(a) Prove that $\frac{(\sin \theta + \cos \theta)^2 - 1}{\cos^2 \theta} \equiv 2 \tan \theta$. (b) Hence solve the equation $\frac{(\sin \theta + \cos \theta)^2 - 1}{\cos^2 \theta} = 5 \tan^3 \theta$ for $-90^\circ < \theta < 90^\circ$.

(a) 2 tan θ · (b) θ = 0°, θ ≈ 32.3°, θ ≈ -32.3°

The functions $f$ and $g$ are defined for all real values of $x$ by $$ f(x) = (3x - 2)^2 + k \quad \text{and} \quad g(x) = 5x - 1, $$ where $k$ is a constant. (a) Given that the range of the function $gf$ is $gf(x) \ge 39$, find the value of $k$. (b) For this value of $k$, determine the range of the function $fg$. (c) The function $h$ is defined for all real values of $x$ and is such that $gh(x) = 35x + 19$. Find an expression for $g^{-1}(x)$ and hence, or otherwise, find an…

(a) k = 8 · (b) fg(x) >= 8 · (c) g^-1(x) = (x+1)/5, h(x) = 7x + 4

The diagram shows the circle with centre $C(-4, 5)$ and radius $\sqrt{20}$ units. The circle intersects the $y$-axis at the points $A$ and $B$. The size of angle $ACB$ is $\theta$ radians. (a) Find the equation of the tangent to the circle at the point $(-6, 9)$. (b) Find the equation of the circle in the form $x^2 + y^2 + ax + by + c = 0$. (c) Find the value of $\theta$ correct to 4 significant figures. (d) Find the perimeter and area of the segment shaded in the diagram.

(a) y = 1/2 x + 12 · (b) x^2 + y^2 + 8x - 10y + 21 = 0 · (c) 0.9273 · (d) Perimeter = 8.15, Area = 1.27

9709/12 Feb/Mar 2024 — Mathematics Solved Past Paper

Q13MMedium-EasyIntegration

Find the exact value of

\int_3^{\infty} \frac{2}{x^2}\,dx

DifficultyMedium-Easy

Worked solution

Approach

Evaluate the improper integral by first rewriting $\frac{2}{x^2}$ as $2x^{-2}$ , finding its antiderivative, and then treating the infinite upper limit as a limit.

Working

\int_3^{\infty} \frac{2}{x^2}\,dx = \int_3^{\infty} 2x^{-2}\,dx

= \left[-2x^{-1}\right]_3^{\infty}

= \left[-\frac{2}{x}\right]_3^{\infty}

Treat the upper limit as a limit:

\lim_{R \to \infty} \left[-\frac{2}{x}\right]_3^R = \lim_{R \to \infty} \left(-\frac{2}{R} + \frac{2}{3}\right) = \frac{2}{3}

Answer

\frac{2}{3}

Final answer

2/3

Detailed explanation

Walkthrough

This is an improper integral because the upper limit is infinite. We cannot substitute $\infty$ directly, so we first find the antiderivative of $\frac{2}{x^2}$ . Since $\frac{2}{x^2} = 2x^{-2}$ , we use the power rule to integrate: the antiderivative is $-2x^{-1} = -\frac{2}{x}$ . Next we evaluate the integral from $3$ to $R$ , where $R$ tends to infinity. Substituting the upper limit $R$ gives $-\frac{2}{R}$ , and substituting the lower limit $3$ gives $-\frac{2}{3}$ . The value of the integral is the upper-limit value minus the lower-limit value: $\left(-\frac{2}{R}\right) - \left(-\frac{2}{3}\right) = -\frac{2}{R} + \frac{2}{3}$ . As $R \to \infty$ , $-\frac{2}{R}$ tends to $0$ , leaving $\frac{2}{3}$ .

Key Takeaways

This question tests the ability to integrate $x^n$ for negative powers and to recognise improper integrals by taking the infinite limit. It also reinforces the useful fact that $\frac{1}{x} \to 0$ as $x \to \infty$ , which makes the upper-limit contribution vanish.

Common Mistakes

Writing the antiderivative of $\frac{2}{x^2}$ as $2\ln x$ or $\frac{2}{x}$ instead of $-\frac{2}{x}$ .
Substituting $\infty$ directly without taking a limit.
Sign errors when subtracting the lower limit: the correct result is $0 - (-\frac{2}{3}) = \frac{2}{3}$ , not $-\frac{2}{3}$ .
Omitting the limit step; the mark scheme requires a clear indication that the upper limit gives $0$ .

Things to Be Careful About

Write an improper integral with an infinite upper bound as a limit before evaluating. In the mark scheme, the integration earns B1, the correct limit/substitution earns M1, and the final value earns A1. Always show the antiderivative, the substitution, and the limit so that all marks are earned.

Techniques used

integrate a power functionevaluate an improper integral by taking a limitsubstitute limits of integration

Similar questions

Q2MediumTrigonometryFunctions

Fig. 2.1

The diagram shows part of the curve with equation $y = k \sin \frac{1}{2}x$ , where $k$ is a positive constant and $x$ is measured in radians. The curve has a minimum point $A$ .

(a)

State the coordinates of $A$ .

1M Similar

(b)

A sequence of transformations is applied to the curve in the following order.

Translation of 2 units in the negative $y$ -direction

Reflection in the $x$ -axis

Find the equation of the new curve and determine the coordinates of the point on the new curve corresponding to $A$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Apply the two transformations in the given order to the equation $y = k \sin \frac{1}{2}x$ . Then track the point $A = (3\pi, -k)$ through the same transformations to find its image.

Working

Step 1: Translation of 2 units in the negative $y$ -direction.

Subtract 2 from $y$ :

y = k \sin \frac{1}{2}x - 2

Step 2: Reflection in the $x$ -axis.

Replace $y$ with $-y$ :

-y = k \sin \frac{1}{2}x - 2

Multiply through by $-1$ :

y = 2 - k \sin \frac{1}{2}x

Coordinates of the image of $A$ :

Point $A$ is $(3\pi, -k)$ .

After translation of 2 units in the negative $y$ -direction:

(3\pi, -k - 2)

After reflection in the $x$ -axis (negate the $y$ -coordinate):

(3\pi, k + 2)

Verification: Substitute $x = 3\pi$ into the new equation:

y = 2 - k \sin \frac{3\pi}{2} = 2 - k(-1) = 2 + k

This confirms the point is $(3\pi, 2 + k)$ .

Answer

Equation: $y = 2 - k \sin \frac{1}{2}x$

Coordinates: $(3\pi, 2 + k)$

Final answer

y = 2 - k sin(1/2 x); (3π, 2 + k)

Detailed explanation

Walkthrough

We apply the transformations in the given order to the original equation $y = k \sin \frac{1}{2}x$ .

First, translate 2 units in the negative $y$ -direction. This means subtracting 2 from the $y$ -value, giving $y = k \sin \frac{1}{2}x - 2$ .

Second, reflect in the $x$ -axis. This means replacing $y$ with $-y$ , giving $-y = k \sin \frac{1}{2}x - 2$ , which simplifies to $y = 2 - k \sin \frac{1}{2}x$ .

For the point corresponding to $A$ : $A$ is at $(3\pi, -k)$ . After translating 2 units down, it moves to $(3\pi, -k - 2)$ . After reflecting in the $x$ -axis, the $y$ -coordinate is negated, giving $(3\pi, k + 2)$ . We can verify this by substituting $x = 3\pi$ into the new equation: $y = 2 - k \sin \frac{3\pi}{2} = 2 - k(-1) = 2 + k$ .

Key Takeaways

A translation of $c$ units in the negative $y$ -direction gives $y = f(x) - c$ .
A reflection in the $x$ -axis replaces $y$ with $-y$ .
When applying multiple transformations, apply them in the given order.
Points transform the same way as the equation: translate coordinates, then reflect.

Common Mistakes

Applying the reflection before the translation (order matters).
Forgetting to negate the entire right-hand side during reflection.
Confusing reflection in the $x$ -axis with reflection in the $y$ -axis.

Things to Be Careful About

The order of transformations is specified and must be followed exactly.
Reflection in the $x$ -axis affects the $y$ -coordinate, not the $x$ -coordinate.
Verify the final point by substituting into the new equation.

Techniques used

apply vertical translationapply reflection in x-axistrack point through transformations

Q35MMedium-EasyIntegration

A curve is such that $\frac{dy}{dx} = 3(4x + 5)^{\frac{1}{2}}$ . It is given that the points $(1, 9)$ and $(5, a)$ lie on the curve.

Find the value of $a$ .

DifficultyMedium-Easy

Worked solution

Approach

Integrate $\frac{dy}{dx}$ to obtain the general equation of the curve, introduce the constant of integration $c$ , then use the known point $(1,9)$ to find $c$ . Finally substitute $x=5$ to evaluate $a$ .

Working

y = \int 3(4x+5)^{\frac{1}{2}} \, dx

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

y = 3 \cdot \frac{(4x+5)^{\frac{3}{2}}}{4 \cdot \frac{3}{2}} + c

Simplify the coefficient:

y = \frac{1}{2}(4x+5)^{\frac{3}{2}} + c

Substitute $x=1$ and $y=9$ :

9 = \frac{1}{2}(4(1)+5)^{\frac{3}{2}} + c

9 = \frac{1}{2}(9)^{\frac{3}{2}} + c = \frac{27}{2} + c

So the constant of integration is

c = 9 - \frac{27}{2} = -\frac{9}{2}

Therefore the curve is

y = \frac{1}{2}(4x+5)^{\frac{3}{2}} - \frac{9}{2}

At $x=5$ , since the point is $(5,a)$ :

a = \frac{1}{2}(4(5)+5)^{\frac{3}{2}} - \frac{9}{2}

a = \frac{1}{2}(25)^{\frac{3}{2}} - \frac{9}{2} = \frac{125}{2} - \frac{9}{2} = 58

Answer

a = 58

Final answer

a = 58

Detailed explanation

Walkthrough

The derivative $\frac{dy}{dx}$ gives the gradient of the curve at every point. To recover $y$ as a function of $x$ , integrate with respect to $x$ . The integrand has the linear form $(4x+5)^{\frac{1}{2}}$ . When integrating something of the form $(ax+b)^n$ , increase the power by $1$ and divide by the new power, and also divide by $a$ because of the chain rule. Here $a=4$ and $n=\frac{1}{2}$ , so

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

Because integration is only defined up to an additive constant, include $+c$ . Use the known point $(1,9)$ : substitute $x=1$ and $y=9$ to form an equation in $c$ . Since $9^{\frac{3}{2}} = 27$ , this gives $9 = \frac{27}{2} + c$ , so $c = -\frac{9}{2}$ . Finally, evaluate at $x=5$ : $4(5)+5=25$ and $25^{\frac{3}{2}}=125$ , so

a = \frac{125}{2} - \frac{9}{2} = 58

Key Takeaways

This question tests the basic reverse-power rule for integration, in particular when the integrand contains a linear expression such as $(4x+5)$ . It also tests the use of a point on the curve to determine the arbitrary constant of integration, and the correct evaluation of fractional powers. The key idea is that integration gives a family of curves, and a given point selects the particular curve.

Common Mistakes

A common mistake is to integrate $(4x+5)^{\frac{1}{2}}$ without dividing by $4$ , giving an incorrect coefficient. Another is to forget the constant $+c$ before substituting the point, which makes it impossible to find the right curve. Students may also evaluate fractional powers incorrectly: $9^{\frac{3}{2}} = 27$ and $25^{\frac{3}{2}} = 125$ , not $9^3$ or $25^3$ . Finally, careless arithmetic when subtracting $\frac{9}{2}$ can produce $a=67$ instead of $58$ .

Things to Be Careful About

Always divide by the coefficient of $x$ inside the bracket when using $\int (ax+b)^n \, dx$ . Introduce $+c$ before using the given point, even though the mark scheme allows it to be omitted immediately after integration. Use exact fractions rather than decimals for $c = -\frac{9}{2}$ . When substituting the second point, use the $x$ -coordinate $5$ , not the $y$ -coordinate, because the point is written as $(5,a)$ .

Techniques used

integrate an expression of the form (ax+b)^ndetermine the constant of integration from a given pointevaluate the resulting function at a given x-coordinate

Similar questions

Q4MediumTrigonometry

(a)

Prove that $\frac{(\sin \theta + \cos \theta)^2 - 1}{\cos^2 \theta} \equiv 2 \tan \theta$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Start from the left-hand side, expand the square, simplify the numerator using $\sin^2\theta + \cos^2\theta = 1$ , and then rewrite the result as $2\tan\theta$ .

Working

\frac{(\sin\theta + \cos\theta)^2 - 1}{\cos^2\theta} = \frac{\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta - 1}{\cos^2\theta}

Using $\sin^2\theta + \cos^2\theta = 1$ :

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

Cancel the common factor $\cos\theta$ (valid wherever the expression is defined):

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

Since $\dfrac{\sin\theta}{\cos\theta} = \tan\theta$ :

= 2\tan\theta

Answer

\frac{(\sin\theta + \cos\theta)^2 - 1}{\cos^2\theta} \equiv 2\tan\theta

Final answer

2 tan θ

Detailed explanation

Walkthrough

Begin with the left-hand side of the identity. Expanding $(\sin\theta + \cos\theta)^2$ gives $\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$ . Subtract $1$ and use $\sin^2\theta + \cos^2\theta = 1$ to replace the sum of the two squared terms by $1$ . The numerator becomes $2\sin\theta\cos\theta$ . Dividing by $\cos^2\theta$ gives $2\dfrac{\sin\theta}{\cos\theta}$ , and using $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ completes the proof.

Key Takeaways

This question tests manipulation of trigonometric expressions: expanding brackets, applying the Pythagorean identity, and using the definition of $\tan\theta$ . It also shows that proving an identity means starting from one side and transforming it until it matches the other side.

Common Mistakes

A common error is forgetting that the $1$ in the numerator came from $\sin^2\theta + \cos^2\theta$ , so students may wrongly simplify $\sin^2\theta + \cos^2\theta - 1$ to zero before expanding. Another common mistake is cancelling a single $\cos\theta$ from one term but not from both terms of the numerator.

Things to Be Careful About

The expression is not defined when $\cos\theta = 0$ , so cancellation by $\cos\theta$ is valid only on the domain where $\cos\theta \neq 0$ . For this identity, the equivalence holds whenever both sides are defined. Show every step so that the mark scheme's method marks are clearly earned.

Techniques used

expand the squared bracketapply the Pythagorean identitysimplify the quotient using the tangent identity

(b)

Hence solve the equation $\frac{(\sin \theta + \cos \theta)^2 - 1}{\cos^2 \theta} = 5 \tan^3 \theta$ for $-90^\circ < \theta < 90^\circ$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the identity from part (a) to replace the left-hand side by $2\tan\theta$ . Rearrange the resulting equation, factorise, solve for $\tan\theta$ , and then find the corresponding angles in $-90^\circ < \theta < 90^\circ$ .

Working

From part (a),

\frac{(\sin\theta + \cos\theta)^2 - 1}{\cos^2\theta} = 2\tan\theta

So the equation becomes

2\tan\theta = 5\tan^3\theta

Bring all terms to one side:

5\tan^3\theta - 2\tan\theta = 0

Factorise:

\tan\theta\left(5\tan^2\theta - 2\right) = 0

Hence either

\tan\theta = 0

or

5\tan^2\theta - 2 = 0 \quad \Rightarrow \quad \tan^2\theta = \frac{2}{5}

so

\tan\theta = \pm\sqrt{\frac{2}{5}}

Now solve for $\theta$ in the interval $-90^\circ < \theta < 90^\circ$ :

\tan\theta = 0 \quad \Rightarrow \quad \theta = 0^\circ

and

\theta = \arctan\left(\sqrt{\frac{2}{5}}\right) \approx 32.3^\circ, \qquad \theta = \arctan\left(-\sqrt{\frac{2}{5}}\right) \approx -32.3^\circ

The tangent function is one-to-one on $-90^\circ < \theta < 90^\circ$ , so these are the only solutions.

Answer

\theta = 0^\circ, \quad \theta \approx 32.3^\circ, \quad \theta \approx -32.3^\circ

Final answer

θ = 0°, θ ≈ 32.3°, θ ≈ -32.3°

Detailed explanation

Walkthrough

Use the identity proved in part (a) to replace the left-hand side of the equation with $2\tan\theta$ . Rearranging gives $5\tan^3\theta - 2\tan\theta = 0$ . Factor out $\tan\theta$ : this is crucial because the factor $\tan\theta$ gives the solution $\theta = 0^\circ$ , which would be lost if we divided by $\tan\theta$ first.

The remaining factor gives $5\tan^2\theta - 2 = 0$ , so $\tan\theta = \pm\sqrt{\frac{2}{5}}$ . Since the interval is $-90^\circ < \theta < 90^\circ$ , inverse tangent gives exactly one angle for each real tangent value: $\theta = \arctan\left(\sqrt{\frac{2}{5}}\right) \approx 32.3^\circ$ and $\theta = \arctan\left(-\sqrt{\frac{2}{5}}\right) \approx -32.3^\circ$ . Together with $0^\circ$ , these are all three solutions.

Key Takeaways

This question shows how a proved identity can be used to simplify a more complicated equation. It also reinforces that factorisation is safer than cancellation, and that the range of inverse tangent helps find all solutions in an interval such as $-90^\circ < \theta < 90^\circ$ .

Common Mistakes

The most common mistake is cancelling $\tan\theta$ from $5\tan^3\theta = 2\tan\theta$ and losing the solution $\tan\theta = 0$ . Another common mistake is giving only $32.3^\circ$ and forgetting the negative angle, or adding $180^\circ$ multiples, which would put the solutions outside the given interval.

Things to Be Careful About

Because the interval is $-90^\circ < \theta < 90^\circ$ , do not add $180^\circ$ periodic solutions. The tangent function has period $180^\circ$ , but the given interval contains only one representative for each value of $\tan\theta$ . The final answer must include $0^\circ$ and both $\pm 32.3^\circ$ , with no extra angles. Show all factoring steps to earn the method mark.

Techniques used

substitute the identity from part (a)rearrange and factorise the trigonometric equationsolve for all possible tangent valuesuse inverse tangent over the given interval

Q56MMediumDifferentiation

A curve has the equation $y = \frac{3}{2x^2 - 5}$ .

Find the equation of the normal to the curve at the point $(2, 1)$ , giving your answer in the form $ax + by + c = 0$ , where $a$ , $b$ and $c$ are integers.

DifficultyMedium

Worked solution

Approach

Differentiate $y = \frac{3}{2x^2 - 5}$ using the chain rule, evaluate the gradient at $x = 2$ , take the negative reciprocal to find the gradient of the normal, then form the equation of the normal through $(2,1)$ and rearrange it into $ax + by + c = 0$ .

Working

Write $y$ as a power of $(2x^2 - 5)$ :

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

Differentiate with respect to $x$ :

\frac{dy}{dx} = 3 \cdot (-1)(2x^2 - 5)^{-2} \cdot 4x = -12x(2x^2 - 5)^{-2}

At the point $(2,1)$ , substitute $x = 2$ :

\frac{dy}{dx}\bigg|_{x=2} = -12(2)\left(2(2)^2 - 5\right)^{-2} = -24(3)^{-2} = -\frac{24}{9} = -\frac{8}{3}

So the gradient of the tangent is $-\frac{8}{3}$ . The gradient of the normal is the negative reciprocal of this:

m_{\text{normal}} = -\frac{1}{-\frac{8}{3}} = \frac{3}{8}

Using the point $(2,1)$ and the normal gradient $\frac{3}{8}$ , the equation of the normal is:

y - 1 = \frac{3}{8}(x - 2)

Multiply through by $8$ :

8y - 8 = 3x - 6

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

3x - 8y + 2 = 0

Answer

3x - 8y + 2 = 0

Final answer

3x - 8y + 2 = 0

Detailed explanation

Walkthrough

We need the equation of the normal to the curve at the point $(2,1)$ . A normal is perpendicular to the tangent, so we first need the gradient of the tangent at that point.

The curve is $y = \frac{3}{2x^2 - 5}$ . To differentiate it, write it as $y = 3(2x^2 - 5)^{-1}$ . Using the chain rule, the derivative is $3 \times (-1)(2x^2 - 5)^{-2} \times 4x = -12x(2x^2 - 5)^{-2}$ . The factor $4x$ comes from differentiating the inside function $2x^2 - 5$ .

Substituting $x = 2$ gives the gradient of the tangent at that point:
$-12(2)(2(2)^2 - 5)^{-2} = -24 \times 3^{-2} = -\frac{24}{9} = -\frac{8}{3}$ .

Since the normal is perpendicular to the tangent, its gradient is the negative reciprocal of $-\frac{8}{3}$ , which is $\frac{3}{8}$ .

Now use the point-slope form of a straight line with the point $(2,1)$ and gradient $\frac{3}{8}$ :

$y - 1 = \frac{3}{8}(x - 2).$

Multiplying by $8$ gives $8y - 8 = 3x - 6$ , and rearranging gives $3x - 8y + 2 = 0$ .

Key Takeaways

To differentiate a quotient of the form $\frac{k}{f(x)}$ , rewrite it as $k(f(x))^{-1}$ and use the chain rule.
The gradient of the tangent at a point is found by substituting the $x$ -coordinate into the derivative.
The gradient of the normal is the negative reciprocal of the gradient of the tangent.
A straight line can be written in the form $y - y_1 = m(x - x_1)$ and then rearranged into $ax + by + c = 0$ .

Common Mistakes

Forgetting the chain rule factor $4x$ when differentiating $2x^2 - 5$ .
Forgetting the negative sign in the derivative, which would give the wrong gradient.
Taking the reciprocal instead of the negative reciprocal when finding the normal gradient.
Substituting only $x = 2$ but not using the point $(2,1)$ when forming the equation of the normal.
Leaving the final answer in a form other than $ax + by + c = 0$ with integer coefficients.

Things to Be Careful About

The derivative must be simplified correctly: $-\frac{24}{9}$ should be reduced to $-\frac{8}{3}$ , though the mark scheme also allows $-2.67$ .
The mark scheme requires an attempt at differentiation before awarding the normal-gradient mark, so the derivative must be shown.
The final answer can be written as a multiple of $3x - 8y + 2 = 0$ , but the question asks for integers, so $3x - 8y + 2 = 0$ is the cleanest form.

Techniques used

differentiate a quotient using the chain ruleevaluate the derivative at a given pointtake the negative reciprocal of a gradientform the equation of a line through a pointrearrange into the form ax + by + c = 0

Similar questions

Q65MMediumSeries

It is given that the coefficient of $x^3$ in the expansion of

(2 + ax)^4(5 - ax)

is 432.

Find the value of the constant $a$ .

DifficultyMedium

Worked solution

Approach

Expand $(2+ax)^4$ only as far as the $x^2$ and $x^3$ terms, since these are the only terms that can produce an $x^3$ term after multiplication by $(5-ax)$ . Then multiply by $(5-ax)$ , collect the coefficient of $x^3$ , equate it to $432$ , and solve for $a$ .

Working

The relevant terms of $(2+ax)^4$ are

\binom{4}{2}2^2(ax)^2 = 24a^2x^2

and

\binom{4}{3}2^1(ax)^3 = 8a^3x^3

The $x^3$ term in $(2+ax)^4(5-ax)$ is obtained by multiplying:

the $x^2$ term by $-ax$ :

24a^2x^2 \times (-ax) = -24a^3x^3

the $x^3$ term by $5$ :

8a^3x^3 \times 5 = 40a^3x^3

So the coefficient of $x^3$ is

-24a^3 + 40a^3 = 16a^3

Equate this coefficient to $432$ :

16a^3 = 432

a^3 = 27

a = 3

Answer

a = 3

Final answer

a = 3

Detailed explanation

Walkthrough

We need the coefficient of $x^3$ in the product $(2+ax)^4(5-ax)$ . The first factor is a binomial raised to the fourth power, so we can use the binomial expansion. However, we do not need the whole expansion: after multiplying by $5-ax$ , only certain terms can contribute to $x^3$ .

First, identify the $x^2$ and $x^3$ terms from $(2+ax)^4$ . The $x^2$ term is $\binom{4}{2}2^2(ax)^2$ , which simplifies to $24a^2x^2$ . The $x^3$ term is $\binom{4}{3}2^1(ax)^3$ , which simplifies to $8a^3x^3$ . These are the only terms that matter because, when multiplied by $5-ax$ , the $x^2$ term can combine with $-ax$ to give $x^3$ , and the $x^3$ term can combine with $5$ to give $x^3$ .

Next, perform those two multiplications. Multiplying $24a^2x^2$ by $-ax$ gives $-24a^3x^3$ . Multiplying $8a^3x^3$ by $5$ gives $40a^3x^3$ . Adding these gives the total coefficient $16a^3$ .

Finally, set $16a^3 = 432$ , divide by $16$ to get $a^3 = 27$ , and take the real cube root to obtain $a = 3$ . Since the cube root is unique for real numbers, there is no second solution.

Key Takeaways

This question tests the binomial expansion of $(a+b)^n$ for a positive integer $n$ , and the ability to select only the terms needed for a particular power in a product. It also reinforces that when multiplying two expansions, the coefficient of a given power comes from combining several contributing products, and that a cubic equation such as $a^3 = 27$ has a unique real solution.

Common Mistakes

Expanding the whole of $(2+ax)^4$ unnecessarily, increasing the chance of arithmetic errors.
Forgetting to multiply the $x^2$ term by $-ax$ , and only using the $x^3$ term multiplied by $5$ .
Sign errors when multiplying by $-ax$ : the product must be $-24a^3x^3$ , not $+24a^3x^3$ .
Incorrect binomial coefficients, e.g. writing $\binom{4}{2}2^2 = 16$ instead of $24$ , or $\binom{4}{3}2 = 4$ instead of $8$ .
Equating the coefficient of $x^3$ in $(2+ax)^4$ itself to $432$ , instead of the coefficient in the full product.

Things to Be Careful About

The coefficient of $x^3$ in the final product is not simply one term; it is the sum of two contributions. Be careful with the sign of $a$ when multiplying by $-ax$ . Also, the final equation is $a^3 = 27$ , not $a^2 = 27$ ; taking the cube root gives the single real value $a = 3$ . The mark scheme requires the answer $a = 3$ only, with no other real solutions.

Techniques used

expand a binomial up to the required termsmultiply by a linear factor to collect a specified coefficientequate the coefficient and solve the resulting equation

Similar questions

Q7MediumQuadratics

The straight line $y = x + 5$ meets the curve $2x^2 + 3y^2 = k$ at a single point $P$ .

(a)

Find the value of the constant $k$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Substitute the line equation into the curve equation to obtain a quadratic in $x$ . Then use the fact that the line meets the curve at a single point, so the quadratic has exactly one solution and its discriminant is zero. Solve this discriminant condition to find $k$ .

Working

Substitute $y = x + 5$ into $2x^2 + 3y^2 = k$ :

2x^2 + 3(x+5)^2 = k

Expand and simplify:

2x^2 + 3(x^2 + 10x + 25) = k

5x^2 + 30x + 75 = k

5x^2 + 30x + 75 - k = 0

For a unique intersection, the discriminant of this quadratic must be zero:

\Delta = b^2 - 4ac = 0

Here $a = 5$ , $b = 30$ , $c = 75 - k$ , so:

30^2 - 4(5)(75 - k) = 0

900 - 20(75 - k) = 0

900 - 1500 + 20k = 0

20k = 600

k = 30

Answer

k = 30

Final answer

k = 30

Detailed explanation

Walkthrough

We are given a line and a curve. To find where they meet, substitute the line equation into the curve equation. This removes one variable and gives a quadratic equation in $x$ . The phrase “meets at a single point” tells us that the line is tangent to the curve, so the quadratic has exactly one repeated root. A quadratic has a repeated root exactly when its discriminant is zero. Therefore, after writing the equation as $5x^2 + 30x + 75 - k = 0$ , we identify $a = 5$ , $b = 30$ , $c = 75 - k$ and set $b^2 - 4ac = 0$ . This simplifies to $900 - 20(75 - k) = 0$ , which gives $k = 30$ .

Key Takeaways

A “single point” intersection between a line and a curve corresponds to a zero discriminant.
Substitution is the standard method for solving a pair of equations where one is linear and one is quadratic.
The discriminant condition can be used to find an unknown constant in the curve equation.

Common Mistakes

Forgetting to set the discriminant equal to zero; some students set it greater than or less than zero.
Expanding $3(x+5)^2$ incorrectly.
Moving $k$ to the wrong side and using $c = 75 + k$ instead of $75 - k$ .
Not simplifying the discriminant fully before solving for $k$ .

Things to Be Careful About

The coefficient of $x^2$ is $5$ , not $2$ , after substitution.
The discriminant must be computed using the coefficients of the quadratic in the form $ax^2 + bx + c = 0$ .
The mark scheme requires explicit substitution and the discriminant equation; an unsupported answer is not sufficient.

Techniques used

substitute linear equation into quadraticform a quadratic equation in one variableapply the zero discriminant conditionsolve for the unknown constant

(b)

Find the coordinates of $P$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the value of $k$ found in part (a) in the quadratic equation obtained by substitution. Solve the resulting quadratic to find the $x$ -coordinate of the single meeting point, then use the line equation to find the $y$ -coordinate.

Working

With $k = 30$ , the equation becomes:

5x^2 + 30x + 75 - 30 = 0

5x^2 + 30x + 45 = 0

Divide by $5$ :

x^2 + 6x + 9 = 0

(x + 3)^2 = 0

So $x = -3$ . Substitute into $y = x + 5$ :

y = -3 + 5 = 2

Answer

P = (-3, 2)

Final answer

P = (-3, 2)

Detailed explanation

Walkthrough

We already found $k = 30$ in part (a). Substitute this value back into the quadratic obtained from the line and curve:

5x^2 + 30x + 75 - 30 = 0

This simplifies to $5x^2 + 30x + 45 = 0$ . Dividing by 5 gives $x^2 + 6x + 9 = 0$ , which is $(x + 3)^2 = 0$ . Therefore the only solution is $x = -3$ . Using the line equation $y = x + 5$ , we get $y = -3 + 5 = 2$ . Hence the point is $(-3, 2)$ .

Key Takeaways

Once the constant is known, the same substitution equation gives the intersection point.
A repeated root corresponds to the single tangent point.
Always use the line equation to find the other coordinate.

Common Mistakes

Using an incorrect value of $k$ from part (a).
Forgetting to divide the quadratic by 5 before factorising.
Stopping after finding $x = -3$ and not finding $y$ .
The mark scheme awards only SC B1 for coordinates without an attempt to solve the quadratic, so working must be shown.

Things to Be Careful About

The root is repeated, so $x = -3$ is the only $x$ -coordinate.
Substitute into the line $y = x + 5$ to find $y$ .
Check the point satisfies the curve: $2(-3)^2 + 3(2)^2 = 18 + 12 = 30 = k$ .

Techniques used

substitute the known constant into the quadratic equationsolve a quadratic equation with a repeated rootdetermine the y-coordinate from the line equation

Q8MediumSeries

(a)

An arithmetic progression is such that its first term is 6 and its tenth term is 19.5.

Find the sum of the first 100 terms of this arithmetic progression.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the formula for the $n$ th term of an arithmetic progression to find the common difference $d$ , then substitute $a = 6$ , $d$ and $n = 100$ into the formula for the sum of the first $n$ terms.

Working

The $n$ th term of an arithmetic progression is

u_n = a + (n-1)d

Given $a = 6$ and $u_{10} = 19.5$ :

6 + 9d = 19.5

Solve for $d$ :

9d = 13.5

d = 1.5

The sum of the first $n$ terms is

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

For $n = 100$ :

S_{100} = \frac{100}{2}\left(2(6) + 99(1.5)\right)

S_{100} = 50(12 + 148.5)

S_{100} = 50 \times 160.5 = 8025

Answer

8025

Final answer

8025

Detailed explanation

Walkthrough

Start with the $n$ th term formula for an arithmetic progression, $u_n = a + (n-1)d$ . Substitute the known first term $a = 6$ and the tenth term $u_{10} = 19.5$ with $n = 10$ . This gives $6 + 9d = 19.5$ , because there are 9 common differences between the first and tenth terms. Solving gives $d = 1.5$ .

Then use the sum formula $S_n = \frac{n}{2}(2a + (n-1)d)$ . Substitute $n = 100$ , $a = 6$ and $d = 1.5$ . The bracket becomes $12 + 99(1.5) = 160.5$ , and multiplying by $50$ gives $8025$ .

Key Takeaways

This question tests the two fundamental arithmetic progression formulae: the $n$ th term and the sum of the first $n$ terms. It also requires careful substitution and arithmetic with decimals.

Common Mistakes

Using $100d$ instead of $99d$ in the sum formula, since there are only 99 differences between 100 terms.
Forgetting to multiply by $\frac{n}{2}$ after finding the bracket.
Substituting the tenth term directly as $u_{10} = 6 + 10d$ instead of $6 + 9d$ .

Things to Be Careful About

Make sure the common difference is found correctly before using the sum formula. The mark scheme requires the substitution into the sum formula to use the candidate's value of $d$ , so keep the working clear. Also check decimal arithmetic carefully: $9 \times 1.5 = 13.5$ and $99 \times 1.5 = 148.5$ .

Techniques used

substitute values into the nth term formulasolve for the common differenceapply the sum formula for an arithmetic progression

(b)

A geometric progression $a_1, a_2, a_3, \dots$ is such that $a_1 = 24$ and the common ratio is $\frac{1}{2}$ .

The sum to infinity of this geometric progression is denoted by $S$ . The sum to infinity of the even-numbered terms (i.e. $a_2, a_4, a_6, \dots$ ) is denoted by $S_E$ .

Find the values of $S$ and $S_E$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the sum to infinity formula $S = \frac{a}{1-r}$ for the full geometric progression. For the even-numbered terms, identify their first term and common ratio, then apply the same formula.

Working

For the full geometric progression, $a = 24$ and $r = \frac{1}{2}$ . Since $|r| < 1$ , the sum to infinity exists:

S = \frac{24}{1 - \frac{1}{2}} = \frac{24}{\frac{1}{2}} = 48

The even-numbered terms are $a_2, a_4, a_6, \dots$ . Their first term is

a_2 = 24\left(\frac{1}{2}\right) = 12

and their common ratio is the square of the original common ratio:

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

Therefore,

S_E = \frac{12}{1 - \frac{1}{4}} = \frac{12}{\frac{3}{4}} = 16

Answer

S = 48, \quad S_E = 16

Final answer

S = 48, S_E = 16

Detailed explanation

Walkthrough

The sum to infinity of a geometric progression is $S = \frac{a}{1-r}$ , valid only when $|r| < 1$ . Here $a = 24$ and $r = \frac{1}{2}$ , so $S = \frac{24}{1 - \frac{1}{2}} = 48$ .

For the even-numbered terms $a_2, a_4, a_6, \dots$ , the first term is $a_2 = 24 \times \frac{1}{2} = 12$ . To find the common ratio, notice that moving from one even term to the next means multiplying by $r$ twice, so the new common ratio is $r^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$ . Then $S_E = \frac{12}{1 - \frac{1}{4}} = \frac{12}{\frac{3}{4}} = 16$ .

Key Takeaways

This question shows how a subsequence of a geometric progression can itself be a geometric progression. It also reinforces the sum to infinity formula and the convergence condition $|r| < 1$ .

Common Mistakes

Using the original common ratio $\frac{1}{2}$ for the even-numbered terms instead of $\frac{1}{4}$ .
Taking the first term of the even-numbered terms as $24$ instead of $a_2 = 12$ .
Applying the sum to infinity formula without checking that the common ratio has magnitude less than 1.

Things to Be Careful About

The mark scheme awards a method mark for attempting the sum to infinity with at least one of the first term and common ratio correct, but only if $|r| < 1$ . Both $\frac{1}{2}$ and $\frac{1}{4}$ satisfy this. Keep the first term and common ratio for the even-numbered subsequence clearly identified before substituting.

Techniques used

apply the sum to infinity formulaidentify the first term and common ratio of the even-numbered termssum a geometric progression to infinity

Q9MediumFunctions

The functions $f$ and $g$ are defined for all real values of $x$ by

f(x) = (3x - 2)^2 + k \quad \text{and} \quad g(x) = 5x - 1,

where $k$ is a constant.

(a)

Given that the range of the function $gf$ is $gf(x) \ge 39$ , find the value of $k$ .

4M Similar

DifficultyMedium

Worked solution

Approach

The composite $gf$ means apply $g$ after $f$ . Since the range of $gf$ is given as $gf(x) \ge 39$ , the minimum value of $gf(x)$ must be $39$ . The expression $(3x-2)^2$ is a square, so its least value is $0$ ; this gives the minimum of $gf$ .

Working

Given:

f(x) = (3x-2)^2 + k, \quad g(x) = 5x - 1

Form $gf(x)$ :

gf(x) = g(f(x)) = 5\left((3x-2)^2 + k\right) - 1

Simplify:

gf(x) = 5(3x-2)^2 + 5k - 1

Since $(3x-2)^2 \ge 0$ , the minimum value of $gf(x)$ is $5k - 1$ . The range is $gf(x) \ge 39$ , so:

5k - 1 = 39

Solve:

5k = 40

k = 8

Answer

k = 8

Final answer

k = 8

Detailed explanation

Walkthrough

First, interpret $gf(x)$ as $g(f(x))$ : replace every $x$ in $g(x)=5x-1$ with $f(x)$ . This gives $5((3x-2)^2+k)-1$ , which simplifies to $5(3x-2)^2+5k-1$ .

The key observation is that $(3x-2)^2$ is a perfect square, so it can never be negative. Its smallest possible value is $0$ , occurring when $3x-2=0$ , i.e. $x=\frac{2}{3}$ . Therefore the smallest value of $gf(x)$ is $5(0)+5k-1=5k-1$ .

The statement "the range of $gf$ is $gf(x)\ge 39$ " tells us that $39$ is exactly the minimum value. So set $5k-1=39$ and solve: $5k=40$ , giving $k=8$ .

Key Takeaways

Composite functions are evaluated from the inside out: $gf(x)=g(f(x))$ .
A squared expression $(ax+b)^2$ has minimum value $0$ .
The lower bound of a range is the minimum value of the function.

Common Mistakes

Using $fg(x)$ instead of $gf(x)$ ; the order matters.
Treating the range condition as $5k-1 \ge 39$ and concluding $k \ge 8$ ; since $39$ is the minimum, equality is required.
Algebraic errors when expanding $5((3x-2)^2+k)-1$ .

Things to Be Careful About

The mark scheme requires equality $5k-1=39$ , not an inequality, to find the exact value of $k$ .
Ensure the coefficient $5$ multiplies both the square and $k$ .

Techniques used

form composite function gfidentify minimum of squared expressionequate minimum to lower bound of rangesolve linear equation for k

(b)

For this value of $k$ , determine the range of the function $fg$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use $k=8$ found in part (a). Form the composite $fg(x)$ by applying $f$ after $g$ , then use the fact that a squared term is always non-negative to identify the minimum value and hence the range.

Working

Given $g(x)=5x-1$ and $f(x)=(3x-2)^2+8$ , form $fg(x)$ :

fg(x) = f(g(x)) = \left(3(5x-1)-2\right)^2 + 8

Simplify inside the square:

3(5x-1)-2 = 15x - 3 - 2 = 15x - 5

So:

fg(x) = (15x - 5)^2 + 8

Since $(15x-5)^2 \ge 0$ , the minimum value of $fg(x)$ is $8$ . Therefore:

fg(x) \ge 8

Answer

fg(x) \ge 8

Final answer

fg(x) >= 8

Detailed explanation

Walkthrough

With $k=8$ , substitute into $f$ : $f(x)=(3x-2)^2+8$ .

Now form $fg(x)=f(g(x))$ . Since $g(x)=5x-1$ , replace $x$ in $f$ by $5x-1$ :

fg(x) = (3(5x-1)-2)^2+8.

Simplify inside the bracket:

3(5x-1)-2 = 15x-5.

Thus $fg(x)=(15x-5)^2+8$ .

The square $(15x-5)^2$ is always at least $0$ , so $fg(x)$ is always at least $8$ . Since $15x-5$ can take every real value as $x$ varies, the square can actually be $0$ , so the range is $fg(x)\ge 8$ .

Key Takeaways

Composition order matters: $fg(x)=f(g(x))$ .
The range of a quadratic in completed-square form is determined by its minimum value.
A square term has minimum $0$ and can attain every non-negative value.

Common Mistakes

Forgetting to add the constant $k$ after squaring.
Writing $fg(x)\ge 0$ instead of including the $+8$ .
Using $>$ instead of $\ge$ ; the minimum is actually attained.

Things to Be Careful About

Use the value of $k$ from part (a), even if it was obtained via an inequality.
The answer must be $\ge 8$ , not $> 8$ , because the square can equal $0$ .

Techniques used

form composite function fgsubstitute known value of kidentify minimum of squared expressionstate range from minimum value

(c)

The function $h$ is defined for all real values of $x$ and is such that $gh(x) = 35x + 19$ .

Find an expression for $g^{-1}(x)$ and hence, or otherwise, find an expression for $h(x)$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Find the inverse of the linear function $g(x)=5x-1$ . Then apply $g^{-1}$ to both sides of $gh(x)=35x+19$ to isolate $h(x)$ .

Working

Let $y = g(x) = 5x - 1$ . Solve for $x$ :

y = 5x - 1

5x = y + 1

x = \frac{y+1}{5}

So:

g^{-1}(x) = \frac{x+1}{5}

Given:

gh(x) = 35x + 19

Apply $g^{-1}$ to both sides:

h(x) = g^{-1}(35x + 19) = \frac{(35x+19)+1}{5}

Simplify:

h(x) = \frac{35x+20}{5} = 7x + 4

Answer

g^{-1}(x) = \frac{x+1}{5}, \quad h(x) = 7x + 4

Final answer

g^-1(x) = (x+1)/5, h(x) = 7x + 4

Detailed explanation

Walkthrough

To find $g^{-1}(x)$ , write $y=5x-1$ and solve for $x$ in terms of $y$ :

y+1=5x \quad\Rightarrow\quad x=\frac{y+1}{5}.

Interchanging the variable names gives $g^{-1}(x)=\frac{x+1}{5}$ .

Now $gh(x)=g(h(x))=35x+19$ . Applying $g^{-1}$ to both sides undoes $g$ on the left:

h(x)=g^{-1}(35x+19).

Substitute $35x+19$ into the inverse:

h(x)=\frac{(35x+19)+1}{5}=\frac{35x+20}{5}=7x+4.

Key Takeaways

The inverse of a linear function $ax+b$ is found by solving $y=ax+b$ for $x$ .
If $gh(x)$ is known, applying $g^{-1}$ isolates $h(x)$ .
Composition and inverse are inverse operations: $g^{-1}(g(h(x)))=h(x)$ .

Common Mistakes

Writing $g^{-1}(x)=5x+1$ instead of $\frac{x+1}{5}$ .
Applying $g^{-1}$ incorrectly, e.g. substituting $x$ rather than $35x+19$ .
Forgetting to state $g^{-1}(x)$ before using it.

Things to Be Careful About

The inverse must be clearly indicated; the mark scheme awards a mark for $g^{-1}(x)=\frac15(x+1)$ .
When simplifying, combine $35x+19$ and $1$ before dividing by $5$ .
The final expression $7x+4$ must be identified as $h(x)$ .

Techniques used

find inverse of linear functionapply inverse to both sides of composite equationsimplify linear expression

Q10MediumCoordinate GeometryTrigonometryCircular Measure

Fig. 10.1

The diagram shows the circle with centre $C(-4, 5)$ and radius $\sqrt{20}$ units. The circle intersects the $y$ -axis at the points $A$ and $B$ . The size of angle $ACB$ is $\theta$ radians.

(a)

Find the equation of the tangent to the circle at the point $(-6, 9)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Find the gradient of the radius connecting the centre $C(-4, 5)$ to the point $(-6, 9)$ . The tangent is perpendicular to this radius, so its gradient is the negative reciprocal. Use the point-slope form to find the equation of the tangent line.

Working

The centre of the circle is $C(-4, 5)$ and the point on the circle is $P(-6, 9)$ . The gradient of the radius $CP$ is:

\begin{aligned} m_{\text{radius}} &= \frac{9 - 5}{-6 - (-4)} \\ &= \frac{4}{-2} \\ &= -2 \end{aligned}

Since the tangent is perpendicular to the radius at the point of contact, the gradient of the tangent $m_{\text{tangent}}$ satisfies $m_{\text{radius}} \times m_{\text{tangent}} = -1$ :

m_{\text{tangent}} = \frac{-1}{-2} = \frac{1}{2}

Using the point-slope form $y - y_1 = m(x - x_1)$ with the point $(-6, 9)$ and gradient $\frac{1}{2}$ :

y - 9 = \frac{1}{2}(x + 6)

Expanding and rearranging:

y - 9 = \frac{1}{2}x + 3

y = \frac{1}{2}x + 12

Answer

y = \frac{1}{2}x + 12

Final answer

y = 1/2 x + 12

Detailed explanation

Walkthrough

First, we identify the coordinates of the centre of the circle $C(-4, 5)$ and the point of tangency $P(-6, 9)$ . We calculate the gradient of the radius $CP$ using the standard gradient formula $\frac{y_2 - y_1}{x_2 - x_1}$ , which gives $-2$ . A key geometric property is that the tangent to a circle is always perpendicular to the radius at the point of contact. Therefore, we find the gradient of the tangent by taking the negative reciprocal of $-2$ , which is $\frac{1}{2}$ . Finally, we use the point-slope form of a straight line equation with the known point and the new gradient to derive the full equation $y = \frac{1}{2}x + 12$ .

Key Takeaways

The gradient of a line through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\frac{y_2 - y_1}{x_2 - x_1}$ .
Perpendicular lines have gradients $m_1$ and $m_2$ such that $m_1 m_2 = -1$ .
The point-slope form $y - y_1 = m(x - x_1)$ is useful for forming line equations when a point and gradient are known.

Common Mistakes

Forgetting the negative reciprocal rule and using the same gradient for the tangent.
Sign errors when calculating the gradient of the radius, especially with negative coordinates like $-6$ and $-4$ .
Algebraic errors when expanding and rearranging the point-slope equation.

Things to Be Careful About

Ensure the gradient calculation uses the correct order of subtraction in the numerator and denominator.
The final equation can be written in any valid form (e.g., $y = \frac{1}{2}x + 12$ , $x - 2y + 24 = 0$ ), but the mark scheme accepts $y = \frac{1}{2}x + 12$ .

Techniques used

calculate gradient of radiusapply perpendicular gradient propertyform equation of tangent using point-slope form

(b)

Find the equation of the circle in the form $x^2 + y^2 + ax + by + c = 0$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the standard form of the equation of a circle $(x - g)^2 + (y - f)^2 = r^2$ with the given centre $(-4, 5)$ and radius $\sqrt{20}$ . Expand the brackets and rearrange all terms to one side to obtain the required general form $x^2 + y^2 + ax + by + c = 0$ .

Working

The centre is $(g, f) = (-4, 5)$ and the radius is $r = \sqrt{20}$ . Substituting into the standard form:

(x - (-4))^2 + (y - 5)^2 = (\sqrt{20})^2

(x + 4)^2 + (y - 5)^2 = 20

Expand the squared binomials:

(x^2 + 8x + 16) + (y^2 - 10y + 25) = 20

Combine like terms:

x^2 + y^2 + 8x - 10y + 41 = 20

Subtract 20 from both sides to set the equation to zero:

x^2 + y^2 + 8x - 10y + 21 = 0

Answer

x^2 + y^2 + 8x - 10y + 21 = 0

Final answer

x^2 + y^2 + 8x - 10y + 21 = 0

Detailed explanation

Walkthrough

The standard equation of a circle with centre $(g, f)$ and radius $r$ is $(x - g)^2 + (y - f)^2 = r^2$ . Substituting $g = -4$ , $f = 5$ , and $r^2 = 20$ gives $(x + 4)^2 + (y - 5)^2 = 20$ . To convert this to the general form $x^2 + y^2 + ax + by + c = 0$ , we expand $(x + 4)^2$ to $x^2 + 8x + 16$ and $(y - 5)^2$ to $y^2 - 10y + 25$ . Adding these and the constant $20$ on the right side gives $x^2 + y^2 + 8x - 10y + 41 = 20$ . Subtracting $20$ from both sides yields the final general form $x^2 + y^2 + 8x - 10y + 21 = 0$ .

Key Takeaways

The standard form of a circle equation is $(x - g)^2 + (y - f)^2 = r^2$ .
Converting to general form requires careful expansion of binomial squares and collecting all terms on one side.

Common Mistakes

Sign errors when expanding $(x + 4)^2$ or $(y - 5)^2$ .
Forgetting to subtract the radius squared from the constant terms when rearranging to equal zero.

Things to Be Careful About

Ensure the radius is squared correctly: $(\sqrt{20})^2 = 20$ , not $20^2$ .
The required form is $x^2 + y^2 + ax + by + c = 0$ , so all terms must be on the left side and the right side must be exactly $0$ .

Techniques used

write standard form equation of circleexpand bracketsrearrange to general form

(c)

Find the value of $\theta$ correct to 4 significant figures.

3M Similar

DifficultyMedium

Worked solution

Approach

To find the angle $\theta$ subtended by the chord $AB$ at the centre, first determine the coordinates of $A$ and $B$ by finding where the circle intersects the $y$ -axis (set $x = 0$ ). Then, use either the right-angled triangle formed by the centre, the midpoint of $AB$ , and one of the points, or the cosine rule in $\triangle ACB$ , to find $\theta$ .

Working

Step 1: Find the coordinates of $A$ and $B$

The circle intersects the $y$ -axis where $x = 0$ . Substitute $x = 0$ into the equation from part (b):

0^2 + y^2 + 8(0) - 10y + 21 = 0

y^2 - 10y + 21 = 0

Factorise the quadratic:

(y - 3)(y - 7) = 0

So $y = 3$ or $y = 7$ . The points are $A(0, 7)$ and $B(0, 3)$ . The length of the chord $AB$ is $7 - 3 = 4$ .

Step 2: Find $\theta$ using the cosine rule

In $\triangle ACB$ , the sides are $AC = \sqrt{20}$ , $BC = \sqrt{20}$ , and $AB = 4$ . Apply the cosine rule:

AB^2 = AC^2 + BC^2 - 2(AC)(BC)\cos\theta

4^2 = 20 + 20 - 2(\sqrt{20})(\sqrt{20})\cos\theta

16 = 40 - 40\cos\theta

40\cos\theta = 24

\cos\theta = \frac{24}{40} = 0.6

Take the inverse cosine to find $\theta$ :

\theta = \cos^{-1}(0.6) \approx 0.927295\ldots

Rounding to 4 significant figures:

\theta = 0.9273

(Alternative method using right-angled triangle: Let $M$ be the midpoint of $AB$ , so $M(0, 5)$ . The distance from $C(-4, 5)$ to $M(0, 5)$ is $4$ . In right-angled $\triangle CMA$ , $\tan(\theta/2) = \frac{AM}{CM} = \frac{2}{4} = 0.5$ . Thus $\theta = 2\tan^{-1}(0.5) = 0.9273$ .)

Answer

\theta = 0.9273

Final answer

0.9273

Detailed explanation

Walkthrough

First, we find where the circle crosses the $y$ -axis by substituting $x = 0$ into the circle's equation. This gives the quadratic $y^2 - 10y + 21 = 0$ , which factorises to $(y - 3)(y - 7) = 0$ , yielding points $A(0, 7)$ and $B(0, 3)$ . The chord length $AB$ is therefore $4$ . To find the angle $\theta$ at the centre, we use the cosine rule in $\triangle ACB$ with sides $\sqrt{20}$ , $\sqrt{20}$ , and $4$ . This gives $\cos\theta = 0.6$ , and taking $\cos^{-1}(0.6)$ yields $\theta \approx 0.9273$ radians. Alternatively, dropping a perpendicular from $C$ to $AB$ bisects the angle, giving $\tan(\theta/2) = 0.5$ , which leads to the same result.

Key Takeaways

Setting $x = 0$ in the circle equation finds the $y$ -intercepts.
The cosine rule is a reliable method for finding angles in triangles when all three side lengths are known.
Inverse trigonometric functions must be used in radian mode to match the question's requirement.

Common Mistakes

Forgetting to square the radius when using the cosine rule ( $AC^2 = 20$ , not $\sqrt{20}$ ).
Using degree mode on the calculator instead of radian mode, giving $\theta \approx 53.1^\circ$ .
Only finding $\theta/2$ and forgetting to multiply by $2$ if using the right-angled triangle method.

Things to Be Careful About

The question asks for $\theta$ correct to 4 significant figures, so $0.9273$ is correct (not $0.927$ or $0.93$ ).
Ensure the calculator is in radian mode, as the angle is specified in radians in the diagram.

Techniques used

find y-intercepts of circleuse right-angled triangle or cosine ruleevaluate inverse trigonometric function

(d)

Find the perimeter and area of the segment shaded in the diagram.

4M Similar

DifficultyMedium

Worked solution

Approach

The shaded segment is bounded by the chord $AB$ and the arc $AB$ . Its perimeter is the sum of the arc length and the chord length. Its area is the area of the sector $ACB$ minus the area of the triangle $ACB$ . Use the formulas $s = r\theta$ for arc length and $A = \frac{1}{2}r^2(\theta - \sin\theta)$ for segment area.

Working

Perimeter of the segment

The arc length $AB$ is given by $s = r\theta$ :

s = \sqrt{20} \times 0.927295\ldots \approx 4.1504

The chord length $AB$ is $4$ (from part c).

The perimeter is:

\text{Perimeter} = s + AB = 4.1504 + 4 = 8.1504

Rounding to 3 significant figures:

\text{Perimeter} \approx 8.15

Area of the segment

The area of the sector $ACB$ is:

A_{\text{sector}} = \frac{1}{2}r^2\theta = \frac{1}{2}(20)(0.927295\ldots) = 9.27295\ldots

The area of the triangle $ACB$ is:

A_{\text{triangle}} = \frac{1}{2}r^2\sin\theta = \frac{1}{2}(20)\sin(0.927295\ldots)

Since $\cos\theta = 0.6$ , we have $\sin\theta = \sqrt{1 - 0.6^2} = 0.8$ :

A_{\text{triangle}} = 10 \times 0.8 = 8.000

The area of the segment is:

A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = 9.27295\ldots - 8.000 = 1.27295\ldots

Rounding to 3 significant figures:

A_{\text{segment}} \approx 1.27

Answer

Perimeter $= 8.15$ , Area $= 1.27$

Final answer

Perimeter = 8.15, Area = 1.27

Detailed explanation

Walkthrough

The perimeter of the shaded segment consists of the curved arc $AB$ and the straight chord $AB$ . We calculate the arc length using $s = r\theta = \sqrt{20} \times 0.9273 \approx 4.150$ . Adding the chord length $AB = 4$ gives a perimeter of $8.15$ . For the area, we use the formula for the area of a circular segment: $A = \frac{1}{2}r^2(\theta - \sin\theta)$ . Substituting $r^2 = 20$ , $\theta = 0.9273$ , and $\sin\theta = 0.8$ (derived from $\cos\theta = 0.6$ ), we get $A = \frac{1}{2}(20)(0.9273 - 0.8) = 10(0.1273) = 1.273 \approx 1.27$ .

Key Takeaways

The perimeter of a circular segment is the arc length plus the chord length.
The area of a circular segment can be found by subtracting the triangle area from the sector area: $\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta = \frac{1}{2}r^2(\theta - \sin\theta)$ .
Exact trigonometric values (like $\sin\theta = 0.8$ when $\cos\theta = 0.6$ ) can simplify calculations and reduce rounding errors.

Common Mistakes

Using $\theta/2$ instead of $\theta$ in the arc length formula.
Forgetting to add the chord length when calculating the perimeter of the segment.
Using the wrong area formula, such as calculating only the sector area or only the triangle area.
Rounding $\theta$ too early in the calculation, which can propagate errors into the final area.

Things to Be Careful About

The mark scheme allows missing or incorrect units for perimeter and area, but it is best practice to include them (e.g., units $^2$ ).
Use the unrounded value of $\theta$ (or the exact value from $\cos^{-1}(0.6)$ ) in intermediate steps to maintain accuracy.
Ensure $\sin\theta$ is calculated correctly; since $\theta$ is in the first quadrant, $\sin\theta$ is positive.

Techniques used

calculate arc length using r*thetasum arc length and chord length for perimeteruse sector minus triangle formula for area

Q11Medium-HardDifferentiationQuadraticsIntegration

Fig. 11.1

The diagram shows the curve with equation $y = 2x^{-\frac{2}{3}} - 3x^{-\frac{1}{3}} + 1$ for $x > 0$ . The curve crosses the $x$ -axis at points $A$ and $B$ and has a minimum point $M$ .

(a)

Find the exact coordinates of $M$ .

4M Similar

DifficultyMedium

Worked solution

Approach

To find the minimum point $M$ , we differentiate $y$ with respect to $x$ , set $\frac{dy}{dx} = 0$ , and solve for $x$ . Then we substitute this $x$ -value back into the original equation to find the $y$ -coordinate.

Working

Differentiate $y = 2x^{-\frac{2}{3}} - 3x^{-\frac{1}{3}} + 1$ :

\begin{aligned} \frac{dy}{dx} &= 2\left(-\frac{2}{3}\right)x^{-\frac{5}{3}} - 3\left(-\frac{1}{3}\right)x^{-\frac{4}{3}} \\ &= -\frac{4}{3}x^{-\frac{5}{3}} + x^{-\frac{4}{3}} \end{aligned}

Set $\frac{dy}{dx} = 0$ to find stationary points:

-\frac{4}{3}x^{-\frac{5}{3}} + x^{-\frac{4}{3}} = 0

Factor out $x^{-\frac{5}{3}}$ (noting $x > 0$ ):

x^{-\frac{5}{3}}\left(-\frac{4}{3} + x^{\frac{1}{3}}\right) = 0

Since $x > 0$ , $x^{-\frac{5}{3}} \neq 0$ , so:

x^{\frac{1}{3}} = \frac{4}{3}

Cube both sides to find $x$ :

x = \left(\frac{4}{3}\right)^3 = \frac{64}{27}

Substitute $x = \frac{64}{27}$ into the original equation to find $y$ . Note that $x^{-\frac{1}{3}} = \frac{3}{4}$ and $x^{-\frac{2}{3}} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$ :

\begin{aligned} y &= 2\left(\frac{9}{16}\right) - 3\left(\frac{3}{4}\right) + 1 \\ &= \frac{9}{8} - \frac{9}{4} + 1 \\ &= \frac{9}{8} - \frac{18}{8} + \frac{8}{8} \\ &= -\frac{1}{8} \end{aligned}

The coordinates of $M$ are $\left(\frac{64}{27}, -\frac{1}{8}\right)$ .

Answer

M = \left(\frac{64}{27}, -\frac{1}{8}\right)

Final answer

(64/27, -1/8)

Detailed explanation

Walkthrough

First, we differentiate the function $y = 2x^{-2/3} - 3x^{-1/3} + 1$ using the power rule. Remember that the derivative of $x^n$ is $nx^{n-1}$ . This gives $dy/dx = -4/3 x^{-5/3} + x^{-4/3}$ . To find the stationary point, we set the derivative to zero. Factoring out the term with the most negative index, $x^{-5/3}$ , leaves us with $-4/3 + x^{1/3} = 0$ . Solving this gives $x^{1/3} = 4/3$ , and cubing both sides yields $x = 64/27$ . Finally, we substitute $x = 64/27$ back into the original equation. Using the fact that $x^{-1/3} = 3/4$ makes the arithmetic much simpler: $y = 2(3/4)^2 - 3(3/4) + 1 = 9/8 - 9/4 + 1 = -1/8$ . Thus, the minimum point $M$ is at $(64/27, -1/8)$ .

Key Takeaways

Differentiating expressions with negative fractional indices requires careful application of the power rule.
Factoring out the term with the most negative index is a reliable way to solve equations involving negative indices.
Substituting back using the simpler form $x^{-1/3}$ avoids messy arithmetic with large fractions.

Common Mistakes

Forgetting to subtract 1 from the index when differentiating (e.g., writing $x^{-2/3}$ instead of $x^{-5/3}$ ).
Incorrectly factoring out negative indices, leading to sign errors.
Making arithmetic errors when substituting $x = 64/27$ back into the original equation.

Things to Be Careful About

The domain is $x > 0$ , so we must ignore any solution $x = 0$ that might appear if we multiply through by positive powers of $x$ .
Ensure you cube both sides correctly: $(4/3)^3 = 64/27$ , not $4/9$ or similar.
The question asks for exact coordinates, so leave fractions as they are; do not convert to decimals unless specified.

Techniques used

differentiate using the power rulefactor out common negative index termsolve for stationary point x-coordinatesubstitute back to find y-coordinate

(b)

Find the area of the region bounded by the curve and the line segment $AB$ .

7M Similar

DifficultyMedium-Hard

Worked solution

Approach

The region is bounded by the curve and the line segment $AB$ on the x-axis. First, find the x-coordinates of $A$ and $B$ by setting $y = 0$ . Recognize the equation as a quadratic in $x^{-1/3}$ to solve for $x$ . Then, integrate the function between these limits. Since the curve is below the x-axis between $A$ and $B$ , the definite integral will be negative; take the absolute value to find the area.

Working

Find the x-intercepts $A$ and $B$ :

Set $y = 0$ :

2x^{-\frac{2}{3}} - 3x^{-\frac{1}{3}} + 1 = 0

Let $a = x^{-\frac{1}{3}}$ . The equation becomes:

2a^2 - 3a + 1 = 0

Factorize:

(2a - 1)(a - 1) = 0

So $a = \frac{1}{2}$ or $a = 1$ .

Substitute back $a = x^{-\frac{1}{3}}$ :

x^{-\frac{1}{3}} = \frac{1}{2} \implies x^{\frac{1}{3}} = 2 \implies x = 8

x^{-\frac{1}{3}} = 1 \implies x = 1

The curve crosses the x-axis at $x = 1$ and $x = 8$ . From the diagram, $A$ is $(1, 0)$ and $B$ is $(8, 0)$ .

Integrate to find the area:

The area is given by $\left|\int_1^8 y \, dx\right|$ . First, find the indefinite integral:

\int \left(2x^{-\frac{2}{3}} - 3x^{-\frac{1}{3}} + 1\right) dx

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

= 6x^{\frac{1}{3}} - \frac{9}{2}x^{\frac{2}{3}} + x

Evaluate from $x = 1$ to $x = 8$ :

At $x = 8$ :

6(8)^{\frac{1}{3}} - \frac{9}{2}(8)^{\frac{2}{3}} + 8 = 6(2) - \frac{9}{2}(4) + 8 = 12 - 18 + 8 = 2

At $x = 1$ :

6(1)^{\frac{1}{3}} - \frac{9}{2}(1)^{\frac{2}{3}} + 1 = 6 - \frac{9}{2} + 1 = 7 - 4.5 = 2.5 = \frac{5}{2}

Definite integral:

\int_1^8 y \, dx = 2 - \frac{5}{2} = -\frac{1}{2} = -0.5

Since the area is below the x-axis, the area is the absolute value:

\text{Area} = \left|-\frac{1}{2}\right| = \frac{1}{2} = 0.5

Answer

\text{Area} = 0.5

Final answer

0.5

Detailed explanation

Walkthrough

To find the area bounded by the curve and the line segment $AB$ , we first need the x-coordinates of $A$ and $B$ , which are the x-intercepts. Setting $y = 0$ gives $2x^{-2/3} - 3x^{-1/3} + 1 = 0$ . By substituting $a = x^{-1/3}$ , this becomes a standard quadratic $2a^2 - 3a + 1 = 0$ , which factors to $(2a-1)(a-1) = 0$ . Solving gives $a = 1/2$ and $a = 1$ , meaning $x^{-1/3} = 1/2 \implies x = 8$ and $x^{-1/3} = 1 \implies x = 1$ . So the limits of integration are 1 and 8.

Next, we integrate the function $y = 2x^{-2/3} - 3x^{-1/3} + 1$ . Using the power rule for integration, $\int x^n dx = x^{n+1}/(n+1)$ , we get $6x^{1/3} - 9/2 x^{2/3} + x$ . Evaluating this at the upper limit $x=8$ gives $6(2) - 4.5(4) + 8 = 2$ . Evaluating at the lower limit $x=1$ gives $6 - 4.5 + 1 = 2.5$ . The definite integral is $2 - 2.5 = -0.5$ . Since the region is below the x-axis (as seen in the diagram and confirmed by the negative value), the area is the absolute value, $0.5$ .

Key Takeaways

Equations with fractional indices can often be solved by substituting a variable to form a quadratic equation.
When finding the area between a curve and the x-axis, always check if the curve is above or below the axis. If it is below, the definite integral will be negative, and you must take the absolute value.
Integration with negative indices follows the same power rule as positive indices, but be careful with the signs and fractions in the denominator.

Common Mistakes

Forgetting to take the absolute value of the definite integral when the region is below the x-axis, resulting in a negative area.
Making errors when integrating negative fractional indices, such as adding 1 to the index incorrectly (e.g., $-2/3 + 1 = 1/3$ , not $-1/3$ ).
Incorrectly solving the quadratic in $x^{-1/3}$ , leading to wrong limits of integration.

Things to Be Careful About

The limits of integration must be in the correct order: lower limit 1, upper limit 8. Reversing them will give the wrong sign.
Ensure you use 'your' limits from your working in the evaluation step, as marked in the scheme.
The question asks for the area, which is always positive. A negative answer indicates the region is below the axis, but the final answer must be positive.

Techniques used

recognize quadratic form to find x-interceptssolve for intercept coordinatesintegrate using power rule for negative indicesevaluate definite integral and take absolute value for area

Mathematics 9709/12 — February/March 2024

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