The complex numbers $z$ and $\omega$ are defined by $z = 1 - \mathbf{i}$ and $\omega = -3 + 3\sqrt{3}\mathbf{i}$. (a) Express $z\omega$ in the form $a + b\mathbf{i}$, where $a$ and $b$ are real and in exact surd form. (b) Express $z$ and $\omega$ in the form $re^{\mathbf{i}\theta}$, where $r > 0$ and $-\pi < \theta \le \pi$. Give the exact values of $r$ and $\theta$ in each case. (c) On an Argand diagram, the points representing $\omega$ and $z\omega$ are $A$ and $B$…

(a) zw = (-3 + 3√3) + (3 + 3√3)i · (b) z = √2 e^(-iπ/4), ω = 6 e^(2iπ/3) · (c) Triangle OAB is isosceles with OA = AB = 6 and is right-angled at A · (d) tan(5π/12) = (√3 + 1)/(√3 - 1)

9709/32 May/June 2024 — Mathematics Solved Past Paper

Q1MediumAlgebra

(a)

Sketch the graph of $y = |x - 2a|$ , where $a$ is a positive constant.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

The function $y = |x - 2a|$ is a modulus function. Its graph is V-shaped with the vertex on the x-axis where the expression inside the modulus is zero. We find the vertex, the y-intercept, and note the slopes of the two linear branches to sketch the graph.

Working

The expression inside the modulus is zero when $x - 2a = 0$ , so $x = 2a$ . At this point, $y = 0$ . Thus, the vertex of the graph is at $(2a, 0)$ .

To find the y-intercept, set $x = 0$ :

y = |0 - 2a| = |-2a| = 2a

since $a$ is a positive constant. The y-intercept is at $(0, 2a)$ .

The graph consists of two straight lines meeting at the vertex:

For $x \geq 2a$ , $y = x - 2a$ (slope $+1$ ).
For $x < 2a$ , $y = -(x - 2a) = -x + 2a$ (slope $-1$ ).

The left arm extends into the second quadrant (negative $x$ values) as $x \to -\infty$ .

Answer

y = |x - 2a| \text{ is a V-shaped graph with vertex at } (2a, 0) \text{ and y-intercept at } (0, 2a).

Final answer

V-shaped graph with vertex at (2a, 0) and y-intercept at (0, 2a)

Detailed explanation

Walkthrough

First, identify the key features of the modulus function $y = |x - 2a|$ . The vertex occurs where the expression inside the absolute value is zero, which is $x = 2a$ . At this point, $y = 0$ , giving the vertex $(2a, 0)$ . Next, find the y-intercept by substituting $x = 0$ , which gives $y = |-2a| = 2a$ (since $a > 0$ ). The graph is composed of two linear rays: $y = x - 2a$ for $x \geq 2a$ and $y = -x + 2a$ for $x < 2a$ . Sketch these as solid straight lines meeting at the vertex, ensuring the left arm extends into the negative $x$ region.

Key Takeaways

The graph of $y = |x - c|$ is V-shaped with its vertex at $(c, 0)$ .
The y-intercept of $y = |x - c|$ (for $c > 0$ ) is at $(0, c)$ .
Modulus graphs consist of straight line segments with slopes $\pm 1$ .

Common Mistakes

Forgetting to label the axes intercepts with the constant $a$ (e.g., writing just $2$ instead of $2a$ ).
Drawing curved lines instead of straight lines for the branches of the modulus graph.
Not extending the left branch into the negative $x$ -axis region.

Things to Be Careful About

The constant $a$ is positive, so $|-2a| = 2a$ .
Both branches must be solid straight lines; construction lines should be dashed or fainter if used.
The vertex must be clearly marked at $x = 2a$ on the x-axis and $y = 2a$ on the y-axis.

Techniques used

identify vertex and y-intercept of modulus functionsketch V-shaped graph with straight line segments

(b)

Solve the inequality $2x - 3a < |x - 2a|$ .

2M Similar

DifficultyMedium

Worked solution

Approach

To solve $2x - 3a < |x - 2a|$ , we consider cases based on the sign of the expression inside the modulus, $x - 2a$ . Alternatively, we can use the fact that if $A < |B|$ , then either $A < 0$ (which automatically satisfies the inequality since $|B| \geq 0$ ) or $A \geq 0$ and $A^2 < B^2$ .

Working

Case 1: $x - 2a < 0$ , i.e., $x < 2a$ .
Here, $|x - 2a| = -(x - 2a) = -x + 2a$ . The inequality becomes:

2x - 3a < -x + 2a

3x < 5a

x < \frac{5}{3}a

Since $\frac{5}{3}a < 2a$ , the condition $x < 2a$ is automatically satisfied. Thus, $x < \frac{5}{3}a$ is a solution in this case.

Case 2: $x - 2a \geq 0$ , i.e., $x \geq 2a$ .
Here, $|x - 2a| = x - 2a$ . The inequality becomes:

2x - 3a < x - 2a

x < a

But we require $x \geq 2a$ . Since $a < 2a$ (as $a > 0$ ), there is no solution in this case.

Combining the cases:
The solution is $x < \frac{5}{3}a$ .

Answer

x < \frac{5}{3}a

Final answer

x < 5a/3

Detailed explanation

Walkthrough

We solve $2x - 3a < |x - 2a|$ by splitting into two cases based on the definition of the modulus function.

Case 1: When $x < 2a$ , the modulus $|x - 2a|$ equals $-(x - 2a) = -x + 2a$ . Substituting this into the inequality gives $2x - 3a < -x + 2a$ , which simplifies to $3x < 5a$ , or $x < \frac{5}{3}a$ . Since $\frac{5}{3}a < 2a$ , this entire region is valid.

Case 2: When $x \geq 2a$ , the modulus $|x - 2a|$ equals $x - 2a$ . The inequality becomes $2x - 3a < x - 2a$ , which simplifies to $x < a$ . However, this contradicts our assumption that $x \geq 2a$ (since $a < 2a$ ). Thus, there is no solution from this case.

The final solution is the union of valid solutions from both cases: $x < \frac{5}{3}a$ .

Key Takeaways

Solving modulus inequalities often requires splitting into cases based on the sign of the expression inside the modulus.
Always check that the solution from each case is consistent with the condition assumed for that case.
Squaring both sides of an inequality involving a modulus can introduce extraneous solutions if the left-hand side is negative.

Common Mistakes

Squaring both sides without ensuring the left-hand side is non-negative, leading to incorrect solutions like $a < x < \frac{5}{3}a$ .
Forgetting to reject solutions that fall outside the assumed range for a particular case.
Writing $x > a$ or $x < a$ as the final answer instead of combining with the case conditions.

Things to Be Careful About

The inequality is strict ( $<$ ), so the final answer must be a strict inequality.
Ensure $a$ is treated as a positive constant throughout; do not substitute a numerical value for $a$ .
The solution $x < \frac{5}{3}a$ must be clearly stated as the final answer.

Techniques used

split modulus inequality into cases based on sign of inner expressionsolve linear inequalitiesfind intersection of solution sets

Q25MMediumAlgebra

Express

\frac{6x^2 - 9x - 16}{2x^2 - 5x - 12}

in partial fractions.

DifficultyMedium

Worked solution

Approach

The numerator and denominator have the same degree, so the fraction is improper. We first write it as a constant plus proper partial fractions, then determine the constants by substituting the roots of the denominator and comparing coefficients.

Working

Factorise the denominator:

2x^2 - 5x - 12 = (2x+3)(x-4)

Since the numerator has degree equal to the denominator, write

\frac{6x^2 - 9x - 16}{(2x+3)(x-4)} = A + \frac{B}{2x+3} + \frac{C}{x-4}

Multiply through by $(2x+3)(x-4)$ :

6x^2 - 9x - 16 = A(2x+3)(x-4) + B(x-4) + C(2x+3)

Compare coefficients of $x^2$ :

6 = 2A \quad \Rightarrow \quad A = 3

Substitute $x = 4$ :

6(4)^2 - 9(4) - 16 = C(2(4)+3)

44 = 11C \quad \Rightarrow \quad C = 4

Substitute $x = -\frac{3}{2}$ :

6\left(-\frac{3}{2}\right)^2 - 9\left(-\frac{3}{2}\right) - 16 = B\left(-\frac{3}{2} - 4\right)

11 = -\frac{11}{2}B \quad \Rightarrow \quad B = -2

Answer

\frac{6x^2 - 9x - 16}{2x^2 - 5x - 12} = 3 - \frac{2}{2x+3} + \frac{4}{x-4}

Final answer

3 - 2/(2x+3) + 4/(x-4)

Detailed explanation

Walkthrough

We start by factorising the denominator. The quadratic $2x^2 - 5x - 12$ factors as $(2x+3)(x-4)$ , which gives us the two distinct linear factors needed for partial fractions. Because the numerator and denominator have the same degree, the fraction is improper, so the partial fraction form must include a constant term $A$ :

\frac{6x^2 - 9x - 16}{(2x+3)(x-4)} = A + \frac{B}{2x+3} + \frac{C}{x-4}

This is the form we state before finding any constants. Next, we multiply both sides by the denominator to clear fractions. The result is an identity in $x$ :

6x^2 - 9x - 16 = A(2x+3)(x-4) + B(x-4) + C(2x+3)

We now find the constants. The constant $A$ is easiest to find by comparing the coefficient of $x^2$ : on the left it is $6$ , and on the right the only $x^2$ term comes from $A(2x+3)(x-4)$ , whose leading term is $2Ax^2$ . Hence $2A=6$ , so $A=3$ .

For $B$ and $C$ , we substitute values of $x$ that make one of the linear factors zero. Substituting $x=4$ makes $x-4=0$ , so the $B$ term disappears and we can solve directly for $C$ :

44 = 11C \quad \Rightarrow \quad C=4

Substituting $x=-\frac{3}{2}$ makes $2x+3=0$ , so the $A$ and $C$ terms vanish and we solve for $B$ :

11 = -\frac{11}{2}B \quad \Rightarrow \quad B=-2

Finally, we substitute these values back into the partial fraction form to write the final answer.

Key Takeaways

This question tests the ability to handle an improper algebraic fraction in partial fractions. The key steps are: factorise the denominator, recognise when a constant term is needed, clear the denominator, and use either substitution of roots or coefficient comparison to find the unknown constants. The method of substituting the roots of the denominator is often the fastest way to find the coefficients of the linear denominators.

Common Mistakes

A common mistake is to omit the constant term $A$ because the numerator and denominator have the same degree. Another is to factorise the denominator incorrectly, for example writing $(x+\frac{3}{2})(x-4)$ instead of $(2x+3)(x-4)$ ; the mark scheme notes that this can limit the marks available. Students may also forget to multiply the $A$ term by the full denominator when clearing fractions, or may make arithmetic errors when substituting $x=-\frac{3}{2}$ .

Things to Be Careful About

Always check that the fraction is proper before choosing the partial fraction form. Since the degrees are equal here, the form must include a constant. When substituting $x=-\frac{3}{2}$ , be careful with the negative signs and fractions. It is also worth checking the final answer by combining the terms back over a common denominator, or by testing a value such as $x=0$ . The mark scheme requires a clear method for finding the constants; an unsupported final answer may not receive full credit.

Techniques used

factorise quadratic denominatorwrite improper fraction in partial fraction formsubstitute roots to find constantscompare coefficients to find constant term

Approach

Take logarithms of both sides to bring the powers down. Expand the brackets, collect the terms in $y$ on one side, factorise, and rearrange into the form $y = mx + c$ . This is the equation of a straight line.

Working

(2y - 1)\ln a = (x - y)\ln b

Expanding:

2y\ln a - \ln a = x\ln b - y\ln b

Collecting $y$ -terms on the left:

2y\ln a + y\ln b = x\ln b + \ln a

Factorising:

y(2\ln a + \ln b) = x\ln b + \ln a

Dividing by the constant coefficient of $y$ :

y = \frac{\ln b}{2\ln a + \ln b}x + \frac{\ln a}{2\ln a + \ln b}

This has the form $y = mx + c$ , where $m = \frac{\ln b}{2\ln a + \ln b}$ and $c = \frac{\ln a}{2\ln a + \ln b}$ are constants. Hence the graph of $y$ against $x$ is a straight line.

Answer

y = \frac{\ln b}{2\ln a + \ln b}x + \frac{\ln a}{2\ln a + \ln b}

This is of the form $y = mx + c$ , so the graph is a straight line.

Final answer

The graph is a straight line, with equation y = (ln b/(2 ln a + ln b))x + (ln a/(2 ln a + ln b)).

Detailed explanation

Walkthrough

The equation has the unknown in the exponents, so direct comparison is not possible. Taking logarithms of both sides brings the powers down as multipliers:

(2y - 1)\ln a = (x - y)\ln b

This is the key step: it converts an exponential equation into a linear equation in $x$ and $y$ . Next, expand the brackets so that the terms involving $y$ can be collected together:

2y\ln a - \ln a = x\ln b - y\ln b

Move the $y$ -terms to one side and the constant terms to the other:

2y\ln a + y\ln b = x\ln b + \ln a

Factorise $y$ out of the left-hand side:

y(2\ln a + \ln b) = x\ln b + \ln a

Finally divide by the constant $2\ln a + \ln b$ :

y = \frac{\ln b}{2\ln a + \ln b}x + \frac{\ln a}{2\ln a + \ln b}

This is exactly the form $y = mx + c$ , with $m$ and $c$ constants, so the graph is a straight line.

Key Takeaways

An exponential equation with two variables can often be turned into a linear equation by taking logarithms.
The coefficients of $y$ must be collected and factorised before the equation can be written in gradient–intercept form.
A relationship of the form $y = mx + c$ always has a straight-line graph.

Common Mistakes

Forgetting to expand the brackets, leading to missing terms when collecting $y$ .
Losing the negative sign on $\ln a$ when moving it across the equation.
Stating that the equation is linear without actually showing it has the form $y = mx + c$ ; the mark scheme requires a clear comparison with the linear form.

Things to Be Careful About

Any base of logarithm can be used, but the same base must be used on both sides.
The quantities $\ln a$ and $\ln b$ are constants because $a$ and $b$ are constants.
Do not condone missing brackets; write $(2y - 1)\ln a$ rather than $2y - 1\ln a$ .

Techniques used

take logarithms of both sidesexpand and collect terms in yfactorise and rearrange into y = mx + c form

(b)

Given that $a = b^3$ , state the equation of the straight line in the form $y = px + q$ , where $p$ and $q$ are rational numbers in their simplest form.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute $a = b^3$ into the logarithmic equation obtained in part (a), then use the laws of logarithms to simplify the coefficient of $y$ and the constant term.

Working

From part (a):

y(2\ln a + \ln b) = x\ln b + \ln a

Since $a = b^3$ , we have $\ln a = 3\ln b$ . Substituting:

y(2(3\ln b) + \ln b) = x\ln b + 3\ln b

y(6\ln b + \ln b) = x\ln b + 3\ln b

y(7\ln b) = (x + 3)\ln b

Dividing by $7\ln b$ :

y = \frac{(x + 3)\ln b}{7\ln b} = \frac{1}{7}x + \frac{3}{7}

This is in the required form $y = px + q$ , with $p = \frac{1}{7}$ and $q = \frac{3}{7}$ .

Answer

y = \frac{1}{7}x + \frac{3}{7}

Final answer

y = (1/7)x + 3/7

Detailed explanation

Walkthrough

We already know from part (a)that

y(2\ln a + \ln b) = x\ln b + \ln a

Use the given relationship $a = b^3$ . Taking logarithms gives $\ln a = 3\ln b$ . Substitute this into the equation:

y(2(3\ln b) + \ln b) = x\ln b + 3\ln b

Simplify:

y(6\ln b + \ln b) = x\ln b + 3\ln b

y(7\ln b) = (x + 3)\ln b

Divide through by $7\ln b$ :

y = \frac{(x + 3)\ln b}{7\ln b} = \frac{1}{7}x + \frac{3}{7}

This is in the required form $y = px + q$ , with $p = \frac{1}{7}$ and $q = \frac{3}{7}$ .

Key Takeaways

A substitution given in the question can be substituted into a previously derived logarithmic equation.
The laws of logarithms allow $\ln(b^3)$ to become $3\ln b$ , simplifying the coefficients.
The final answer must be written in the form $y = px + q$ , not as a single fraction.

Common Mistakes

Writing the final answer as $y = \frac{x+3}{7}$ : this is equivalent but not in the required form $y = px + q$ , and is not accepted by the mark scheme.
Forgetting to substitute $\ln a = 3\ln b$ into both the coefficient of $y$ and the constant term.
Dividing incorrectly by the coefficient of $y$ ; every term on the right must be divided by $7\ln b$ .

Things to Be Careful About

The mark scheme accepts $y = \frac{x}{7} + \frac{3}{7}$ as well as $y = \frac{1}{7}x + \frac{3}{7}$ .
If using the alternative exponential method, equate the exponents after writing $a$ as $b^3$ ; this gives $6y - 3 = x - y$ , leading to the same answer.
Keep the logarithms consistent: since $a = b^3$ , $\ln a = 3\ln b$ for any valid base.

Techniques used

substitute the given relationship between constantsapply laws of logarithmssimplify into gradient-intercept form

Q46MMediumDifferentiation

The equation of a curve is $ye^{2x} + y^2e^x = 6$ .

Find the gradient of the curve at the point where $y = 1$ .

DifficultyMedium

Worked solution

Approach

This is an implicit differentiation problem. First substitute $y=1$ into the curve equation and solve the resulting quadratic in $e^x$ to find the value of $x$ at the required point. Then differentiate the whole equation implicitly with respect to $x$ , using the product rule on $ye^{2x}$ and $y^2e^x$ . Finally substitute the known values of $y$ and $e^x$ and solve for $\frac{dy}{dx}$ .

Working

Substitute $y=1$ into the curve equation:

e^{2x}+e^x=6

e^{2x}+e^x-6=0

Let $u=e^x$ . Then:

u^2+u-6=0

(u+3)(u-2)=0

Since $u=e^x>0$ , the only valid solution is $u=2$ , so $e^x=2$ and $x=\ln 2$ .

Now differentiate the equation implicitly with respect to $x$ .

For the first term:

\frac{d}{dx}\left(ye^{2x}\right)=2ye^{2x}+e^{2x}\frac{dy}{dx}

For the second term:

\frac{d}{dx}\left(y^2e^x\right)=y^2e^x+2ye^x\frac{dy}{dx}

The derivative of the right-hand side is zero, so:

2ye^{2x}+e^{2x}\frac{dy}{dx}+y^2e^x+2ye^x\frac{dy}{dx}=0

At the required point, $y=1$ , $e^x=2$ and $e^{2x}=4$ . Substitute these values:

2(1)(4)+4\frac{dy}{dx}+(1)^2(2)+2(1)(2)\frac{dy}{dx}=0

8+4\frac{dy}{dx}+2+4\frac{dy}{dx}=0

8\frac{dy}{dx}=-10

\frac{dy}{dx}=-\frac{5}{4}

Answer

\frac{dy}{dx}=-\frac{5}{4}

The gradient of the curve at the point where $y=1$ is $-\frac{5}{4}$ .

Final answer

-5/4

Detailed explanation

Walkthrough

We are told the curve is defined implicitly by an equation involving both $y$ and $x$ , and we need the gradient $\frac{dy}{dx}$ at a specific $y$ -value. The first step is to find the corresponding $x$ -value, or more usefully the value of $e^x$ , by substituting $y=1$ .

Substituting $y=1$ gives $e^{2x}+e^x=6$ , which is a quadratic in $e^x$ . Writing $u=e^x$ turns it into $u^2+u-6=0$ . Factorising gives $(u+3)(u-2)=0$ , so $u=-3$ or $u=2$ . Since $e^x$ is always positive, $u=-3$ is impossible, so $e^x=2$ . This is important because the derivatives below need $e^x$ and $e^{2x}$ .

Next, differentiate the whole equation with respect to $x$ . Because $y$ is a function of $x$ , we must use implicit differentiation. The term $ye^{2x}$ is a product of $y$ and $e^{2x}$ , so the product rule gives $2ye^{2x}+e^{2x}\frac{dy}{dx}$ . The term $y^2e^x$ is also a product; differentiating $y^2$ gives $2y\frac{dy}{dx}$ , so the result is $y^2e^x+2ye^x\frac{dy}{dx}$ . The right-hand side is the constant $6$ , whose derivative is $0$ .

Now substitute $y=1$ , $e^x=2$ and $e^{2x}=4$ into the differentiated equation. This gives $8+4\frac{dy}{dx}+2+4\frac{dy}{dx}=0$ , so $8\frac{dy}{dx}=-10$ and $\frac{dy}{dx}=-\frac{5}{4}$ .

Key Takeaways

This question tests implicit differentiation, the product rule, and the ability to solve an equation that is quadratic in $e^x$ . It also shows that when a curve is given implicitly, we can find the gradient at a point without first making $y$ the subject, by differentiating both sides with respect to $x$ and then substituting coordinates.

Common Mistakes

Forgetting that $e^x>0$ and accepting $e^x=-3$ as a solution.
Applying the product rule incorrectly to $ye^{2x}$ or $y^2e^x$ , especially forgetting the factor $\frac{dy}{dx}$ when differentiating $y$ or $y^2$ .
Forgetting that the derivative of the constant $6$ is $0$ .
Substituting $y=1$ but using the wrong value for $e^x$ , or using $x=1$ instead of $e^x=2$ .
Giving a decimal rounded to $-1.25$ but not exact; the mark scheme requires exact $-\frac{5}{4}$ or $-1.25$ from correct working.

Things to Be Careful About

The negative root $e^x=-3$ must be rejected because an exponential function is never negative.
When substituting, $e^{2x}=(e^x)^2=4$ , not $2x$ .
The derivative of $ye^{2x}$ has two terms because of the product rule; the derivative of $y^2e^x$ also has two terms.
The mark scheme awards the final answer only from correct working, so show the substitution step clearly.
Either keep the equation in its original form or divide through by $e^x$ before differentiating; both methods are valid, but the substitution step must be shown.

Techniques used

solve a quadratic in e^xdifferentiate implicitly using the product rulesubstitute the point to find the gradient

Approach

Let $f(x)=e^{2x}-5-\cos 3x$ . A root of the equation is a zero of $f$ . Evaluate $f$ at $x=0.7$ and $x=0.8$ ; if the values have opposite signs, continuity of $f$ guarantees at least one root in the interval.

Working

At $x=0.7$ :

f(0.7)=e^{1.4}-5-\cos(2.1)

=4.05520\ldots - 5 - (-0.50485\ldots)

= -0.43995\ldots < 0

At $x=0.8$ :

f(0.8)=e^{1.6}-5-\cos(2.4)

=4.95303\ldots - 5 - (-0.73739\ldots)

=0.69043\ldots > 0

Since $f(0.7)<0$ and $f(0.8)>0$ , and $f$ is continuous, there is a sign change in $(0.7,0.8)$ . Hence the root lies in $0.7<x<0.8$ .

Answer

The root lies in the interval $0.7<x<0.8$ .

Final answer

The root lies in the interval 0.7 < x < 0.8

Detailed explanation

Walkthrough

Define $f(x)=e^{2x}-5-\cos 3x$ . The equation $e^{2x}=5+\cos 3x$ is equivalent to $f(x)=0$ . We choose two points, $0.7$ and $0.8$ , and evaluate $f$ . At $0.7$ , $f$ is negative; at $0.8$ , $f$ is positive. Because $f$ is continuous, a continuous function cannot pass from negative to positive without crossing zero. Therefore there is a root in between. Since the question states there is only one root, this is the unique root.

Key Takeaways

A sign change of a continuous function across an interval is sufficient to prove the existence of a root. Always check that trigonometric values are calculated in radians when working with numerical methods.

Common Mistakes

Working in degrees instead of radians; this gives different values and can falsely show no sign change.
Only evaluating one endpoint or not comparing the signs clearly.
Forgetting that a sign change proves at least one root, not uniqueness; here uniqueness is given.

Things to Be Careful About

Use radians. Keep enough decimal places to see the sign clearly. The sign-change argument requires the function to be continuous on the interval; here it is.

Techniques used

evaluate the function at interval endpointscompare signs to detect a sign changeapply continuity to conclude a root lies in the interval

(b)

Show that if a sequence of values in the interval $0.7 < x < 0.8$ given by the iterative formula

x_{n+1} = \frac{1}{2}\ln(5 + \cos 3x_n)

converges then it converges to the root of the equation in part (a).

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

Start from the original equation and rearrange it into the form $x = \frac{1}{2}\ln(5+\cos 3x)$ . If the sequence converges to a limit $L$ , then the recurrence tends to $L=F(L)$ , which is exactly the rearranged equation.

Working

Given

e^{2x} = 5 + \cos 3x

take natural logarithms of both sides:

2x = \ln(5+\cos 3x)

so

x = \frac{1}{2}\ln(5+\cos 3x)

If $x_n \to L$ as $n \to \infty$ , then in the recurrence

x_{n+1} = \frac{1}{2}\ln(5+\cos 3x_n)

both sides tend to $L$ , giving

L = \frac{1}{2}\ln(5+\cos 3L)

which is equivalent to

e^{2L} = 5 + \cos 3L

Thus $L$ is a root of the original equation.

Answer

Any convergent limit of the sequence satisfies $e^{2L}=5+\cos 3L$ , so it converges to the root.

Final answer

The limit L satisfies e^(2L) = 5 + cos 3L, so it is the root.

Detailed explanation

Walkthrough

The iterative formula is obtained by solving the original equation for $x$ . Taking logs of both sides gives $2x=\ln(5+\cos 3x)$ , so $x=\frac{1}{2}\ln(5+\cos 3x)$ . If the sequence converges to $L$ , then as $n$ grows, $x_{n+1}$ and $x_n$ both approach $L$ . Substituting $L$ into the recurrence gives $L=F(L)$ . This fixed-point equation is exactly the rearranged original equation, so $L$ is a root. Since the equation has only one root, any convergent sequence must converge to that root.

Key Takeaways

An iterative formula is often a rearrangement of the original equation into the form $x=F(x)$ . A convergent sequence must converge to a fixed point of $F$ , which is a root of the original equation.

Common Mistakes

Not taking logs correctly, especially forgetting the factor of 2.
Assuming convergence without justification; the question only asks to show that if it converges, the limit is the root.
Reversing the rearrangement and not arriving at the iterative formula.

Things to Be Careful About

The natural logarithm is defined because $5+\cos 3x>0$ . Use radians for $\cos 3x$ . In the limit, continuity of $\ln$ and $\cos$ allows the limit to be passed inside the functions.

Techniques used

rearrange the equation into fixed-point formtake natural logarithmstake limits of a convergent sequence

(c)

Use this iterative formula to determine the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the iterative formula $x_{n+1}=\frac{1}{2}\ln(5+\cos 3x_n)$ starting from a value in the interval, for example $x_0=0.7$ . Continue until successive iterates agree to the required accuracy, recording each to 5 decimal places.

Working

Starting with $x_0=0.70000$ :

$n$	$x_n$
0	0.70000
1	0.75150
2	0.73719
3	0.74105
4	0.74000
5	0.74028

The last two iterates are $0.74000$ and $0.74028$ . Both round to $0.740$ to 3 decimal places, so the sequence has converged to this accuracy.

Answer

x \approx 0.740

Final answer

0.740

Detailed explanation

Walkthrough

Choose a starting value in the interval, say $0.7$ . Substitute into the formula to get $x_1=0.75150$ . Repeat the substitution. The values initially move around but settle near $0.740$ . After $x_4=0.74000$ and $x_5=0.74028$ , both are $0.740$ when rounded to 3 decimal places, so we can state the root is $0.740$ correct to 3 decimal places. Recording each iteration to 5 decimal places lets the reader see the convergence and justify the final rounding.

Key Takeaways

Iterative methods produce a sequence of approximations. To claim an answer correct to a given number of decimal places, successive iterates should agree when rounded to that accuracy. Recording to extra decimal places helps justify the final rounding.

Common Mistakes

Working in degrees; the sequence will not converge to $0.740$ .
Rounding each iteration too early, which can change the convergence.
Stopping after one or two iterations without showing enough iterates to justify 3 decimal places.
Quoting $0.7402$ or $0.7403$ instead of the required 3 decimal places.

Things to Be Careful About

Use radians. Keep at least 5 decimal places in intermediate results. To justify $0.740$ to 3 decimal places, note that both $0.74000$ and $0.74028$ lie in the interval $(0.7395,0.7405)$ , so any value in that interval rounds to $0.740$ . Small differences in the final decimal place are acceptable if a calculator gives slightly different last digits.

Techniques used

iterate the fixed-point formularound intermediate values to 5 decimal placescompare successive iterates to establish accuracy

Q6MediumDifferentiationIntegration

Fig. 6

The diagram shows the curve $y = xe^{-ax}$ , where $a$ is a positive constant, and its maximum point $M$ .

(a)

Find the exact coordinates of $M$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate $y = xe^{-ax}$ using the product rule, set $\frac{dy}{dx} = 0$ to find the $x$ -coordinate of the maximum point $M$ , then substitute back to find the $y$ -coordinate.

Working

\begin{aligned} y &= xe^{-ax} \end{aligned}

Apply the product rule with $u = x$ and $v = e^{-ax}$ :

\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}(x) \cdot e^{-ax} + x \cdot \frac{d}{dx}(e^{-ax}) \\ &= 1 \cdot e^{-ax} + x \cdot (-a)e^{-ax} \\ &= e^{-ax} - axe^{-ax} \\ &= e^{-ax}(1 - ax) \end{aligned}

Set $\frac{dy}{dx} = 0$ :

e^{-ax}(1 - ax) = 0

Since $e^{-ax} \neq 0$ for all real $x$ , we have:

1 - ax = 0 \implies x = \frac{1}{a}

Substitute $x = \frac{1}{a}$ into $y = xe^{-ax}$ :

y = \frac{1}{a} \cdot e^{-a \cdot \frac{1}{a}} = \frac{1}{a} \cdot e^{-1} = \frac{1}{ae}

Answer

M = \left(\frac{1}{a}, \frac{1}{ae}\right)

Final answer

M = (1/a, 1/(ae))

Detailed explanation

Walkthrough

We are given the curve $y = xe^{-ax}$ and need to find its maximum point $M$ .

Step 1: Differentiate using the product rule. The function is a product of two factors: $x$ and $e^{-ax}$ . We apply the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$ with $u = x$ and $v = e^{-ax}$ . The derivative of $x$ is $1$ , and the derivative of $e^{-ax}$ requires the chain rule, giving $-ae^{-ax}$ . This yields $\frac{dy}{dx} = e^{-ax} - axe^{-ax} = e^{-ax}(1 - ax)$ .

Step 2: Find the critical point. At a maximum, the gradient is zero, so we set $\frac{dy}{dx} = 0$ . Since $e^{-ax}$ is always positive and never zero, the factor $(1 - ax)$ must be zero, giving $x = \frac{1}{a}$ .

Step 3: Find the $y$ -coordinate. Substitute $x = \frac{1}{a}$ back into the original equation: $y = \frac{1}{a} \cdot e^{-a \cdot (1/a)} = \frac{1}{a} \cdot e^{-1} = \frac{1}{ae}$ .

Key Takeaways

The product rule is essential when differentiating a product of algebraic and exponential functions.
The chain rule must be applied correctly to composite exponentials like $e^{-ax}$ , giving a factor of $-a$ .
At a turning point, $\frac{dy}{dx} = 0$ , and since $e^{-ax}$ is never zero, we only need to solve the algebraic factor.

Common Mistakes

Forgetting the product rule and treating $xe^{-ax}$ as if it were just $e^{-ax}$ or just $x$ .
Incorrectly differentiating $e^{-ax}$ as $e^{-ax}$ without the chain rule factor of $-a$ .
Using a specific numerical value for $a$ instead of keeping it as a parameter.
Forgetting to find the $y$ -coordinate after finding $x$ .

Things to Be Careful About

Always keep $a$ as a parameter throughout; do not substitute a numerical value.
Verify that the critical point is indeed a maximum (the curve rises then falls as shown in Fig. 6, confirming $x = 1/a$ gives a maximum).
The answer must be in exact form with no decimal approximations.

Techniques used

apply product ruledifferentiate exponential composite functionsolve dy/dx = 0 for critical pointevaluate y-coordinate at maximum

(b)

Find the exact value of

\int_0^{\frac{2}{a}} xe^{-ax}\,dx

5M Similar

DifficultyMedium

Worked solution

Approach

Use integration by parts with $u = x$ and $dv = e^{-ax}\,dx$ to evaluate the integral, then apply the limits $0$ and $\frac{2}{a}$ .

Working

We need to evaluate:

\int_0^{\frac{2}{a}} xe^{-ax}\,dx

Apply integration by parts: $\int u\,dv = uv - \int v\,du$ , with $u = x$ and $dv = e^{-ax}\,dx$ .

Then $du = dx$ and $v = -\frac{1}{a}e^{-ax}$ .

\begin{aligned} \int xe^{-ax}\,dx &= x\left(-\frac{1}{a}e^{-ax}\right) - \int \left(-\frac{1}{a}e^{-ax}\right)\,dx \\ &= -\frac{x}{a}e^{-ax} + \frac{1}{a}\int e^{-ax}\,dx \\ &= -\frac{x}{a}e^{-ax} + \frac{1}{a}\left(-\frac{1}{a}e^{-ax}\right) \\ &= -\frac{x}{a}e^{-ax} - \frac{1}{a^2}e^{-ax} \end{aligned}

Now apply the limits from $0$ to $\frac{2}{a}$ :

\left[-\frac{x}{a}e^{-ax} - \frac{1}{a^2}e^{-ax}\right]_0^{\frac{2}{a}}

At the upper limit $x = \frac{2}{a}$ :

-\frac{\frac{2}{a}}{a}e^{-a \cdot \frac{2}{a}} - \frac{1}{a^2}e^{-a \cdot \frac{2}{a}} = -\frac{2}{a^2}e^{-2} - \frac{1}{a^2}e^{-2} = -\frac{3}{a^2}e^{-2}

At the lower limit $x = 0$ :

-\frac{0}{a}e^{0} - \frac{1}{a^2}e^{0} = 0 - \frac{1}{a^2} = -\frac{1}{a^2}

Subtract lower limit from upper limit:

-\frac{3}{a^2}e^{-2} - \left(-\frac{1}{a^2}\right) = \frac{1}{a^2} - \frac{3}{a^2}e^{-2} = \frac{1}{a^2}\left(1 - 3e^{-2}\right)

Answer

\frac{1}{a^2}\left(1 - 3e^{-2}\right)

Final answer

(1/a^2)(1 - 3e^(-2))

Detailed explanation

Walkthrough

We need to evaluate the definite integral $\int_0^{2/a} xe^{-ax}\,dx$ .

Step 1: Integration by parts. The integrand is a product of $x$ (algebraic) and $e^{-ax}$ (exponential). We choose $u = x$ so that $du = dx$ simplifies the integral, and $dv = e^{-ax}\,dx$ so that $v = -\frac{1}{a}e^{-ax}$ . Applying $\int u\,dv = uv - \int v\,du$ gives $-\frac{x}{a}e^{-ax} + \frac{1}{a}\int e^{-ax}\,dx$ .

Step 2: Complete the integration. The remaining integral $\int e^{-ax}\,dx = -\frac{1}{a}e^{-ax}$ , so the full antiderivative is $-\frac{x}{a}e^{-ax} - \frac{1}{a^2}e^{-ax}$ .

Step 3: Evaluate at the upper limit $x = 2/a$ . Substituting gives $-\frac{2}{a^2}e^{-2} - \frac{1}{a^2}e^{-2} = -\frac{3}{a^2}e^{-2}$ .

Step 4: Evaluate at the lower limit $x = 0$ . Substituting gives $0 - \frac{1}{a^2}e^{0} = -\frac{1}{a^2}$ .

Step 5: Subtract. Upper minus lower: $-\frac{3}{a^2}e^{-2} + \frac{1}{a^2} = \frac{1}{a^2}(1 - 3e^{-2})$ .

Key Takeaways

Integration by parts is the standard technique for integrals of the form $\int x \cdot e^{kx}\,dx$ .
When choosing $u$ and $dv$ in integration by parts, let $u$ be the algebraic factor so it simplifies upon differentiation.
Careful substitution of limits is essential; the lower limit $x = 0$ gives a non-zero contribution from the $e^{-ax}$ term.

Common Mistakes

Incorrect sign in the integration by parts formula (forgetting the minus sign in $uv - \int v\,du$ ).
Forgetting to include $dx$ in the $dv$ term.
Using a numerical value for $a$ instead of keeping it as a parameter.
Sign errors when evaluating the lower limit: at $x = 0$ , the term $-\frac{1}{a^2}e^{0} = -\frac{1}{a^2}$ , not zero.
Forgetting to subtract the lower limit evaluation from the upper limit evaluation.

Things to Be Careful About

Keep $a$ as a parameter throughout; do not substitute a numerical value.
The upper limit is $\frac{2}{a}$ , so $-ax$ at the upper limit becomes $-a \cdot \frac{2}{a} = -2$ , giving $e^{-2}$ .
The final answer can be written as $\frac{1}{a^2}(1 - 3e^{-2})$ or equivalently $\frac{e^2 - 3}{a^2 e^2}$ .
Always show the full working for integration by parts; unsupported answers do not earn marks.

Techniques used

apply integration by partsintegrate exponential composite functionevaluate definite integral with given limits

Q7Medium-HardTrigonometryIntegration

(a)

Show that $\cos^4 \theta - \sin^4 \theta \equiv \cos 2\theta$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Factorise the left-hand side as a difference of two squares, then use the Pythagorean identity and the double-angle formula for cosine.

Working

\begin{aligned} \cos^4 \theta - \sin^4 \theta &= (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) \\ &= (\cos^2 \theta - \sin^2 \theta)(1) \\ &= \cos^2 \theta - \sin^2 \theta \\ &= \cos 2\theta \end{aligned}

Answer

\cos^4 \theta - \sin^4 \theta \equiv \cos 2\theta

Final answer

cos^4 θ - sin^4 θ ≡ cos 2θ

Detailed explanation

Walkthrough

The left-hand side is $\cos^4 \theta - \sin^4 \theta$ . This is a difference of two squares because $\cos^4 \theta = (\cos^2 \theta)^2$ and $\sin^4 \theta = (\sin^2 \theta)^2$ . Factorising gives $(\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta)$ . The second factor is exactly $1$ by the Pythagorean identity $\cos^2 \theta + \sin^2 \theta = 1$ . The first factor is the double-angle formula $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ . Hence the identity follows.

Key Takeaways

This question tests the ability to recognise a difference of two squares and to combine the Pythagorean identity with the double-angle formula for cosine. It also emphasises that a proof of an identity must show the steps leading from one side to the other.

Common Mistakes

A common mistake is not recognising that $\cos^4 \theta$ and $\sin^4 \theta$ are squares of $\cos^2 \theta$ and $\sin^2 \theta$ . Another is forgetting that $\cos^2 \theta + \sin^2 \theta = 1$ . Since the answer is given, the working must be shown clearly; an unsupported statement of the identity would not receive full credit.

Things to Be Careful About

Be careful to factorise before simplifying. Also remember the exact double-angle formula: $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ . The mark scheme requires the factorisation step and the use of the Pythagorean identity to be visible.

Techniques used

factorise difference of two squaresapply Pythagorean identityapply double-angle formula for cosine

(b)

Hence find the exact value of

\int_{-\frac{1}{8}\pi}^{\frac{1}{8}\pi} (\cos^4 \theta - \sin^4 \theta + 4\sin^2 \theta \cos^2 \theta)\,d\theta

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Use the identity from part (a) and the double-angle formula $\sin 2\theta = 2\sin\theta\cos\theta$ to simplify the integrand, then integrate term by term and evaluate at the limits.

Working

From part (a),

\cos^4 \theta - \sin^4 \theta = \cos 2\theta.

Also,

4\sin^2\theta\cos^2\theta = (2\sin\theta\cos\theta)^2 = \sin^2 2\theta.

So the integral becomes

\int_{-\pi/8}^{\pi/8} \left(\cos 2\theta + \sin^2 2\theta\right)\,d\theta.

Now integrate:

\int \cos 2\theta\,d\theta = \frac{1}{2}\sin 2\theta,

and

\int \sin^2 2\theta\,d\theta = \int \frac{1 - \cos 4\theta}{2}\,d\theta = \frac{\theta}{2} - \frac{1}{8}\sin 4\theta.

Hence an antiderivative is

F(\theta) = \frac{1}{2}\sin 2\theta + \frac{\theta}{2} - \frac{1}{8}\sin 4\theta.

Evaluate at the limits:

\begin{aligned} F\left(\frac{\pi}{8}\right) &= \frac{1}{2}\cdot\frac{\sqrt{2}}{2} + \frac{\pi}{16} - \frac{1}{8}\cdot 1 \\ &= \frac{\sqrt{2}}{4} + \frac{\pi}{16} - \frac{1}{8}, \\ F\left(-\frac{\pi}{8}\right) &= -\frac{\sqrt{2}}{4} - \frac{\pi}{16} + \frac{1}{8}. \end{aligned}

Therefore

\begin{aligned} \int_{-\pi/8}^{\pi/8} \left(\cos 2\theta + \sin^2 2\theta\right)\,d\theta &= F\left(\frac{\pi}{8}\right) - F\left(-\frac{\pi}{8}\right) \\ &= \left(\frac{\sqrt{2}}{4} + \frac{\pi}{16} - \frac{1}{8}\right) - \left(-\frac{\sqrt{2}}{4} - \frac{\pi}{16} + \frac{1}{8}\right) \\ &= \frac{\sqrt{2}}{2} + \frac{\pi}{8} - \frac{1}{4}. \end{aligned}

Answer

\frac{\sqrt{2}}{2} + \frac{\pi}{8} - \frac{1}{4}

Final answer

sqrt(2)/2 + pi/8 - 1/4

Detailed explanation

Walkthrough

The integrand is $\cos^4 \theta - \sin^4 \theta + 4\sin^2\theta\cos^2\theta$ . From part (a), the first two terms simplify to $\cos 2\theta$ . The term $4\sin^2\theta\cos^2\theta$ can be written as $(2\sin\theta\cos\theta)^2$ , and since $2\sin\theta\cos\theta = \sin 2\theta$ , this is $\sin^2 2\theta$ . So the integrand becomes $\cos 2\theta + \sin^2 2\theta$ .

To integrate $\sin^2 2\theta$ , use the double-angle identity $\sin^2 2\theta = \frac{1 - \cos 4\theta}{2}$ . Then integrate term by term:

\int \cos 2\theta\,d\theta = \frac{1}{2}\sin 2\theta,

\int \frac{1 - \cos 4\theta}{2}\,d\theta = \frac{\theta}{2} - \frac{1}{8}\sin 4\theta.

So an antiderivative is $F(\theta) = \frac{1}{2}\sin 2\theta + \frac{\theta}{2} - \frac{1}{8}\sin 4\theta$ . Evaluate at $\pm\frac{\pi}{8}$ and subtract. Since all terms in $F(\theta)$ are odd functions, $F(-\pi/8) = -F(\pi/8)$ , so the integral equals $2F(\pi/8)$ . This gives $\frac{\sqrt{2}}{2} + \frac{\pi}{8} - \frac{1}{4}$ .

Key Takeaways

This question combines the identity from part (a) with the double-angle formula for $\sin 2\theta$ , then uses the double-angle identity again to integrate $\sin^2 2\theta$ . It also tests exact evaluation of trigonometric functions at special angles and careful handling of symmetric limits.

Common Mistakes

A common mistake is forgetting to use part (a) and trying to integrate the fourth powers directly. Another is failing to rewrite $4\sin^2\theta\cos^2\theta$ as $\sin^2 2\theta$ . When integrating $\sin^2 2\theta$ , it is essential to use $\sin^2 2\theta = \frac{1 - \cos 4\theta}{2}$ ; integrating it as if it were $\sin 2\theta$ gives the wrong sign and factor. Substituting the negative limit also frequently causes sign errors.

Things to Be Careful About

Remember the exact values $\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ and $\sin\frac{\pi}{2} = 1$ . When substituting $\theta = -\frac{\pi}{8}$ , both $\sin 2\theta$ and $\sin 4\theta$ change sign, and the $\frac{\theta}{2}$ term also changes sign. The mark scheme allows a mixture of $x$ and $\theta$ in the antiderivative, but the final exact value must be obtained from correct working.

Techniques used

rewrite integrand using part (a) identityuse double-angle formula for sin^2integrate cos and sin termsevaluate definite integral at symmetric limits

Q8MediumVectors

The points $A$ , $B$ and $C$ have position vectors $\vec{OA} = -2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ , $\vec{OB} = 5\mathbf{i} + 2\mathbf{j}$ and $\vec{OC} = 8\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}$ , where $O$ is the origin. The line $l_1$ passes through $B$ and $C$ .

(a)

Find a vector equation for $l_1$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The line $l_1$ passes through $B$ and $C$ , so a direction vector is $\vec{BC} = \vec{OC} - \vec{OB}$ . A vector equation can then be written as $\mathbf{r} = \vec{OB} + \lambda \vec{BC}$ (or using $C$ as the base point with any scalar multiple of the direction vector).

Working

Use $B(5, 2, 0)$ and $C(8, 5, -3)$ :

\vec{BC} = \begin{pmatrix} 8 - 5 \\ 5 - 2 \\ -3 - 0 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -3 \end{pmatrix}

This can be simplified by dividing by $3$ :

\vec{BC} = 3\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}

so a direction vector is $\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ . Hence a vector equation for $l_1$ is

\mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}

or equivalently

\mathbf{r} = 5\mathbf{i} + 2\mathbf{j} + \lambda(\mathbf{i} + \mathbf{j} - \mathbf{k}).

Answer

\mathbf{r} = 5\mathbf{i} + 2\mathbf{j} + \lambda(\mathbf{i} + \mathbf{j} - \mathbf{k})

Final answer

r = 5i + 2j + λ(i + j - k)

Detailed explanation

Walkthrough

A line in 3D is determined by a base point and a direction vector. Since $l_1$ passes through $B$ and $C$ , the vector from $B$ to $C$ is a natural direction vector: $\vec{BC} = \vec{OC} - \vec{OB}$ . We subtract the components of $B$ from the components of $C$ to get $(3, 3, -3)$ . Any scalar multiple of a direction vector is also a direction vector, so dividing by $3$ gives the simpler vector $(1, 1, -1)$ . Finally, we write $\mathbf{r} = \vec{OB} + \lambda(1, 1, -1)$ , which means: start at $B$ and move any number of direction vectors along the line.

Key Takeaways

A vector equation of a line needs one position vector of a point on the line and one direction vector.
The direction vector can be scaled by any non-zero constant.
Position vectors are measured from the origin, so $\vec{BC} = \vec{OC} - \vec{OB}$ .

Common Mistakes

Writing the equation as $l_1 = \dots$ instead of $\mathbf{r} = \dots$ or $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \dots$ ; the mark scheme requires the vector form.
Using the wrong order for the direction vector, e.g. $B - C$ instead of $C - B$ . This is actually acceptable if used consistently, but it changes the sign of the parameter.
Forgetting the parameter $\lambda$ and writing only a point.

Things to Be Careful About

The direction vector can be left as $(3, 3, -3)$ or simplified to $(1, 1, -1)$ ; both are accepted.
Any point on the line can be used as the base point, so using $C$ instead of $B$ is also correct.
Keep the same order of components in all vectors.

Techniques used

subtract position vectors to find a direction vectorwrite the vector equation of a line through a given pointsimplify a direction vector by a common scalar

(b)

The line $l_2$ has equation $\mathbf{r} = -2\mathbf{i} + \mathbf{j} + 4\mathbf{k} + \mu(3\mathbf{i} + \mathbf{j} - 2\mathbf{k})$ .

Find the coordinates of the point of intersection of $l_1$ and $l_2$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

At the intersection, the position vectors on $l_1$ and $l_2$ are equal. Write both lines in component form using different parameters, equate the $\mathbf{i}$ , $\mathbf{j}$ and $\mathbf{k}$ components, then solve the resulting simultaneous equations.

Working

From part (a), $l_1$ can be written as

\mathbf{r} = \begin{pmatrix} 5 + \lambda \\ 2 + \lambda \\ -\lambda \end{pmatrix}

and $l_2$ is

\mathbf{r} = \begin{pmatrix} -2 + 3\mu \\ 1 + \mu \\ 4 - 2\mu \end{pmatrix}.

Equating components gives

\begin{aligned} 5 + \lambda &= -2 + 3\mu \\ 2 + \lambda &= 1 + \mu \\ -\lambda &= 4 - 2\mu \end{aligned}

From the second equation,

\lambda = \mu - 1.

Substitute into the first equation:

5 + (\mu - 1) = -2 + 3\mu

4 + \mu = -2 + 3\mu

6 = 2\mu

\mu = 3

Therefore $\lambda = \mu - 1 = 2$ . Check the third component:

-\lambda = -2, \qquad 4 - 2\mu = 4 - 6 = -2

so the equations are consistent. Substitute $\lambda = 2$ into $l_1$ :

\mathbf{r} = \begin{pmatrix} 5 + 2 \\ 2 + 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \\ -2 \end{pmatrix}.

Answer

(7, 4, -2)

Final answer

(7, 4, -2)

Detailed explanation

Walkthrough

To find where two lines meet, we set their position vectors equal. Because the two lines have different parameters, call the parameter on $l_1$ $\lambda$ and the parameter on $l_2$ $\mu$ . Writing both lines in component form gives three equations: one for each of $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ . We only need two of these equations to solve for the two unknowns $\lambda$ and $\mu$ ; the third equation acts as a consistency check. Here, from the $\mathbf{j}$ -component we get $\lambda = \mu - 1$ . Substituting into the $\mathbf{i}$ -component gives $\mu = 3$ , so $\lambda = 2$ . Checking the $\mathbf{k}$ -component confirms the point is on both lines. Substituting either parameter value gives the intersection point $(7, 4, -2)$ .

Key Takeaways

Intersection of two lines: equate the vector equations.
Use different parameters for different lines.
Two component equations are enough to solve for two unknowns; the third checks consistency.

Common Mistakes

Using the same parameter for both lines; this is not valid and would not earn the first mark.
Equating only one component and trying to solve.
Losing signs when equating the $\mathbf{k}$ -component.

Things to Be Careful About

The mark scheme says there is no need to check the third equation because the question implies the lines intersect, but checking is a good safeguard.
If the direction vector of $l_1$ is left as $(3, 3, -3)$ , the parameter values will be different but the final point is the same.
The final answer may be written as coordinates or as a position vector.

Techniques used

equate components of two vector line equationssolve simultaneous linear equations for the two parameterssubstitute a parameter value to find the intersection point

(c)

The point $D$ on $l_2$ is such that $AB = BD$ .

Find the position vector of $D$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Point $D$ lies on $l_2$ , so write its position vector in terms of the parameter $\mu$ . Compute $AB$ using the distance formula, then set $BD^2 = AB^2$ , solve for $\mu$ , and substitute back to find $D$ .

Working

For $A(-2, 1, 4)$ and $B(5, 2, 0)$ ,

\vec{AB} = \begin{pmatrix} 5 - (-2) \\ 2 - 1 \\ 0 - 4 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \\ -4 \end{pmatrix}

so

AB^2 = 7^2 + 1^2 + (-4)^2 = 49 + 1 + 16 = 66.

Since $D$ lies on $l_2$ , write

\vec{OD} = \begin{pmatrix} -2 + 3\mu \\ 1 + \mu \\ 4 - 2\mu \end{pmatrix}.

Then

\vec{BD} = \vec{OD} - \vec{OB} = \begin{pmatrix} -7 + 3\mu \\ -1 + \mu \\ 4 - 2\mu \end{pmatrix}.

Set $BD^2 = AB^2$ :

(-7 + 3\mu)^2 + (-1 + \mu)^2 + (4 - 2\mu)^2 = 66.

Expand:

(9\mu^2 - 42\mu + 49) + (\mu^2 - 2\mu + 1) + (4\mu^2 - 16\mu + 16) = 66

14\mu^2 - 60\mu + 66 = 66

14\mu^2 - 60\mu = 0

2\mu(7\mu - 30) = 0.

Thus $\mu = 0$ or $\mu = \frac{30}{7}$ . The value $\mu = 0$ gives $D = A$ , the trivial solution; the required point is $\mu = \frac{30}{7}$ . Then

\vec{OD} = \begin{pmatrix} -2 + 3\cdot\frac{30}{7} \\ 1 + \frac{30}{7} \\ 4 - 2\cdot\frac{30}{7} \end{pmatrix} = \begin{pmatrix} \frac{76}{7} \\ \frac{37}{7} \\ -\frac{32}{7} \end{pmatrix}.

Answer

\vec{OD} = \frac{76}{7}\mathbf{i} + \frac{37}{7}\mathbf{j} - \frac{32}{7}\mathbf{k}

Final answer

OD = (76/7)i + (37/7)j - (32/7)k

Detailed explanation

Walkthrough

We need a point $D$ on $l_2$ for which the distance from $B$ to $D$ equals the distance from $A$ to $B$ . First compute $AB$ : $\vec{AB} = B - A = (7, 1, -4)$ , so $AB^2 = 66$ . Next, write $D$ using the parameter $\mu$ on $l_2$ : $D = (-2+3\mu, 1+\mu, 4-2\mu)$ . Then $\vec{BD} = D - B = (-7+3\mu, -1+\mu, 4-2\mu)$ . Setting $BD^2 = AB^2$ gives a quadratic in $\mu$ . Expanding and simplifying gives $2\mu(7\mu - 30) = 0$ , so $\mu = 0$ or $\mu = \frac{30}{7}$ . The value $\mu = 0$ gives $D = A$ , which is the trivial point where $BD = BA = AB$ ; the intended point is the other one. Substituting $\mu = \frac{30}{7}$ gives the position vector $\left(\frac{76}{7}, \frac{37}{7}, -\frac{32}{7}\right)$ .

Key Takeaways

Distances can be compared using squared distances to avoid square roots.
A point on a line can be expressed in terms of a single parameter.
A quadratic equation may produce two solutions; the context or wording selects the required one.

Common Mistakes

Expanding the squared components without first writing the correct squared-bracket equation; the mark scheme requires the correct form before expansion.
Forgetting that $\mu = 0$ is also a solution and must be rejected or ignored.
Giving the answer as coordinates rather than a position vector; the final answer must be a vector.

Things to Be Careful About

A sign error in $\vec{AB}$ is condoned by the mark scheme because the squared length is unchanged, but it is safer to compute it correctly.
Show the equation $(-7+3\mu)^2 + (-1+\mu)^2 + (4-2\mu)^2 = 66$ before expanding.
The final vector has fractions; do not round them.

Techniques used

compute the distance between two points using vector componentsexpress a point on a line using a parameterform and solve a quadratic equation from a distance equalityselect the non-trivial solution

Q9MediumComplex Numbers

The complex numbers $z$ and $\omega$ are defined by $z = 1 - \mathbf{i}$ and $\omega = -3 + 3\sqrt{3}\mathbf{i}$ .

(a)

Express $z\omega$ in the form $a + b\mathbf{i}$ , where $a$ and $b$ are real and in exact surd form.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

Multiply $z$ and $\omega$ directly using the distributive law, then simplify using $i^2 = -1$ and collect the real and imaginary parts.

Working

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

Answer

z\omega = (-3 + 3\sqrt{3}) + (3 + 3\sqrt{3})i

Final answer

zw = (-3 + 3√3) + (3 + 3√3)i

Detailed explanation

Walkthrough

We need to multiply $z = 1 - i$ and $\omega = -3 + 3\sqrt{3}i$ . The standard approach is to treat $i$ like any other algebraic symbol and apply the distributive law (or 'FOIL' for two binomials).

First, we multiply each term:

$1 \cdot (-3) = -3$
$1 \cdot 3\sqrt{3}i = 3\sqrt{3}i$
$-i \cdot (-3) = 3i$
$-i \cdot 3\sqrt{3}i = -3\sqrt{3}i^2$

The last term involves $i^2$ , which equals $-1$ . So $-3\sqrt{3}i^2 = -3\sqrt{3}(-1) = 3\sqrt{3}$ .

Now collect the real and imaginary parts:

Real: $-3 + 3\sqrt{3}$
Imaginary: $3\sqrt{3}i + 3i = (3 + 3\sqrt{3})i$

So $z\omega = (-3 + 3\sqrt{3}) + (3 + 3\sqrt{3})i$ .

Key Takeaways

To multiply complex numbers in Cartesian form, expand the product and use the rule $i^2 = -1$ to simplify.
Always group the real parts together and the imaginary parts together at the end.

Common Mistakes

Forgetting to apply $i^2 = -1$ and leaving $i^2$ in the answer.
Mixing up the signs when collecting the real and imaginary parts (especially the sign change from $-3\sqrt{3}i^2$ to $+3\sqrt{3}$ ).
Writing the answer as two separate numbers instead of as a single complex number in the form $a + bi$ .

Things to Be Careful About

The coefficient of $i$ must be in brackets, e.g. $(3 + 3\sqrt{3})i$ , to avoid ambiguity.
Watch for sign errors: both imaginary terms $3\sqrt{3}i$ and $3i$ are positive, while the standalone real term $-3$ is negative.

Techniques used

expand product using distributive lawsimplify using i^2 = -1collect real and imaginary parts

(b)

Express $z$ and $\omega$ in the form $re^{\mathbf{i}\theta}$ , where $r > 0$ and $-\pi < \theta \le \pi$ . Give the exact values of $r$ and $\theta$ in each case.

4M Similar

DifficultyMedium

Worked solution

Approach

For each complex number, compute the modulus using $|w| = \sqrt{\text{Re}(w)^2 + \text{Im}(w)^2}$ , and the argument using the inverse tangent, taking care to identify the correct quadrant so that $\theta$ lies in $(-\pi, \pi]$ .

Working

For $z = 1 - i$ :

|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}

Since $z$ lies in the fourth quadrant (positive real part, negative imaginary part), its argument is negative:

\arg(z) = -\frac{\pi}{4}

Hence $z = \sqrt{2}e^{-i\pi/4}$ .

For $\omega = -3 + 3\sqrt{3}i$ :

|\omega| = \sqrt{(-3)^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6

Since $\omega$ lies in the second quadrant (negative real part, positive imaginary part), and

\tan(\arg\omega) = \frac{3\sqrt{3}}{-3} = -\sqrt{3}

the reference angle is $\frac{\pi}{3}$ , so

\arg(\omega) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}

Hence $\omega = 6e^{2i\pi/3}$ .

Answer

z = \sqrt{2}e^{-i\pi/4}, \quad \omega = 6e^{2i\pi/3}

Final answer

z = √2 e^(-iπ/4), ω = 6 e^(2iπ/3)

Detailed explanation

Walkthrough

To express a complex number $w = a + bi$ in polar form $re^{i\theta}$ , we need two pieces of information:

The modulus $r = |w| = \sqrt{a^2 + b^2}$ , which gives the distance from the origin.
The argument $\theta = \arg(w)$ , which gives the angle measured counterclockwise from the positive real axis.

Computing $|z|$ :
For $z = 1 - i$ , we have $\text{Re}(z) = 1$ and $\text{Im}(z) = -1$ . Using the Pythagorean formula:

|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}.

Computing $\arg(z)$ :
The point $z = 1 - i$ lies in the fourth quadrant of the Argand diagram. The tangent of the angle to the positive real axis is $\frac{-1}{1} = -1$ . The reference angle is $\frac{\pi}{4}$ . Since we are in the fourth quadrant, the argument is negative: $\arg(z) = -\frac{\pi}{4}$ .

Computing $|\omega|$ :
For $\omega = -3 + 3\sqrt{3}i$ , we have $\text{Re}(\omega) = -3$ and $\text{Im}(\omega) = 3\sqrt{3}$ . So:

|\omega| = \sqrt{(-3)^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6.

Computing $\arg(\omega)$ :
The point $\omega$ lies in the second quadrant (real part negative, imaginary part positive). The tangent of the angle to the positive real axis is $\frac{3\sqrt{3}}{-3} = -\sqrt{3}$ . The reference angle is $\frac{\pi}{3}$ since $\tan(\pi/3) = \sqrt{3}$ . In the second quadrant, the argument is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$ .

Key Takeaways

The modulus is always non-negative and is computed via the Pythagorean formula $|w| = \sqrt{a^2 + b^2}$ .
The argument is the angle from the positive real axis, measured counterclockwise, and must lie in $(-\pi, \pi]$ (the principal range).
When computing the argument, always identify the quadrant first to determine the sign and the correct reference angle.

Common Mistakes

Computing $\arg(\omega) = -\frac{\pi}{3}$ instead of $\frac{2\pi}{3}$ (forgetting that the second quadrant gives a positive angle greater than $\frac{\pi}{2}$ ).
Forgetting that the principal range is $(-\pi, \pi]$ , so $-\pi < \theta \leq \pi$ (not $[0, 2\pi)$ ).
Using a calculator to compute the argument and getting a decimal approximation instead of the exact value.

Things to Be Careful About

The argument is in radians, not degrees. $\frac{\pi}{4}$ is $45°$ and $\frac{2\pi}{3}$ is $120°$ .
The argument of a real positive number is $0$ , of a real negative number is $\pi$ , of a purely imaginary positive number is $\frac{\pi}{2}$ , and of a purely imaginary negative number is $-\frac{\pi}{2}$ .

Techniques used

compute modulus using Pythagorean formuladetermine the quadrant of a complex numbercompute the argument using inverse tangent and reference angleexpress in exponential polar form re^{iθ}

(c)

On an Argand diagram, the points representing $\omega$ and $z\omega$ are $A$ and $B$ respectively.

Prove that $OAB$ is an isosceles right-angled triangle, where $O$ is the origin.

2M Similar

DifficultyMedium

Worked solution

Approach

Identify the points $O$ , $A$ , $B$ on the Argand diagram. Compute $|OA| = |\omega|$ and $|AB| = |z\omega - \omega|$ , then use the argument difference to find the angle at $O$ . Combine these to show the triangle is isosceles and right-angled.

Working

The point $A$ represents $\omega$ and the point $B$ represents $z\omega$ .

|OA| = |\omega| = 6

The displacement from $A$ to $B$ is:

B - A = z\omega - \omega = \omega(z - 1) = \omega(1 - i - 1) = -i\omega

-i\omega = -i(-3 + 3\sqrt{3}i) = 3i - 3\sqrt{3}i^2 = 3\sqrt{3} + 3i

|AB| = |3\sqrt{3} + 3i| = \sqrt{(3\sqrt{3})^2 + 3^2} = \sqrt{27 + 9} = \sqrt{36} = 6

So $|OA| = |AB| = 6$ , and triangle $OAB$ is isosceles, with the two equal sides $OA$ and $AB$ meeting at $A$ .

The angle at $O$ is the difference of the arguments:

\angle AOB = \arg(\omega) - \arg(z\omega) = \arg(\omega) - (\arg(z) + \arg(\omega)) = -\arg(z) = \frac{\pi}{4}

Since the triangle is isosceles with equal sides $OA$ and $AB$ , the base angles at $O$ and $B$ are equal. The angle at $B$ is also $\frac{\pi}{4}$ , so the angle at $A$ is

\angle OAB = \pi - \frac{\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2}.

Hence triangle $OAB$ is right-angled at $A$ .

Answer

The triangle $OAB$ is isosceles with $|OA| = |AB| = 6$ and is right-angled at $A$ .

Final answer

Triangle OAB is isosceles with |OA| = |AB| = 6 and is right-angled at A

Detailed explanation

Walkthrough

On the Argand diagram, the complex numbers $\omega$ and $z\omega$ are represented by points $A$ and $B$ , and $O$ is the origin (representing $0$ ). The lengths of the sides of triangle $OAB$ are the moduli of the corresponding vectors, and the angles can be read off from the arguments.

Length $OA$ :
$OA$ is the vector from $O$ to $A$ , which is the complex number $\omega$ itself. From part (b), $|\omega| = 6$ , so $|OA| = 6$ .

Length $AB$ :
$AB$ is the vector from $A$ to $B$ , which equals $B - A = z\omega - \omega$ . We can factor this as $\omega(z - 1)$ . Since $z = 1 - i$ , we have $z - 1 = -i$ , so $AB = -i\omega$ , and $|AB| = |-i| \cdot |\omega| = 1 \cdot 6 = 6$ .

Isosceles property:
Since $|OA| = |AB| = 6$ , the triangle has two equal sides sharing the vertex $A$ . This means it is isosceles, and the base is $OB$ . The base angles (at $O$ and $B$ ) are equal.

Angle at $O$ :
The angle $\angle AOB$ is the angle between the vectors $OA$ and $OB$ , which is the difference of their arguments:

\angle AOB = \arg(\omega) - \arg(z\omega) = -\arg(z) = \frac{\pi}{4}.

Right-angled property:
Since the triangle is isosceles with equal sides $OA$ and $AB$ , the base angles at $O$ and $B$ are equal. So the angle at $B$ is also $\frac{\pi}{4}$ . The sum of angles in a triangle is $\pi$ , so the angle at $A$ is $\pi - \frac{\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2}$ . Therefore, the triangle is right-angled at $A$ .

Verification by Pythagoras (alternative):
$|OB| = |z\omega| = |z|\,|\omega| = \sqrt{2} \cdot 6 = 6\sqrt{2}$ , so $|OA|^2 + |AB|^2 = 36 + 36 = 72 = (6\sqrt{2})^2 = |OB|^2$ . By the converse of Pythagoras' theorem, the triangle is right-angled at $A$ , confirming our conclusion.

Geometric picture (described, not drawn):
$A$ lies in the second quadrant at $\omega = -3 + 3\sqrt{3}i$ , and $B$ lies in the first quadrant at $z\omega = (-3 + 3\sqrt{3}) + (3 + 3\sqrt{3})i \approx 2.196 + 8.196i$ . The vector $AB = -i\omega$ represents a quarter-turn rotation of $OA$ about $A$ (multiplication by $-i$ rotates by $-\pi/2$ ), which explains why $|OA| = |AB|$ and the angle at $A$ is $\pi/2$ .

Key Takeaways

On an Argand diagram, the vector from the origin to a point $P$ representing $w$ is $w$ itself, with length $|w|$ .
The vector from $P_1$ to $P_2$ is the difference $w_2 - w_1$ , with length $|w_2 - w_1|$ .
The angle subtended at the origin by two points is the difference of their arguments.
For an isosceles triangle, the base angles are equal.
Multiplication by $-i$ rotates a vector by $-\pi/2$ (a quarter turn clockwise), preserving the modulus.

Common Mistakes

Forgetting that $AB$ is the modulus of $z\omega - \omega$ , not of $z\omega$ or $\omega$ alone.
Computing the angle at $O$ as the sum of arguments instead of the difference.
Concluding that the right angle is at $O$ or $B$ instead of $A$ .
Reversing the sign in $\arg(\omega) - \arg(z\omega)$ and obtaining a negative angle (the magnitude is what matters).

Things to Be Careful About

The displacement $AB$ is the vector from $A$ to $B$ , not from $O$ to $B$ (which is $OB = z\omega$ ).
The two equal sides $OA$ and $AB$ meet at $A$ , so $A$ is the apex and $OB$ is the base. The base angles are at $O$ and $B$ .
The mark scheme notes that the diagram is not required for this proof — it can be done entirely algebraically.

Techniques used

compute |AB| as the modulus of the displacement zω − ωcompare moduli to establish the isosceles propertycompute the angle at O using the difference of argumentsapply the isosceles triangle angle property to find the right angle

(d)

Using your answers to part (b), prove that

\tan \frac{5}{12}\pi = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

3M Similar

DifficultyMedium

Worked solution

Approach

Use the property that the argument of a product is the sum of arguments: $\arg(z\omega) = \arg(z) + \arg(\omega)$ . From part (b), this gives $\arg(z\omega) = \frac{5\pi}{12}$ . Then from part (a), $\tan(\arg(z\omega)) = \frac{3 + 3\sqrt{3}}{-3 + 3\sqrt{3}}$ . Equate the two and simplify.

Working

From part (b):

\arg(z) = -\frac{\pi}{4}, \quad \arg(\omega) = \frac{2\pi}{3}

By the addition rule for arguments:

\arg(z\omega) = \arg(z) + \arg(\omega) = -\frac{\pi}{4} + \frac{2\pi}{3} = \frac{-3\pi + 8\pi}{12} = \frac{5\pi}{12}

From part (a), $z\omega = (-3 + 3\sqrt{3}) + (3 + 3\sqrt{3})i$ , which lies in the first quadrant (since $-3 + 3\sqrt{3} = 3(\sqrt{3} - 1) > 0$ and $3 + 3\sqrt{3} > 0$ ). Therefore:

\tan(\arg(z\omega)) = \frac{3 + 3\sqrt{3}}{-3 + 3\sqrt{3}}

Equating the two expressions:

\tan\left(\frac{5\pi}{12}\right) = \frac{3 + 3\sqrt{3}}{-3 + 3\sqrt{3}}

Factor $3$ from the numerator and denominator:

\tan\left(\frac{5\pi}{12}\right) = \frac{3(1 + \sqrt{3})}{3(\sqrt{3} - 1)} = \frac{1 + \sqrt{3}}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

Answer

\tan\frac{5\pi}{12} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

Final answer

tan(5π/12) = (√3 + 1)/(√3 - 1)

Detailed explanation

Walkthrough

This part connects the polar form of complex numbers (from part b) to the Cartesian form (from part a) to prove a trigonometric identity.

Step 1: Find $\arg(z\omega)$ using the addition rule.
A key property of complex numbers in polar form is that the argument of a product equals the sum of the arguments (modulo $2\pi$ ): $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$ .

From part (b), $\arg(z) = -\pi/4$ and $\arg(\omega) = 2\pi/3$ . So:

\arg(z\omega) = -\frac{\pi}{4} + \frac{2\pi}{3}.

To add these, we need a common denominator. The least common multiple of $4$ and $3$ is $12$ , so:

\arg(z\omega) = \frac{-3\pi}{12} + \frac{8\pi}{12} = \frac{5\pi}{12}.

Step 2: Find $\tan(\arg(z\omega))$ from the Cartesian form.
From part (a), $z\omega = (-3 + 3\sqrt{3}) + (3 + 3\sqrt{3})i$ . For a complex number $a + bi$ (with $a \neq 0$ ), the tangent of its argument is $b/a$ . We check: $-3 + 3\sqrt{3} = 3(\sqrt{3} - 1) > 0$ and $3 + 3\sqrt{3} > 0$ , so $z\omega$ is in the first quadrant (consistent with $\arg = 5\pi/12 = 75°$ ). Therefore:

\tan(\arg(z\omega)) = \frac{3 + 3\sqrt{3}}{-3 + 3\sqrt{3}}.

Step 3: Equate and simplify.
Since $\arg(z\omega) = 5\pi/12$ from both expressions:

\tan\left(\frac{5\pi}{12}\right) = \frac{3 + 3\sqrt{3}}{-3 + 3\sqrt{3}}.

Factor out $3$ from the numerator and denominator:

= \frac{3(1 + \sqrt{3})}{3(\sqrt{3} - 1)} = \frac{1 + \sqrt{3}}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}.

Verification (optional):
We can verify using the compound angle formula: $5\pi/12 = \pi/4 + \pi/3$ , so

\tan\left(\frac{5\pi}{12}\right) = \tan\left(\frac{\pi}{4} + \frac{\pi}{3}\right) = \frac{\tan(\pi/4) + \tan(\pi/3)}{1 - \tan(\pi/4)\tan(\pi/3)} = \frac{1 + \sqrt{3}}{1 - \sqrt{3}}.

Multiplying numerator and denominator by $-1$ :

= \frac{-(1 + \sqrt{3})}{-(1 - \sqrt{3})} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}.

This confirms the result.

Key Takeaways

The argument of a product is the sum of the arguments: $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$ .
For a complex number $a + bi$ in the first or second quadrant, $\tan(\arg) = b/a$ (not $a/b$ !).
Trigonometric identities can be derived by combining the polar and Cartesian forms of complex numbers.

Common Mistakes

Forgetting to add a common denominator when adding $\arg(z)$ and $\arg(\omega)$ , leading to a wrong value for $\arg(z\omega)$ .
Using the wrong formula for $\tan(\arg)$ . It is $\frac{\text{imaginary}}{\text{real}}$ , not $\frac{\text{real}}{\text{imaginary}}$ .
Confusing the sign in the denominator of $-3 + 3\sqrt{3}$ (e.g. writing $\sqrt{3} - 1$ directly instead of starting from $3 + 3\sqrt{3}$ over $-3 + 3\sqrt{3}$ ).

Things to Be Careful About

The mark scheme emphasises that the student must show where $5\pi/12$ comes from. Simply stating $\arg(z\omega) = 5\pi/12$ without showing the addition of the angles is not enough.
The mark scheme also emphasises that the unsimplified form $\frac{3 + 3\sqrt{3}}{-3 + 3\sqrt{3}}$ must appear, not just the simplified $\frac{1 + \sqrt{3}}{\sqrt{3} - 1}$ , to demonstrate the connection to part (a).
Verify that the point $z\omega$ lies in the first quadrant before using $\tan(\arg) = b/a$ directly, since the formula assumes the correct quadrant.

Techniques used

apply the addition of arguments property arg(z1z2) = arg(z1) + arg(z2)compute tan of the argument from the Cartesian formsimplify a surd fraction by factoring

Q10Medium-HardDifferentiationDifferential EquationsIntegration

(a)

By writing $y = \sec^3 \theta$ as $\frac{1}{\cos^3 \theta}$ , show that

\frac{dy}{d\theta} = 3\sin \theta \sec^4 \theta

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write $\sec^3 \theta$ as $(\cos \theta)^{-3}$ and differentiate using the chain rule. The derivative of $\cos \theta$ is $-\sin \theta$ , so the two negative signs cancel.

Working

Let

y = \sec^3 \theta = (\cos \theta)^{-3}

Using the chain rule:

\frac{dy}{d\theta} = -3(\cos \theta)^{-4} \cdot \frac{d}{d\theta}(\cos \theta)

= -3(\cos \theta)^{-4}(-\sin \theta)

= 3\sin \theta (\cos \theta)^{-4}

= 3\sin \theta \sec^4 \theta

Answer

\frac{dy}{d\theta} = 3\sin \theta \sec^4 \theta

Final answer

dy/dθ = 3 sin θ sec^4 θ

Detailed explanation

Walkthrough

We are asked to differentiate $y = \sec^3 \theta$ . The function is a composite: the outer function is a power of the secant, and the inner function is $\cos \theta$ . Writing it as $(\cos \theta)^{-3}$ makes the power explicit. Differentiate the outer power first, bringing down $-3$ and reducing the power to $-4$ , then multiply by the derivative of $\cos \theta$ , which is $-\sin \theta$ . The two minus signs cancel, giving $3\sin \theta (\cos \theta)^{-4}$ , which is the same as $3\sin \theta \sec^4 \theta$ .

Key Takeaways

This question tests the chain rule on a trigonometric composite. It is often easier to rewrite reciprocal trigonometric functions as powers of $\cos \theta$ before differentiating. It also reinforces the equivalence between $(\cos \theta)^{-4}$ and $\sec^4 \theta$ .

Common Mistakes

A common error is forgetting the minus sign from the derivative of $\cos \theta$ , which would give $-3\sin \theta \sec^4 \theta$ . Another common error is differentiating $\sec^3 \theta$ as if it were $3\sec^2 \theta$ , omitting the inner derivative entirely.

Things to Be Careful About

The mark scheme requires the signs to be shown clearly. It also requires the final answer to be expressed in the given form $3\sin \theta \sec^4 \theta$ , not left as $3\sin \theta (\cos \theta)^{-4}$ .

Techniques used

rewrite secant as a reciprocal power of cosineapply the chain rulesimplify using secant notation

(b)

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

(x^2 + 9)\sin \theta \frac{d\theta}{dx} = (x + 3)\cos^4 \theta

It is given that $x = 3$ when $\theta = \frac{1}{3}\pi$ .

Solve the differential equation to find the value of $\cos \theta$ when $x = 0$ . Give your answer correct to 3 significant figures.

8M Similar

DifficultyMedium-Hard

Worked solution

Approach

Separate the variables so that all $\theta$ terms are on one side and all $x$ terms on the other. Use the result from part (a) to integrate the $\theta$ side, split the $x$ side into a logarithmic part and an arctangent part, then use the initial condition to find the constant. Finally substitute $x = 0$ and solve for $\cos \theta$ .

Working

Separate the variables:

\frac{\sin \theta}{\cos^4 \theta}\,d\theta = \frac{x+3}{x^2+9}\,dx

Multiply both sides by 3 so that the left-hand side is ready to use $\frac{d}{d\theta}(\sec^3 \theta) = 3\sin\theta\sec^4\theta$ :

\int \frac{3\sin \theta}{\cos^4 \theta}\,d\theta = \int \frac{3x+9}{x^2+9}\,dx

Integrate the left-hand side:

\int \frac{3\sin \theta}{\cos^4 \theta}\,d\theta = \sec^3 \theta + A

Split and integrate the right-hand side:

\int \frac{3x+9}{x^2+9}\,dx = \int \frac{3x}{x^2+9}\,dx + \int \frac{9}{x^2+9}\,dx

= \frac{3}{2}\ln(x^2+9) + 3\tan^{-1}\left(\frac{x}{3}\right) + B

Therefore:

\sec^3 \theta = \frac{3}{2}\ln(x^2+9) + 3\tan^{-1}\left(\frac{x}{3}\right) + C

Use $x = 3$ , $\theta = \frac{1}{3}\pi$ :

\sec^3\left(\frac{\pi}{3}\right) = \left(\frac{1}{\cos(\pi/3)}\right)^3 = 2^3 = 8

8 = \frac{3}{2}\ln(18) + 3\tan^{-1}(1) + C = \frac{3}{2}\ln 18 + \frac{3\pi}{4} + C

So:

C = 8 - \frac{3}{2}\ln 18 - \frac{3\pi}{4}

Thus:

\sec^3 \theta = \frac{3}{2}\ln(x^2+9) + 3\tan^{-1}\left(\frac{x}{3}\right) + 8 - \frac{3}{2}\ln 18 - \frac{3\pi}{4}

At $x = 0$ :

\sec^3 \theta = \frac{3}{2}\ln 9 + 0 + 8 - \frac{3}{2}\ln 18 - \frac{3\pi}{4}

= 8 - \frac{3\pi}{4} + \frac{3}{2}\ln\left(\frac{9}{18}\right) = 8 - \frac{3\pi}{4} - \frac{3}{2}\ln 2

Numerically:

\sec^3 \theta \approx 4.60408

Therefore:

\cos \theta = \left(\frac{1}{\sec^3 \theta}\right)^{1/3} \approx 0.601

Answer

\cos \theta = 0.601 \quad (3\text{ s.f.})

Final answer

cos θ = 0.601 (3 s.f.)

Detailed explanation

Walkthrough

First separate the variables so that every $\theta$ is on one side and every $x$ is on the other. The left-hand side is $\frac{\sin \theta}{\cos^4 \theta}$ . From part (a), the derivative of $\sec^3 \theta$ is $3\sin \theta \sec^4 \theta$ , which equals $\frac{3\sin \theta}{\cos^4 \theta}$ . Therefore multiplying both sides by 3 makes the left-hand side integrate directly to $\sec^3 \theta$ .

On the right-hand side, split $\frac{3x+9}{x^2+9}$ into $\frac{3x}{x^2+9} + \frac{9}{x^2+9}$ . The first term is of the form $k\frac{f'(x)}{f(x)}$ , so it integrates to $\frac{3}{2}\ln(x^2+9)$ . The second term is a standard arctangent integral: $\int \frac{9}{x^2+9}\,dx = 3\tan^{-1}\left(\frac{x}{3}\right)$ .

After integrating, add the constant of integration. Use $x = 3$ , $\theta = \frac{\pi}{3}$ to find the constant. At this point $\sec^3\left(\frac{\pi}{3}\right) = 8$ , $\ln(3^2+9) = \ln 18$ , and $\tan^{-1}(1) = \frac{\pi}{4}$ . Solving gives the constant.

Finally substitute $x = 0$ . Then $\ln(9)$ and $\tan^{-1}(0) = 0$ appear. Simplify using $\ln 9 - \ln 18 = \ln\frac{1}{2} = -\ln 2$ , evaluate numerically, and take the cube root of the reciprocal to find $\cos \theta$ .

Key Takeaways

This question combines separable differential equations with several integration techniques. It shows why recognising $\int \frac{kf'(x)}{f(x)}\,dx = k\ln|f(x)|$ and $\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)$ is essential. It also reinforces using an initial condition to determine the arbitrary constant in a first-order differential equation.

Common Mistakes

A common mistake is forgetting to multiply both sides by 3, which makes the left-hand integral harder to recognise. Another is omitting the constant of integration, or using the initial condition before the constant is introduced. Some candidates also confuse $\sec^3 \theta$ with $\sec \theta^3$ , or incorrectly take the cube root when solving for $\cos \theta$ .

Things to Be Careful About

Remember that $\sec^3 \theta = \frac{1}{\cos^3 \theta}$ , so $\cos \theta = \left(\frac{1}{\sec^3 \theta}\right)^{1/3}$ . Use radians throughout. The mark scheme accepts the constant in the form $8 - \frac{3}{2}\ln 18 - \frac{3\pi}{4}$ , and the final answer must be given to 3 significant figures, so $0.601$ is required.

Techniques used

separate the variablesuse part (a) to integrate the trigonometric sidesplit the rational integrand into logarithmic and arctangent partsapply the initial condition to determine the constantsolve for cos theta

Mathematics 9709/32 — May/June 2024

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