6 \times 36 = 216

and the coefficient of $x$ is:

\binom{4}{1}6^{3}(-1) = 4 \times 216 \times (-1) = -864

So the total coefficient of $x^2$ is $270a^2 + 216$ , and the total coefficient of $x$ is $405a - 864$ .

The condition gives:

270a^2 + 216 = 6(405a - 864)

Simplify:

270a^2 + 216 = 2430a - 5184

270a^2 - 2430a + 5400 = 0

Divide by 6:

45a^2 - 405a + 900 = 0

Factorise:

45(a-4)(a-5) = 0

Therefore

a = 4 \quad \text{or} \quad a = 5

Answer

a = 4 \text{ or } a = 5

Final answer

a = 4 or a = 5

Detailed explanation

Walkthrough

This problem asks about binomial coefficients in the sum of two expansions. We only need the terms in $x$ and $x^2$ , not the full expansions.

For $(3+ax)^5$ , the term containing $x^r$ is $\binom{5}{r}3^{5-r}(ax)^r$ . Putting $r=2$ gives $\binom{5}{2}3^3a^2 = 10 \times 27a^2 = 270a^2$ . Putting $r=1$ gives $\binom{5}{1}3^4a = 5 \times 81a = 405a$ .

For $(6-x)^4$ , write the second term as $-x$ . The $x^2$ coefficient is $\binom{4}{2}6^2(-1)^2 = 6 \times 36 = 216$ . The $x$ coefficient is $\binom{4}{1}6^3(-1) = 4 \times 216 \times (-1) = -864$ . The minus sign matters because the second term is negative.

Now add the coefficients: the total $x^2$ coefficient is $270a^2+216$ , and the total $x$ coefficient is $405a-864$ . The condition says the first is six times the second, so $270a^2+216 = 6(405a-864)$ . Expand and simplify: $270a^2 - 2430a + 5400 = 0$ , divide by 6 to get $45a^2 - 405a + 900 = 0$ , then factorise as $45(a-4)(a-5)=0$ . Hence $a=4$ or $a=5$ .

Key Takeaways

The binomial theorem lets you extract one coefficient without expanding fully: the coefficient of $x^r$ in $(p+q)^n$ is $\binom{n}{r}p^{n-r}$ .
When the second term in a binomial is negative, include $(-1)^r$ in the coefficient.
Coefficients from separate expansions add when the expansions are added together.
A coefficient condition produces a quadratic equation in $a$ ; solving it gives the possible values.

Common Mistakes

Forgetting the binomial coefficient $\binom{n}{r}$ .
Making a sign error with $(6-x)^4$ , so the coefficient of $x$ is written as positive instead of $-864$ .
Setting up the condition incorrectly, e.g. writing $6(270a^2+216) = 405a-864$ instead of $270a^2+216 = 6(405a-864)$ .
Not simplifying the combinations, which the mark scheme guidance requires.
Losing one of the two solutions when solving the quadratic.

Things to Be Careful About

The mark scheme expects the combinations to be evaluated, so write $10 \times 27$ and $6 \times 36$ rather than leaving $\binom{5}{2}$ or $\binom{4}{2}$ unsimplified.
The coefficient of $x$ in $(6-x)^4$ is negative because of the $(-1)$ factor.
If using the quadratic formula, a full substitution must be shown; factorisation is simpler here.
The final answer must include both possible values, $a=4$ and $a=5$ .

Techniques used

extract binomial coefficients for x and x^2combine coefficients from two expansionsformulate a quadratic equation from a given conditionfactorise and solve the quadratic

Approach

Divide the equation by 4 so that the coefficient of $x^2$ is 1, then complete the square on $x^2 - x$ . Rearrange to isolate the square term and take square roots.

Working

4x^2 - 4x - 1 = 0

Divide by 4:

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

Complete the square on $x^2 - x$ :

x^2 - x = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4}

Therefore the equation becomes:

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

\left(x - \frac{1}{2}\right)^2 - \frac{1}{2} = 0

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

Taking square roots:

x - \frac{1}{2} = \pm \sqrt{\frac{1}{2}}

x = \frac{1}{2} \pm \frac{1}{\sqrt{2}}

Equivalently,

x = \frac{1 \pm \sqrt{2}}{2}

Answer

x = \frac{1}{2} \pm \sqrt{\frac{1}{2}}

or

x = \frac{1 \pm \sqrt{2}}{2}

Final answer

x = (1 ± √2)/2

Detailed explanation

Walkthrough

We solve $4x^2 - 4x - 1 = 0$ by completing the square. First divide by 4 so the coefficient of $x^2$ becomes 1. Then write $x^2 - x$ as $\left(x - \frac{1}{2}\right)^2 - \frac{1}{4}$ . Substituting back gives $\left(x - \frac{1}{2}\right)^2 - \frac{1}{2} = 0$ , so $\left(x - \frac{1}{2}\right)^2 = \frac{1}{2}$ . Taking square roots gives $x = \frac{1}{2} \pm \sqrt{\frac{1}{2}}$ , equivalent to $\frac{1 \pm \sqrt{2}}{2}$ .

Key Takeaways

Completing the square requires a coefficient of 1 on $x^2$ . For $x^2 + bx$ , use $\left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4}$ . After isolating the square, remember the $\pm$ sign when taking square roots.

Common Mistakes

Dividing only some terms by 4 instead of the whole equation.
Forgetting the $\pm$ sign when taking square roots.
Not simplifying $\sqrt{\frac{1}{2}}$ correctly.
Giving a decimal answer instead of the exact surd form.

Things to Be Careful About

The mark scheme requires the completed-square method to be shown; a correct answer by another method without this only earns a partial credit mark. Keep the answer exact.

Techniques used

divide by the leading coefficientcomplete the squaretake square roots

(b)

Hence solve the equation $4\tan\theta = 4 + \frac{1}{\tan\theta}$ for $0^\circ < \theta < 180^\circ$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Convert the equation into a quadratic in $\tan\theta$ , use the solutions from part (a), then find the angles in $0^\circ < \theta < 180^\circ$ using inverse tangent and the period of $\tan$ .

Working

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

Let $x = \tan\theta$ . Multiply by $x$ (valid because $\tan\theta \neq 0$ in the given interval):

4x = 4 + \frac{1}{x}

4x^2 = 4x + 1

4x^2 - 4x - 1 = 0

From part (a):

x = \frac{1 \pm \sqrt{2}}{2}

Therefore

\tan\theta = \frac{1 + \sqrt{2}}{2} \quad \text{or} \quad \tan\theta = \frac{1 - \sqrt{2}}{2}

Numerically,

\frac{1 + \sqrt{2}}{2} \approx 1.2071, \qquad \frac{1 - \sqrt{2}}{2} \approx -0.2071

For $\tan\theta = 1.2071$ :

\theta = \tan^{-1}(1.2071) \approx 50.4^\circ

Since $\tan\theta > 0$ and $0^\circ < \theta < 180^\circ$ , the only solution from this value is $50.4^\circ$ .

For $\tan\theta = -0.2071$ :

\tan^{-1}(-0.2071) \approx -11.7^\circ

Add $180^\circ$ to bring it into the interval:

\theta = 180^\circ - 11.7^\circ = 168.3^\circ

There are no other solutions in the interval.

Answer

\theta = 50.4^\circ, \ 168.3^\circ \quad (0^\circ < \theta < 180^\circ)

Final answer

θ = 50.4°, 168.3°

Detailed explanation

Walkthrough

The equation is quadratic in $\tan\theta$ . Let $x = \tan\theta$ . Multiplying by $x$ gives $4x^2 - 4x - 1 = 0$ , exactly the quadratic solved in part (a). Hence $x = \frac{1 \pm \sqrt{2}}{2}$ . Now solve $\tan\theta = \frac{1 + \sqrt{2}}{2}$ and $\tan\theta = \frac{1 - \sqrt{2}}{2}$ . The first value is positive, so its inverse tangent gives the first-quadrant angle $50.4^\circ$ . The second value is negative; its principal inverse tangent is about $-11.7^\circ$ , so add $180^\circ$ to obtain $168.3^\circ$ . Each value of $\tan\theta$ has exactly one solution in $0^\circ < \theta < 180^\circ$ , so there are no extra answers.

Key Takeaways

An equation can be rewritten as a quadratic in a trigonometric function, allowing earlier algebra to be reused.
To solve $\tan\theta = k$ , find the principal value with $\tan^{-1}$ and then add or subtract multiples of $180^\circ$ to place solutions in the required interval.
The sign of $k$ tells you which quadrant to consider.

Common Mistakes

Forgetting to multiply both sides by $\tan\theta$ or making a sign error.
Giving only the positive-tangent solution and missing the negative-tangent solution.
Giving answers in radians when the question asks for degrees.
Writing final answers without showing the substitution $x = \tan\theta$ .

Things to Be Careful About

The interval is $0^\circ < \theta < 180^\circ$ , so $\tan\theta$ is never zero at the endpoints and is undefined only at $90^\circ$ . When $\tan\theta$ is negative, the solution lies in the second quadrant, so use $\theta = 180^\circ + \tan^{-1}(k)$ with the negative principal value, or equivalently $180^\circ - \tan^{-1}(|k|)$ . The mark scheme requires a clear method; answers only, or answers only in radians, do not earn full marks.

Techniques used

set tan theta equal to xreuse the quadratic solution from part (a)apply inverse tangentadjust the angle using the period of tangent

Q45MMediumIntegration

Fig. 4.1

The diagram shows part of the curve $y = x^2 - \frac{1}{x^2}$ . The shaded region is bounded by the curve, the line $x = 2$ and the $x$ -axis.

Find the volume formed when the shaded region is rotated through $360^\circ$ about the $x$ -axis, giving your answer correct to 2 decimal places.

DifficultyMedium

Worked solution

Approach

To find the volume formed when the shaded region is rotated about the x-axis, we use the formula $V = \pi \int y^2 \, dx$ . First, we must determine the lower limit of integration by finding where the curve intersects the x-axis. Then we set up the integral, expand the integrand, integrate term by term, and evaluate between the limits.

Working

Find the x-intercept (lower limit):

y = 0 \implies x^2 - \frac{1}{x^2} = 0

x^4 = 1 \implies x = 1 \quad (\text{since } x > 0)

Set up the volume integral with limits $x = 1$ to $x = 2$ :

V = \pi \int_1^2 \left(x^2 - \frac{1}{x^2}\right)^2 dx

Expand the integrand:

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

Integrate term by term:

V = \pi \left[ \frac{x^5}{5} - 2x - \frac{1}{3x^3} \right]_1^2

Evaluate at the upper limit $x = 2$ :

\frac{2^5}{5} - 2(2) - \frac{1}{3(2^3)} = \frac{32}{5} - 4 - \frac{1}{24} = \frac{283}{120}

Evaluate at the lower limit $x = 1$ :

\frac{1^5}{5} - 2(1) - \frac{1}{3(1^3)} = \frac{1}{5} - 2 - \frac{1}{3} = -\frac{32}{15}

Subtract the lower limit value from the upper limit value:

V = \pi \left( \frac{283}{120} - \left(-\frac{32}{15}\right) \right) = \pi \left( \frac{283}{120} + \frac{256}{120} \right) = \frac{539\pi}{120}

Calculate the numerical value:

V = \frac{539\pi}{120} \approx 14.11

Answer

14.11

Final answer

14.11

Detailed explanation

Walkthrough

First, we identify the boundaries of the shaded region. The right boundary is given as $x = 2$ . The bottom boundary is the x-axis ( $y = 0$ ). To find the left boundary, we find where the curve $y = x^2 - 1/x^2$ crosses the x-axis by setting $y = 0$ . This gives $x^2 = 1/x^2$ , so $x^4 = 1$ . Since we are in the first quadrant, $x = 1$ . Thus, the limits of integration are from $x = 1$ to $x = 2$ .

Next, we apply the volume of revolution formula about the x-axis: $V = \pi \int y^2 \, dx$ . Substituting the curve equation, we get $V = \pi \int_1^2 (x^2 - 1/x^2)^2 \, dx$ .

We expand the squared binomial: $(x^2 - x^{-2})^2 = x^4 - 2x^2(x^{-2}) + x^{-4} = x^4 - 2 + x^{-4}$ . This is much easier to integrate than the original form.

We integrate each term using the power rule: $\int x^4 \, dx = x^5/5$ , $\int -2 \, dx = -2x$ , and $\int x^{-4} \, dx = x^{-3}/-3 = -1/(3x^3)$ . This gives the antiderivative $\frac{x^5}{5} - 2x - \frac{1}{3x^3}$ .

We then evaluate this antiderivative at the upper limit ( $x = 2$ ) and subtract the value at the lower limit ( $x = 1$ ). Careful fraction arithmetic is needed here: at $x=2$ , we get $32/5 - 4 - 1/24 = 283/120$ . At $x=1$ , we get $1/5 - 2 - 1/3 = -32/15 = -256/120$ . The difference is $(283 + 256)/120 = 539/120$ . Multiplying by $\pi$ gives the final volume.

Key Takeaways

The volume of a solid of revolution about the x-axis is given by $V = \pi \int y^2 \, dx$ .
Limits of integration must be determined from the geometry of the region, often by finding x-intercepts or intersections with given lines.
Expanding binomials before integrating can simplify the process, especially when dealing with negative exponents.

Common Mistakes

Using $x = 0$ as the lower limit instead of the actual x-intercept $x = 1$ . The function is undefined at $x = 0$ , and the shaded region does not extend to the y-axis.
Forgetting to square the $y$ expression in the volume formula, integrating $x^2 - 1/x^2$ instead of $(x^2 - 1/x^2)^2$ .
Sign errors when expanding $(a - b)^2$ , such as writing $x^4 + 1/x^4$ instead of $x^4 - 2 + 1/x^4$ .
Arithmetic errors when evaluating and subtracting the fractional values at the limits.

Things to Be Careful About

Always check the domain of the function. $y = x^2 - 1/x^2$ has a vertical asymptote at $x = 0$ , so the lower limit cannot be $0$ .
Ensure all fractions are converted to a common denominator before adding or subtracting to avoid calculation errors.
The final answer must be given to 2 decimal places as requested, so compute $539\pi/120$ and round appropriately to $14.11$ .

Techniques used

find x-intercept to determine lower limitset up volume of revolution integralexpand binomial and integrate term by termevaluate definite integral with limits

Approach

The shaded region $BCD$ is bounded by the line segment $CD$ , the line segment $CB$ , and the circular arc $BD$ . We need to find the lengths of these three boundaries. We use the right-angled triangle $ACD$ to find the angle $\angle CAD$ and the length $CD$ , then use the arc length formula for $BD$ .

Working

Since $CD$ is the perpendicular bisector of $AB$ , $C$ is the midpoint of $AB$ . Given the radius $AB = 10$ cm, we have $AC = CB = 5$ cm.

In right-angled triangle $ACD$ :

Hypotenuse $AD = 10$ cm (radius)
Adjacent side $AC = 5$ cm

Using the cosine ratio:

\cos(\angle CAD) = \frac{AC}{AD} = \frac{5}{10} = \frac{1}{2}

\angle CAD = 60^\circ = \frac{\pi}{3} \text{ radians}

Using the sine ratio to find $CD$ :

CD = AD \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} = \sqrt{75} \text{ cm}

The arc length $BD$ is given by $r\theta$ :

\text{Arc } BD = 10 \times \frac{\pi}{3} = \frac{10\pi}{3} \text{ cm}

The perimeter of the shaded region is the sum of $CD$ , $CB$ , and arc $BD$ :

\text{Perimeter} = CD + CB + \text{Arc } BD = 5\sqrt{3} + 5 + \frac{10\pi}{3}

\text{Perimeter} \approx 8.660 + 5 + 10.472 = 24.132 \text{ cm}

\text{Perimeter} = 24.1 \text{ cm} \text{ (to 1 d.p.)}

Answer

24.1 \text{ cm}

Final answer

24.1 cm

Detailed explanation

Walkthrough

First, identify the boundaries of the shaded region $BCD$ : the straight line $CD$ , the straight line $CB$ , and the curved arc $BD$ . Since $CD$ is the perpendicular bisector of $AB$ , point $C$ is exactly halfway along $AB$ , making $AC = CB = 5$ cm.

Next, focus on the right-angled triangle $ACD$ . We know the hypotenuse $AD$ is the radius of the circle ( $10$ cm) and the adjacent side $AC$ is $5$ cm. Using the cosine ratio ( $\cos \theta = \text{adjacent} / \text{hypotenuse}$ ), we find $\cos(\angle CAD) = 1/2$ , which immediately tells us $\angle CAD = 60^\circ$ or $\pi/3$ radians. This is a standard exact trigonometric value.

Then, find the length of $CD$ using the sine ratio ( $\sin 60^\circ = \sqrt{3}/2$ ), giving $CD = 5\sqrt{3}$ cm. Alternatively, Pythagoras' theorem ( $10^2 - 5^2 = 75$ ) yields $\sqrt{75} = 5\sqrt{3}$ cm.

With the angle $\angle CAD = \pi/3$ radians, calculate the arc length $BD$ using the formula $\text{arc} = r\theta = 10 \times \pi/3$ . Finally, add the three boundary lengths together ( $5\sqrt{3} + 5 + 10\pi/3$ ) and evaluate to 1 decimal place.

Key Takeaways

Recognising that a perpendicular bisector creates a midpoint and right-angled triangles.
Using exact trigonometric values (like $\cos 60^\circ = 1/2$ ) to find angles and side lengths in right-angled triangles.
Applying the arc length formula $s = r\theta$ with $\theta$ in radians.
Carefully identifying all boundaries of a composite shaded region when calculating perimeter.

Common Mistakes

Forgetting to include the arc length in the perimeter and only summing the straight lines.
Using degrees instead of radians in the arc length formula $s = r\theta$ .
Miscalculating $5\sqrt{3}$ or $10\pi/3$ during the final addition.

Things to Be Careful About

Ensure the angle is converted to radians before using the arc length formula.
The perimeter includes the curved arc $BD$ , not the straight line $AB$ or $AD$ .
Round only the final answer to 1 decimal place, not intermediate values.

Techniques used

use right-angled triangle to find angle and sidecalculate arc length using r\thetasum lengths for perimeter

(b)

Find the area of the shaded region $BCD$ , giving your answer correct to 1 decimal place.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The shaded region $BCD$ is the difference between the area of the circular sector $ABD$ and the area of the right-angled triangle $ACD$ . We use the sector area formula and the triangle area formula with the values found in part (a).

Working

The area of sector $ABD$ with radius $r = 10$ cm and angle $\theta = \frac{\pi}{3}$ radians is:

\text{Area of sector } ABD = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 10^2 \times \frac{\pi}{3} = \frac{50\pi}{3} \text{ cm}^2

The area of right-angled triangle $ACD$ with base $AC = 5$ cm and height $CD = 5\sqrt{3}$ cm is:

\text{Area of triangle } ACD = \frac{1}{2} \times AC \times CD = \frac{1}{2} \times 5 \times 5\sqrt{3} = \frac{25\sqrt{3}}{2} \text{ cm}^2

The area of the shaded region $BCD$ is the sector area minus the triangle area:

\text{Area} = \frac{50\pi}{3} - \frac{25\sqrt{3}}{2}

\text{Area} \approx 52.360 - 21.651 = 30.709 \text{ cm}^2

\text{Area} = 30.7 \text{ cm}^2 \text{ (to 1 d.p.)}

Answer

30.7 \text{ cm}^2

Final answer

30.7 cm^2

Detailed explanation

Walkthrough

To find the area of the shaded region $BCD$ , observe that it is formed by taking the entire circular sector $ABD$ and removing the right-angled triangle $ACD$ from it.

First, calculate the area of sector $ABD$ using the formula $\text{Area} = \frac{1}{2}r^2\theta$ . We know $r = 10$ cm and $\theta = \pi/3$ radians (from part a). This gives $\frac{1}{2} \times 100 \times \frac{\pi}{3} = \frac{50\pi}{3}$ cm $^2$ .

Next, calculate the area of triangle $ACD$ . Since it is a right-angled triangle at $C$ , the area is simply $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times CD$ . Using $AC = 5$ cm and $CD = 5\sqrt{3}$ cm, the area is $\frac{25\sqrt{3}}{2}$ cm $^2$ .

Subtract the triangle area from the sector area to get the shaded area: $\frac{50\pi}{3} - \frac{25\sqrt{3}}{2} \approx 30.7$ cm $^2$ .

Key Takeaways

Recognising composite shapes as the difference between simpler shapes (sector minus triangle).
Applying the sector area formula $\frac{1}{2}r^2\theta$ correctly with $\theta$ in radians.
Using the standard triangle area formula for right-angled triangles.

Common Mistakes

Using the wrong angle for the sector area (e.g., using $30^\circ$ instead of $60^\circ$ ).
Forgetting to convert degrees to radians before using the sector area formula.
Calculating the area of triangle $ABD$ instead of triangle $ACD$ .

Things to Be Careful About

Ensure consistent use of radians for circular measure formulas.
The shaded region is specifically bounded by $CD$ , $CB$ , and arc $BD$ , which perfectly matches sector $ABD$ minus triangle $ACD$ .

Techniques used

calculate sector area using \frac{1}{2}r^2\thetacalculate triangle area using \frac{1}{2}bhsubtract triangle area from sector area

Q6MediumSeries

Each year, on her birthday, Ananya receives some money from each of her parents.

On Ananya's first birthday, her father gives her $10. Every subsequent year, her father gives her $5 more than he gave her the previous year.

On Ananya's first birthday, her mother also gives her $10. Every subsequent year, her mother gives her 20% more than she gave her the previous year.

(a)

Show that on Ananya's eleventh birthday she receives more from her mother than from her father.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

On the $n$ -th birthday, the father's gift follows an arithmetic progression with first term $a=10$ and common difference $d=5$ . The mother's gift follows a geometric progression with first term $a=10$ and common ratio $r=1.2$ . Apply the appropriate $n$ -th term formulas with $n=11$ and compare the results.

Working

Father's amount on the 11th birthday:

F_{11} = 10 + (11-1)\times 5 = 10 + 50 = 60

Mother's amount on the 11th birthday:

M_{11} = 10 \times 1.2^{11-1} = 10 \times 1.2^{10}

10 \times 1.2^{10} = 61.92 \text{ (to 2 d.p.)}

Since $61.92 > 60$ , Ananya receives more from her mother on her eleventh birthday.

Answer

Father gives $60 and mother gives $61.92, so the mother's gift is larger.

Final answer

Mother gives 61.92 and father gives 60, so mother gives more.

Detailed explanation

Walkthrough

On each birthday, we need an individual gift amount. The father's gifts form an arithmetic progression because a fixed amount of $5 is added every year. The mother's gifts form a geometric progression because the amount is multiplied by the same factor every year: a 20% increase means multiplying by $1.2$ .

For the eleventh birthday, $n=11$ is used in each nth-term formula. The father's calculation is:

10 + (11-1)\times 5 = 60

The mother's calculation is:

10 \times 1.2^{10} = 61.92

Since 61.92 exceeds 60, the mother's gift is larger. The key is to remember that the eleventh term uses 10 increments or 10 multiplications after the first birthday.

Key Takeaways

Recognise arithmetic and geometric sequences from word-based descriptions.
Use $a_n = a + (n-1)d$ for an arithmetic sequence and $a_n = ar^{n-1}$ for a geometric sequence.
An increase of 20% corresponds to a common ratio of $1.2$ , not $0.2$ .

Common Mistakes

Using $n=10$ instead of $n=11$ .
Raising $1.2$ to the power $11$ instead of $10$ .
Stopping after finding one amount without comparing the two values.

Things to Be Careful About

The first birthday corresponds to $n=1$ .
The father's gift after 10 years of increases is $10 + 10 \times 5$ , and the mother's after 10 years of growth is $10 \times 1.2^{10}$ .
Values may be given to 2 significant figures, but showing both exact forms first avoids rounding errors.

Techniques used

apply nth term formula for an arithmetic progressionapply nth term formula for a geometric progressioncompare numerical values

(b)

Find the total amount of money Ananya receives up to and including her eighteenth birthday.

5M Similar

DifficultyMedium

Worked solution

Approach

We need the total amount over 18 birthdays, so we sum the first 18 terms of the father's arithmetic progression and the mother's geometric progression, then add the two totals.

Use

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

for an AP and

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

for a GP.

Working

Father's total over 18 birthdays:

S_{18} = \frac{18}{2}\left(2\times 10 + (18-1)\times 5\right) = 9(20 + 85) = 9 \times 105 = 945

Mother's total over 18 birthdays:

Using $1.2^{18} = 26.6233$ ,

S_{18} = \frac{10(1 - 1.2^{18})}{1 - 1.2} = \frac{10(1 - 26.6233)}{-0.2} = 1281.17 \text{ (to 2 d.p.)}

Total received up to and including the eighteenth birthday:

945 + 1281.17 = 2226.17

Answer

The total amount is $2226.17.

Final answer

2226.17

Detailed explanation

Walkthrough

We need the total amount across 18 birthdays, so we sum the first 18 terms of each sequence.

For the father's arithmetic sequence, the sum formula is:

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

With $a=10$ , $d=5$ , and $n=18$ :

S_{18} = \frac{18}{2}(20 + 85) = 9 \times 105 = 945

For the mother's geometric sequence, the sum formula is:

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

With $a=10$ , $r=1.2$ , and $n=18$ :

S_{18} = \frac{10(1 - 1.2^{18})}{1 - 1.2} = 1281.17

Because $1.2^{18} \approx 26.6233$ , both the numerator and denominator are negative, which gives the positive sum. Finally, add the two totals:

945 + 1281.17 = 2226.17

So the total is $2226.17.

Key Takeaways

Sum an arithmetic progression with $S_n = \frac{n}{2}(2a + (n-1)d)$ .
Sum a geometric progression with $S_n = \frac{a(1-r^n)}{1-r}$ .
The phrase "up to and including her eighteenth birthday" means the sum of the first 18 terms, not 17.

Common Mistakes

Summing only 17 terms for the mother or father.
Forgetting to add the father's and mother's totals together.
Using the GP sum formula with the subtraction in the wrong order, causing a sign error.
Reporting $S_{17}$ as the total; the mark scheme notes to ignore it as an extra answer.

Things to Be Careful About

Rounding $1.2^{18}$ too early can change the last digits; keep enough precision before the final answer.
When $r > 1$ , the formula $\frac{a(1-r^n)}{1-r}$ still works because both numerator and denominator are negative.
The final total should be given to at least 3 significant figures; $2226.17 is acceptable.

Techniques used

apply sum formula for an arithmetic progressionapply sum formula for a geometric progressioncombine the two totals

Q7MediumCoordinate Geometry

In the parallelogram $ABCD$ , the coordinates of $A$ are $(3, 7)$ , the coordinates of $B$ are $(6, p)$ and the coordinates of $D$ are $(1, p)$ . It is given that the gradient of $AB$ is $-\frac{2}{3}$ .

(a)

Find the value of $p$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the gradient formula for the line through $A(3,7)$ and $B(6,p)$ , set it equal to $-\frac{2}{3}$ , and solve for $p$ .

Working

The gradient of $AB$ is

\frac{p-7}{6-3} = \frac{p-7}{3}

Given that this equals $-\frac{2}{3}$ , we have

\frac{p-7}{3} = -\frac{2}{3}

Multiply both sides by $3$ :

p - 7 = -2

Hence

p = 5

Answer

p = 5

Final answer

p = 5

Detailed explanation

Walkthrough

The gradient of a line through $(x_1,y_1)$ and $(x_2,y_2)$ is

\frac{y_2-y_1}{x_2-x_1}

Take $A=(3,7)$ and $B=(6,p)$ . The vertical change from $A$ to $B$ is $p-7$ , and the horizontal change is $6-3=3$ . We are told that the gradient is $-\frac{2}{3}$ , so we set up

\frac{p-7}{3} = -\frac{2}{3}

Multiplying both sides by $3$ gives $p-7=-2$ , so $p=5$ . This missing coordinate will be needed for the rest of the question.

Key Takeaways

This part tests the definition of gradient: it is the ratio of the change in $y$ to the change in $x$ . It also reinforces that if one endpoint is unknown, the gradient formula can be used to solve for the unknown coordinate.

Common Mistakes

Reversing only one of $x$ and $y$ when applying the gradient formula, which produces the wrong sign.
Omitting the method line. The mark scheme requires the gradient equation to be seen explicitly, not just the final answer.
Making a sign error when moving the constant across the equation.

Things to Be Careful About

The vertical difference should be on top and the horizontal difference on the bottom. Since $A(3,7)$ and $B(6,p)$ have the same form, both $(p-7)/(6-3)$ and $(7-p)/(3-6)$ are acceptable, but mixing the two orders gives the wrong sign.

Techniques used

use the gradient formula between two pointsset the gradient equal to a given valuesolve a linear equation

(b)

Find the coordinates of $C$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the fact that in a parallelogram $ABCD$ , opposite sides are equal and parallel, so $\overrightarrow{AD} = \overrightarrow{BC}$ . Add this displacement to $B$ to find $C$ .

Working

The displacement from $A$ to $D$ is

\overrightarrow{AD} = D - A = (1-3,\, p-7) = (-2,\, p-7)

Since $\overrightarrow{BC} = \overrightarrow{AD}$ ,

C = B + \overrightarrow{AD} = (6-2,\, p + (p-7)) = (4,\, 2p-7)

Using $p=5$ from part (a):

C = (4,\, 2(5)-7) = (4,\, 3)

Answer

C = (4,3)

Final answer

C = (4, 3)

Detailed explanation

Walkthrough

In a parallelogram with vertices in order $A, B, C, D$ , the side $BC$ is parallel and equal to the side $AD$ . Therefore the displacement from $A$ to $D$ is the same as the displacement from $B$ to $C$ .

First find this displacement:

\overrightarrow{AD} = D - A = (1-3,\, p-7) = (-2,\, p-7)

Starting at $B=(6,p)$ and applying the same displacement gives

C = (6,p) + (-2,\, p-7) = (4,\, 2p-7)

Substituting the value $p=5$ found in part (a):

C = (4,\, 2(5)-7) = (4,\, 3)

Alternatively, using the fact that the diagonals of a parallelogram bisect each other, $A+C=B+D$ , which gives the same result $C=B+D-A$ .

Key Takeaways

This part shows how vector displacements can be used to locate missing vertices of a parallelogram. It also demonstrates the important follow-through idea: the $y$ -coordinate of $C$ depends on the value of $p$ found earlier.

Common Mistakes

Getting the order of vertices wrong and using the wrong side relation, for example writing $\overrightarrow{AB} = \overrightarrow{CD}$ instead of $\overrightarrow{AD} = \overrightarrow{BC}$ .
Forgetting that $y$ -coordinate of $D$ is $\text{their }p$ , so the vertical displacement is $p-7$ rather than $7-p$ .
Simplifying $2p-7$ incorrectly when $p=5$ .
Giving multiple possible coordinates for $C$ ; the mark scheme specifically warns that extra solutions lose both marks.

Things to Be Careful About

Keep the vertices in the order $A, B, C, D$ . When applying $C = B + (D-A)$ , be careful to subtract coordinates in the same order for both $x$ and $y$ . If you already found $p=5$ , substitute it only at the end so the general expression $C=(4,\, 2p-7)$ is clear.

Techniques used

use the parallelogram property of equal displacement vectorsadd the displacement vector to a pointsimplify coordinate expressions

(c)

Find the area of the triangle formed by the perpendicular bisector of $AB$ and the $x$ - and $y$ -axes.

5M Similar

DifficultyMedium

Worked solution

Approach

Find the midpoint of $AB$ and the gradient of the perpendicular bisector. Use the point-slope form to write its equation, find its intercepts with the axes, then compute the area of the resulting right-angled triangle.

Working

With $p=5$ , $B=(6,5)$ . The midpoint $M$ of $AB$ is

M = \left(\frac{3+6}{2},\, \frac{7+5}{2}\right) = \left(\frac{9}{2},\, 6\right)

The gradient of $AB$ is $-\frac{2}{3}$ , so the gradient of the perpendicular bisector is

m_\perp = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}

The perpendicular bisector passes through $M$ , so its equation is

y - 6 = \frac{3}{2}\left(x - \frac{9}{2}\right)

Simplifying to gradient-intercept form:

y = \frac{3}{2}x - \frac{27}{4} + 6 = \frac{3}{2}x - \frac{3}{4}

On the $y$ -axis, $x=0$ , giving

y = -\frac{3}{4}

so the intercept is $\left(0,\, -\frac{3}{4}\right)$ .

On the $x$ -axis, $y=0$ :

\frac{3}{2}x - \frac{3}{4} = 0 \quad \Rightarrow \quad x = \frac{1}{2}

so the intercept is $\left(\frac{1}{2},\, 0\right)$ .

The triangle with the axes has vertices $(0,0)$ , $\left(\frac{1}{2},0\right)$ and $\left(0,-\frac{3}{4}\right)$ . Its legs are $\frac{1}{2}$ and $\frac{3}{4}$ , so its area is

\frac{1}{2} \times \frac{1}{2} \times \frac{3}{4} = \frac{3}{16}

Answer

\frac{3}{16} \text{ square units} \quad (0.1875)

Final answer

3/16 square units (0.1875)

Detailed explanation

Walkthrough

After part (a), $p=5$ , so $B=(6,5)$ .

The perpendicular bisector of $AB$ is the line through the midpoint of $AB$ that is perpendicular to $AB$ . First find the midpoint:

M = \left(\frac{3+6}{2},\, \frac{7+5}{2}\right) = \left(\frac{9}{2},\, 6\right)

Next, because the original gradient is $-\frac{2}{3}$ , the perpendicular gradient is the negative reciprocal:

m_\perp = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}

Use the point-slope form through $M$ :

y - 6 = \frac{3}{2}\left(x - \frac{9}{2}\right)

Simplify to $y = \frac{3}{2}x - \frac{3}{4}$ . This line cuts the axes at two points: on the $y$ -axis, $x=0$ gives $y=-\frac{3}{4}$ ; on the $x$ -axis, $y=0$ gives $x=\frac{1}{2}$ . Together with the origin, these two points form a right-angled triangle with legs $\frac{1}{2}$ along the $x$ -axis and $\frac{3}{4}$ along the $y$ -axis. Its area is

\frac{1}{2} \times \frac{1}{2} \times \frac{3}{4} = \frac{3}{16} = 0.1875

The line lies below the $x$ -axis between $x=0$ and $x=\frac{1}{2}$ , so the ordinary triangle-area formula is the most direct method.

Key Takeaways

This part brings together several coordinate geometry ideas: midpoint, perpendicular gradients, the equation of a line, intercepts, and area of a triangle formed by a line with the coordinate axes.

Common Mistakes

Using the original gradient $-\frac{2}{3}$ instead of its negative reciprocal $\frac{3}{2}$ .
Using the wrong midpoint, e.g. averaging $A$ and $D$ or $B$ and $D$ instead of $A$ and $B$ .
Writing the line equation without using the midpoint and perpendicular gradient; the mark scheme requires both to be used.
Forgetting that the intercept on the $y$ -axis is negative, and then using a negative length in the area formula. Take absolute values of the intercept distances.
Simplifying $\frac{3}{2}\left(x - \frac{9}{2}\right)$ incorrectly; carefully expand to $\frac{3}{2}x - \frac{27}{4}$ .

Things to Be Careful About

The perpendicular bisector of $AB$ must pass through the midpoint of $AB$ , so the equation must use that exact midpoint. When finding the triangle area, the two axes meet at right angles at the origin, so the area is simply half the product of the two intercept distances. If using integration, integrate the absolute value of the line between $x=0$ and $x=\frac{1}{2}$ ; the mark scheme allows either method.

Techniques used

find the midpoint of a segmentuse negative reciprocal to find a perpendicular gradientwrite the equation of the perpendicular bisectorfind the x- and y-intercepts of a linecompute the area of a right triangle

Q8Medium-EasyDifferentiation

The equation of a curve is $y = x^3 + ax^2 + bx + 5$ . The curve has a stationary point at $(1, 9)$ .

(a)

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

5M Similar

DifficultyMedium-Easy

Worked solution

Approach

The curve passes through $(1,9)$ , so substitute these coordinates into the equation. Since $(1,9)$ is a stationary point, the gradient there is zero: differentiate the curve and set $\frac{dy}{dx}=0$ at $x=1$ . Solve the resulting linear simultaneous equations for $a$ and $b$ .

Working

Substitute $(1,9)$ into $y = x^3 + ax^2 + bx + 5$ :

9 = 1^3 + a(1)^2 + b(1) + 5

Therefore

9 = 1 + a + b + 5

a + b = 3

Differentiate the curve:

\frac{dy}{dx} = 3x^2 + 2ax + b

At the stationary point, the gradient is zero:

0 = 3(1)^2 + 2a(1) + b

0 = 3 + 2a + b

So

2a + b = -3

Subtract the first equation from the second:

(2a + b) - (a + b) = -3 - 3

a = -6

Then, using $a + b = 3$ :

-6 + b = 3

b = 9

Answer

a = -6, \quad b = 9

Final answer

a = -6, b = 9

Detailed explanation

Walkthrough

The point $(1,9)$ gives two conditions. Because it lies on the curve, substituting $x=1$ and $y=9$ must satisfy the equation; this gives one linear equation in $a$ and $b$ . Because it is a stationary point, the gradient $\frac{dy}{dx}$ must be zero at $x=1$ . Differentiating $y$ gives $\frac{dy}{dx}=3x^2+2ax+b$ , and setting this equal to zero at $x=1$ produces the second equation. The two equations are then solved simultaneously. The mark scheme awards B1 for each correct condition set up (the point on the curve and the derivative), M1 for using the zero gradient, DM1 for solving the two equations, and A1 for the correct final values.

Key Takeaways

This question tests the fundamental meaning of a stationary point: at a stationary point the derivative is zero, while the curve itself also passes through the given point. It also tests differentiation of powers of $x$ and solving linear simultaneous equations. A point on a curve and a stationary point each supply one independent condition, so two unknowns can be found.

Common Mistakes

Substituting only the $x$ -coordinate but not using $y=9$ .
Forgetting to set the derivative equal to zero at the stationary point.
Making an arithmetic slip when solving the simultaneous equations.
Misreading the equation $9 = 1 + a + b + 5$ as $a + b = 9$ rather than $a+b=3$ .

The mark scheme also notes that the final A1 is only available if the working is correct with no wrong working (WWW).

Things to Be Careful About

Check that the values $a=-6$ and $b=9$ satisfy both equations: $a+b=3$ and $2a+b=-3$ . Keep the derivative notation $\frac{dy}{dx}$ clear, and do not confuse the stationary point condition with a point simply lying on the curve.

Techniques used

differentiate polynomial term by termsubstitute given point into curve equationset derivative equal to zero at stationary pointsolve simultaneous linear equations

(b)

Find the coordinates of the other stationary point.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the values $a=-6$ and $b=9$ to write the derivative. Stationary points occur where $\frac{dy}{dx}=0$ . Solve the resulting quadratic to find both $x$ -coordinates, then find the $y$ -coordinate of the stationary point that is not $(1,9)$ .

Working

With $a=-6$ and $b=9$ ,

\frac{dy}{dx} = 3x^2 - 12x + 9

At stationary points:

3x^2 - 12x + 9 = 0

Divide by 3:

x^2 - 4x + 3 = 0

Factorise:

(x-1)(x-3) = 0

So $x=1$ or $x=3$ . The point $x=1$ is the given stationary point, so the other stationary point has $x=3$ .

Find its $y$ -coordinate:

y = (3)^3 - 6(3)^2 + 9(3) + 5

y = 27 - 54 + 27 + 5 = 5

Answer

(3, 5)

Final answer

(3, 5)

Detailed explanation

Walkthrough

Once $a$ and $b$ are known, the original curve is fully determined. Substitute them into the derivative to obtain $\frac{dy}{dx}=3x^2-12x+9$ . At every stationary point the derivative is zero, so solve $3x^2-12x+9=0$ . Dividing by 3 and factorising gives $x=1$ and $x=3$ . Since $x=1$ is the already-known stationary point from the question, the other one is at $x=3$ . Substitute $x=3$ into the curve equation to get $y=5$ . The mark scheme gives the first M1 for setting a correct derivative to zero, DM1 for solving the three-term quadratic, and A1 for the final point.

Key Takeaways

Stationary points are found by solving $\frac{dy}{dx}=0$ . A cubic can have up to two stationary points, and one of them was already given in the question. Factorising a quadratic derivative is a common way to locate stationary points.

Common Mistakes

Giving only $x=3$ without computing $y=5$ .
Including $(1,9)$ as a second answer; the mark scheme says it may be ignored if it appears.
Incorrectly factorising $x^2-4x+3$ .
Substituting $x=3$ into the derivative instead of the curve equation to find $y$ .

Things to Be Careful About

Make sure the quadratic is set equal to zero before solving. Check the final coordinates by substituting back: at $(3,5)$ , the derivative is $27-36+9=0$ . The mark scheme notes the final A1 is only dependent on the first M1, so solving must follow from a correct derivative expression.

Techniques used

set derivative equal to zerofactorise quadratic equationevaluate curve at stationary x-coordinate

(c)

A point $P$ is moving along part of the curve in such a way that the $y$ -coordinate of $P$ is increasing at a constant rate of 6 units per second.

Find the rate at which the $x$ -coordinate of $P$ is increasing when $x = 5$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the chain rule

\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}

to relate the two rates of change. First evaluate $\frac{dy}{dx}$ at $x=5$ , then solve for $\frac{dx}{dt}$ using the given $\frac{dy}{dt}=6$ .

Working

From part (a), the derivative is

\frac{dy}{dx} = 3x^2 - 12x + 9

At $x=5$ :

\frac{dy}{dx} = 3(5)^2 - 12(5) + 9

= 75 - 60 + 9 = 24

Given $\frac{dy}{dt}=6$ and using the chain rule,

6 = 24\cdot\frac{dx}{dt}

Therefore

\frac{dx}{dt} = \frac{6}{24} = \frac{1}{4}

Answer

\frac{dx}{dt} = \frac{1}{4}

The $x$ -coordinate is increasing at $\frac{1}{4}$ units per second.

Final answer

dx/dt = 1/4

Detailed explanation

Walkthrough

This part is a related-rate problem. The quantity we know is how fast $y$ changes with time, and we want how fast $x$ changes with time. These are connected through the slope $\frac{dy}{dx}$ : a change in time produces a change in $x$ , which in turn produces a change in $y$ . The chain rule states $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$ . First evaluate the slope at the instant when $x=5$ , using $\frac{dy}{dx}=3x^2-12x+9$ , which gives 24. Then substitute the known $\frac{dy}{dt}=6$ : $6=24\frac{dx}{dt}$ , so $\frac{dx}{dt}=\frac{6}{24}=\frac{1}{4}$ . This means one unit of time increases $y$ by 6 units; since the slope is 24, $x$ needs to increase by only $1/4$ unit.

Key Takeaways

Rates of change with respect to time are connected by the chain rule. Evaluating a derivative at the relevant value gives the instantaneous slope, which is the conversion factor between $\frac{dy}{dt}$ and $\frac{dx}{dt}$ . This question also reinforces that the derivative is an instantaneous rate.

Common Mistakes

Writing the chain rule upside down, e.g. $\frac{dx}{dt}=\frac{dy}{dx}\cdot\frac{dy}{dt}$ .
Substituting $x=5$ into $y$ rather than into $\frac{dy}{dx}$ .
Simplifying $\frac{6}{24}$ incorrectly.
Forgetting to state the units (units per second) in the final answer.

Things to Be Careful About

The mark scheme accepts equivalent statements such as $24=6\cdot\frac{dt}{dx}$ ; the important step is linking the slope 24, the 6, and $\frac{dx}{dt}$ . Always check that a larger slope means a smaller $\frac{dx}{dt}$ for a fixed $\frac{dy}{dt}$ . Use the derivative from part (a), not the original curve, when finding the rate at $x=5$ .

Techniques used

evaluate derivative at given x-coordinateapply chain rule to relate ratessolve for required rate

Q9MediumFunctionsTrigonometry

Functions $\mathrm{f}$ and $\mathrm{g}$ are defined as follows.

\begin{aligned} \mathrm{f}(x) &= \cos x \text{ for } 0 \leqslant x \leqslant \pi \\ \mathrm{g}(x) &= 3\cos(x - \pi) + 2 \text{ for } \pi \leqslant x \leqslant 2\pi \end{aligned}

(a)

Describe fully the transformations that have been combined to transform the graph of $y = \mathrm{f}(x)$ to the graph of $y = \mathrm{g}(x)$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Express $g(x)$ in terms of transformations applied to $f(x) = \cos x$ . Start with $f(x)$ , apply a horizontal translation, a vertical stretch, and a vertical translation to obtain $g(x) = 3\cos(x - \pi) + 2$ .

Working

Starting from $y = f(x) = \cos x$ :

\begin{aligned} y &= \cos x \\ \xrightarrow{\text{translate } \binom{\pi}{0}} y &= \cos(x - \pi) \\ \xrightarrow{\text{stretch factor 3, } y\text{-direction}} y &= 3\cos(x - \pi) \\ \xrightarrow{\text{translate } \binom{0}{2}} y &= 3\cos(x - \pi) + 2 = g(x) \end{aligned}

The combined transformations are:

A translation by the vector $\begin{pmatrix} \pi \\ 0 \end{pmatrix}$ (shift right by $\pi$ ),
A stretch with scale factor $3$ in the $y$ -direction,
A translation by the vector $\begin{pmatrix} 0 \\ 2 \end{pmatrix}$ (shift up by $2$ ).

Equivalently, the last two steps can be combined as a single translation by $\begin{pmatrix} \pi \\ 2 \end{pmatrix}$ applied after the stretch.

Answer

The graph of $y = \mathrm{f}(x)$ is transformed by:

a translation by $\begin{pmatrix} \pi \\ 0 \end{pmatrix}$ ,
a stretch with scale factor $3$ in the $y$ -direction,
a translation by $\begin{pmatrix} 0 \\ 2 \end{pmatrix}$ .

Final answer

Translation by (π, 0), stretch factor 3 in y-direction, translation by (0, 2)

Detailed explanation

Walkthrough

We need to describe how to transform $y = \cos x$ into $y = 3\cos(x - \pi) + 2$ . Work from the inside out:

Horizontal translation: The argument $(x - \pi)$ means the graph shifts right by $\pi$ . This gives $y = \cos(x - \pi)$ .
Vertical stretch: The factor $3$ multiplying the cosine means every $y$ -value is multiplied by $3$ . This is a stretch with scale factor $3$ in the $y$ -direction, giving $y = 3\cos(x - \pi)$ .
Vertical translation: The $+ 2$ at the end shifts the graph up by $2$ units, giving $y = 3\cos(x - \pi) + 2$ .

Note that the translation vector $\begin{pmatrix} \pi \\ 2 \end{pmatrix}$ can be stated as a single combined translation applied after the stretch, since the horizontal shift does not interact with the vertical stretch.

Key Takeaways

When describing transformations from $y = f(x)$ to $y = af(bx - c) + d$ :

Replace $x$ with $(x - c/b)$ : horizontal translation by $c/b$ .
Multiply by $a$ : vertical stretch by factor $a$ .
Add $d$ : vertical translation by $d$ .
The order matters: horizontal translations and stretches are applied first, then vertical stretches, then vertical translations.

Common Mistakes

Giving two or more stretches (e.g., splitting the vertical stretch incorrectly) — the mark scheme gives B0 for extra stretches.
Not stating 'translation' or 'shift' when giving vectors — only B1 is awarded if the vectors are correct but the transformation type is not named.
Incorrect order: if the translation and stretch are not ordered correctly, the maximum is 3/4.

Things to Be Careful About

The domain also changes from $[0, \pi]$ to $[\pi, 2\pi]$ , which is consistent with the horizontal translation by $\pi$ . This is not a separate transformation but a consequence of the horizontal shift.
The mark scheme accepts the translation split as $\begin{pmatrix} \pi \\ 0 \end{pmatrix}$ before the stretch and $\begin{pmatrix} 0 \\ 2 \end{pmatrix}$ after, or as a single vector $\begin{pmatrix} \pi \\ 2 \end{pmatrix}$ after the stretch.

Techniques used

identify horizontal translationidentify vertical stretchidentify vertical translation

(b)

On the given axes, sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{g}(x)$ .

Fig. 9.1

4M Similar

DifficultyMedium

Worked solution

Approach

Sketch $y = \mathrm{f}(x) = \cos x$ for $0 \leqslant x \leqslant \pi$ , and $y = \mathrm{g}(x) = 3\cos(x - \pi) + 2$ for $\pi \leqslant x \leqslant 2\pi$ . Evaluate at endpoints and use knowledge of the cosine curve shape.

Working

Graph of $\mathrm{f}(x) = \cos x$ , $0 \leqslant x \leqslant \pi$ :

\begin{aligned} \mathrm{f}(0) &= \cos 0 = 1 \\ \mathrm{f}\left(\frac{\pi}{2}\right) &= \cos\frac{\pi}{2} = 0 \\ \mathrm{f}(\pi) &= \cos \pi = -1 \end{aligned}

This is a decreasing curve from $(0, 1)$ to $(\pi, -1)$ with zero gradient at both endpoints.

Graph of $\mathrm{g}(x) = 3\cos(x - \pi) + 2$ , $\pi \leqslant x \leqslant 2\pi$ :

\begin{aligned} \mathrm{g}(\pi) &= 3\cos(0) + 2 = 5 \\ \mathrm{g}\left(\frac{3\pi}{2}\right) &= 3\cos\left(\frac{\pi}{2}\right) + 2 = 2 \\ \mathrm{g}(2\pi) &= 3\cos(\pi) + 2 = -1 \end{aligned}

This is a decreasing curve from $(\pi, 5)$ to $(2\pi, -1)$ with zero gradient at both endpoints.

Answer

Sketch the cosine curve from $(0, 1)$ to $(\pi, -1)$ for $\mathrm{f}(x)$ , and the curve from $(\pi, 5)$ to $(2\pi, -1)$ for $\mathrm{g}(x)$ , both with zero gradients at their endpoints.

Final answer

f(x): curve from (0,1) to (π,-1); g(x): curve from (π,5) to (2π,-1), both decreasing with zero gradients at endpoints

Detailed explanation

Walkthrough

We sketch both functions on the same axes, noting their domains and key values.

For $\mathrm{f}(x) = \cos x$ , $0 \leqslant x \leqslant \pi$ :
This is the standard cosine curve over its first half-period. At $x = 0$ , $\cos 0 = 1$ . At $x = \pi/2$ , $\cos(\pi/2) = 0$ . At $x = \pi$ , $\cos \pi = -1$ . The curve is monotonically decreasing on this interval, with zero gradient (horizontal tangent) at both endpoints $x = 0$ and $x = \pi$ .

For $\mathrm{g}(x) = 3\cos(x - \pi) + 2$ , $\pi \leqslant x \leqslant 2\pi$ :
Let $u = x - \pi$ , so as $x$ goes from $\pi$ to $2\pi$ , $u$ goes from $0$ to $\pi$ . Then $\mathrm{g}(x) = 3\cos u + 2$ .

At $x = \pi$ ( $u = 0$ ): $3\cos 0 + 2 = 5$ .
At $x = 3\pi/2$ ( $u = \pi/2$ ): $3\cos(\pi/2) + 2 = 2$ .
At $x = 2\pi$ ( $u = \pi$ ): $3\cos \pi + 2 = -1$ .

This is also a decreasing curve, from $(\pi, 5)$ to $(2\pi, -1)$ , with zero gradients at both endpoints.

Key Takeaways

When sketching trigonometric functions with restricted domains:

Always evaluate at the endpoints of the domain.
Check monotonicity to determine whether the curve is increasing or decreasing.
Trigonometric curves have zero gradient at their peaks and troughs (endpoints of half-periods).
The domain restriction means the curve should not be drawn outside the given interval.

Common Mistakes

Drawing the curve beyond the specified domain — the mark scheme requires the domain to be clearly shown.
Not having zero gradient at the endpoints of the domain — sketches must be curves with zero gradient at the ends.
Getting the range of $\mathrm{g}(x)$ wrong — it should be from $-1$ to $5$ , not from $-3$ to $3$ .
Drawing $\mathrm{g}(x)$ as an increasing curve — it is decreasing on $[\pi, 2\pi]$ .

Things to Be Careful About

The sketch of $\mathrm{f}(x)$ must have domain $0 \leqslant x \leqslant \pi$ and range $[-1, 1]$ .
The sketch of $\mathrm{g}(x)$ must have domain $\pi \leqslant x \leqslant 2\pi$ and range $[-1, 5]$ .
Both curves must have zero gradient at their domain endpoints (horizontal tangents at $(0,1)$ , $(\pi,-1)$ for $\mathrm{f}$ , and at $(\pi,5)$ , $(2\pi,-1)$ for $\mathrm{g}$ ).

Techniques used

evaluate function at domain endpointsidentify monotonicitysketch trigonometric curve with correct domain and range

(c)

Find $\mathrm{g}^{-1}\mathrm{f}\left(\frac{1}{3}\pi\right)$ .

4M Similar

DifficultyMedium

Worked solution

Approach

First evaluate $\mathrm{f}\left(\frac{\pi}{3}\right)$ , then find $\mathrm{g}^{-1}$ of that result. To find $\mathrm{g}^{-1}$ , swap $x$ and $y$ in $y = \mathrm{g}(x)$ and solve for $y$ , ensuring the correct branch of $\cos^{-1}$ is used given the domain of $\mathrm{g}$ .

Working

Step 1: Evaluate $\mathrm{f}\left(\frac{\pi}{3}\right)$

\mathrm{f}\left(\frac{\pi}{3}\right) = \cos\frac{\pi}{3} = \frac{1}{2}

So we need $\mathrm{g}^{-1}\left(\frac{1}{2}\right)$ .

Step 2: Find $\mathrm{g}^{-1}(x)$

Let $y = \mathrm{g}(x) = 3\cos(x - \pi) + 2$ , where $\pi \leqslant x \leqslant 2\pi$ .

\begin{aligned} y &= 3\cos(x - \pi) + 2 \\ y - 2 &= 3\cos(x - \pi) \\ \cos(x - \pi) &= \frac{y - 2}{3} \end{aligned}

Since $\pi \leqslant x \leqslant 2\pi$ , we have $0 \leqslant x - \pi \leqslant \pi$ , so $x - \pi = \cos^{-1}\left(\frac{y - 2}{3}\right)$ .

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

Therefore:

\mathrm{g}^{-1}(x) = \cos^{-1}\left(\frac{x - 2}{3}\right) + \pi

Step 3: Evaluate $\mathrm{g}^{-1}\left(\frac{1}{2}\right)$

\mathrm{g}^{-1}\left(\frac{1}{2}\right) = \cos^{-1}\left(\frac{\frac{1}{2} - 2}{3}\right) + \pi = \cos^{-1}\left(\frac{-\frac{3}{2}}{3}\right) + \pi = \cos^{-1}\left(-\frac{1}{2}\right) + \pi

Since $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ :

\mathrm{g}^{-1}\left(\frac{1}{2}\right) = \frac{2\pi}{3} + \pi = \frac{5\pi}{3}

Answer

\mathrm{g}^{-1}\mathrm{f}\left(\frac{\pi}{3}\right) = \frac{5\pi}{3}

Final answer

5π/3

Detailed explanation

Walkthrough

We need to compute $\mathrm{g}^{-1}(\mathrm{f}(\pi/3))$ , which is a composition where we first apply $\mathrm{f}$ , then apply $\mathrm{g}^{-1}$ .

Step 1: Compute $\mathrm{f}(\pi/3) = \cos(\pi/3) = 1/2$ . This is a standard exact trigonometric value.

Step 2: Find $\mathrm{g}^{-1}(x)$ . Start with $y = 3\cos(x - \pi) + 2$ and solve for $x$ in terms of $y$ :

Subtract 2: $y - 2 = 3\cos(x - \pi)$
Divide by 3: $\cos(x - \pi) = (y - 2)/3$
Apply $\cos^{-1}$ : $x - \pi = \cos^{-1}((y - 2)/3)$
Add $\pi$ : $x = \cos^{-1}((y - 2)/3) + \pi$

The key point is that since the domain of $\mathrm{g}$ is $\pi \leqslant x \leqslant 2\pi$ , we have $0 \leqslant x - \pi \leqslant \pi$ , which is exactly the range of $\cos^{-1}$ , so we don't need to worry about multiple branches.

Step 3: Substitute $x = 1/2$ into $\mathrm{g}^{-1}$ :

$(1/2 - 2)/3 = (-3/2)/3 = -1/2$
$\cos^{-1}(-1/2) = 2\pi/3$ (since $\cos(2\pi/3) = -1/2$ and $2\pi/3 \in [0, \pi]$ )
$2\pi/3 + \pi = 5\pi/3$

Check: $5\pi/3$ is in the domain $[\pi, 2\pi]$ of $\mathrm{g}$ , so this is valid.

Key Takeaways

When finding $\mathrm{g}^{-1}(\mathrm{f}(a))$ :

Always evaluate the inner function first.
When finding the inverse, ensure the domain restriction is used to select the correct branch of the inverse trigonometric function.
Always verify that the final answer lies within the domain of the original function $\mathrm{g}$ .

Common Mistakes

Forgetting to add $\pi$ when finding $\mathrm{g}^{-1}$ — the inverse is $\cos^{-1}((x-2)/3) + \pi$ , not just $\cos^{-1}((x-2)/3)$ .
Using the wrong value for $\cos^{-1}(-1/2)$ — it is $2\pi/3$ , not $-\pi/3$ or $4\pi/3$ .
Not checking that the answer $5\pi/3$ lies in the domain $[\pi, 2\pi]$ of $\mathrm{g}$ .
Sign errors when computing $(1/2 - 2)/3$ — this equals $-1/2$ , not $1/2$ .

Things to Be Careful About

The range of $\mathrm{g}$ is $[-1, 5]$ , and $1/2$ is within this range, so $\mathrm{g}^{-1}(1/2)$ is defined.
The domain of $\mathrm{g}^{-1}$ is $[-1, 5]$ (the range of $\mathrm{g}$ ), and the range of $\mathrm{g}^{-1}$ is $[\pi, 2\pi]$ (the domain of $\mathrm{g}$ ).
$\cos^{-1}$ always returns values in $[0, \pi]$ , so $\mathrm{g}^{-1}(x) = \cos^{-1}((x-2)/3) + \pi$ returns values in $[\pi, 2\pi]$ , which matches the domain of $\mathrm{g}$ .

Techniques used

evaluate inner function at given valuefind inverse function algebraicallysubstitute into inverse expression

(d)

Explain why the composite function $\mathrm{fg}$ cannot be formed.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

For the composite function $\mathrm{fg}$ to be defined, the range of $\mathrm{g}$ must be a subset of the domain of $\mathrm{f}$ . Find the range of $\mathrm{g}$ and compare it with the domain of $\mathrm{f}$ .

Working

Domain and range of $\mathrm{f}$ :

Domain: $0 \leqslant x \leqslant \pi$ , i.e., $[0, \pi]$ .
Range: $\mathrm{f}(x) = \cos x$ on $[0, \pi]$ gives range $[-1, 1]$ .

Domain and range of $\mathrm{g}$ :

Domain: $\pi \leqslant x \leqslant 2\pi$ , i.e., $[\pi, 2\pi]$ .
Range: From part (b), $\mathrm{g}(x)$ decreases from $5$ to $-1$ on $[\pi, 2\pi]$ , so the range is $[-1, 5]$ .

Condition for $\mathrm{fg}$ to be formed:
For $\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))$ to be defined, every value in the range of $\mathrm{g}$ must lie within the domain of $\mathrm{f}$ .

The range of $\mathrm{g}$ is $[-1, 5]$ . The domain of $\mathrm{f}$ is $[0, \pi] \approx [0, 3.14]$ .

Since $5 > \pi$ , the value $5$ (and other values in $(\pi, 5]$ ) is in the range of $\mathrm{g}$ but not in the domain of $\mathrm{f}$ . For example, $\mathrm{g}(\pi) = 5$ , and $\mathrm{f}(5)$ is undefined because $5 > \pi$ .

Therefore, the domain of $\mathrm{f}$ does not include the whole of the range of $\mathrm{g}$ , so $\mathrm{fg}$ cannot be formed.

Answer

The domain of $\mathrm{f}$ is $[0, \pi]$ and the range of $\mathrm{g}$ is $[-1, 5]$ . Since $5 > \pi$ , the range of $\mathrm{g}$ is not entirely contained within the domain of $\mathrm{f}$ , so $\mathrm{fg}$ cannot be formed.

Final answer

The range of g is [-1, 5] and the domain of f is [0, π]; since 5 > π, the range of g is not a subset of the domain of f, so fg cannot be formed.

Detailed explanation

Walkthrough

The composite function $\mathrm{fg}$ means $\mathrm{f}(\mathrm{g}(x))$ . For this to be defined for any $x$ , we need:

$x$ must be in the domain of $\mathrm{g}$ .
$\mathrm{g}(x)$ must be in the domain of $\mathrm{f}$ .

This means the range of $\mathrm{g}$ must be a subset of the domain of $\mathrm{f}$ .

Range of $\mathrm{g}$ : From part (b), $\mathrm{g}(x) = 3\cos(x - \pi) + 2$ on $[\pi, 2\pi]$ . At $x = \pi$ , $\mathrm{g}(\pi) = 5$ . At $x = 2\pi$ , $\mathrm{g}(2\pi) = -1$ . Since $\mathrm{g}$ is continuous and decreasing on this interval, the range is $[-1, 5]$ .

Domain of $\mathrm{f}$ : $\mathrm{f}(x) = \cos x$ is defined for $0 \leqslant x \leqslant \pi$ , so the domain is $[0, \pi] \approx [0, 3.14159]$ .

Comparison: The range of $\mathrm{g}$ includes values like $4$ , $5$ , etc., which are greater than $\pi$ . For example, $\mathrm{g}(\pi) = 5$ , but $\mathrm{f}(5)$ is undefined because $5 \notin [0, \pi]$ . Therefore, $\mathrm{fg}(\pi) = \mathrm{f}(\mathrm{g}(\pi)) = \mathrm{f}(5)$ is undefined, and the composite $\mathrm{fg}$ cannot be formed.

Key Takeaways

For a composite function $\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))$ to be defined:

The range of the inner function $\mathrm{g}$ must be a subset of the domain of the outer function $\mathrm{f}$ .
Always check this condition by comparing the range of $\mathrm{g}$ with the domain of $\mathrm{f}$ .

Common Mistakes

Saying 'the domain of $\mathrm{f}$ does not include the range of $\mathrm{g}$ ' without specifying what the range of $\mathrm{g}$ actually is — the mark scheme requires mentioning both the range of $\mathrm{g}$ and the domain of $\mathrm{f}$ .
Confusing $\mathrm{fg}$ with $\mathrm{gf}$ — $\mathrm{gf}(x) = \mathrm{g}(\mathrm{f}(x))$ requires the range of $\mathrm{f}$ to be in the domain of $\mathrm{g}$ , which is $[-1, 1]$ vs $[\pi, 2\pi]$ ; this also fails, but the question asks about $\mathrm{fg}$ .
Not giving a specific reason or counterexample — the mark scheme accepts either a general statement about range/domain mismatch or a specific value showing the failure.

Things to Be Careful About

The mark scheme requires mentioning both 'range of $\mathrm{g}$ ' and 'domain of $\mathrm{f}$ ' — just saying 'the domains don't match' is not sufficient.
$\pi \approx 3.14$ , so any value in the range of $\mathrm{g}$ that is greater than $\pi$ (i.e., values in $(\pi, 5]$ ) is outside the domain of $\mathrm{f}$ .
The alternative method in the mark scheme accepts showing a specific value: for example, $x = \pi$ gives $\mathrm{g}(\pi) = 5$ , and $5 > \pi$ , so $\mathrm{f}(5)$ is undefined.

Techniques used

determine range of inner functiondetermine domain of outer functioncompare range and domain for composition validity

Q10Medium-HardCoordinate GeometryQuadratics

The equation of a circle is $x^2 + y^2 + 4x - 8y - 12 = 0$ .

(a)

Find an equation of the tangent to the circle at the point $(2, 8)$ , giving your answer in the form $ax + by + c = 0$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Complete the square to find the centre of the circle, then use the fact that the tangent at a point is perpendicular to the radius to the point. Write the tangent in the form $ax+by+c=0$ .

Working

The circle is $x^2 + y^2 + 4x - 8y - 12 = 0$ . Complete the square in $x$ and $y$ :

(x+2)^2 - 4 + (y-4)^2 - 16 - 12 = 0

(x+2)^2 + (y-4)^2 = 32

So the centre is $(-2, 4)$ .

The gradient of the radius from $(-2,4)$ to the point $(2,8)$ is

\frac{8-4}{2-(-2)} = \frac{4}{4} = 1

The tangent is perpendicular to this radius, so its gradient is the negative reciprocal:

m_{\text{tangent}} = -1

Using $y - y_1 = m(x - x_1)$ with the point $(2,8)$ :

y - 8 = -1(x - 2)

y - 8 = -x + 2

x + y - 10 = 0

Answer

x + y - 10 = 0

Final answer

x + y - 10 = 0

Detailed explanation

Walkthrough

We are given the circle $x^2 + y^2 + 4x - 8y - 12 = 0$ and need the tangent at the point $(2,8)$ . The radius to a point on a circle is perpendicular to the tangent at that point, so the tangent is found once the centre is known. Completing the square gives $(x+2)^2 + (y-4)^2 = 32$ , so the centre is $(-2,4)$ . The gradient of the radius joining $(-2,4)$ and $(2,8)$ is $\frac{8-4}{2-(-2)} = 1$ . The tangent must have gradient $-1$ , the negative reciprocal of $1$ . Using $y-y_1=m(x-x_1)$ with $(2,8)$ gives $y-8=-1(x-2)$ , which simplifies to $x+y-10=0$ .

Key Takeaways

This question tests completing the square to find the centre of a circle, using the perpendicular relationship between a radius and a tangent, and writing the equation of a straight line in the required general form $ax+by+c=0$ .

Common Mistakes

Using the wrong centre because of a sign error when completing the square.
Taking the gradient of the radius as the gradient of the tangent instead of using its negative reciprocal.
Leaving the tangent as $y = -x + 10$ instead of the required $ax+by+c=0$ form.
Giving an answer without working; the mark scheme allows a maximum of 2 marks for an unsupported answer.

Things to Be Careful About

The centre is $(-2,4)$ , not $(2,-4)$ .
The tangent gradient is $-1$ , not $1$ .
The final equation must be in the form $ax+by+c=0$ ; either $x+y-10=0$ or $-x-y+10=0$ is accepted.

Techniques used

complete the square to locate the centre of a circlecalculate the gradient of a radiustake the negative reciprocal to find the tangent gradientwrite the tangent line in general form

(b)

Given that the line $x + 3y = k$ does not intersect the circle, show that $k^2 - 20k - 220 > 0$ .

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Substitute the line $x+3y=k$ into the circle equation to obtain a quadratic in $y$ . Since the line does not intersect the circle, this quadratic has no real roots, so its discriminant is negative. Simplify the resulting inequality to reach the required form.

Working

From $x + 3y = k$ , write $x = k - 3y$ . Substitute into the circle equation:

(k-3y)^2 + y^2 + 4(k-3y) - 8y - 12 = 0

Expand and collect like terms:

k^2 - 6ky + 9y^2 + y^2 + 4k - 12y - 8y - 12 = 0

10y^2 - (6k + 20)y + (k^2 + 4k - 12) = 0

For no intersection, this quadratic in $y$ has no real roots, so its discriminant is negative:

(6k+20)^2 - 4(10)(k^2 + 4k - 12) < 0

Expand the discriminant:

36k^2 + 240k + 400 - 40k^2 - 160k + 480 < 0

-4k^2 + 80k + 880 < 0

Divide by $-4$ and reverse the inequality:

k^2 - 20k - 220 > 0

Answer

k^2 - 20k - 220 > 0

Final answer

k^2 - 20k - 220 > 0

Detailed explanation

Walkthrough

We are told that the line $x+3y=k$ does not intersect the circle $x^2+y^2+4x-8y-12=0$ . To test intersection, substitute the line into the circle equation. Rewriting the line as $x=k-3y$ gives a quadratic in $y$ ; its roots are the $y$ -coordinates of any intersection points. If the line does not intersect the circle, this quadratic has no real roots, so its discriminant is less than zero. Expanding $(k-3y)^2$ , $4(k-3y)$ and collecting like terms gives $10y^2-(6k+20)y+(k^2+4k-12)=0$ . The discriminant is $(6k+20)^2-4(10)(k^2+4k-12)$ , which simplifies to $-4k^2+80k+880$ . Setting this less than zero and dividing by $-4$ reverses the inequality, giving $k^2-20k-220>0$ .

Key Takeaways

A line and a circle intersect when the quadratic equation obtained by substituting the line into the circle has real roots. No intersection corresponds to a negative discriminant. This question combines the algebra of lines and circles with the quadratic discriminant condition.

Common Mistakes

Substituting incorrectly, for example replacing $x$ by $k+3y$ instead of $k-3y$ .
Forgetting to substitute into every term, including $x^2$ and $4x$ .
Making sign errors when expanding $(k-3y)^2$ .
Using the wrong discriminant condition: no intersection means discriminant $<0$ , not $>0$ .
Dividing by a negative number without reversing the inequality.

Things to Be Careful About

The discriminant must be set less than zero for no real roots; it is greater than zero for two intersections and equal to zero for a tangent.
When simplifying $-4k^2+80k+880<0$ , dividing by $-4$ reverses the inequality sign.
The mark scheme also accepts the distance-from-centre method, but whichever method is used, the final inequality must be obtained with full working.

Techniques used

rewrite the line equation in terms of one variablesubstitute the line equation into the circle equationexpand to form a quadratic equationapply the discriminant condition for no real rootssimplify the resulting inequality

Mathematics 9709/15 — May/June 2025

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