The equation of a curve is $2y^2 + 3xy + x = x^2$. (a) Show that $$ \frac{dy}{dx} = \frac{2x - 3y - 1}{4y + 3x}. $$ (b) Hence show that the curve does not have a tangent that is parallel to the $x$-axis.

(a) dy/dx = (2x - 3y - 1)/(4y + 3x) · (b) No tangent parallel to the x-axis exists.

Let $f(x) = \frac{36a^2}{(2a + x)(2a - x)(5a - 2x)}$, where $a$ is a positive constant. (a) Express $f(x)$ in partial fractions. (b) Hence find the exact value of $\int_{-a}^{a} f(x)\,dx$, giving your answer in the form $p\ln q + r\ln s$ where $p$ and $r$ are integers and $q$ and $s$ are prime numbers.

(a) 1/(2a+x) + 9/(2a-x) - 16/(5a-2x) · (b) 18 ln 3 - 8 ln 7

9709/32 Feb/Mar 2024 — Mathematics Solved Past Paper

Q13MMedium-EasyAlgebra

Find the quotient and remainder when $x^4 - 3x^3 + 9x^2 - 12x + 27$ is divided by $x^2 + 5$ .

DifficultyMedium-Easy

Worked solution

Approach

Perform polynomial long division of $x^4 - 3x^3 + 9x^2 - 12x + 27$ by $x^2 + 5$ to find the quotient and remainder.

Working

Divide the leading term of the dividend by the leading term of the divisor:

\frac{x^4}{x^2} = x^2

Multiply the divisor by $x^2$ and subtract from the dividend:

\begin{aligned} (x^4 - 3x^3 + 9x^2 - 12x + 27) - x^2(x^2 + 5) &= x^4 - 3x^3 + 9x^2 - 12x + 27 - x^4 - 5x^2 \\ &= -3x^3 + 4x^2 - 12x + 27 \end{aligned}

Divide the leading term of the new expression by the leading term of the divisor:

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

Multiply the divisor by $-3x$ and subtract:

\begin{aligned} (-3x^3 + 4x^2 - 12x + 27) - (-3x)(x^2 + 5) &= -3x^3 + 4x^2 - 12x + 27 + 3x^3 + 15x \\ &= 4x^2 + 3x + 27 \end{aligned}

Divide the leading term by the leading term of the divisor:

\frac{4x^2}{x^2} = 4

Multiply the divisor by $4$ and subtract:

\begin{aligned} (4x^2 + 3x + 27) - 4(x^2 + 5) &= 4x^2 + 3x + 27 - 4x^2 - 20 \\ &= 3x + 7 \end{aligned}

The degree of the remainder $3x + 7$ is less than the degree of the divisor $x^2 + 5$ , so the division is complete.

Answer

\text{Quotient} = x^2 - 3x + 4, \quad \text{Remainder} = 3x + 7

Final answer

Quotient: x^2 - 3x + 4, Remainder: 3x + 7

Detailed explanation

Walkthrough

We are asked to divide the polynomial $x^4 - 3x^3 + 9x^2 - 12x + 27$ by $x^2 + 5$ . Since the divisor is a quadratic, we use polynomial long division.

First, we divide the highest degree term of the dividend, $x^4$ , by the highest degree term of the divisor, $x^2$ , to get the first term of the quotient: $x^2$ . We then multiply the entire divisor by $x^2$ to get $x^4 + 5x^2$ , and subtract this from the dividend. This gives $-3x^3 + 4x^2 - 12x + 27$ .

Next, we take the new leading term $-3x^3$ and divide it by $x^2$ to get $-3x$ . Multiplying the divisor by $-3x$ gives $-3x^3 - 15x$ . Subtracting this from the current expression leaves $4x^2 + 3x + 27$ .

Finally, we divide $4x^2$ by $x^2$ to get $4$ . Multiplying the divisor by $4$ gives $4x^2 + 20$ . Subtracting this from the current expression leaves $3x + 7$ .

Since the degree of the remainder ( $1$ ) is less than the degree of the divisor ( $2$ ), we stop. The quotient is $x^2 - 3x + 4$ and the remainder is $3x + 7$ .

Key Takeaways

Polynomial long division works similarly to numerical long division, matching terms by descending degree.
The remainder must always have a degree strictly less than the divisor.
Missing terms in the dividend (like an $x^3$ term if it were absent) should be represented with a coefficient of $0$ to keep columns aligned.

Common Mistakes

Forgetting to subtract the entire product, leading to sign errors (e.g., subtracting $-15x$ as $-15x$ instead of $+15x$ ).
Misaligning terms of different degrees during subtraction.
Stopping the division too early or continuing until the remainder is $0$ when it shouldn't be.

Things to Be Careful About

Ensure all terms of the dividend are included, even if their coefficient is zero (e.g., if dividing by $x^2+5$ , the $x$ term in the dividend must be accounted for).
The final answer should clearly state both the quotient and the remainder, as the question asks for both.

Techniques used

perform polynomial long divisionidentify quotient and remainder

Approach

Rewrite $\sqrt{4-x}$ as $2\left(1-\frac{x}{4}\right)^{1/2}$ and expand using the binomial series for $(1+u)^n$ with $n=\frac{1}{2}$ and $u=-\frac{x}{4}$ , up to the term in $u^2$ . Then multiply by $(2x-5)$ and collect the coefficient of $x^2$ .

Working

First expand $(4-x)^{1/2}$ :

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

Using $(1+u)^{1/2}=1+\frac{1}{2}u+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}u^2+\cdots$ with $u=-\frac{x}{4}$ :

\begin{aligned} \left(1-\frac{x}{4}\right)^{1/2} &=1+\frac{1}{2}\left(-\frac{x}{4}\right)+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}\left(-\frac{x}{4}\right)^2+\cdots \\ &=1-\frac{x}{8}-\frac{x^2}{128}+\cdots \end{aligned}

Therefore

(4-x)^{1/2}=2\left(1-\frac{x}{8}-\frac{x^2}{128}+\cdots\right) =2-\frac{x}{4}-\frac{x^2}{64}+\cdots

Now multiply by $(2x-5)$ :

(2x-5)(4-x)^{1/2} =(2x-5)\left(2-\frac{x}{4}-\frac{x^2}{64}+\cdots\right)

The terms in $x^2$ are

2x\left(-\frac{x}{4}\right)+(-5)\left(-\frac{x^2}{64}\right) =-\frac{1}{2}x^2+\frac{5}{64}x^2 =-\frac{27}{64}x^2

Hence the coefficient of $x^2$ is $-\frac{27}{64}$ .

Answer

-\frac{27}{64}

Final answer

-27/64

Detailed explanation

Walkthrough

Start by rewriting the expression so that the binomial expansion can be applied. Since $\sqrt{4-x}=(4-x)^{1/2}=4^{1/2}\left(1-\frac{x}{4}\right)^{1/2}$ , we factor out $4^{1/2}=2$ to get the standard form $2(1+u)^{1/2}$ with $u=-\frac{x}{4}$ . This is necessary because the binomial series for $(1+u)^n$ is only directly usable when the first term is $1$ .

Next expand $(1+u)^{1/2}$ up to the $u^2$ term using

(1+u)^{1/2}=1+\frac{1}{2}u+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}u^2+\cdots

Substituting $u=-\frac{x}{4}$ gives the linear term $-\frac{x}{8}$ and the quadratic term $-\frac{x^2}{128}$ . Multiplying by the factor $2$ from $4^{1/2}$ gives $(4-x)^{1/2}=2-\frac{x}{4}-\frac{x^2}{64}+\cdots$ .

Then multiply this expansion by $(2x-5)$ . The only ways to obtain an $x^2$ term are $2x$ times the $x$ -term of the expansion and $-5$ times the $x^2$ -term of the expansion. Collecting these gives $-\frac{1}{2}x^2+\frac{5}{64}x^2=-\frac{27}{64}x^2$ , so the coefficient is $-\frac{27}{64}$ .

Key Takeaways

To expand $\sqrt{4-x}$ , rewrite it as $2\left(1-\frac{x}{4}\right)^{1/2}$ .
The binomial expansion for rational $n$ requires the bracket to be of the form $(1+u)^n$ .
When multiplying expansions, identify all contributions to the required power.
The coefficient is the number multiplying $x^2$ in the final expansion.

Common Mistakes

Forgetting to multiply by $4^{1/2}=2$ after rewriting the bracket.
Using $(1+u)^{1/2}$ but substituting $u=-\frac{x}{4}$ incorrectly, especially sign errors in the $x^2$ term.
Only using one of the two contributions to $x^2$ when multiplying by $(2x-5)$ .
Giving the $x^2$ term of $(4-x)^{1/2}$ instead of the coefficient in the full product.

Things to Be Careful About

The binomial coefficient for $u^2$ is $\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}$ , not $\frac{1}{2}\left(-\frac{1}{2}\right)$ alone.
The expansion is only valid for $|x|<4$ , so any terms beyond $x^2$ should not be used for the coefficient.
The mark scheme accepts unsimplified forms such as $2\times\left(-\frac{1}{4}\right)+(-5)\times\left(-\frac{1}{64}\right)$ ; showing the unsimplified terms can earn method marks.

Techniques used

rewrite the expression in the form (1+u)^napply the binomial expansion for rational nmultiply by a linear factor and collect coefficientsidentify the coefficient of a specified power

(b)

State the set of values of $x$ for which the expansion in part (a) is valid.

1M Similar

Q3MediumComplex Numbers

It is given that $z = -\sqrt{3} + i$ .

(a)

Express $z^2$ in the form $re^{i\theta}$ , where $r > 0$ and $-\pi < \theta \leq \pi$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Square $z$ in Cartesian form, then convert the result to polar (exponential) form by finding its modulus and argument, ensuring the argument lies in $-\pi < \theta \leq \pi$ .

Working

\begin{aligned} z^2 &= (-\sqrt{3} + i)^2 \\ &= (-\sqrt{3})^2 + 2(-\sqrt{3})(i) + i^2 \\ &= 3 - 2\sqrt{3}\,i - 1 \\ &= 2 - 2\sqrt{3}\,i \end{aligned}

Modulus:

r = |z^2| = \sqrt{2^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4

Argument: since the real part $2 > 0$ and the imaginary part $-2\sqrt{3} < 0$ , the point lies in the fourth quadrant.

\tan\theta = \frac{-2\sqrt{3}}{2} = -\sqrt{3}

\theta = -\frac{\pi}{3}

This already satisfies $-\pi < \theta \leq \pi$ .

Answer

z^2 = 4e^{-\frac{\pi}{3}i}

Final answer

z^2 = 4e^{-iπ/3}

Detailed explanation

Walkthrough

We start by squaring $z$ directly in Cartesian form. Using the binomial expansion, $(-\sqrt{3} + i)^2$ gives three terms: $(-\sqrt{3})^2 = 3$ , twice the product $2(-\sqrt{3})(i) = -2\sqrt{3}i$ , and $i^2 = -1$ . Combining these yields $z^2 = 2 - 2\sqrt{3}i$ .

To write this in the form $re^{i\theta}$ , we need the modulus $r$ and the argument $\theta$ . The modulus is found from the Pythagorean formula: the square root of the sum of the squares of the real and imaginary parts. With real part $2$ and imaginary part $-2\sqrt{3}$ , we have $r = \sqrt{4 + 12} = 4$ .

For the argument, the reference angle has $\tan = |{-2\sqrt{3}}|/|2| = \sqrt{3}$ , giving a reference angle of $\pi/3$ . Because the real part is positive and the imaginary part is negative, the angle lies in the fourth quadrant, so $\theta = -\pi/3$ . This value is already inside the required range $-\pi < \theta \leq \pi$ , so no adjustment is needed.

Key Takeaways

Squaring a complex number in Cartesian form is straightforward binomial expansion.
The modulus is always non-negative and computed from the real and imaginary parts.
The argument must be assigned to the correct quadrant based on the signs of both parts.
The principal range $-\pi < \theta \leq \pi$ excludes $\pi$ but includes $-\pi$ (equivalent to $\pi$ ).

Common Mistakes

Forgetting that $i^2 = -1$ when squaring, leading to $z^2 = 3 + 2\sqrt{3}i + 1 = 4 + 2\sqrt{3}i$ — a wrong answer.
Taking $\tan^{-1}(-\sqrt{3})$ and reporting $-\pi/3$ without checking the quadrant (here it happens to be correct, but in general this can fail).
Reporting the argument as $\pi/3$ instead of $-\pi/3$ because of sign carelessness.

Things to Be Careful About

Always state the quadrant explicitly when determining an argument from a $\tan$ ratio.
Do not give a decimal approximation for the argument — keep it in terms of $\pi$ .

Techniques used

compute square of complex number in Cartesian formfind modulus via Pythagorean rulefind argument using inverse tangent and quadrant checkadjust argument into required principal range

(b)

The complex number $\omega$ is such that $z^2\omega$ is real and $\left| \frac{z^2}{\omega} \right| = 12$ .

Find the two possible values of $\omega$ , giving your answers in the form $Re^{i\alpha}$ , where $R > 0$ and $-\pi < \alpha \leq \pi$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Write $\omega = Re^{i\alpha}$ and use the polar-form rules: $\arg(z^2\omega) = \arg z^2 + \alpha$ and $|z^2/\omega| = |z^2|/R$ . The condition that $z^2\omega$ is real forces the sum of arguments to be $0$ or $\pm\pi$ , while the modulus equation fixes $R$ .

Working

From part (a), $z^2 = 4e^{-\frac{\pi}{3}i}$ , so $|z^2| = 4$ and $\arg z^2 = -\frac{\pi}{3}$ .

Let $\omega = Re^{i\alpha}$ .

Condition 1 — modulus:

\left|\frac{z^2}{\omega}\right| = \frac{|z^2|}{R} = \frac{4}{R} = 12

R = \frac{4}{12} = \frac{1}{3}

Condition 2 — argument (product is real):

\arg(z^2\omega) = \arg z^2 + \alpha = -\frac{\pi}{3} + \alpha \in \{0, \pi, -\pi\}

If $-\frac{\pi}{3} + \alpha = 0$ , then $\alpha = \frac{\pi}{3}$ (valid, in range).
If $-\frac{\pi}{3} + \alpha = \pi$ , then $\alpha = \frac{4\pi}{3}$ (not in range; equivalent to $-\frac{2\pi}{3}$ ).
If $-\frac{\pi}{3} + \alpha = -\pi$ , then $\alpha = -\frac{2\pi}{3}$ (valid, in range).

So the two admissible arguments are $\alpha = \frac{\pi}{3}$ and $\alpha = -\frac{2\pi}{3}$ .

Answer

\omega = \frac{1}{3}e^{\frac{\pi}{3}i} \quad \text{or} \quad \omega = \frac{1}{3}e^{-\frac{2\pi}{3}i}

Final answer

ω = (1/3)e^{iπ/3} or ω = (1/3)e^{-i2π/3}

Detailed explanation

Walkthrough

We are told $z^2\omega$ is real and $|z^2/\omega| = 12$ . Writing $\omega$ in polar form, $\omega = Re^{i\alpha}$ , lets us translate both conditions into simple polar equations.

For the modulus, the rule is $|A/B| = |A|/|B|$ . We know $|z^2| = 4$ from part (a), so $4/R = 12$ , giving $R = 1/3$ . This alone fixes the magnitude of $\omega$ .

For the argument, the rule is $\arg(AB) = \arg A + \arg B$ . The product $z^2\omega$ being real means its argument is $0$ , $\pi$ , or $-\pi$ (all three describe the positive or negative real axis). So we need

-\frac{\pi}{3} + \alpha \in \{0, \pi, -\pi\}.

This gives three candidate values of $\alpha$ : $\pi/3$ , $4\pi/3$ , and $-2\pi/3$ . Because $4\pi/3$ lies outside the range $-\pi < \alpha \leq \pi$ , only two values are valid: $\alpha = \pi/3$ and $\alpha = -2\pi/3$ .

Combining with $R = 1/3$ yields the two possible values of $\omega$ .

Key Takeaways

A complex number is real iff its argument is $0$ , $\pi$ , or $-\pi$ (equivalently, $k\pi$ for integer $k$ ).
Modulus is multiplicative: $|AB| = |A|\,|B|$ and $|A/B| = |A|/|B|$ .
Argument is additive: $\arg(AB) = \arg A + \arg B$ (modulo $2\pi$ ).
After solving, always check the final arguments lie in the requested principal range.

Common Mistakes

Forgetting to include both $\alpha = 0$ and $\alpha = \pm\pi$ as conditions for a real product, leading to only one value of $\omega$ .
Stating the result with $\alpha = 4\pi/3$ without reducing to its principal-range equivalent $-2\pi/3$ .
Confusing the modulus equation: writing $R = 4 \times 12$ instead of $R = 4/12$ .
Using degrees instead of radians, or mixing the two.

Things to Be Careful About

The argument equation $-\pi/3 + \alpha = 0, \pm\pi$ has three candidate solutions, but only those inside the principal range $-\pi < \alpha \leq \pi$ are accepted.
The question explicitly asks for the answer in the form $Re^{i\alpha}$ with $R > 0$ , so do not give answers in Cartesian form.

Techniques used

use argument-addition rule for a product to be realuse modulus rule for a quotientset up two simultaneous conditions on argument and modulus of ωsolve for the modulus and the two admissible arguments of ω

Q44MMediumLogarithmic and Exponential Functions

The positive numbers $p$ and $q$ are such that

\ln\left(\frac{p}{q}\right) = a \quad \text{and} \quad \ln(q^2p) = b.

Express $\ln(p^7q)$ in terms of $a$ and $b$ .

DifficultyMedium

Worked solution

Approach

Apply the laws of logarithms to rewrite the two given equations as linear equations in $\ln p$ and $\ln q$ . Solve these simultaneous equations, then use the power law for logarithms to form $\ln(p^7 q)$ .

Working

From $\ln\left(\frac{p}{q}\right) = a$ :

\ln p - \ln q = a. \qquad (1)

From $\ln(q^2 p) = b$ :

\ln q^2 + \ln p = b

\ln p + 2\ln q = b. \qquad (2)

From (1), $\ln p =a + \ln q$ . Substitute into (2):

(a + \ln q) + 2\ln q = b

3\ln q = b - a

\ln q = \frac{b-a}{3}.

Then

\ln p =a + \frac{b-a}{3} = \frac{3a + b - a}{3} = \frac{2a+b}{3}.

Now

\ln(p^7 q) = 7\ln p + \ln q

= 7\left(\frac{2a+b}{3}\right) + \frac{b-a}{3}

= \frac{14a+7b+b-a}{3} = \frac{13a+8b}{3}.

Answer

\ln(p^7 q) = \frac{13a+8b}{3}.

Final answer

(13a + 8b)/3

Detailed explanation

Walkthrough

Start with the first equation. The logarithm of a quotient is the difference of the logarithms, so $\ln\left(\frac{p}{q}\right) = \ln p - \ln q$ . This gives one linear equation in the two unknown logarithms. Similarly, the logarithm of a product is the sum of the logarithms, and $\ln(q^2) = 2\ln q$ , so the second equation becomes $\ln p + 2\ln q = b$ . These are simultaneous linear equations in $\ln p$ and $\ln q$ . Solve by substituting $\ln p =a + \ln q$ from the first equation into the second; this gives $\ln q = \frac{b-a}{3}$ and then $\ln p = \frac{2a+b}{3}$ . Finally, use the power law: $\ln(p^7 q) = 7\ln p + \ln q$ . Substitute the two values and simplify to get $\frac{13a+8b}{3}$ .

Key Takeaways

This question tests the laws of logarithms: the quotient law, the product law, and the power law. It also shows that logarithms convert multiplicative relationships into linear equations, making simultaneous equation techniques available.

Common Mistakes

Forgetting that $\ln\left(\frac{p}{q}\right) = \ln p - \ln q$ , not $\ln p / \ln q$ .
Forgetting that $\ln(q^2 p) = \ln q^2 + \ln p = 2\ln q + \ln p$ .
Making sign errors when solving the simultaneous equations, especially using $a - b$ instead of $b - a$ .
If using the alternative method of writing $\ln(p^7 q)$ as a linear combination of $a$ and $b$ , do not leave exponentials in the working; the mark scheme requires all exponentials to be removed.

Things to Be Careful About

The variables $p$ and $q$ are positive, so all logarithms are defined; there are no domain issues.
Keep all work in terms of $\ln p$ and $\ln q$ until the final substitution; do not introduce decimals or approximate values.
Simplify the final fraction carefully: $14a+7b+b-a = 13a+8b$ .
Show the rewritten equations, the substitution, and the final simplification to earn the method marks (M1) and answer mark (A1).

Techniques used

apply logarithm laws to rewrite products and quotientssolve simultaneous linear equationscombine logarithmic terms and simplify

Approach

Interpret each modulus inequality as a region on the Argand diagram, then find their intersection and shade it.

Working

First inequality: $|z - 4 - 2i| \leq 3$

This represents all points $z$ whose distance from the point $4 + 2i$ is at most 3. Geometrically, this is the closed disk (interior and boundary) of a circle with centre $(4, 2)$ and radius 3.

(x - 4)^2 + (y - 2)^2 \leq 9

Second inequality: $|z| \geq |10 - z|$

Let $z = x + iy$ . Squaring both sides:

|z|^2 \geq |10 - z|^2

x^2 + y^2 \geq (10 - x)^2 + y^2

x^2 \geq 100 - 20x + x^2

20x \geq 100

x \geq 5

This is the half-plane to the right of (and including) the vertical line $\text{Re}(z) = 5$ .

Intersection:

The required region is where both conditions hold: inside or on the circle of radius 3 centred at $(4, 2)$ , and with $\text{Re}(z) \geq 5$ .

The circle crosses the line $x = 5$ :

(5 - 4)^2 + (y - 2)^2 = 9

1 + (y - 2)^2 = 9

(y - 2)^2 = 8

y = 2 \pm 2\sqrt{2}

So the circle intersects the line at $(5,\; 2 - 2\sqrt{2})$ and $(5,\; 2 + 2\sqrt{2})$ .

The shaded region is the segment of the circle to the right of $x = 5$ .

Answer

The region is the part of the disk $(x-4)^2 + (y-2)^2 \leq 9$ with $x \geq 5$ , shaded between the vertical line $\text{Re}(z) = 5$ and the right arc of the circle centred at $4 + 2i$ with radius 3.

Final answer

The shaded region is the segment of the circle $|z - 4 - 2i| = 3$ (centre $4+2i$ , radius 3) to the right of the line $\text{Re}(z) = 5$ .

Detailed explanation

Walkthrough

Step 1: Interpret the first inequality.

The expression $|z - 4 - 2i| \leq 3$ asks for all complex numbers $z$ whose distance from the point $4 + 2i$ is at most 3. On an Argand diagram, this is the filled circle (disk) with centre $(4, 2)$ and radius 3. The boundary circle has equation $(x-4)^2 + (y-2)^2 = 9$ .

Step 2: Interpret the second inequality.

The expression $|z| \geq |10 - z|$ compares the distance of $z$ from the origin with the distance of $z$ from $10$ . Points equidistant from $0$ and $10$ lie on the perpendicular bisector of the segment joining them, which is the vertical line $x = 5$ . Points closer to $10$ than to the origin satisfy $|z| \geq |10 - z|$ , which is the half-plane $x \geq 5$ .

Algebraically, squaring both sides of $|z|^2 \geq |10-z|^2$ with $z = x+iy$ gives $x^2 + y^2 \geq (10-x)^2 + y^2$ , which simplifies to $x \geq 5$ .

Step 3: Find the intersection.

The region satisfying both inequalities is the part of the disk that lies to the right of (or on) the line $x = 5$ . The circle intersects this line at $y = 2 \pm 2\sqrt{2}$ , giving the points $(5,\; 2 - 2\sqrt{2})$ and $(5,\; 2 + 2\sqrt{2})$ .

Step 4: Shade the region.

The shaded region is the circular segment bounded by the chord on $x = 5$ (from $y = 2 - 2\sqrt{2}$ to $y = 2 + 2\sqrt{2}$ ) and the right arc of the circle.

Key Takeaways

$|z - a| \leq r$ represents a closed disk of radius $r$ centred at $a$ on the Argand diagram.
$|z - a| \geq |z - b|$ represents the half-plane on the side of the perpendicular bisector of $ab$ that is closer to $b$ .
When multiple modulus inequalities are given, interpret each geometrically and find the intersection of the regions.

Common Mistakes

Forgetting that $|z| \geq |10 - z|$ gives $x \geq 5$ (the region to the right), not $x \leq 5$ .
Drawing the circle with the wrong centre or radius.
Shading the wrong side of the line or the wrong part of the circle.
Not showing the line $\text{Re}(z) = 5$ clearly on the diagram.

Things to Be Careful About

The inequality $|z - 4 - 2i| \leq 3$ includes the boundary (closed disk), so the circle itself is part of the region.
The line $\text{Re}(z) = 5$ is included ( $x \geq 5$ , not $x > 5$ ).
The centre of the circle is at $(4, 2)$ , not at the origin. Mark it clearly on the diagram.
The radius is 3, so the circle extends from $x = 1$ to $x = 7$ and from $y = -1$ to $y = 5$ .

Techniques used

interpret modulus inequality as circular regionrewrite second modulus inequality as linear inequalityidentify intersection of regionsshade correct region on Argand diagram

(b)

Find the greatest value of $\arg z$ for points in this region.

2M Similar

DifficultyMedium

Worked solution

Approach

The argument $\arg z$ is the angle that the line from the origin to $z$ makes with the positive real axis. To maximise this angle for points in the shaded region, we need the point in the region that is 'highest' relative to its horizontal distance from the origin.

The shaded region is bounded by the line $x = 5$ on the left and the circular arc on the right. The greatest argument occurs at the upper intersection point of the line $x = 5$ and the circle, since moving along the circular arc to the right decreases the argument (the arc curves away from the origin's line of sight).

Working

The upper intersection point is at $x = 5$ and $y = 2 + 2\sqrt{2}$ .

At this point, $z = 5 + (2 + 2\sqrt{2})i$ .

The argument is:

\arg z = \tan^{-1}\left(\frac{2 + 2\sqrt{2}}{5}\right)

Calculating:

2 + 2\sqrt{2} \approx 2 + 2(1.4142) = 4.8284

\frac{4.8284}{5} = 0.9657

\arg z = \tan^{-1}(0.9657) \approx 0.768 \text{ radians}

Converting to degrees:

0.768 \times \frac{180}{\pi} \approx 44.0^\circ

Answer

\arg z = \tan^{-1}\left(\frac{2 + 2\sqrt{2}}{5}\right) \approx 0.768 \text{ radians} \approx 44.0^\circ

Final answer

0.768 radians (or 44.0°)

Detailed explanation

Walkthrough

Step 1: Identify where the maximum argument occurs.

The argument $\arg z = \tan^{-1}(y/x)$ for a point $(x, y)$ in the first quadrant. To maximise this, we want $y/x$ to be as large as possible within the shaded region.

The shaded region is bounded on the left by $x = 5$ and on the right by the circular arc. Along the line $x = 5$ , the argument increases as $y$ increases, reaching its maximum at the upper intersection point $(5,\; 2 + 2\sqrt{2})$ .

Along the circular arc (where $x > 5$ ), as we move from the upper intersection point to the right, $x$ increases and $y$ decreases, so $y/x$ decreases and the argument decreases. Therefore, the maximum argument is at $(5,\; 2 + 2\sqrt{2})$ .

Step 2: Compute the argument.

At $z = 5 + (2 + 2\sqrt{2})i$ :

\arg z = \tan^{-1}\left(\frac{2 + 2\sqrt{2}}{5}\right)

Numerically: $2 + 2\sqrt{2} \approx 4.8284$ , so $\frac{4.8284}{5} \approx 0.9657$ .

\arg z \approx \tan^{-1}(0.9657) \approx 0.768 \text{ rad} \approx 44.0^\circ

Key Takeaways

The maximum argument in a region bounded by a line and a curve often occurs at a boundary intersection point.
When comparing points in a region, check both boundary segments to find the extremum.
Always verify that the candidate point is actually in the feasible region.

Common Mistakes

Assuming the maximum argument occurs at a tangent from the origin to the circle. The tangent point $(\approx 1.2,\; 3.1)$ lies outside the region $x \geq 5$ , so it is not valid.
Forgetting to check both boundaries (the line and the arc) when searching for the maximum argument.
Computing the wrong intersection point (using the lower one instead of the upper one).

Things to Be Careful About

The tangent from the origin to the full circle gives an argument of about $68.7^\circ$ , but the tangent point has $x \approx 1.2 < 5$ , so it is outside the shaded region. This is a common trap.
The answer must be in the correct units (radians or degrees) as specified or clearly stated.
$2\sqrt{2}$ can be written as $\sqrt{8}$ or computed as $\sqrt{3^2 - 1^2}$ from the geometry (radius 3, horizontal distance from centre to line is 1).

Techniques used

identify that greatest argument occurs at circle-line intersectioncalculate argument using arctangentcompute numerical value in radians and degrees

Q6MediumDifferentiation

The equation of a curve is $2y^2 + 3xy + x = x^2$ .

(a)

Show that

\frac{dy}{dx} = \frac{2x - 3y - 1}{4y + 3x}.

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate both sides of the curve equation implicitly with respect to $x$ , treating $y$ as a function of $x$ . Use the product rule for the $3xy$ term, then collect the terms containing $\frac{dy}{dx}$ on one side and solve.

Working

Differentiate each term:

\begin{aligned} \frac{d}{dx}(2y^2) &= 4y\frac{dy}{dx} \\ \frac{d}{dx}(3xy) &= 3y + 3x\frac{dy}{dx} \\ \frac{d}{dx}(x) &= 1 \\ \frac{d}{dx}(x^2) &= 2x \end{aligned}

So the differentiated equation is

4y\frac{dy}{dx} + 3y + 3x\frac{dy}{dx} + 1 = 2x.

Collect the $\frac{dy}{dx}$ terms:

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

Hence

\frac{dy}{dx} = \frac{2x - 3y - 1}{4y + 3x}.

Answer

\frac{dy}{dx} = \frac{2x - 3y - 1}{4y + 3x}

Final answer

dy/dx = (2x - 3y - 1)/(4y + 3x)

Detailed explanation

Walkthrough

We need to differentiate the equation $2y^2 + 3xy + x = x^2$ because $y$ is defined implicitly as a function of $x$ . Differentiate each term with respect to $x$ .

For $2y^2$ , use the chain rule: the derivative is $4y\frac{dy}{dx}$ . For $3xy$ , use the product rule: $3y + 3x\frac{dy}{dx}$ . The derivative of $x$ is $1$ , and the derivative of $x^2$ is $2x$ .

Putting these together gives

4y\frac{dy}{dx} + 3y + 3x\frac{dy}{dx} + 1 = 2x.

Now collect the terms containing $\frac{dy}{dx}$ on one side:

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

Finally divide by $4y + 3x$ to obtain the required expression.

Key Takeaways

Implicit differentiation treats $y$ as a function of $x$ , so every $y$ -term needs a $\frac{dy}{dx}$ factor. The product rule is needed for $3xy$ . Once differentiated, collect derivative terms and solve algebraically.

Common Mistakes

Forgetting the $\frac{dy}{dx}$ factor when differentiating $2y^2$ .
Differentiating $3xy$ as just $3y$ , omitting the $3x\frac{dy}{dx}$ term.
Moving terms with the wrong sign.
Not writing a complete equation such as $(4y + 3x)\frac{dy}{dx} = 2x - 3y - 1$ ; since the answer is given, the intermediate equation must be shown.

Things to Be Careful About

The mark scheme requires seeing $4y\frac{dy}{dx} + 3x\frac{dy}{dx} = 2x - 3y - 1$ or equivalent. The derivative of the term $x$ is $1$ , and the right-hand side $x^2$ differentiates to $2x$ . Keep $\frac{dy}{dx}$ attached to every term involving $y$ .

Techniques used

differentiate implicitlyapply the product rulecollect derivative termssolve for dy/dx

(b)

Hence show that the curve does not have a tangent that is parallel to the $x$ -axis.

3M Similar

DifficultyMedium

Worked solution

Approach

A tangent parallel to the $x$ -axis is horizontal, so its gradient is zero. Set $\frac{dy}{dx}=0$ , which means the numerator is zero. Use this relation to eliminate one variable from the curve equation, obtaining a quadratic in the other variable. Show this quadratic has no real roots by its discriminant.

Working

For a horizontal tangent,

\frac{dy}{dx} = 0 \quad \Rightarrow \quad 2x - 3y - 1 = 0.

Thus

3y = 2x - 1, \quad y = \frac{2x - 1}{3}.

Substitute into $2y^2 + 3xy + x = x^2$ :

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

Simplify:

\frac{2}{9}(2x - 1)^2 + x(2x - 1) + x = x^2.

Since $x(2x - 1) + x = 2x^2$ , this becomes

\frac{2}{9}(4x^2 - 4x + 1) + 2x^2 = x^2.

Multiply by $9$ :

2(4x^2 - 4x + 1) + 18x^2 = 9x^2,

8x^2 - 8x + 2 + 18x^2 = 9x^2,

17x^2 - 8x + 2 = 0.

Its discriminant is

(-8)^2 - 4(17)(2) = 64 - 136 = -72 < 0.

Therefore the quadratic has no real roots, so there is no real point on the curve where $\frac{dy}{dx}=0$ . Hence the curve has no tangent parallel to the $x$ -axis.

Answer

No tangent parallel to the $x$ -axis exists.

Final answer

No tangent parallel to the x-axis exists.

Detailed explanation

Walkthrough

A tangent parallel to the $x$ -axis is horizontal, so its gradient is zero. From part (a), $\frac{dy}{dx}=0$ requires the numerator to be zero:

2x - 3y - 1 = 0.

This gives $3y = 2x - 1$ , so $y = \frac{2x - 1}{3}$ . Substitute this into the curve equation to eliminate $y$ :

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

Simplify. The term $3x\left(\frac{2x-1}{3}\right) + x$ becomes $x(2x-1)+x = 2x^2$ . Thus

\frac{2}{9}(2x-1)^2 + 2x^2 = x^2.

Multiply by $9$ and expand:

2(4x^2 - 4x + 1) + 18x^2 = 9x^2,

so

17x^2 - 8x + 2 = 0.

The discriminant is

(-8)^2 - 4(17)(2) = 64 - 136 = -72 < 0.

A negative discriminant means there are no real values of $x$ satisfying the equation. Therefore no point on the curve has $\frac{dy}{dx}=0$ , so no tangent is parallel to the $x$ -axis.

Key Takeaways

A horizontal tangent is found by setting the derivative equal to zero. For an implicit derivative, this means setting the numerator to zero. Substituting the resulting relation into the original curve gives an equation in one variable; the discriminant can prove that no real solution exists.

Common Mistakes

Setting the denominator $4y+3x$ equal to zero instead of the numerator.
Substituting incorrectly or losing the $+x$ term when simplifying.
Making expansion errors with $\left(\frac{2x-1}{3}\right)^2$ .
Concluding no tangent without showing the discriminant is negative or that the quadratic has no real roots.

Things to Be Careful About

The mark scheme expects the numerator to be equated to zero, a correct equation in one variable, and a clear conclusion of no real roots. If you solve the quadratic and obtain complex roots, you must still state that there are no real roots. Keep the algebra tidy; multiplying by $9$ early avoids fractions.

Techniques used

set derivative numerator to zerosubstitute linear relation into curveform quadratic in one variableapply discriminant test

Q7MediumDifferentiationLogarithmic and Exponential FunctionsNumerical Solution of Equations

Fig. 7.1

The diagram shows the curve $y = xe^{2x} - 5x$ and its minimum point $M$ , where $x = \alpha$ .

(a)

Show that $\alpha$ satisfies the equation

\alpha = \frac{1}{2}\ln\left(\frac{5}{1 + 2\alpha}\right).

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate $y = xe^{2x} - 5x$ using the product rule, set $\frac{dy}{dx} = 0$ at the minimum point $x = \alpha$ , and rearrange the resulting equation to obtain the required form.

Working

y = xe^{2x} - 5x

Apply the product rule to $xe^{2x}$ :

\frac{dy}{dx} = (1)e^{2x} + x(2e^{2x}) - 5

\frac{dy}{dx} = e^{2x} + 2xe^{2x} - 5

At the minimum point $M$ , $x = \alpha$ and $\frac{dy}{dx} = 0$ :

e^{2\alpha} + 2\alpha e^{2\alpha} - 5 = 0

Factor out $e^{2\alpha}$ :

e^{2\alpha}(1 + 2\alpha) = 5

e^{2\alpha} = \frac{5}{1 + 2\alpha}

Take the natural logarithm of both sides:

\ln\left(e^{2\alpha}\right) = \ln\left(\frac{5}{1 + 2\alpha}\right)

2\alpha = \ln\left(\frac{5}{1 + 2\alpha}\right)

Divide by 2:

\alpha = \frac{1}{2}\ln\left(\frac{5}{1 + 2\alpha}\right)

Answer

\alpha = \frac{1}{2}\ln\left(\frac{5}{1 + 2\alpha}\right)

Final answer

α = ½ln(5 / (1 + 2α))

Detailed explanation

Walkthrough

We are given the curve $y = xe^{2x} - 5x$ with a minimum point $M$ at $x = \alpha$ . To find the condition satisfied by $\alpha$ , we differentiate $y$ with respect to $x$ and set the derivative to zero.

First, we apply the product rule to the term $xe^{2x}$ . Let $u = x$ and $v = e^{2x}$ . Then $u' = 1$ and $v' = 2e^{2x}$ . The derivative is $u'v + uv' = e^{2x} + 2xe^{2x}$ . Differentiating $-5x$ gives $-5$ . So $\frac{dy}{dx} = e^{2x} + 2xe^{2x} - 5$ .

At a minimum point, the gradient is zero. Substituting $x = \alpha$ and $\frac{dy}{dx} = 0$ :

e^{2\alpha} + 2\alpha e^{2\alpha} - 5 = 0

We factor $e^{2\alpha}$ from the first two terms:

e^{2\alpha}(1 + 2\alpha) = 5

Isolating the exponential term:

e^{2\alpha} = \frac{5}{1 + 2\alpha}

To solve for $\alpha$ , we take the natural logarithm of both sides. Since $\ln(e^{2\alpha}) = 2\alpha$ , we get:

2\alpha = \ln\left(\frac{5}{1 + 2\alpha}\right)

Dividing by 2 yields the required equation:

\alpha = \frac{1}{2}\ln\left(\frac{5}{1 + 2\alpha}\right)

Key Takeaways

The product rule is essential when differentiating a product of two functions of $x$ .
At stationary points (minima or maxima), the first derivative equals zero.
Exponential equations can be solved by taking logarithms, using the property $\ln(e^k) = k$ .

Common Mistakes

Forgetting the product rule and incorrectly differentiating $xe^{2x}$ as $e^{2x} + x \cdot 2$ or $e^{2x}$ .
Failing to factor $e^{2\alpha}$ correctly when rearranging.
Not substituting $\alpha$ for $x$ before equating to zero, or not rearranging into the exact form requested.

Things to Be Careful About

Ensure the derivative is fully simplified before setting to zero.
When taking logarithms, remember that $\ln(e^{2\alpha}) = 2\alpha$ , not just $\alpha$ .
The final equation must be expressed entirely in terms of $\alpha$ , not $x$ .

Techniques used

apply product rule to differentiatedifferentiate exponential functionequate derivative to zero for minimum pointrearrange using logarithms

(b)

Verify by calculation that $\alpha$ lies between 0.4 and 0.5.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Evaluate the derivative $\frac{dy}{dx} = e^{2x}(1 + 2x) - 5$ at $x = 0.4$ and $x = 0.5$ . If the values have opposite signs, by the Intermediate Value Theorem, there is a root of $\frac{dy}{dx} = 0$ between these values, meaning $\alpha$ lies between 0.4 and 0.5.

Working

\frac{dy}{dx} = e^{2x}(1 + 2x) - 5

At $x = 0.4$ :

\frac{dy}{dx} = e^{0.8}(1 + 0.8) - 5 = 1.8e^{0.8} - 5

\approx 1.8(2.22554) - 5 = 4.00597 - 5 = -0.99403

At $x = 0.5$ :

\frac{dy}{dx} = e^{1.0}(1 + 1.0) - 5 = 2e^1 - 5

\approx 2(2.71828) - 5 = 5.43656 - 5 = 0.43656

Since $\frac{dy}{dx}$ changes sign from negative ( $-0.994$ ) at $x = 0.4$ to positive ( $0.437$ ) at $x = 0.5$ , and $\frac{dy}{dx}$ is continuous, there must be a value of $x$ between $0.4$ and $0.5$ where $\frac{dy}{dx} = 0$ .

This value is $\alpha$ , so $\alpha$ lies between $0.4$ and $0.5$ .

Answer

0.4 < \alpha < 0.5

Final answer

0.4 < α < 0.5

Detailed explanation

Walkthrough

We need to show that the $x$ -coordinate of the minimum point, $\alpha$ , lies between $0.4$ and $0.5$ . Since $\alpha$ is a root of $\frac{dy}{dx} = 0$ , we can evaluate the derivative at the boundary values $x = 0.4$ and $x = 0.5$ .

Using the derivative from part (a):

\frac{dy}{dx} = e^{2x}(1 + 2x) - 5

Substituting $x = 0.4$ :

\frac{dy}{dx} = e^{0.8}(1 + 0.8) - 5 = 1.8e^{0.8} - 5 \approx -0.994

Substituting $x = 0.5$ :

\frac{dy}{dx} = e^{1.0}(1 + 1.0) - 5 = 2e^1 - 5 \approx 0.437

The derivative is negative at $x = 0.4$ and positive at $x = 0.5$ . Because the derivative is a continuous function, it must cross zero somewhere between $x = 0.4$ and $x = 0.5$ . This zero corresponds to the minimum point $M$ , so $\alpha$ must lie in the interval $(0.4, 0.5)$ .

Key Takeaways

A sign change in a continuous function guarantees a root between the two points (Intermediate Value Theorem).
Evaluating the derivative at two points is a standard method to locate stationary points.

Common Mistakes

Calculating only one value instead of both.
Forgetting to state that the derivative changes sign and thus a root exists between the values.
Rounding errors that lead to incorrect sign conclusions.

Things to Be Careful About

Ensure calculations are accurate to at least 3 decimal places to clearly show the sign change.
Explicitly state that the derivative is continuous, which justifies the sign change argument.

Techniques used

evaluate derivative at boundary valuesapply sign change argument

(c)

Use an iterative formula based on the equation in part (a) to determine $\alpha$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the iterative formula derived in part (a): $\alpha_{n+1} = \frac{1}{2}\ln\left(\frac{5}{1 + 2\alpha_n}\right)$ . Start with an initial value (e.g., $\alpha_1 = 0.4$ ) and iterate, recording each result to 4 decimal places, until the value stabilizes to 2 decimal places.

Working

Iterative formula:

\alpha_{n+1} = \frac{1}{2}\ln\left(\frac{5}{1 + 2\alpha_n}\right)

Let $\alpha_1 = 0.4$ :

\alpha_2 = \frac{1}{2}\ln\left(\frac{5}{1 + 2(0.4)}\right) = \frac{1}{2}\ln\left(\frac{5}{1.8}\right) = 0.5108

\alpha_3 = \frac{1}{2}\ln\left(\frac{5}{1 + 2(0.5108)}\right) = \frac{1}{2}\ln\left(\frac{5}{2.0216}\right) = 0.4528

\alpha_4 = \frac{1}{2}\ln\left(\frac{5}{1 + 2(0.4528)}\right) = \frac{1}{2}\ln\left(\frac{5}{1.9056}\right) = 0.4823

\alpha_5 = \frac{1}{2}\ln\left(\frac{5}{1 + 2(0.4823)}\right) = \frac{1}{2}\ln\left(\frac{5}{1.9646}\right) = 0.4670

\alpha_6 = \frac{1}{2}\ln\left(\frac{5}{1 + 2(0.4670)}\right) = \frac{1}{2}\ln\left(\frac{5}{1.9340}\right) = 0.4749

\alpha_7 = \frac{1}{2}\ln\left(\frac{5}{1 + 2(0.4749)}\right) = \frac{1}{2}\ln\left(\frac{5}{1.9498}\right) = 0.4707

\alpha_8 = \frac{1}{2}\ln\left(\frac{5}{1 + 2(0.4707)}\right) = \frac{1}{2}\ln\left(\frac{5}{1.9414}\right) = 0.4730

The values are oscillating around $0.47$ . To confirm $\alpha = 0.47$ correct to 2 decimal places, we check for a sign change in $f(x) = \frac{1}{2}\ln\left(\frac{5}{1 + 2x}\right) - x$ between $0.465$ and $0.475$ :

At $x = 0.465$ :

\frac{1}{2}\ln\left(\frac{5}{1 + 2(0.465)}\right) - 0.465 = \frac{1}{2}\ln\left(\frac{5}{1.93}\right) - 0.465 = 0.4727 - 0.465 = 0.0077 > 0

At $x = 0.475$ :

\frac{1}{2}\ln\left(\frac{5}{1 + 2(0.475)}\right) - 0.475 = \frac{1}{2}\ln\left(\frac{5}{1.95}\right) - 0.475 = 0.4682 - 0.475 = -0.0068 < 0

Since there is a sign change between $0.465$ and $0.475$ , $\alpha = 0.47$ correct to 2 decimal places.

Answer

\alpha = 0.47

Final answer

0.47

Detailed explanation

Walkthrough

We use the iterative formula $\alpha_{n+1} = \frac{1}{2}\ln\left(\frac{5}{1 + 2\alpha_n}\right)$ to find $\alpha$ . Starting with $\alpha_1 = 0.4$ , we substitute this value into the right-hand side to get $\alpha_2$ , then substitute $\alpha_2$ to get $\alpha_3$ , and so on.

Calculations:

$\alpha_1 = 0.4000$
$\alpha_2 = \frac{1}{2}\ln\left(\frac{5}{1.8}\right) = 0.5108$
$\alpha_3 = \frac{1}{2}\ln\left(\frac{5}{2.0216}\right) = 0.4528$
$\alpha_4 = \frac{1}{2}\ln\left(\frac{5}{1.9056}\right) = 0.4823$
$\alpha_5 = \frac{1}{2}\ln\left(\frac{5}{1.9646}\right) = 0.4670$
$\alpha_6 = \frac{1}{2}\ln\left(\frac{5}{1.9340}\right) = 0.4749$
$\alpha_7 = \frac{1}{2}\ln\left(\frac{5}{1.9498}\right) = 0.4707$
$\alpha_8 = \frac{1}{2}\ln\left(\frac{5}{1.9414}\right) = 0.4730$

The sequence is oscillating and converging towards a value around $0.47$ . To confirm that $\alpha = 0.47$ correct to 2 decimal places, we verify that the root lies in the interval $[0.465, 0.475)$ . We define $f(x) = \frac{1}{2}\ln\left(\frac{5}{1 + 2x}\right) - x$ and check the signs at the boundaries:

$f(0.465) \approx 0.0077 > 0$
$f(0.475) \approx -0.0068 < 0$

The sign change confirms the root is in this interval, so $\alpha = 0.47$ to 2 decimal places.

Key Takeaways

Iterative methods can find roots of equations to a desired accuracy.
Showing a sign change in an interval of width $0.01$ centered on the answer is the standard way to justify a 2 decimal place answer.
Iterations must be carried out to sufficient precision (4 d.p. here) to avoid rounding errors.

Common Mistakes

Using the wrong iterative formula or making arithmetic errors in early iterations.
Stopping the iteration too early without verifying convergence.
Not showing sufficient iterations or the sign change argument to justify the final answer.

Things to Be Careful About

Always carry at least 4 decimal places during iterations to ensure the final answer is accurate.
Explicitly show the sign change at $0.465$ and $0.475$ to justify the 2 decimal place answer; simply stating the iteration values is not enough.

Techniques used

set up iterative formulaiterate to convergenceverify accuracy to required decimal places

Q8MediumTrigonometry

(a)

Express $3\sin x + 2\sqrt{2}\cos\left(x + \frac{1}{4}\pi\right)$ in the form $R\sin(x + \alpha)$ , where $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$ . State the exact value of $R$ and give $\alpha$ correct to 3 decimal places.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Expand $\cos\left(x+\frac{1}{4}\pi\right)$ using the compound angle formula, collect the terms into $a\sin x + b\cos x$ , and then write the result as $R\sin(x+\alpha)$ .

Working

Expand the cosine term:

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

Multiply by $2\sqrt{2}$ :

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

Add $3\sin x$ :

3\sin x + 2\sqrt{2}\cos\left(x+\frac{1}{4}\pi\right) = \sin x + 2\cos x

Now write $\sin x + 2\cos x = R\sin(x+\alpha)$ . Expanding the right-hand side:

R\sin(x+\alpha) = R\sin x\cos\alpha + R\cos x\sin\alpha

Matching coefficients gives $R\cos\alpha = 1$ and $R\sin\alpha = 2$ . Therefore

R = \sqrt{1^2 + 2^2} = \sqrt{5}

and

\tan\alpha = \frac{2}{1} = 2

\alpha = \tan^{-1} 2 = 1.1071487\ldots

To 3 decimal places, $\alpha = 1.107$ .

Answer

3\sin x + 2\sqrt{2}\cos\left(x+\frac{1}{4}\pi\right) = \sqrt{5}\sin(x + 1.107)

with $R = \sqrt{5}$ and $\alpha = 1.107$ .

Final answer

R = sqrt(5), alpha = 1.107

Detailed explanation

Walkthrough

Start by expanding $\cos\left(x+\frac{1}{4}\pi\right)$ using the compound angle formula. Because $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ , the shifted cosine becomes $\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x$ . Multiplying by $2\sqrt{2}$ gives $2\cos x - 2\sin x$ . Adding $3\sin x$ gives $\sin x + 2\cos x$ .

Next, compare this with $R\sin(x+\alpha) = R\sin x\cos\alpha + R\cos x\sin\alpha$ . The coefficient of $\sin x$ is $R\cos\alpha$ and the coefficient of $\cos x$ is $R\sin\alpha$ , so $R\cos\alpha = 1$ and $R\sin\alpha = 2$ . Squaring and adding gives $R^2 = 1^2 + 2^2 = 5$ , so $R = \sqrt{5}$ . Dividing the two equations gives $\tan\alpha = 2$ , so $\alpha = \tan^{-1}2 \approx 1.107$ radians. Since both $R\cos\alpha$ and $R\sin\alpha$ are positive, $\alpha$ lies in the first quadrant, matching the required range $0 < \alpha < \frac{1}{2}\pi$ .

Key Takeaways

The compound angle formula for $\cos(A+B)$ is $\cos A\cos B - \sin A\sin B$ .
To write $a\sin x + b\cos x$ as $R\sin(x+\alpha)$ , use $R = \sqrt{a^2+b^2}$ and $\tan\alpha = \frac{b}{a}$ , choosing the quadrant from the signs of $a$ and $b$ .
Exact values should be given exactly where requested; approximations are only for the final decimal answer.

Common Mistakes

Forgetting the minus sign in $\cos(A+B)$ .
Using $\tan\alpha = \frac{a}{b}$ instead of $\frac{b}{a}$ .
Choosing the wrong quadrant for $\alpha$ ; here both coefficients are positive, so $\alpha$ is acute.
Giving $R$ as a decimal instead of the exact value $\sqrt{5}$ .

Things to Be Careful About

$\alpha$ must be in radians because the question asks for 3 decimal places and the interval is in radians; giving $63.435°$ is treated as a misread.
The mark scheme requires the correct expansion before the $R$ and $\alpha$ values are accepted.
Check that $0 < \alpha < \frac{1}{2}\pi$ ; here it is satisfied.

Techniques used

expand a compound angle cosinecollect sine and cosine termscompute R from coefficientsdetermine phase shift with arctangent

(b)

Hence solve the equation

6\sin\frac{1}{2}\theta + 4\sqrt{2}\cos\left(\frac{1}{2}\theta + \frac{1}{4}\pi\right) = 3

for $-4\pi < \theta < 4\pi$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use the result from part (a) with $x = \frac{1}{2}\theta$ . The left-hand side is exactly twice the expression in part (a), so the equation becomes a sine equation. Solve for $u = \frac{1}{2}\theta + \alpha$ , then list all solutions in the interval $-4\pi < \theta < 4\pi$ .

Working

From part (a),

3\sin x + 2\sqrt{2}\cos\left(x+\frac{1}{4}\pi\right) = \sqrt{5}\sin(x+\alpha), \quad \alpha = 1.1071487\ldots

Putting $x = \frac{1}{2}\theta$ and multiplying by 2 gives

6\sin\frac{\theta}{2} + 4\sqrt{2}\cos\left(\frac{\theta}{2}+\frac{\pi}{4}\right) = 2\sqrt{5}\sin\left(\frac{\theta}{2}+\alpha\right)

So the equation is

2\sqrt{5}\sin\left(\frac{\theta}{2}+\alpha\right) = 3

\sin\left(\frac{\theta}{2}+\alpha\right) = \frac{3}{2\sqrt{5}} = \frac{1.5}{\sqrt{5}}

Let $u = \frac{\theta}{2}+\alpha$ . The principal value is

u = \sin^{-1}\left(\frac{1.5}{\sqrt{5}}\right) = 0.7353\ldots

Since $-4\pi < \theta < 4\pi$ , we have $-2\pi < \frac{\theta}{2} < 2\pi$ , so

-2\pi + \alpha < u < 2\pi + \alpha

which is approximately $-5.176 < u < 7.390$ .

The general solutions of $\sin u = k$ are

u = \sin^{-1} k + 2n\pi \quad \text{or} \quad u = \pi - \sin^{-1} k + 2n\pi

Therefore

u = 0.7353 + 2n\pi \quad \text{or} \quad u = 2.4063 + 2n\pi

Converting back, $\theta = 2(u - \alpha)$ :

\theta = 2(0.7353 - 1.10715 + 2n\pi) = -0.7437 + 4n\pi

or

\theta = 2(2.4063 - 1.10715 + 2n\pi) = 2.5983 + 4n\pi

For the first family, $n=0$ gives $-0.74$ and $n=1$ gives $11.82$ ; $n=-1$ gives $-13.31$ , outside the interval. For the second family, $n=0$ gives $2.60$ and $n=-1$ gives $-9.97$ ; $n=1$ gives $15.16$ , outside the interval. These are the only integer values of $n$ that keep $\theta$ in the interval.

Answer

The solutions in the interval are

\theta = -9.97, -0.74, 2.60, 11.8

Final answer

theta = -9.97, -0.74, 2.60, 11.8

Detailed explanation

Walkthrough

The left-hand side of the equation is exactly twice the expression from part (a) with $x=\frac{1}{2}\theta$ . Therefore it can be replaced by $2\sqrt{5}\sin\left(\frac{\theta}{2}+\alpha\right)$ . Dividing by $2\sqrt{5}$ gives $\sin\left(\frac{\theta}{2}+\alpha\right)=\frac{3}{2\sqrt{5}}$ .

Let $u=\frac{\theta}{2}+\alpha$ . The principal solution is $u=\sin^{-1}\left(\frac{1.5}{\sqrt{5}}\right)\approx0.7353$ . Because sine is periodic with period $2\pi$ , the general solutions are $u=0.7353+2n\pi$ and $u=\pi-0.7353+2n\pi=2.4063+2n\pi$ .

Now convert back: $\theta=2(u-\alpha)$ . This gives $\theta=-0.7437+4n\pi$ from the first family and $\theta=2.5983+4n\pi$ from the second. The interval $-4\pi<\theta<4\pi$ is approximately $-12.566<\theta<12.566$ . Testing integer values of $n$ : first family gives $-0.74$ and $11.82$ ; second family gives $2.60$ and $-9.97$ . These are the four required solutions.

Key Takeaways

The R-form result can be used directly to solve equations of the form $a\sin x+b\cos x=c$ .
The general solution of $\sin u=k$ has two families: $u=\sin^{-1}k+2n\pi$ and $u=\pi-\sin^{-1}k+2n\pi$ .
When an interval is given, substitute integer values of $n$ and discard solutions outside the interval.

Common Mistakes

Forgetting the second family $\pi-u$ when solving $\sin u=k$ .
Forgetting to add $2n\pi$ and therefore missing solutions.
Forgetting to convert the interval for $\theta$ into an interval for $u$ or $\frac{\theta}{2}$ .
Using degrees instead of radians; the mark scheme treats degree answers as a misread.
Giving only one or two solutions instead of all four in the interval.

Things to Be Careful About

The interval is strict, $-4\pi<\theta<4\pi$ , but none of the four solutions lies exactly on an endpoint.
Answers outside the interval should be ignored, not counted.
If a rounded value of $\alpha$ is used, the final answers may differ slightly in the last decimal place; this is acceptable.
The mark scheme says that if $1.11°$ is used (degrees), the first accuracy mark is withheld, but later accuracy marks may still be allowed.

Techniques used

scale the R-form resultisolate a sine functionapply inverse sine to find principal valuegenerate all solutions using periodicityrestrict solutions to the given interval

Q9MediumVectors

Relative to the origin $O$ , the position vectors of the points $A$ , $B$ and $C$ are given by

\vec{OA} = 5i - 2j + k, \quad \vec{OB} = 8i + 2j - 6k \quad \text{and} \quad \vec{OC} = 3i + 4j - 7k.

(a)

Show that $OABC$ is a rectangle.

4M Similar

DifficultyMedium

Worked solution

Approach

We use vectors. To show that $OABC$ is a rectangle, we show that one pair of adjacent sides is perpendicular and that both pairs of opposite sides are parallel and equal in length.

Working

Let

\vec{OA} = 5\mathbf{i} - 2\mathbf{j} + \mathbf{k}, \quad \vec{OB} = 8\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}, \quad \vec{OC} = 3\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}.

Then the sides of quadrilateral $OABC$ are:

\vec{AB} = \vec{OB} - \vec{OA} = (8\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}) - (5\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = 3\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}.

So $\vec{AB} = \vec{OC}$ , showing that side $AB$ is parallel and equal to side $OC$ .

Also,

\vec{BC} = \vec{OC} - \vec{OB} = (3\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}) - (8\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}) = -5\mathbf{i} + 2\mathbf{j} - \mathbf{k} = -\vec{OA},

so $\vec{BC}$ is parallel and equal in length to $\vec{OA}$ .

Now check a pair of adjacent sides:

\vec{OA} \cdot \vec{AB} = (5)(3) + (-2)(4) + (1)(-7) = 15 - 8 - 7 = 0.

Thus $\vec{OA}$ is perpendicular to $\vec{AB}$ . Since opposite sides are parallel and equal, $OABC$ is a parallelogram; with a right angle at $A$ , it is a rectangle.

Answer

$OABC$ is a rectangle because its adjacent sides are perpendicular and its opposite sides are parallel and equal in length.

Final answer

OABC is a rectangle because adjacent sides are perpendicular and opposite sides are parallel and equal in length.

Detailed explanation

Walkthrough

We start from the position vectors of $A$ , $B$ and $C$ relative to the origin $O$ . To show that $OABC$ is a rectangle, we first show that it is a parallelogram, then check that one of its angles is a right angle.

Compute the side vectors from the vertex order $O \to A \to B \to C$ . In vector form, $\vec{AB} = \vec{OB} - \vec{OA}$ , and subtraction gives $3\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}$ , which is exactly $\vec{OC}$ . Hence $AB$ is parallel to $OC$ and has the same length. Similarly, $\vec{BC} = \vec{OC} - \vec{OB} = -5\mathbf{i} + 2\mathbf{j} - \mathbf{k} = -\vec{OA}$ , so $BC$ is parallel to $OA$ and has the same length.

Next, use the scalar product to test whether adjacent sides meet at a right angle. Since $\vec{AB} = \vec{OC}$ , the side $AB$ has the same vector as $\vec{OC}$ . The scalar product $\vec{OA} \cdot \vec{AB} = 15 - 8 - 7 = 0$ , so the two adjacent sides $OA$ and $AB$ are perpendicular. A parallelogram with a right angle is a rectangle.

Key Takeaways

A rectangle is a parallelogram with a right angle. To prove a quadrilateral is a rectangle using vectors, show both pairs of opposite sides are parallel and equal, and use a zero scalar product to show one pair of adjacent sides is perpendicular. Vector subtraction is used to find side vectors from position vectors.

Common Mistakes

Using the given position vectors $\vec{OA}$ , $\vec{OB}$ , $\vec{OC}$ as if they were the four sides of the rectangle, instead of computing the actual side vectors of $OABC$ .
Omitting one of the two pairs of opposite sides when proving parallelism.
Only showing that lengths are equal without showing that the sides are parallel.
Sign errors when subtracting position vectors, e.g. writing $\vec{AB} = \vec{OA} - \vec{OB}$ .

Things to Be Careful About

The correct subtraction is $\vec{AB} = \vec{OB} - \vec{OA}$ .
A scalar product of zero means the vectors are perpendicular; to earn the mark, show the numerator $15 - 8 - 7$ , so the denominator is irrelevant here.
Showing $\vec{AB} = \vec{OC}$ is enough to establish both parallel and equal in length for that pair; confirming $\vec{BC} = -\vec{OA}$ handles the other pair.
If you only show equal lengths for the opposite sides, you have not yet proved they are parallel, which is needed for the rectangle conclusion.

Techniques used

subtract position vectors to find side vectorsapply the scalar product to test perpendicularitycompare opposite side vectors to show parallelism and equal length

(b)

Use a scalar product to find the acute angle between the diagonals of $OABC$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

We need the two diagonals of the parallelogram $OABC$ . Since $O$ is the origin and the vertices are $O,A,B,C$ , the diagonals are $\vec{OB}$ and $\vec{AC}$ . We use the scalar product formula

\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta.

Working

First find the diagonal $\vec{AC}$ :

\vec{AC} = \vec{OC} - \vec{OA} = (3\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}) - (5\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = -2\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}.

The other diagonal is

\vec{OB} = 8\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}.

Their scalar product is

\vec{AC} \cdot \vec{OB} = (-2)(8) + (6)(2) + (-8)(-6) = -16 + 12 + 48 = 44.

Their magnitudes are

|\vec{AC}| = \sqrt{(-2)^2 + 6^2 + (-8)^2} = \sqrt{4 + 36 + 64} = \sqrt{104}

and

|\vec{OB}| = \sqrt{8^2 + 2^2 + (-6)^2} = \sqrt{64 + 4 + 36} = \sqrt{104}.

Therefore,

\cos \theta = \frac{\vec{AC} \cdot \vec{OB}}{|\vec{AC}| |\vec{OB}|} = \frac{44}{\sqrt{104} \cdot \sqrt{104}} = \frac{44}{104} = \frac{11}{26}.

Since this dot product is positive, $\theta$ is the acute angle:

\theta = \cos^{-1}\left(\frac{11}{26}\right) = 65.0^\circ.

This is $1.13$ radians.

Answer

The acute angle between the diagonals is $65.0^\circ$ .

Final answer

65.0°

Detailed explanation

Walkthrough

In the quadrilateral $OABC$ , with vertices in the order $O, A, B, C$ , the two diagonals are $OB$ and $AC$ . We need the acute angle between them.

First find the vector for diagonal $AC$ by subtracting position vectors: $\vec{AC} = \vec{OC} - \vec{OA} = -2\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}$ . The other diagonal is $\vec{OB} = 8\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}$ .

Apply the scalar product formula

\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta.

The scalar product of the diagonals is $(-2)(8) + (6)(2) + (-8)(-6) = 44$ . The magnitudes are both $\sqrt{104}$ , so

\cos \theta = \frac{44}{\sqrt{104} \cdot \sqrt{104}} = \frac{44}{104} = \frac{11}{26}.

Because this value is positive, the angle found by $\cos^{-1}(11/26)$ is already the acute angle between the diagonals: $65.0^\circ$ (or $1.13$ radians).

Key Takeaways

This question tests the scalar product definition of the angle between two vectors, finding diagonal vectors by vector subtraction, and computing vector magnitudes. It also highlights that if the scalar product is positive, the angle from the cosine formula is acute, while a negative scalar product would need the supplement $180^\circ - \theta$ .

Common Mistakes

Using non-diagonal vectors such as $\vec{OA}$ and $\vec{OB}$ as the diagonals, or using four sides instead of diagonals.
Forgetting to divide by the product of the moduli; the cosine formula requires both magnitudes in the denominator.
Arithmetic sign mistakes in the scalar product, especially with the negative components.
If a negative scalar product occurs, reporting the obtuse angle instead of converting it to the acute angle.

Things to Be Careful About

The diagonals of $OABC$ are $\vec{OB}$ and $\vec{AC} = \vec{OC} - \vec{OA}$ , not $\vec{OA}$ and $\vec{BC}$ .
Here the scalar product is $44 > 0$ , so the resulting angle is acute. If it were negative, the acute angle would be $180^\circ$ minus the computed angle.
Simplify $44/104$ to $11/26$ ; the answer is accepted as $65.0^\circ$ or $1.13$ radians.
In the mark scheme, the vector for $AC$ may be written with either sign; using the opposite sign would change the sign of the scalar product, so choose the direction that gives the required acute angle.

Techniques used

find a diagonal vector by subtractioncompute the scalar product of the two diagonalscalculate magnitudes and apply the cosine formulareport the required acute angle

Q10Medium-HardAlgebraIntegrationLogarithmic and Exponential Functions

Let $f(x) = \frac{36a^2}{(2a + x)(2a - x)(5a - 2x)}$ , where $a$ is a positive constant.

(a)

Express $f(x)$ in partial fractions.

5M Similar

DifficultyMedium

Worked solution

Approach

Write the rational function in the standard partial fraction form for three distinct linear factors. Multiply through by the common denominator, then substitute values of $x$ that make each factor zero to find the coefficients.

Working

Let

\frac{36a^2}{(2a+x)(2a-x)(5a-2x)} = \frac{A}{2a+x} + \frac{B}{2a-x} + \frac{C}{5a-2x}

Multiplying by the denominator:

36a^2 = A(2a-x)(5a-2x) + B(2a+x)(5a-2x) + C(2a+x)(2a-x)

Set $x = -2a$ :

36a^2 = A(4a)(9a) = 36a^2 A \quad \Rightarrow \quad A = 1

Set $x = 2a$ :

36a^2 = B(4a)(a) = 4a^2 B \quad \Rightarrow \quad B = 9

Set $x = \frac{5a}{2}$ :

36a^2 = C\left(\frac{9a}{2}\right)\left(-\frac{a}{2}\right) = -\frac{9a^2}{4}C \quad \Rightarrow \quad C = -16

Answer

f(x) = \frac{1}{2a+x} + \frac{9}{2a-x} - \frac{16}{5a-2x}

Final answer

1/(2a+x) + 9/(2a-x) - 16/(5a-2x)

Detailed explanation

Walkthrough

We have a rational function whose denominator is a product of three distinct linear factors. The first step is to write the correct partial fraction form:

\frac{36a^2}{(2a+x)(2a-x)(5a-2x)} = \frac{A}{2a+x} + \frac{B}{2a-x} + \frac{C}{5a-2x}

This form is guaranteed because each factor is linear and none is repeated. The mark scheme gives a mark for stating this form before any coefficients are found.

Next, multiply both sides by the denominator. This clears the fractions and gives an identity valid for all $x$ :

36a^2 = A(2a-x)(5a-2x) + B(2a+x)(5a-2x) + C(2a+x)(2a-x)

Now choose values of $x$ that make one factor zero. This isolates one coefficient at a time. Setting $x=-2a$ makes the factor $2a+x$ zero, so only the $A$ term survives; this gives $A=1$ . Setting $x=2a$ makes $2a-x$ zero, giving $B=9$ . Setting $x=\frac{5a}{2}$ makes $5a-2x$ zero, giving $C=-16$ . The substitution method is quick and avoids expanding and comparing coefficients.

Key Takeaways

A rational function with three distinct linear factors decomposes into three simple fractions. Substituting the roots of the linear factors is an efficient way to find coefficients. The coefficients may depend on parameters such as $a$ , but here they are constants.

Common Mistakes

Using an incorrect form, such as combining the first two factors into one quadratic denominator. The mark scheme allows at most 2 marks if only $C=-16$ is found with the wrong form.
Not showing a valid method for finding a coefficient; the method mark requires more than just writing the final answer.
Sign errors when substituting $x=\frac{5a}{2}$ , because one of the remaining factors is negative.
Forgetting to multiply the whole numerator by the missing factors when clearing denominators.

Things to Be Careful About

The factors are distinct, so no repeated-factor term such as $\frac{B}{(2a-x)^2}$ is needed.
When substituting a value that makes a factor zero, that factor must be omitted from the corresponding product; otherwise the term becomes zero and the coefficient cannot be found.
Keep $a$ symbolic throughout; do not assign a numerical value to $a$ unless the question asks for it.

Techniques used

state the partial fraction form with distinct linear denominatorsmultiply through by the common denominatorsubstitute values of x to determine coefficients

(b)

Hence find the exact value of $\int_{-a}^{a} f(x)\,dx$ , giving your answer in the form $p\ln q + r\ln s$ where $p$ and $r$ are integers and $q$ and $s$ are prime numbers.

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Integrate each partial fraction term separately using the rule for $\frac{1}{ax+b}$ . Then evaluate the definite integral by substituting $x=a$ and $x=-a$ , and simplify the logarithms using the laws of logarithms.

Working

From part (a),

f(x) = \frac{1}{2a+x} + \frac{9}{2a-x} - \frac{16}{5a-2x}

Therefore

\int f(x)\,dx = \ln|2a+x| - 9\ln|2a-x| + 8\ln|5a-2x| + c

Let $F(x)$ be this antiderivative. At $x = a$ :

F(a) = \ln(3a) - 9\ln(a) + 8\ln(3a) = 9\ln(3a) - 9\ln(a)

At $x = -a$ :

F(-a) = \ln(a) - 9\ln(3a) + 8\ln(7a)

Thus

\begin{aligned} \int_{-a}^{a} f(x)\,dx &= F(a) - F(-a) \\ &= 9\ln(3a) - 9\ln(a) - \ln(a) + 9\ln(3a) - 8\ln(7a) \\ &= 18\ln(3a) - 10\ln(a) - 8\ln(7a) \end{aligned}

Using $\ln(3a) = \ln 3 + \ln a$ and $\ln(7a) = \ln 7 + \ln a$ :

18\ln(3a) - 10\ln(a) - 8\ln(7a) = 18\ln 3 - 8\ln 7

Answer

\int_{-a}^{a} f(x)\,dx = 18\ln 3 - 8\ln 7

So $p = 18$ , $q = 3$ , $r = -8$ and $s = 7$ .

Final answer

18 ln 3 - 8 ln 7

Detailed explanation

Walkthrough

Start from the partial fraction decomposition in part (a). Integrate each term separately. For $\frac{1}{2a+x}$ , the derivative of the denominator is $1$ , so the integral is $\ln|2a+x|$ . For $\frac{9}{2a-x}$ , the derivative of the denominator is $-1$ , so the integral is $-9\ln|2a-x|$ . For $-\frac{16}{5a-2x}$ , the derivative of the denominator is $-2$ , so the integral is $+8\ln|5a-2x|$ .

Define $F(x)$ as the antiderivative. The definite integral is $F(a)-F(-a)$ . Since $a>0$ and $x$ lies between $-a$ and $a$ , all of $2a+x$ , $2a-x$ and $5a-2x$ are positive, so the modulus signs can be removed. Substitute the limits:

F(a)=9\ln(3a)-9\ln(a), \qquad F(-a)=\ln(a)-9\ln(3a)+8\ln(7a)

Then subtract carefully:

F(a)-F(-a)=18\ln(3a)-10\ln(a)-8\ln(7a)

Finally use $\ln(3a)=\ln 3+\ln a$ and $\ln(7a)=\ln 7+\ln a$ . The $\ln a$ terms cancel, leaving $18\ln 3-8\ln 7$ . This matches $p\ln q+r\ln s$ with $p=18$ , $q=3$ , $r=-8$ , $s=7$ .

Key Takeaways

A partial fraction decomposition converts a complicated rational integrand into simple reciprocal-linear terms that integrate to logarithms. Definite integrals are evaluated by substituting the upper limit then subtracting the lower limit. Laws of logarithms are often needed to simplify the result into the requested form.

Common Mistakes

Forgetting the factor from the chain rule when integrating $-\frac{16}{5a-2x}$ ; the correct term is $+8\ln|5a-2x|$ , not $-16\ln|5a-2x|$ or $-8\ln|5a-2x|$ .
Subtracting $F(-a)$ incorrectly. Every term of $F(-a)$ must be subtracted, which changes several signs.
Assigning a numerical value to $a$ . The mark scheme awards at most 3 marks in part (b) if a value is assigned to $a$ .
Writing logarithms of negative numbers; the mark scheme gives A0 if the solution involves logarithms of negative numbers. Here all arguments are positive on the interval.
Stopping before the $\ln a$ terms cancel; the final answer must be independent of $a$ and in the form $p\ln q+r\ln s$ with prime bases.

Things to Be Careful About

The order of limits is important: evaluate $F(a)-F(-a)$ , not $F(-a)-F(a)$ .
When combining $\ln(3a)$ terms, note that $\ln(3a)=\ln 3+\ln a$ ; the $\ln a$ contributions must cancel exactly.
The coefficient of $\ln(3a)$ is $1+8=9$ at the upper limit and $-9$ at the lower limit, giving $18\ln(3a)$ in total.
The answer should be expressed with prime bases: $q=3$ and $s=7$ .

Techniques used

integrate reciprocal-linear termsevaluate definite integral using antiderivative at limitsapply laws of logarithms to simplify and remove a

Q119MMedium-HardDifferential EquationsIntegration

The variables $y$ and $\theta$ satisfy the differential equation

(1 + y)(1 + \cos 2\theta)\frac{dy}{d\theta} = e^{3y}.

It is given that $y = 0$ when $\theta = \frac{1}{4}\pi$ .

Solve the differential equation and find the exact value of $\tan\theta$ when $y = 1$ .

DifficultyMedium-Hard

Worked solution

Approach

Separate the variables so that all $y$ terms are on one side and all $\theta$ terms on the other. Integrate the left side using integration by parts and the right side using the identity $1 + \cos 2\theta = 2\cos^2\theta$ . Use the given initial condition to fix the constant, then substitute $y = 1$ and solve for $\tan\theta$ .

Working

Separate variables:

\int (1 + y)e^{-3y}\,dy = \int \frac{1}{1 + \cos 2\theta}\,d\theta

Integrate the left side by parts with $u = 1 + y$ , $dv = e^{-3y}\,dy$ , so $du = dy$ and $v = -\frac{1}{3}e^{-3y}$ :

\int (1 + y)e^{-3y}\,dy = -\frac{1}{3}(1 + y)e^{-3y} + \frac{1}{3}\int e^{-3y}\,dy

= -\frac{1}{3}(1 + y)e^{-3y} - \frac{1}{9}e^{-3y} + C

For the right side, use $1 + \cos 2\theta = 2\cos^2\theta$ :

\int \frac{1}{1 + \cos 2\theta}\,d\theta = \int \frac{1}{2\cos^2\theta}\,d\theta = \int \frac{1}{2}\sec^2\theta\,d\theta = \frac{1}{2}\tan\theta + D

Combine the constants into one constant $C$ :

-\frac{1}{3}(1 + y)e^{-3y} - \frac{1}{9}e^{-3y} + C = \frac{1}{2}\tan\theta

Use $y = 0$ , $\theta = \frac{\pi}{4}$ . Since $\tan\frac{\pi}{4} = 1$ and $e^0 = 1$ :

-\frac{1}{3} - \frac{1}{9} + C = \frac{1}{2}

-\frac{4}{9} + C = \frac{1}{2}

C = \frac{1}{2} + \frac{4}{9} = \frac{17}{18}

So the particular solution is:

-\frac{1}{3}(1 + y)e^{-3y} - \frac{1}{9}e^{-3y} + \frac{17}{18} = \frac{1}{2}\tan\theta

Now set $y = 1$ :

-\frac{2}{3}e^{-3} - \frac{1}{9}e^{-3} + \frac{17}{18} = \frac{1}{2}\tan\theta

-\frac{7}{9}e^{-3} + \frac{17}{18} = \frac{1}{2}\tan\theta

Multiply by 2:

\tan\theta = \frac{17}{9} - \frac{14}{9}e^{-3}

Answer

\tan\theta = \frac{17}{9} - \frac{14}{9}e^{-3}

Final answer

tan theta = 17/9 - 14/(9e^3)

Detailed explanation

Walkthrough

This is a separable first-order differential equation. The first step is to rearrange it so that every $y$ appears with $dy$ and every $\theta$ appears with $d\theta$ . Dividing both sides by $e^{3y}$ and by $1 + \cos 2\theta$ gives

\int (1 + y)e^{-3y}\,dy = \int \frac{1}{1 + \cos 2\theta}\,d\theta.

The left-hand side is a product of a linear factor and an exponential, so integration by parts is the natural tool. Let $u = 1 + y$ and $dv = e^{-3y}\,dy$ . Then $du = dy$ and $v = -\frac{1}{3}e^{-3y}$ . The formula $\int u\,dv = uv - \int v\,du$ gives

-\frac{1}{3}(1 + y)e^{-3y} + \frac{1}{3}\int e^{-3y}\,dy = -\frac{1}{3}(1 + y)e^{-3y} - \frac{1}{9}e^{-3y} + C.

For the right-hand side, the denominator $1 + \cos 2\theta$ is simplified using the double-angle identity $1 + \cos 2\theta = 2\cos^2\theta$ . This turns the integrand into $\frac{1}{2}\sec^2\theta$ , whose integral is $\frac{1}{2}\tan\theta$ .

After adding the constant of integration, the general solution is

-\frac{1}{3}(1 + y)e^{-3y} - \frac{1}{9}e^{-3y} + C = \frac{1}{2}\tan\theta.

The initial condition $y = 0$ when $\theta = \frac{\pi}{4}$ is used to find $C$ . Since $e^0 = 1$ and $\tan\frac{\pi}{4} = 1$ , substituting gives

-\frac{1}{3} - \frac{1}{9} + C = \frac{1}{2},

so $C = \frac{17}{18}$ .

Finally, substitute $y = 1$ into the particular solution:

-\frac{2}{3}e^{-3} - \frac{1}{9}e^{-3} + \frac{17}{18} = \frac{1}{2}\tan\theta.

Combining the exponential terms gives $-\frac{7}{9}e^{-3}$ , so

\frac{1}{2}\tan\theta = \frac{17}{18} - \frac{7}{9}e^{-3}.

Multiplying by 2 yields the exact value

\tan\theta = \frac{17}{9} - \frac{14}{9}e^{-3}.

Key Takeaways

This question combines several core skills: separating variables in a differential equation, integrating a product using integration by parts, using a trigonometric identity to make an integral tractable, and using an initial condition to determine an arbitrary constant. It also tests careful algebraic manipulation with fractions and exponentials. After solving, a student should be comfortable recognising when a first-order differential equation is separable and executing each integration technique accurately.

Common Mistakes

A common mistake is separating variables incorrectly, for example writing $\int (1 + y)e^{3y}\,dy$ instead of $\int (1 + y)e^{-3y}\,dy$ . In integration by parts, sign errors frequently occur in the second term: the integral of $-\frac{1}{3}e^{-3y}$ is $-\frac{1}{9}e^{-3y}$ , not $+\frac{1}{9}e^{-3y}$ . Another common error is forgetting the constant of integration or placing it inconsistently. On the right-hand side, students sometimes fail to use $1 + \cos 2\theta = 2\cos^2\theta$ and attempt a more complicated integration. The mark scheme also requires an exact final answer; a decimal approximation is not accepted, and using $x$ instead of $\theta$ loses the final mark.

Things to Be Careful About

Be consistent about which side carries the constant of integration. Here it is convenient to write $+C$ on the left-hand side, giving $C = \frac{17}{18}$ ; if it is put on the right-hand side, the sign of $C$ changes. When substituting $y = 1$ , remember that $e^{-3}$ is a common factor: $-\frac{2}{3}e^{-3} - \frac{1}{9}e^{-3} = -\frac{7}{9}e^{-3}$ . Finally, multiply by 2 correctly to obtain $\frac{17}{9} - \frac{14}{9}e^{-3}$ , and leave the answer in exact form with $\theta$ as the variable.

Techniques used

separate variablesintegrate by partsapply double-angle identityuse initial condition to find constantsolve for tan theta

Mathematics 9709/32 — February/March 2024

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