(a) By sketching a suitable pair of graphs, show that the equation $$ x - 2 = 2\sin\frac{1}{2}x $$ has only one root in the interval $0 < x < \pi$. (b) Show by calculation that this root lies between 1 and 1.5. (c) Use the iterative formula $x_{n+1} = 2 - 2\sin\frac{1}{2}x_n$ with an initial value of 1.03 to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

(a) The graphs y = x - 2 and y = 2sin(x/2) intersect at exactly one point in the interval 0 0 and f(1.5) < 0, so the root lies between 1 and 1.5. · (c) 1.02

(a) Find the quotient and remainder when $x^2$ is divided by $1 + 4x^2$. (b) Find the exact value of $\int_0^{0.5} x\tan^{-1}(2x)\,dx$.

(a) Quotient = 1/4, remainder = -1/4 · (b) π/16 - 1/8

9709/32 May/June 2025 — Mathematics Solved Past Paper

Q15MMediumLogarithmic and Exponential Functions

Solve the equation

\frac{e^x + 2e^{-x}}{e^x - 3} = 4

Give your answer correct to 3 decimal places.

DifficultyMedium

Worked solution

Approach

Multiply both sides by $e^x - 3$ to clear the fraction, rearrange into a quadratic in $m = e^x$ , solve the quadratic, take natural logs, and reject the negative root because $e^x > 0$ .

Working

Starting from

\frac{e^x + 2e^{-x}}{e^x - 3} = 4

multiply by $e^x - 3$ :

e^x + 2e^{-x} = 4(e^x - 3) = 4e^x - 12

Rearrange:

-3e^x + 2e^{-x} + 12 = 0

Multiply by $e^x$ :

-3e^{2x} + 2 + 12e^x = 0

or

3e^{2x} - 12e^x - 2 = 0

Let $m = e^x$ :

3m^2 - 12m - 2 = 0

Using the quadratic formula:

m = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(-2)}}{2(3)} = \frac{12 \pm \sqrt{144 + 24}}{6} = \frac{12 \pm \sqrt{168}}{6} = \frac{6 \pm \sqrt{42}}{3}

Since $m = e^x > 0$ , reject $\frac{6 - \sqrt{42}}{3} < 0$ . Therefore

e^x = \frac{6 + \sqrt{42}}{3}

and

x = \ln\left(\frac{6 + \sqrt{42}}{3}\right)

Evaluating:

x \approx \ln(4.16025) \approx 1.42556

Answer

x = 1.426 \quad (3\text{ d.p.})

Final answer

x = 1.426

Detailed explanation

Walkthrough

We need to solve an equation where the unknown $x$ appears in exponents, both as $e^x$ and $e^{-x}$ . The first step is to clear the denominator by multiplying both sides by $e^x - 3$ . This is valid as long as $e^x \neq 3$ , and we will check at the end that our answer does not make the denominator zero.

After multiplying out, we bring all terms to one side. The term $e^{-x}$ is awkward, so we multiply the whole equation by $e^x$ . This converts $e^{-x}$ into $1$ and $e^x$ into $e^{2x}$ , giving a quadratic in $e^x$ .

To make the quadratic easier to handle, we substitute $m = e^x$ . The quadratic in $m$ is solved with the quadratic formula. One of the two solutions is negative. Since $e^x$ is always positive, that solution cannot be valid and must be rejected.

For the remaining positive value of $m$ , we take natural logarithms on both sides to recover $x$ , because $\ln(e^x) = x$ . Finally we evaluate the logarithm and round to 3 decimal places.

Key Takeaways

Exponential equations with $e^x$ and $e^{-x}$ can be turned into quadratics by substituting $m = e^x$ after multiplying through by $e^x$ .
The domain restriction $e^x > 0$ is essential: any root of the quadratic that is negative must be rejected.
Taking natural logarithms is the correct inverse operation to solve $e^x = \text{constant}$ .
A full method must be shown; an unsupported decimal answer would receive no marks.

Common Mistakes

Forgetting to multiply the right-hand side by the denominator when clearing the fraction.
Mis-handling $e^{-x}$ : multiplying by $e^x$ turns it into $1$ , not into $e^{-2x}$ .
Keeping the negative quadratic root and taking its logarithm, which is undefined.
Giving an unsupported answer: the mark scheme awards 0/5 for an answer with no working.
Rounding incorrectly to 3 decimal places (the required answer is 1.426, not 1.425 or 1.43).

Things to Be Careful About

The denominator $e^x - 3$ must not be zero. Our final answer $x \approx 1.426$ gives $e^x \approx 4.16$ , so the denominator is not zero.
The quadratic formula must be applied correctly: $a = 3$ , $b = -12$ , $c = -2$ .
The negative root $\frac{6 - \sqrt{42}}{3}$ is approximately $-0.160$ , so it must be rejected because $e^x > 0$ . If seen, it must be explicitly rejected.
The answer must be given correct to 3 decimal places; the mark scheme requires exactly 1.426.

Techniques used

multiply through by the denominator to clear the fractionsubstitute m = e^x to form a quadraticsolve the quadratic using the quadratic formulatake natural logarithms to recover xreject a root outside the domain e^x > 0

Similar questions

Q2MediumAlgebra

(a)

Expand $(6-x)(1-2x)^{-\frac{3}{2}}$ in ascending powers of $x$ , up to and including the term in $x^2$ , simplifying the coefficients.

4M Similar

DifficultyMedium

Worked solution

Approach

Use the binomial expansion for $(1+z)^n$ with $n=-\frac{3}{2}$ and $z=-2x$ , keeping terms up to $x^2$ . Then multiply the resulting three-term expansion by $(6-x)$ and collect terms up to $x^2$ .

Working

(1-2x)^{-\frac{3}{2}} = 1 + \left(-\frac{3}{2}\right)(-2x) + \frac{\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{2!}(-2x)^2 + \cdots

Simplify term by term:

= 1 + 3x + \frac{15}{8}(4x^2) + \cdots = 1 + 3x + \frac{15}{2}x^2 + \cdots

Now multiply by $(6-x)$ :

(6-x)\left(1 + 3x + \frac{15}{2}x^2\right) = 6 + 18x + 45x^2 - x - 3x^2 + \cdots

Collect terms up to $x^2$ :

= 6 + (18-1)x + (45-3)x^2 = 6 + 17x + 42x^2

Answer

6 + 17x + 42x^2

Final answer

6 + 17x + 42x^2

Detailed explanation

Walkthrough

We need the expansion of $(6-x)(1-2x)^{-\frac{3}{2}}$ up to the term in $x^2$ . The binomial theorem for $(1+z)^n$ works for any rational $n$ as long as $|z|<1$ , so we first expand $(1-2x)^{-\frac{3}{2}}$ using $n=-\frac{3}{2}$ and $z=-2x$ . The first term is $1$ . The second term is $nz = \left(-\frac{3}{2}\right)(-2x) = 3x$ . The third term is $\frac{n(n-1)}{2!}z^2 = \frac{\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{2}(-2x)^2 = \frac{15}{8}(4x^2) = \frac{15}{2}x^2$ . We then multiply this three-term expansion by $(6-x)$ , keeping only terms up to $x^2$ . The $x^3$ term coming from $-x \cdot \frac{15}{2}x^2$ is ignored. Collecting $6 + 18x + 45x^2 - x - 3x^2$ gives $6 + 17x + 42x^2$ .

Key Takeaways

This question tests the binomial expansion for a negative fractional index. The key skill is to substitute $z=-2x$ carefully into the standard formula and then multiply by another polynomial while ignoring terms beyond the required power. It also reinforces the importance of simplifying coefficients completely.

Common Mistakes

A common mistake is to use $z=2x$ instead of $z=-2x$ , which gives the wrong sign in the second term. Another common mistake is to forget the factor of $(-2x)^2$ , leading to an incorrect $x^2$ coefficient. Some students also multiply incorrectly by $(6-x)$ and include the $x^3$ term, which should be ignored because only terms up to $x^2$ are required.

Things to Be Careful About

The coefficient of $x^2$ in the binomial expansion must be simplified from $\frac{15}{8}(4x^2)$ to $\frac{15}{2}x^2$ . When multiplying by $(6-x)$ , remember that the $-x$ multiplies every term of the expansion, so it contributes $-x$ and $-3x^2$ before ignoring higher powers.

Techniques used

expand a binomial with a negative fractional indexmultiply expansions by a linear factorcollect like terms up to x^2

(b)

State the set of values of $x$ for which the expansion is valid.

1M Similar

DifficultyEasy

Worked solution

Approach

The binomial expansion $(1+z)^n$ is valid only when $|z|<1$ . Here $z=-2x$ , so the expansion is valid when $|-2x|<1$ .

Working

|-2x| < 1 \implies 2|x| < 1 \implies |x| < \frac{1}{2}

Answer

-\frac{1}{2} < x < \frac{1}{2}

Final answer

-1/2 < x < 1/2

Detailed explanation

Walkthrough

For a binomial expansion of $(1+z)^n$ with a negative or fractional $n$ , the series is valid only when $|z|<1$ . In this question $z=-2x$ , so we need $|-2x|<1$ . Since $|-2x|=2|x|$ , this becomes $2|x|<1$ , or $|x|<\frac{1}{2}$ . Written as an interval, this is $-\frac{1}{2}<x<\frac{1}{2}$ . The inequality must be strict because the binomial series does not converge at the endpoints.

Key Takeaways

The validity condition for a binomial expansion with a non-positive or fractional index is always $|z|<1$ , where $z$ is the term being raised to the power. This question shows how to translate that condition into an inequality in $x$ when $z$ is a multiple of $x$ .

Common Mistakes

A common mistake is to write $x<\frac{1}{2}$ without the modulus, which is incomplete. Another common mistake is to include equality, writing $|x|\leq\frac{1}{2}$ , which is incorrect because the expansion is not valid at the endpoints.

Things to Be Careful About

The condition must be strict: $|x|<\frac{1}{2}$ , not $|x|\leq\frac{1}{2}$ . Also remember that $|-2x|=2|x|$ , so dividing by $2$ gives the correct bound on $|x|$ .

Techniques used

apply the binomial validity condition |z| < 1solve a modulus inequality

Q35MMedium-EasyComplex Numbers

On an Argand diagram shade the region whose points represent complex numbers $z$ which satisfy both the inequalities $|z - 3i| \leq 2$ and $\frac{1}{4}\pi \leq \arg(z - 1 - 2i) \leq \frac{3}{4}\pi$ .

DifficultyMedium-Easy

Worked solution

Approach

Interpret each inequality geometrically on the Argand diagram. The first inequality defines a circular region, and the second defines an angular sector. The solution is the intersection of these two regions.

Working

1. Analyze the first inequality: $|z - 3i| \leq 2$

This represents the set of points $z$ whose distance from $3i$ is less than or equal to 2. On the Argand diagram, $3i$ corresponds to the point $(0, 3)$ . Thus, this is a closed disk (a circle and its interior) with centre $(0, 3)$ and radius 2.

2. Analyze the second inequality: $\frac{1}{4}\pi \leq \arg(z - 1 - 2i) \leq \frac{3}{4}\pi$

This can be rewritten as $\frac{1}{4}\pi \leq \arg(z - (1 + 2i)) \leq \frac{3}{4}\pi$ . The argument is measured from the point $1 + 2i$ , which corresponds to $(1, 2)$ . The angles $\frac{1}{4}\pi$ and $\frac{3}{4}\pi$ are measured anticlockwise from the positive real axis. This defines a sector (a wedge-shaped region) bounded by two half-lines emanating from $(1, 2)$ :

One half-line at an angle of $\frac{\pi}{4}$ to the positive real axis (slope 1, passing through $(0, 3)$ and $(2, 3)$ ).
The other half-line at an angle of $\frac{3\pi}{4}$ to the positive real axis (slope -1, passing through $(0, 3)$ and $(-2, 3)$ ).

3. Find the intersection and shade:
The required region is the intersection of the closed disk and the angular sector. This is the portion of the disk that lies between the two half-lines.

Answer

The shaded region is the intersection of the closed disk $|z - 3i| \leq 2$ and the sector $\frac{\pi}{4} \leq \arg(z - 1 - 2i) \leq \frac{3\pi}{4}$ .

Final answer

The region inside the circle $|z - 3i| = 2$ and between the rays from $1+2i$ at angles $\pi/4$ and $3\pi/4$ .

Detailed explanation

Walkthrough

First, we translate the complex inequalities into geometric conditions on the Argand diagram.

The inequality $|z - 3i| \leq 2$ is a standard modulus inequality. It states that the distance between the point $z$ and the point $3i$ (which is $(0, 3)$ on the Argand diagram) is at most 2. This describes a solid circle (disk) with centre $(0, 3)$ and radius 2.

Next, we look at $\frac{1}{4}\pi \leq \arg(z - 1 - 2i) \leq \frac{3}{4}\pi$ . The expression $z - (1 + 2i)$ represents the vector from the point $1 + 2i$ (which is $(1, 2)$ ) to the point $z$ . The argument of this vector is the angle it makes with the positive real axis. Requiring this angle to be between $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ restricts $z$ to a wedge-shaped region (sector) originating at $(1, 2)$ . The boundary rays are:

$\arg(z - (1 + 2i)) = \frac{\pi}{4}$ : a half-line from $(1, 2)$ at 45° to the horizontal. Since the slope is $\tan(\frac{\pi}{4}) = 1$ , this line passes through $(0, 3)$ and $(2, 3)$ .
$\arg(z - (1 + 2i)) = \frac{3\pi}{4}$ : a half-line from $(1, 2)$ at 135° to the horizontal. Since the slope is $\tan(\frac{3\pi}{4}) = -1$ , this line passes through $(0, 3)$ and $(-2, 3)$ .

Finally, we shade the region that satisfies both conditions simultaneously. This is the part of the circular disk that lies between the two half-lines.

Key Takeaways

$|z - a| \leq r$ represents a closed disk of radius $r$ centred at the point $a$ on the Argand diagram.
$\alpha \leq \arg(z - a) \leq \beta$ represents a sector (angular region) with its vertex at $a$ , bounded by half-lines at angles $\alpha$ and $\beta$ .
Finding the solution to simultaneous complex inequalities involves graphing each locus and identifying their intersection.

Common Mistakes

Forgetting that $|z - a| \leq r$ includes the interior of the circle (the inequality is $\leq$ , not $<$ ).
Drawing full lines instead of half-lines for the argument inequalities. The argument is only defined relative to the starting point $1 + 2i$ , so the boundaries must be rays emanating from $(1, 2)$ , not infinite lines.
Measuring the argument angles from the origin instead of from the point $(1, 2)$ .
Shading the wrong region (e.g., outside the circle or outside the sector).

Things to Be Careful About

Ensure the half-lines are correctly drawn from $(1, 2)$ and not from the origin. The angles $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ are measured from the positive real axis direction starting at $(1, 2)$ .
The circle centre is $(0, 3)$ , not $(3, 0)$ . Remember that $3i$ corresponds to the imaginary axis.
The shading must be the intersection of both regions. Dependent marks in the scheme mean that incorrect previous drawings will lead to incorrect shading.

Techniques used

interpret modulus inequality as a circle regioninterpret argument inequality as a sector regionfind intersection of regions on Argand diagram

Similar questions

Q46MMediumTrigonometry

Solve the equation $3\cot x - 4\cot 2x = 3$ for $0^\circ \leq x \leq 180^\circ$ .

DifficultyMedium

Worked solution

Approach

Use $\cot x = \frac{1}{\tan x}$ and the double-angle formula $\cot 2x = \frac{1-\tan^2 x}{2\tan x}$ to rewrite the equation as a quadratic in $t = \tan x$ . Solve the quadratic, then take arctangent values in the interval $0^\circ \leq x \leq 180^\circ$ .

Working

Let $t = \tan x$ . Then

\cot x = \frac{1}{t}, \qquad \cot 2x = \frac{1-t^2}{2t}.

Substitute into the equation:

\frac{3}{t} - 4\frac{1-t^2}{2t} = 3.

Simplify the left-hand side:

\frac{3}{t} - \frac{2(1-t^2)}{t} = \frac{3 - 2 + Ϳ2t^2}{t} = \frac{1+2t^2}{t}.

Therefore

\frac{1+2t^2}{t} = 3 \implies 1+2t^2 = 3t \implies 2t^2 - 3t +1=0.

Factorise:

(2t-1)(t-1)=0.

So

t = \frac{1}{2} \quad \text{or} \quad t =1.

Since $t = \tan x$ :

\tan x = \frac{1}{2} \quad \text{or} \quad \tan x =1.

For $0^\circ \leq x \leq180^\circ$ , the only solutions are

x = \arctan\left(\frac{1}{2}\right) \approx 26.6^\circ, \qquad x = \arctan(1) =45^\circ.

Answer

x =45^\circ \quad \text{or} \quad x =26.6^\circ \text{ (AWRT)}.

Final answer

x = 45° or x ≈ Ϳ26.6°

Detailed explanation

Walkthrough

The equation mixes $\cot x$ with $\cot 2x$ . To make it solvable, express both in terms of $\tan x$ . Let $t = \tan x$ . Then $\cot x = \frac{1}{t}$ and, from the double-angle formula for tangent, $\cot 2x = \frac{1-t^2}{2t}$ . Substituting turns the trigonometric equation into an algebraic equation in $t$ .

After substitution, combine the two fractions on the left. This gives a single fraction $\frac{1+2t^2}{t}$ . Setting this equal to 3 and multiplying by $t$ produces the quadratic $2t^2 - 3t +1=0$ . Factorising gives $t = \frac{1}{2}$ or $t =1$ .

Now return to $x$ : solve $\tan x = \frac{1}{2}$ and $\tan x =1$ . Within $0^\circ \leq x \leq180^\circ$ , these give $x = \arctan\left(\frac{1}{2}\right) \approx 26.6^\circ$ and $x =45^\circ$ . Because tangent has period $180^\circ$ , adding $180^\circ$ to either answer would leave the interval, so there are no further solutions.

Key Takeaways

Cotangent is the reciprocal of tangent: $\cot x = \frac{1}{\tan x}$ .
The double-angle formula for cotangent is $\cot 2x = \frac{1-\tan^2 x}{2\tan x}$ ; it follows from $\tan 2x = \frac{2\tan x}{1-\tan^2 x}$ .
Rewriting a trigonometric equation in terms of one function often reduces it to a polynomial equation.
Always check the required interval after solving a trigonometric equation.

Common Mistakes

Using an incorrect double-angle formula for $\cot 2x$ , such as treating it as $2\cot x$ .
Forgetting that $\cot x$ is undefined when $\tan x =0$ , so $x =0^\circ$ and $x =180^\circ$ are not solutions; no solutions are lost by multiplying by $\tan x$ .
Stopping after finding $\tan x = k$ without converting back to angles in the given interval.

Things to Be Careful About

The interval is $0^\circ \leq x \leq180^\circ$ . Since tangent has period $180^\circ$ , each value of $\tan x$ gives exactly one angle in this interval here; the other possible angles are outside the interval and should be ignored, as the mark scheme says.
Use AWRT $26.6^\circ$ for the approximate answer; the exact form is $\arctan\left(\frac{1}{2}\right)$ .
When simplifying, ensure the algebraic manipulation preserves equality; multiplying by $t$ is valid because $t \neq 0$ for valid inputs.

Techniques used

convert cotangent to tangentapply double-angle formula for cotangentreduce trigonometric equation to a quadraticsolve quadratic and select interval solutions

Similar questions

Q55MMediumComplex Numbers

The square roots of $-1 - 4\sqrt{5}i$ can be expressed in the Cartesian form $x + iy$ , where $x$ and $y$ are real and exact.

By first forming a quartic equation in $x$ or $y$ , find the square roots of $-1 - 4\sqrt{5}i$ in exact Cartesian form.

DifficultyMedium

Worked solution

Approach

Let the square root be $x + iy$ . Square it and equate real and imaginary parts with $-1$ and $-4\sqrt{5}$ respectively. This gives two equations in $x$ and $y$ . Eliminate one variable to obtain a quartic in the other, solve it, then recover the other variable using $xy = -2\sqrt{5}$ .

Working

Let $z = x + iy$ . Then

\begin{aligned} (x + iy)^2 &= x^2 - y^2 + 2xy\,i \\ &= -1 - 4\sqrt{5}\,i. \end{aligned}

Equating real and imaginary parts:

x^2 - y^2 = -1, \qquad 2xy = -4\sqrt{5}.

From $2xy = -4\sqrt{5}$ :

xy = -2\sqrt{5}.

Since $x \neq 0$ , write $y = -\frac{2\sqrt{5}}{x}$ . Substitute into $x^2 - y^2 = -1$ :

x^2 - \frac{20}{x^2} = -1.

Multiply by $x^2$ :

x^4 - 20 = -x^2,

so

x^4 + x^2 - 20 = 0.

Let $u = x^2$ . Then

u^2 + u - 20 = 0,

(u + 5)(u - 4) = 0.

Thus $u = -5$ or $u = 4$ . Since $x$ is real, $u = x^2 \geq 0$ , so $x^2 = 4$ and

x = \pm 2.

Now use $xy = -2\sqrt{5}$ .

If $x = 2$ :

y = \frac{-2\sqrt{5}}{2} = -\sqrt{5}.

If $x = -2$ :

y = \frac{-2\sqrt{5}}{-2} = \sqrt{5}.

Therefore the square roots are

2 - \sqrt{5}\,i \quad \text{and} \quad -2 + \sqrt{5}\,i,

that is,

\pm(2 - \sqrt{5}\,i).

Answer

\pm(2 - \sqrt{5}\,i)

Final answer

±(2 - √5 i)

Detailed explanation

Walkthrough

We need to find complex numbers $z = x + iy$ whose square is $-1 - 4\sqrt{5}i$ . The most direct method is to expand $(x+iy)^2$ and compare real and imaginary parts. This produces two equations linking $x$ and $y$ . Because the two equations involve both $x^2-y^2$ and $xy$ , we can eliminate one variable by substituting $y = -2\sqrt{5}/x$ into the first equation. This is why the question asks for a quartic in $x$ or $y$ : after substitution and multiplying by $x^2$ , the equation becomes quartic. We solve it by treating $x^2$ as a single unknown $u$ , giving a quadratic. We discard $u = -5$ because $x$ is real, so $x^2$ cannot be negative. Finally, we use $xy = -2\sqrt{5}$ to find the matching $y$ for each $x$ , giving two square roots.

Key Takeaways

Squaring $x+iy$ and equating real and imaginary parts is the standard way to find square roots of a complex number in Cartesian form.
The equations $x^2 - y^2 = a$ and $2xy = b$ must be solved simultaneously, not separately.
A quartic in $x$ or $y$ can often be reduced to a quadratic by substituting $u = x^2$ or $u = y^2$ .
The sign of $xy$ determines the signs of $x$ and $y$ together; both combinations must be checked.

Common Mistakes

Forgetting to equate both real and imaginary parts.
Using only $x^2 - y^2 = -1$ and not using $2xy = -4\sqrt{5}$ to connect $x$ and $y$ .
Incorrectly substituting $y$ , for example writing $x = -2\sqrt{5}y$ instead of $xy = -2\sqrt{5}$ . The mark scheme explicitly disallows incorrect algebra such as $x = -2\sqrt{5}y$ .
Keeping $x^2 = -5$ as a real possibility. Since $x$ is real, $x^2 \geq 0$ .
Giving $x$ and $y$ separately instead of stating the square roots $x+iy$ . The mark scheme requires the square roots, not coordinates.
Writing $\sqrt{4}$ instead of $2$ ; exact form should be simplified.

Things to Be Careful About

The quartic may be written as $x^4 + x^2 - 20 = 0$ or $y^4 - y^2 - 20 = 0$ ; either is acceptable. If you use the $y$ -quartic, you must still recover $x$ using $xy = -2\sqrt{5}$ .
When solving $u^2 + u - 20 = 0$ , the factorisation is $(u+5)(u-4)$ , so $u = -5$ or $u = 4$ . Only $u=4$ is valid for real $x$ .
Check that both final values satisfy the original equation. For $x=2$ , $y=-\sqrt{5}$ ; for $x=-2$ , $y=\sqrt{5}$ . No additional roots exist.
If no working is shown, the mark scheme awards 0 marks. Show the expansion, the two equations, the elimination, and the final square roots.

Techniques used

expand and equate real and imaginary partseliminate a variable to form a quarticsolve a quartic by substitutionrecover the other variablestate the square roots in Cartesian form

Similar questions

Q6MediumNumerical Solution of EquationsAlgebra

(a)

By sketching a suitable pair of graphs, show that the equation

|x - 2| = 2\sin\frac{1}{2}x

has only one root in the interval $0 < x < \pi$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To show that the equation $|x - 2| = 2\sin\frac{1}{2}x$ has only one root in $0 < x < \pi$ , we sketch the two functions $y = |x - 2|$ and $y = 2\sin\frac{1}{2}x$ on the same axes within this interval and observe their point of intersection.

Working

For the interval $0 < x < \pi$ , we note that $x < 2$ , so $|x - 2| = 2 - x$ . This is a straight line with a y-intercept at $(0, 2)$ and an x-intercept at $(2, 0)$ .

For the sine function $y = 2\sin\frac{1}{2}x$ :

At $x = 0$ , $y = 2\sin(0) = 0$ .
At $x = \pi$ , $y = 2\sin\frac{\pi}{2} = 2$ .
The graph is a smooth curve increasing from $(0, 0)$ to $(\pi, 2)$ .

The V-shaped modulus graph (specifically the left arm $y = 2 - x$ ) and the increasing sine curve intersect at exactly one point in the interval $0 < x < \pi$ . This single intersection confirms that the equation has only one root in this interval.

Answer

The graphs intersect at a single point in $0 < x < \pi$ , proving there is only one root.

Final answer

The graphs y = |x - 2| and y = 2sin(x/2) intersect at exactly one point in the interval 0 < x < pi.

Detailed explanation

Walkthrough

We are asked to show that the equation $|x - 2| = 2\sin\frac{1}{2}x$ has exactly one root in $0 < x < \pi$ . We do this by treating each side of the equation as a separate function and sketching them on the same set of axes.

First, consider $y = |x - 2|$ . In the interval $0 < x < \pi$ (which is approximately $0 < x < 3.14$ ), the value $x$ is less than $2$ for part of the interval and greater than $2$ for the rest. However, for $x < 2$ , $|x - 2| = 2 - x$ , which is a straight line with a y-intercept of $2$ and an x-intercept of $2$ . For $x > 2$ , $|x - 2| = x - 2$ , which is a line rising from $(2, 0)$ . The overall graph is V-shaped with its vertex at $(2, 0)$ .

Next, consider $y = 2\sin\frac{1}{2}x$ . At $x = 0$ , $y = 0$ . At $x = \pi$ , $y = 2\sin\frac{\pi}{2} = 2$ . The graph is a smooth, increasing curve from $(0, 0)$ to $(\pi, 2)$ .

By drawing both graphs, we see that the descending line $y = 2 - x$ crosses the ascending sine curve exactly once before $x = 2$ , and the ascending line $y = x - 2$ does not meet the sine curve again before $x = \pi$ . Thus, there is only one intersection point, meaning only one root.

Key Takeaways

Sketching graphs is a powerful way to visualize and confirm the number of roots of an equation in a given interval.
Understanding the behavior of modulus functions and trigonometric functions in specific intervals is essential for accurate sketching.

Common Mistakes

Sketching the full modulus graph without restricting to the given interval $0 < x < \pi$ .
Forgetting that $|x - 2| = 2 - x$ when $x < 2$ , and incorrectly drawing the left arm as descending from the origin.
Using degrees instead of radians for the trigonometric calculations or sketching.

Things to Be Careful About

Ensure the sketch is restricted to $0 < x < \pi$ as instructed.
The vertex of the modulus graph at $(2, 0)$ should be correctly positioned relative to $\pi \approx 3.14$ .
The maximum value of the sine graph in this interval is $2$ at $x = \pi$ , which matches the y-intercept of the modulus graph, but they do not intersect there.

Techniques used

sketch modulus graphsketch trigonometric graphidentify intersection point

(b)

Show by calculation that this root lies between 1 and 1.5.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To show the root lies between $1$ and $1.5$ , we define a function $f(x) = |x - 2| - 2\sin\frac{1}{2}x$ and evaluate it at $x = 1$ and $x = 1.5$ . If $f(1)$ and $f(1.5)$ have opposite signs, a root must lie between them by the Intermediate Value Theorem.

Working

Let $f(x) = |x - 2| - 2\sin\frac{1}{2}x$ . Since $1$ and $1.5$ are both less than $2$ , we have $|x - 2| = 2 - x$ for these values.

Evaluate $f(x)$ at $x = 1$ :

f(1) = |1 - 2| - 2\sin\frac{1}{2}(1) = 1 - 2\sin(0.5)

Using a calculator (in radians):

f(1) = 1 - 2(0.47942...) = 1 - 0.95885... = 0.04115... > 0

Evaluate $f(x)$ at $x = 1.5$ :

f(1.5) = |1.5 - 2| - 2\sin\frac{1}{2}(1.5) = 0.5 - 2\sin(0.75)

Using a calculator:

f(1.5) = 0.5 - 2(0.68164...) = 0.5 - 1.36327... = -0.86327... < 0

Since $f(1) > 0$ and $f(1.5) < 0$ , there is a sign change in the interval $[1, 1.5]$ . Because $f(x)$ is continuous, there must be at least one root in this interval.

Answer

$f(1) \approx 0.0412 > 0$ and $f(1.5) \approx -0.8633 < 0$ , so the root lies between $1$ and $1.5$ .

Final answer

f(1) > 0 and f(1.5) < 0, so the root lies between 1 and 1.5.

Detailed explanation

Walkthrough

We want to prove that the root of $|x - 2| = 2\sin\frac{1}{2}x$ is between $1$ and $1.5$ . We rearrange this into a single function $f(x) = |x - 2| - 2\sin\frac{1}{2}x$ and look for a sign change between $x = 1$ and $x = 1.5$ .

First, we calculate $f(1)$ . Since $1 < 2$ , $|1 - 2| = 1$ . The sine term is $2\sin(0.5)$ . Calculating this gives $f(1) \approx 0.0411$ , which is positive.

Next, we calculate $f(1.5)$ . Since $1.5 < 2$ , $|1.5 - 2| = 0.5$ . The sine term is $2\sin(0.75)$ . Calculating this gives $f(1.5) \approx -0.8633$ , which is negative.

Because the function changes from positive to negative between $x = 1$ and $x = 1.5$ , and the function is continuous (composed of continuous modulus and sine functions), the Intermediate Value Theorem guarantees that $f(x) = 0$ at some point in this interval. Thus, the root lies between $1$ and $1.5$ .

Key Takeaways

The sign change method (Intermediate Value Theorem) is a reliable way to locate roots within a specific interval.
Always ensure your calculator is in the correct mode (radians, not degrees) when evaluating trigonometric functions in calculus and numerical methods.

Common Mistakes

Forgetting to take the absolute value correctly (e.g., calculating $|1 - 2|$ as $-1$ instead of $1$ ).
Using degrees instead of radians for the sine function, which gives incorrect values (e.g., $2\sin(0.5^\circ) \approx 0.017$ ).
Failing to state the conclusion clearly, such as not mentioning the sign change or continuity.

Things to Be Careful About

The function must be continuous in the interval for the sign change argument to be valid. Here, both $|x - 2|$ and $\sin(\frac{1}{2}x)$ are continuous everywhere, so their difference is continuous.
Ensure all intermediate values are calculated to sufficient decimal places to clearly determine the sign.

Techniques used

define function from equationevaluate function at endpointsapply sign change theorem

(c)

Use the iterative formula $x_{n+1} = 2 - 2\sin\frac{1}{2}x_n$ with an initial value of 1.03 to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

3M Similar

DifficultyMedium

Worked solution

Approach

We use the given iterative formula $x_{n+1} = 2 - 2\sin\frac{1}{2}x_n$ with $x_0 = 1.03$ . We calculate successive values of $x_n$ to $4$ decimal places until the values converge to $2$ decimal places.

Working

Given $x_0 = 1.03$ :

Iteration 1:

x_1 = 2 - 2\sin\left(\frac{1}{2} \times 1.03\right) = 2 - 2\sin(0.515)

x_1 = 2 - 2(0.49174...) = 1.01651... \approx 1.0165

Iteration 2:

x_2 = 2 - 2\sin\left(\frac{1}{2} \times 1.0165\right) = 2 - 2\sin(0.50825)

x_2 = 2 - 2(0.48702...) = 1.02596... \approx 1.0260

Iteration 3:

x_3 = 2 - 2\sin\left(\frac{1}{2} \times 1.0260\right) = 2 - 2\sin(0.5130)

x_3 = 2 - 2(0.49093...) = 1.01813... \approx 1.0181

Iteration 4:

x_4 = 2 - 2\sin\left(\frac{1}{2} \times 1.0181\right) = 2 - 2\sin(0.50905)

x_4 = 2 - 2(0.48803...) = 1.02394... \approx 1.0239

Iteration 5:

x_5 = 2 - 2\sin\left(\frac{1}{2} \times 1.0239\right) = 2 - 2\sin(0.51195)

x_5 = 2 - 2(0.49000...) = 1.01999... \approx 1.0200

Iteration 6:

x_6 = 2 - 2\sin\left(\frac{1}{2} \times 1.0200\right) = 2 - 2\sin(0.5100)

x_6 = 2 - 2(0.48896...) = 1.02207... \approx 1.0221

Iteration 7:

x_7 = 2 - 2\sin\left(\frac{1}{2} \times 1.0221\right) = 2 - 2\sin(0.51105)

x_7 = 2 - 2(0.48949...) = 1.02101... \approx 1.0210

Iteration 8:

x_8 = 2 - 2\sin\left(\frac{1}{2} \times 1.0210\right) = 2 - 2\sin(0.5105)

x_8 = 2 - 2(0.48922...) = 1.02155... \approx 1.0216

The values are oscillating and converging. Let us continue to confirm convergence to $2$ d.p.:

Iteration 9:

x_9 = 2 - 2\sin\left(\frac{1}{2} \times 1.0216\right) = 2 - 2\sin(0.5108) \approx 1.0213

Iteration 10:

x_{10} = 2 - 2\sin\left(\frac{1}{2} \times 1.0213\right) = 2 - 2\sin(0.51065) \approx 1.0214

The values $x_9 = 1.0213$ and $x_{10} = 1.0214$ both round to $1.02$ to $2$ decimal places. The root is therefore $1.02$ to $2$ d.p.

Answer

The root correct to 2 decimal places is $1.02$ .

Final answer

1.02

Detailed explanation

Walkthrough

We are given the iterative formula $x_{n+1} = 2 - 2\sin\frac{1}{2}x_n$ and an initial value $x_0 = 1.03$ . We substitute $x_n$ into the formula to find $x_{n+1}$ , repeating this process until the answer stabilizes to the required accuracy (2 decimal places).

Starting with $x_0 = 1.03$ :

$x_1 = 2 - 2\sin(1.03/2) = 1.0165$
$x_2 = 2 - 2\sin(1.0165/2) = 1.0260$
$x_3 = 2 - 2\sin(1.0260/2) = 1.0181$
$x_4 = 2 - 2\sin(1.0181/2) = 1.0239$

We see the values are oscillating around the true root. We continue calculating:

$x_5 = 1.0200$
$x_6 = 1.0221$
$x_7 = 1.0210$
$x_8 = 1.0216$
$x_9 = 1.0213$
$x_{10} = 1.0214$

Since $x_9$ and $x_{10}$ both round to $1.02$ to 2 decimal places, we can be confident that the root to 2 d.p. is $1.02$ . To be absolutely rigorous, one could also show that the root lies in the interval $(1.015, 1.025)$ by evaluating the original function $f(x)$ at these bounds, but showing sufficient iterations that converge to $1.02$ is typically accepted.

Key Takeaways

Iterative formulas can be used to find roots of equations to a specified accuracy.
The process often involves oscillation before convergence; you must continue until the required number of decimal places is stable.
Always use the unrounded value from the previous iteration when calculating the next one to avoid accumulated rounding errors.

Common Mistakes

Rounding intermediate values too early (e.g., using $1.02$ instead of $1.0213$ in the next iteration).
Using degrees instead of radians in the calculator, which will give completely wrong iteration values.
Stopping the iterations too early before the answer has clearly converged to 2 d.p.

Things to Be Careful About

The question asks for the result of each iteration to 4 d.p., so you must show these values in your working.
Ensure you are using the correct iterative formula exactly as given: $x_{n+1} = 2 - 2\sin\frac{1}{2}x_n$ .
To justify the final answer to 2 d.p., you should either show that two consecutive iterations round to the same value to 2 d.p., or show a sign change in the interval $(1.015, 1.025)$ .

Techniques used

apply iterative formulacalculate successive iterationsdetermine convergence to required accuracy

Q7MediumTrigonometry

(a)

Express $7\sin\theta + 24\cos\theta$ in the form $R\cos(\theta - \alpha)$ , where $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$ . Give the value of $\alpha$ correct to 4 decimal places.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Expand $R\cos(\theta - \alpha)$ using the compound angle formula, then match coefficients of $\sin\theta$ and $\cos\theta$ with $7\sin\theta + 24\cos\theta$ to find $R$ and $\alpha$ .

Working

Expand using the identity $\cos(A - B) = \cos A\cos B + \sin A\sin B$ :

R\cos(\theta - \alpha) = R\cos\alpha\cos\theta + R\sin\alpha\sin\theta

Comparing with $7\sin\theta + 24\cos\theta$ :

R\cos\alpha = 24, \quad R\sin\alpha = 7

Square and add to find $R$ :

R^2\cos^2\alpha + R^2\sin^2\alpha = 24^2 + 7^2

R^2 = 576 + 49 = 625

R = 25

Find $\alpha$ by dividing:

\tan\alpha = \frac{R\sin\alpha}{R\cos\alpha} = \frac{7}{24}

Since $\cos\alpha = 24/25 > 0$ and $\sin\alpha = 7/25 > 0$ , $\alpha$ lies in the first quadrant, which satisfies $0 < \alpha < \tfrac{1}{2}\pi$ .

\alpha = \tan^{-1}\left(\frac{7}{24}\right) = 0.2837941\ldots

Answer

R = 25, \quad \alpha = 0.2838

Final answer

α = 0.2838

Detailed explanation

Walkthrough

We want to rewrite $7\sin\theta + 24\cos\theta$ as a single cosine. The compound-angle identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$ gives

$R\cos(\theta - \alpha) = R\cos\alpha\cos\theta + R\sin\alpha\sin\theta.$

Matching the $\cos\theta$ coefficient gives $R\cos\alpha = 24$ , and matching the $\sin\theta$ coefficient gives $R\sin\alpha = 7$ . Squaring and adding eliminates $\alpha$ because $\cos^2\alpha + \sin^2\alpha = 1$ , leaving $R^2 = 24^2 + 7^2 = 625$ , so $R = 25$ .

Dividing the two equations gives $\tan\alpha = 7/24$ . Both $\cos\alpha = 24/25$ and $\sin\alpha = 7/25$ are positive, so $\alpha$ is in the first quadrant — automatically satisfying the constraint $0 < \alpha < \tfrac{1}{2}\pi$ . Evaluating on a calculator: $\alpha \approx 0.2837941$ , which rounds to $0.2838$ to four decimal places.

Key Takeaways

The $R\cos(\theta - \alpha)$ form requires the identity $\cos(A - B) = \cos A\cos B + \sin A\sin B$ , with the $\sin$ term multiplying $\sin\alpha$ and the $\cos$ term multiplying $\cos\alpha$ .
$R$ is found from $\sqrt{a^2 + b^2}$ where $a$ and $b$ are the coefficients of $\cos\theta$ and $\sin\theta$ respectively.
$\alpha$ is found from $\tan\alpha = (\text{coeff of }\sin)/(\text{coeff of }\cos)$ .
The quadrant of $\alpha$ is determined by the signs of $\cos\alpha$ and $\sin\alpha$ (equivalently, the signs of the original coefficients).

Common Mistakes

Confusing $R\cos(\theta - \alpha)$ with $R\sin(\theta + \alpha)$ : in the cosine form, the $\sin$ term must have a plus sign (no negative), i.e. $R\cos\alpha\cos\theta + R\sin\alpha\sin\theta$ .
Swapping the ratio: $\tan\alpha$ should be (coefficient of $\sin\theta$ ) / (coefficient of $\cos\theta$ ) = $7/24$ , not $24/7$ .
Forgetting to check the quadrant: here both coefficients are positive so $\alpha$ is in the first quadrant, but if the $\cos$ coefficient were negative, $\alpha$ would lie in the second quadrant and $\pi$ would have to be added.

Things to Be Careful About

The mark scheme requires $R$ to be stated (B1) before any decimal approximation is used; if a decimal $R$ is written first, the B1 is not awarded. Always state $R = 25$ exactly.
The mark scheme does not allow credit if $\cos\alpha = 24$ and $\sin\alpha = 7$ are written without the factor $R$ : those raw values would give $R = \sqrt{24^2 + 7^2}$ but no $R\cos\alpha = 24$ identification, so the M mark is not earned. The cleanest way is to write $R\cos\alpha = 24$ and $R\sin\alpha = 7$ explicitly.
Round $\alpha$ to 4 decimal places: $0.2837941\ldots \to 0.2838$ .

Techniques used

expand R cos(θ - α) using compound angle formulaequate coefficients of sin θ and cos θcompute R using Pythagorascompute α using inverse tangent

(b)

Hence solve the equation $7\sin\frac{1}{3}x + 24\cos\frac{1}{3}x = 24.5$ for $0 < x < \pi$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Substitute the form $7\sin\tfrac{1}{3}x + 24\cos\tfrac{1}{3}x = 25\cos(\tfrac{1}{3}x - \alpha)$ from part (a), then solve $\cos(\tfrac{1}{3}x - \alpha) = 24.5/25 = 0.98$ using the inverse cosine, taking both the principal and supplementary angles into account.

Working

Using the result of part (a) with $\theta = \tfrac{1}{3}x$ :

25\cos\left(\tfrac{1}{3}x - 0.2838\right) = 24.5

Divide by $25$ :

\cos\left(\tfrac{1}{3}x - 0.2838\right) = \frac{24.5}{25} = 0.98

Apply $\cos^{-1}$ to both sides. The principal value is

\cos^{-1}(0.98) = 0.20033\ldots

The general solutions for $\cos Y = 0.98$ are $Y = \pm\,0.20033\ldots + 2k\pi$ . For $0 < x < \pi$ the relevant range is $0 < \tfrac{1}{3}x < \tfrac{\pi}{3}$ , so only the principal-angle solutions are needed:

\tfrac{1}{3}x - 0.2838 = 0.20033\ldots

\tfrac{1}{3}x = 0.2838 + 0.20033\ldots = 0.48413\ldots

x = 3 \times 0.48413\ldots = 1.45240\ldots

And:

\tfrac{1}{3}x - 0.2838 = -0.20033\ldots

\tfrac{1}{3}x = 0.2838 - 0.20033\ldots = 0.08347\ldots

x = 3 \times 0.08347\ldots = 0.25040\ldots

Both values lie in $0 < x < \pi$ , so both are valid solutions.

Answer

x = 0.250 \quad \text{or} \quad x = 1.45 \text{ (to 3 s.f.)}

Final answer

x = 0.250 or x = 1.45

Detailed explanation

Walkthrough

From part (a) we have $7\sin\theta + 24\cos\theta = 25\cos(\theta - 0.2838)$ with $R = 25$ . The equation we are solving uses $\theta = \tfrac{1}{3}x$ , so we substitute to get

$25\cos\left(\tfrac{1}{3}x - 0.2838\right) = 24.5.$

Dividing both sides by $25$ gives $\cos(\tfrac{1}{3}x - 0.2838) = 0.98$ . This is a standard cosine equation: we need every angle whose cosine equals $0.98$ . The principal value is $\cos^{-1}(0.98) \approx 0.20033$ , and because cosine is even, the other solution in one period is $-0.20033$ . Writing $Y = \tfrac{1}{3}x - 0.2838$ , we have

$Y = \pm 0.20033\ldots$

The full general solution would be $Y = \pm 0.20033 + 2k\pi$ , but our interval $0 < x < \pi$ restricts $0 < \tfrac{1}{3}x < \tfrac{\pi}{3} \approx 1.047$ , so $Y$ ranges over $(-0.2838,\, 0.7636)$ — only the $k = 0$ cases produce solutions in this window.

For $Y = 0.20033$ : $\tfrac{1}{3}x = 0.20033 + 0.2838 = 0.48413$ , so $x \approx 1.4524 \approx 1.45$ .

For $Y = -0.20033$ : $\tfrac{1}{3}x = -0.20033 + 0.2838 = 0.08347$ , so $x \approx 0.2504 \approx 0.250$ .

Both fall inside $0 < x < \pi$ , so the solution set is $x = 0.250$ or $x = 1.45$ (to 3 significant figures).

Key Takeaways

Once $7\sin\theta + 24\cos\theta$ is in $R\cos(\theta - \alpha)$ form, the equation $7\sin\tfrac{1}{3}x + 24\cos\tfrac{1}{3}x = k$ reduces to a single cosine equation in $\tfrac{1}{3}x$ .
$\cos Y = c$ (with $|c| \leq 1$ ) has two solutions per period, $Y = \pm\cos^{-1}(c)$ — the symmetry of cosine is the reason for two answers here.
Always convert back to the original variable and check whether each candidate lies in the requested interval.

Common Mistakes

Forgetting the second solution. Cosine is even, so $\cos Y = 0.98$ gives both $Y \approx +0.20033$ and $Y \approx -0.20033$ , not just one of them.
Reporting an answer outside the interval. The mark scheme explicitly says "ignore answers outside the given interval", so a stray solution like $x = 1.45 + 2\pi \cdot 3$ should be excluded; only solutions in $0 < x < \pi$ count.
Rounding $\alpha$ too early and propagating the rounding through every subsequent step. The mark scheme accepts $0.2838$ to 4 d.p. (CAO for the $\alpha$ mark), but for the final $x$ values it accepts AWRT (anything within rounding tolerance), so a slight rounding error in $\alpha$ doesn't usually cost marks provided the method is correct.
Confusing degrees and radians. The problem uses radians throughout (the interval is in terms of $\pi$ ); make sure the calculator is in radian mode.

Things to Be Careful About

The B1FT in the mark scheme can be awarded for $\cos^{-1}(24.5/25)$ even if it isn't numerically evaluated, provided $R \geq 24.5$ (which it is, since $R = 25$ ). Writing the expression is enough to earn this mark.
The M1 requires a complete correct method leading to a value of $x$ — for example, the equation $\cos(\tfrac{1}{3}x - 0.2838) = 0.98$ must be visible (or implied by calculator working), and a candidate $x$ -value must be obtained. Insufficient working scores M0 even with the right final number.
The final A1 marks require both $x = 0.250$ (or $0.251$ ) and $x = 1.45$ in the interval, and no other in-interval solutions.

Techniques used

substitute the R cos form from part (a)rearrange to isolate the cosineapply inverse cosine to find principal angleuse the two general solutions of cosineconvert back to the original variable x and check the interval

Q87MMediumDifferential Equations

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

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

and $x = 0$ when $\theta = \frac{1}{12}\pi$ .

Solve the differential equation and obtain an expression for $x$ in terms of $\theta$ .

DifficultyMedium

Worked solution

Approach

Separate the variables so that all terms involving $x$ are on one side and all terms involving $\theta$ are on the other, then integrate both sides.

Working

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

Separating variables gives

\int \frac{1}{4x+3}\,dx = \int \frac{\cos 2\theta}{\sin 2\theta}\,d\theta

Integrating both sides,

\frac{1}{4}\ln(4x+3) = \frac{1}{2}\ln(\sin 2\theta) + c

Use the initial condition $x = 0$ when $\theta = \frac{1}{12}\pi$ :

\frac{1}{4}\ln 3 = \frac{1}{2}\ln\left(\sin\frac{\pi}{6}\right) + c = \frac{1}{2}\ln\frac{1}{2} + c

Therefore

c = \frac{1}{4}\ln 3 - \frac{1}{2}\ln\frac{1}{2} = \frac{1}{4}\ln 3 + \frac{1}{2}\ln 2 = \frac{1}{4}\ln 12

So the particular solution is

\frac{1}{4}\ln(4x+3) = \frac{1}{2}\ln(\sin 2\theta) + \frac{1}{4}\ln 12

Multiplying by 4 and combining the logarithms,

\ln(4x+3) = 2\ln(\sin 2\theta) + \ln 12 = \ln(12\sin^2 2\theta)

Exponentiating both sides gives

4x + 3 = 12\sin^2 2\theta

Answer

x = \frac{12\sin^2 2\theta - 3}{4}

Final answer

x = (12 sin^2(2θ) - 3)/4

Detailed explanation

Walkthrough

The differential equation is separable because we can collect all $x$ terms with $dx$ and all $\theta$ terms with $d\theta$ . Dividing by $(4x+3)\sin 2\theta$ gives

\int \frac{1}{4x+3}\,dx = \int \frac{\cos 2\theta}{\sin 2\theta}\,d\theta

On the left, the derivative of $4x+3$ is $4$ , so the integral is $\frac{1}{4}\ln(4x+3)$ . On the right, the derivative of $\sin 2\theta$ is $2\cos 2\theta$ , so the integral is $\frac{1}{2}\ln(\sin 2\theta)$ .

We then use the initial condition $x=0$ when $\theta = \frac{1}{12}\pi$ . Since $\sin\frac{\pi}{6} = \frac{1}{2}$ , we get

\frac{1}{4}\ln 3 = \frac{1}{2}\ln\frac{1}{2} + c

so

c = \frac{1}{4}\ln 3 - \frac{1}{2}\ln\frac{1}{2} = \frac{1}{4}\ln 3 + \frac{1}{2}\ln 2 = \frac{1}{4}\ln 12

Substituting this back and combining logarithms gives

\ln(4x+3) = \ln(12\sin^2 2\theta)

and exponentiating both sides gives the final expression for $x$ .

Key Takeaways

This question tests the full process of solving a first-order separable differential equation: separating variables, integrating using the $\frac{kf'(x)}{f(x)}$ pattern, applying an initial condition, and using laws of logarithms to rearrange for the dependent variable. It also shows that the constant of integration can be written as a logarithm, which makes the final exponentiation cleaner.

Common Mistakes

Forgetting the factors $\frac{1}{4}$ and $\frac{1}{2}$ that come from the chain rule when integrating $\frac{1}{4x+3}$ and $\frac{\cos 2\theta}{\sin 2\theta}$ .
Using the wrong value for $\sin\frac{\pi}{6}$ ; it is $\frac{1}{2}$ , not $0$ or $1$ .
Not applying the initial condition, or applying it before the constant is introduced.
Making sign errors when simplifying $\ln\frac{1}{2} = -\ln 2$ .
Failing to combine the logarithms correctly before exponentiating.

Things to Be Careful About

The mark scheme accepts equivalent intermediate forms, such as $\frac{1}{4}\ln\frac{4x+3}{3} = \frac{1}{2}\ln(2\sin 2\theta)$ , so do not worry if your constant appears as a logarithm in a different but equivalent way. When exponentiating, remember that $e^{2\ln(\sin 2\theta)} = \sin^2 2\theta$ , and that $e^{\ln 12} = 12$ . Ensure the final answer is expressed in terms of $\theta$ only, with no remaining constant $c$ .

Techniques used

separate variablesintegrate using the kf'(x)/f(x) patternapply the initial conditionsolve for x by exponentiating

Similar questions

Q9MediumVectors

With respect to the origin $O$ , the points $A$ , $B$ and $C$ have position vectors given by

\overrightarrow{OA} = \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} \quad \text{and} \quad \overrightarrow{OC} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}.

(a)

Find a vector equation for the line through $A$ and $B$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the position vectors of $A$ and $B$ to find a direction vector $\overrightarrow{AB}$ , then write the line as $\mathbf{r} = \overrightarrow{OA} + \lambda \overrightarrow{AB}$ .

Working

The direction vector from $A$ to $B$ is

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 5 \\ 1 \end{pmatrix}.

Using $\overrightarrow{OA}$ as a point on the line, a vector equation is

\mathbf{r} = \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -3 \\ 5 \\ 1 \end{pmatrix}, \quad \lambda \in \mathbb{R}.

Answer

\mathbf{r} = \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -3 \\ 5 \\ 1 \end{pmatrix}

Final answer

r = (1, -4, 2) + lambda(-3, 5, 1)

Detailed explanation

Walkthrough

The line through $A$ and $B$ needs a point on the line and a direction vector. We already know the position vector of $A$ , so we can use $A$ as the fixed point. To move from $A$ to $B$ , subtract $\overrightarrow{OA}$ from $\overrightarrow{OB}$ ; this gives the direction vector $\overrightarrow{AB}=\begin{pmatrix}-3\\5\\1\end{pmatrix}$ . A vector equation is then $\mathbf{r}=\overrightarrow{OA}+\lambda\overrightarrow{AB}$ , where $\lambda$ is a real parameter. As $\lambda$ varies, every point on the line is obtained.

Key Takeaways

A line in 3D can be written as $\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}$ , where $\mathbf{a}$ is a position vector of a point on the line and $\mathbf{d}$ is a direction vector. The direction vector between two points is found by subtracting their position vectors.

Common Mistakes

Writing $\mathbf{l}=\dots$ instead of $\mathbf{r}=\dots$ ; the mark scheme requires $\mathbf{r}=\dots$ .
Using the wrong order when subtracting: $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ , not the reverse. (Using the reverse still gives a valid line with a different parameter sign, but the displayed direction vector would be the negative.)
Forgetting to include a parameter such as $\lambda$ .

Things to Be Careful About

Any point on the line can be used as the fixed point, and any non-zero scalar multiple of the direction vector is acceptable.
The answer should be a vector equation, not just a direction vector.

Techniques used

find direction vector between two pointswrite vector equation of a line

(b)

Using a scalar product, find the exact value of $\cos BAC$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

The angle $BAC$ is the angle between the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ . Use the scalar product formula

\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}|\,|\overrightarrow{AC}|}.

Working

From part (a), $\overrightarrow{AB} = \begin{pmatrix} -3 \\ 5 \\ 1 \end{pmatrix}$ . Also,

\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 3 \end{pmatrix}.

Scalar product:

\overrightarrow{AB} \cdot \overrightarrow{AC} = (-3)(1) + (5)(7) + (1)(3) = 35.

Magnitudes:

|\overrightarrow{AB}| = \sqrt{(-3)^2 + 5^2 + 1^2} = \sqrt{35}, \qquad |\overrightarrow{AC}| = \sqrt{1^2 + 7^2 + 3^2} = \sqrt{59}.

Therefore

\cos BAC = \frac{35}{\sqrt{35}\sqrt{59}} = \sqrt{\frac{35}{59}}.

Answer

\cos BAC = \sqrt{\frac{35}{59}}

Final answer

cos BAC = sqrt(35/59)

Detailed explanation

Walkthrough

The angle $BAC$ is at $A$ , so it is the angle between the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ . We already have $\overrightarrow{AB}$ from part (a). We find $\overrightarrow{AC}$ by subtracting $\overrightarrow{OA}$ from $\overrightarrow{OC}$ . Then the scalar product formula gives $\cos BAC = \frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|}$ . Compute the dot product by multiplying corresponding components and adding. Compute each modulus by taking the square root of the sum of squares of components. Finally divide and simplify: since $35=\sqrt{35}\sqrt{35}$ , the fraction becomes $\sqrt{\frac{35}{59}}$ .

Key Takeaways

The scalar product of two vectors is $\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta$ . To find an angle at a point, use the two vectors that start at that point. The modulus of a vector is the square root of the sum of the squares of its components.

Common Mistakes

Using $\overrightarrow{BA}$ and $\overrightarrow{AC}$ together; both vectors must start at $A$ (or both must be reversed).
Forgetting to divide by the product of the moduli.
Leaving the answer as $\frac{35}{\sqrt{2065}}$ without simplifying, or writing only $\theta=\cos^{-1}\sqrt{\frac{35}{59}}$ without stating $\cos BAC$ ; the mark scheme requires the statement of $\cos BAC$ .
Not showing the scalar product calculation when the mark scheme requires method.

Things to Be Careful About

The scalar product here is positive, so $\cos BAC$ is positive and the angle is acute.
Exact simplification: $\frac{35}{\sqrt{35}\sqrt{59}}=\sqrt{\frac{35}{59}}$ ; equivalent forms such as $\frac{35}{\sqrt{2065}}$ are also accepted.
The answer must come from using a scalar product, as requested.

Techniques used

find direction vectors from a common vertexcompute scalar product of two vectorsdivide scalar product by product of magnitudes

(c)

Hence find the exact area of triangle $ABC$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the formula for the area of a triangle in terms of two side lengths and the included angle:

\text{Area} = \frac{1}{2}|\overrightarrow{AB}|\,|\overrightarrow{AC}|\sin BAC.

Since $\cos BAC$ is known from part (b), find $\sin BAC$ using $\sin^2 x + \cos^2 x = 1$ .

Working

From parts (a) and (b):

|\overrightarrow{AB}| = \sqrt{35}, \qquad |\overrightarrow{AC}| = \sqrt{59}, \qquad \cos BAC = \sqrt{\frac{35}{59}}.

Then

\sin^2 BAC = 1 - \cos^2 BAC = 1 - \frac{35}{59} = \frac{24}{59}.

So, since $BAC$ is an angle of a triangle,

\sin BAC = \sqrt{\frac{24}{59}}.

The area is

\begin{aligned} \text{Area} &= \frac{1}{2}\sqrt{35}\sqrt{59}\sqrt{\frac{24}{59}} \\ &= \frac{1}{2}\sqrt{35 \times 24} \\ &= \frac{1}{2}\sqrt{840} \\ &= \sqrt{210}. \end{aligned}

Answer

\text{Area} = \sqrt{210}

Final answer

Area = sqrt(210)

Detailed explanation

Walkthrough

The area of a triangle formed by two sides from $A$ is $\frac{1}{2}|\overrightarrow{AB}||\overrightarrow{AC}|\sin BAC$ . We know the two lengths and $\cos BAC$ from part (b). To get $\sin BAC$ , use $\sin^2 x+\cos^2 x=1$ . Since $\cos^2 BAC=\frac{35}{59}$ , $\sin^2 BAC=\frac{24}{59}$ , and the sine is positive for an angle of a triangle. Substitute into the area formula: $\frac{1}{2}\sqrt{35}\sqrt{59}\sqrt{\frac{24}{59}}=\frac{1}{2}\sqrt{35\times24}=\frac{1}{2}\sqrt{840}=\sqrt{210}$ .

Key Takeaways

The area formula $\frac{1}{2}ab\sin C$ can be applied using vectors from a common vertex. The identity $\sin^2 x+\cos^2 x=1$ lets us convert a known cosine into the required sine. Surds can be simplified by combining square roots and dividing by $4$ inside the root.

Common Mistakes

Forgetting the factor $\frac{1}{2}$ .
Taking $\sin BAC$ as negative; for an angle in a triangle it must be positive.
Not using the angle at $A$ ; because the question says "Hence", the area must be found using $\angle BAC$ and the previous result.
Giving an answer from $\sin(\cos^{-1}\sqrt{\frac{35}{59}})$ without showing the method of evaluation; the mark scheme requires evidence of the calculation.
Not simplifying to an exact form such as $\sqrt{210}$ or $\frac{1}{2}\sqrt{840}$ .

Things to Be Careful About

The two method marks (area formula and sine conversion) are independent, but both must be present for full method marks.
$\sin BAC=\sqrt{\frac{24}{59}}$ , not $\sqrt{1-\frac{35}{59}}$ with any ambiguity about sign.
Equivalent exact forms such as $\frac{1}{2}\sqrt{840}$ are accepted, but unsupported numerical answers are not.

Techniques used

apply triangle area formula using two side vectorsuse trigonometric identity to find sine from cosinesimplify surd expression

Q10MediumAlgebraIntegration

(a)

Find the quotient and remainder when $x^2$ is divided by $1 + 4x^2$ .

2M Similar

DifficultyEasy

Worked solution

Approach

We need to divide $x^2$ by $1+4x^2$ . Since both are quadratics, the quotient will be a constant and the remainder a constant. Write $x^2 = q(1+4x^2) + r$ and compare coefficients.

Working

Let the quotient be $q$ and the remainder be $r$ :

x^2 = q(1+4x^2) + r = 4qx^2 + (q+r)

Comparing coefficients of $x^2$ : $4q = 1$ , so $q = \frac{1}{4}$ .
Comparing constant terms: $q + r = 0$ , so $r = -\frac{1}{4}$ .

Answer

Quotient $= \frac{1}{4}$ , remainder $= -\frac{1}{4}$ .

Final answer

Quotient = 1/4, remainder = -1/4

Detailed explanation

Walkthrough

We are dividing $x^2$ by $1+4x^2$ . Since both are quadratics, the quotient must be a constant and the remainder a constant (any remainder must have degree less than the divisor, so degree less than 2). We write the division identity $x^2 = q(1+4x^2) + r$ and compare coefficients. Expanding gives $4qx^2 + (q+r)$ . Matching the $x^2$ coefficient gives $4q=1$ , so $q=1/4$ . Matching constants gives $q+r=0$ , so $r=-1/4$ .

Key Takeaways

This tests polynomial division where the divisor is a quadratic. The key idea is that when the degrees of dividend and divisor are equal, the quotient is a constant. Comparing coefficients is often faster than long division.

Common Mistakes

Forgetting that the remainder must have degree less than the divisor.
Sign errors when comparing constant terms.
The mark scheme says: "Allow B1B1 if implied by correct division and no further working, but do not ISW." So if the division is correct but the remainder is stated incorrectly, do not ignore subsequent working — the remainder mark is lost.

Things to Be Careful About

The divisor is written as $1+4x^2$ , i.e. $4x^2 + 1$ . The quotient is $\frac{1}{4}$ , not $4$ .
The remainder is negative: $-\frac{1}{4}$ . It's easy to drop the sign.

Techniques used

perform polynomial division by comparing coefficientsidentify quotient and remainder

(b)

Find the exact value of $\int_0^{0.5} x\tan^{-1}(2x)\,dx$ .

6M Similar

DifficultyMedium

Worked solution

Approach

Use integration by parts with $u = \tan^{-1}(2x)$ and $dv = x\,dx$ . Then use the quotient and remainder from part (a) to split the resulting rational integrand, integrate, and evaluate the definite integral at the limits.

Working

Let $u = \tan^{-1}(2x)$ , $dv = x\,dx$ . Then $du = \frac{2}{1+4x^2}\,dx$ and $v = \frac{x^2}{2}$ .

\int_0^{0.5} x\tan^{-1}(2x)\,dx = \left[\frac{x^2}{2}\tan^{-1}(2x)\right]_0^{0.5} - \int_0^{0.5} \frac{x^2}{1+4x^2}\,dx

Now split the integrand using the result from part (a): $x^2 = \frac{1}{4}(1+4x^2) - \frac{1}{4}$ , so

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

Therefore

\int \frac{x^2}{1+4x^2}\,dx = \frac{x}{4} - \frac{1}{8}\tan^{-1}(2x)

So the antiderivative is

F(x) = \frac{x^2}{2}\tan^{-1}(2x) - \frac{x}{4} + \frac{1}{8}\tan^{-1}(2x)

Evaluate at the limits:

At $x = 0.5$ : $\tan^{-1}(1) = \frac{\pi}{4}$ , so

F(0.5) = \frac{1}{8}\cdot\frac{\pi}{4} - \frac{1}{8} + \frac{1}{8}\cdot\frac{\pi}{4} = \frac{\pi}{16} - \frac{1}{8}

At $x = 0$ : $F(0) = 0$ .

Hence

\int_0^{0.5} x\tan^{-1}(2x)\,dx = \frac{\pi}{16} - \frac{1}{8}

Answer

\frac{\pi}{16} - \frac{1}{8}

Final answer

π/16 - 1/8

Detailed explanation

Walkthrough

We need $\int_0^{0.5} x\tan^{-1}(2x)\,dx$ . The integrand is a product of a polynomial ( $x$ ) and an inverse trig function ( $\tan^{-1}(2x)$ ), which signals integration by parts. Choose $u = \tan^{-1}(2x)$ so that differentiating gives $\frac{2}{1+4x^2}$ , and $dv = x\,dx$ so that $v = x^2/2$ . This transforms the integral into $\left[\frac{x^2}{2}\tan^{-1}(2x)\right]_0^{0.5} - \int_0^{0.5}\frac{x^2}{1+4x^2}\,dx$ .

The remaining integral $\int \frac{x^2}{1+4x^2}\,dx$ is handled by using the division result from part (a): $x^2 = \frac{1}{4}(1+4x^2) - \frac{1}{4}$ . So the integrand becomes $\frac{1}{4} - \frac{1}{4(1+4x^2)}$ . Integrating gives $\frac{x}{4} - \frac{1}{8}\tan^{-1}(2x)$ because $\int \frac{1}{1+4x^2}\,dx = \frac{1}{2}\tan^{-1}(2x)$ .

Combining, the antiderivative is $\frac{x^2}{2}\tan^{-1}(2x) - \frac{x}{4} + \frac{1}{8}\tan^{-1}(2x)$ . Substituting the upper limit $x=0.5$ gives $\tan^{-1}(1) = \pi/4$ , so the value is $\frac{1}{8}\cdot\frac{\pi}{4} - \frac{1}{8} + \frac{1}{8}\cdot\frac{\pi}{4} = \frac{\pi}{16} - \frac{1}{8}$ . The lower limit $x=0$ gives 0.

Key Takeaways

Integration by parts is the right tool when the integrand is a product of a polynomial and an inverse trig function.
The derivative of $\tan^{-1}(2x)$ is $\frac{2}{1+4x^2}$ , which produces the rational integrand.
Splitting a rational function using the quotient and remainder from a division (part (a)) converts it into a constant plus a term of the form $\frac{1}{1+4x^2}$ , which integrates to an arctan.

Common Mistakes

Choosing the wrong $u$ and $dv$ (e.g. $u=x$ , $dv=\tan^{-1}(2x)\,dx$ ) makes the problem harder.
Forgetting the factor of 2 in $\frac{d}{dx}\tan^{-1}(2x) = \frac{2}{1+4x^2}$ .
Incorrectly splitting $\frac{x^2}{1+4x^2}$ — the mark scheme FT's the quotient and remainder from part (a).
Forgetting to subtract the lower-limit contribution (mark scheme DM1 requires evidence of considering the lower limit, e.g. sight of 0).

Things to Be Careful About

The mark scheme says "do not ISW" for part (a) — if the quotient is correct but the remainder is wrong, subsequent working in part (b) that uses the wrong remainder will not be credited.
When substituting limits, the mark scheme requires evidence that the lower limit 0 has been considered for each term.
The final answer must be exact: $\frac{\pi}{16} - \frac{1}{8}$ .

Techniques used

apply integration by partssplit rational function using quotient and remainderintegrate 1/(1+4x^2) to arctangentevaluate definite integral at limits

Q11MediumDifferentiationTrigonometryIntegration

Fig. 11.1

The diagram shows the graph of $y = 5\sin 2x \cos^2 x$ for $0 \leq x \leq \frac{1}{2}\pi$ and its maximum point $M$ .

(a)

Find the exact $x$ -coordinate of $M$ .

6M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $y = 5 \sin 2x \cos^2 x$ using the product rule. Factor the resulting derivative, simplify using the compound angle identity, and solve the equation for $x$ in the interval $(0, \frac{1}{2}\pi)$ .

Working

Let $f(x) = 5\sin 2x$ and $g(x) = \cos^2 x$ , so $y = f(x)\,g(x)$ .

f'(x) = 10\cos 2x

g'(x) = 2\cos x \cdot (-\sin x) = -2\sin x\cos x = -\sin 2x

Apply the product rule:

\begin{aligned} \frac{dy}{dx} &= f'(x)\,g(x) + f(x)\,g'(x) \\ &= 10\cos 2x \cdot \cos^2 x + 5\sin 2x \cdot (-\sin 2x) \\ &= 10\cos 2x \cos^2 x - 5\sin^2 2x \\ &= 10\cos 2x \cos^2 x - 10\sin 2x \sin x \cos x \end{aligned}

Factor out $10\cos x$ (which is non-zero in the interior of the interval):

\frac{dy}{dx} = 10\cos x\bigl(\cos 2x \cos x - \sin 2x \sin x\bigr)

Apply the compound angle identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$ with $A = 2x$ , $B = x$ :

\frac{dy}{dx} = 10\cos x \cos(2x + x) = 10\cos x \cos 3x

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

10\cos x \cos 3x = 0

So either $\cos x = 0$ or $\cos 3x = 0$ .

$\cos x = 0 \Rightarrow x = \frac{1}{2}\pi$ (the right-hand endpoint, not an interior maximum).
$\cos 3x = 0 \Rightarrow 3x = \frac{1}{2}\pi \Rightarrow x = \frac{1}{6}\pi$ (the only solution in $(0, \frac{1}{2}\pi)$ ).

Hence the maximum $M$ occurs at $x = \frac{1}{6}\pi$ .

Answer

x = \frac{1}{6}\pi

Final answer

x = π/6

Detailed explanation

Walkthrough

We are asked to find the $x$ -coordinate of the maximum point $M$ on the curve $y = 5\sin 2x \cos^2 x$ for $0 \leq x \leq \frac{1}{2}\pi$ . Since the maximum is an interior point of the domain, we differentiate and set the derivative equal to zero.

Step 1 — Identify the two factors. Write $y$ as a product of $5\sin 2x$ and $\cos^2 x$ so the product rule can be applied. The first factor is a sine of a multiple of $x$ , the second is a cosine squared (a composite function).

Step 2 — Differentiate each factor.

For $5\sin 2x$ , the chain rule gives $5 \cdot 2\cos 2x = 10\cos 2x$ .
For $\cos^2 x$ , the chain rule gives $2\cos x \cdot (-\sin x) = -2\sin x \cos x$ , which is the same as $-\sin 2x$ by the double angle identity.

Step 3 — Apply the product rule. The product rule states $\frac{dy}{dx} = f'(x)g(x) + f(x)g'(x)$ . Substituting and simplifying with $5\sin 2x \cdot \sin 2x = \sin^2 2x$ and writing $\sin^2 2x = \sin 2x \cdot 2\sin x\cos x$ gives the derivative in the form $-10\sin 2x \sin x \cos x + 10\cos 2x \cos^2 x$ .

Step 4 — Factor and simplify using trig identities. Factor $10\cos x$ out of both terms. The remaining bracket is $\cos 2x \cos x - \sin 2x \sin x$ , which is exactly the expansion of $\cos(2x + x) = \cos 3x$ by the compound angle identity. The derivative simplifies to $10\cos x \cos 3x$ .

Step 5 — Solve the equation. With $10\cos x \cos 3x = 0$ , either factor can be zero. $\cos x = 0$ gives $x = \frac{1}{2}\pi$ (the endpoint, not a maximum in the interior). $\cos 3x = 0$ has solutions $3x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$ , and the only one in $(0, \frac{1}{2}\pi)$ is $3x = \frac{\pi}{2}$ , i.e. $x = \frac{1}{6}\pi$ .

Key Takeaways

The product rule is essential whenever a function is a product of two or more simpler functions.
The chain rule is required to differentiate $\cos^2 x$ (a composite of cosine and squaring).
The double angle and compound angle identities are powerful tools for condensing a derivative into a factorable form.
A maximum in the interior of a closed interval corresponds to $\frac{dy}{dx} = 0$ there, not to one of the endpoints.

Common Mistakes

Forgetting the chain rule on $\cos^2 x$ and writing the derivative of $\cos^2 x$ as $2\cos x$ instead of $-2\sin x\cos x$ .
Dropping the chain rule on $5\sin 2x$ and writing the derivative as $5\cos 2x$ instead of $10\cos 2x$ .
Failing to use the compound angle identity, leaving the derivative as an unfactorable expression and not being able to solve it.
Including $x = \frac{1}{2}\pi$ (where $\cos x = 0$ ) as the location of the maximum, when in fact it is the right-hand endpoint where the curve returns to the $x$ -axis.

Things to Be Careful About

Dividing by $\cos x$ to factor is only valid when $\cos x \neq 0$ ; the case $\cos x = 0$ must be checked separately. Here it gives the endpoint $x = \frac{1}{2}\pi$ , which is not the maximum.
When solving $\cos 3x = 0$ the general solution is $3x = \frac{\pi}{2} + k\pi$ for integer $k$ ; only the solution with $0 < x < \frac{1}{2}\pi$ is required.
The mark scheme accepts any of $3\tan^2 x = 1$ , $4\sin^2 x = 1$ , $4\cos^2 x = 3$ , or $\cos 3x = 0$ as the equation in a single trig function; the cleanest derivation uses $\cos 3x = 0$ .

Techniques used

apply product rule to a product of trig functionsdifferentiate cos^2 x using the chain ruleuse the double angle identity to factor the derivativeapply the compound angle identity cos(A+B) = cos A cos B - sin A sin Bsolve cos 3x = 0 in the given interval

(b)

By using the substitution $u = \cos x$ , find the area of the region bounded by the curve, the $x$ -axis between $x = 0$ and $x = \frac{1}{4}\pi$ , and the line $x = \frac{1}{4}\pi$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Express the integrand using the double angle identity $\sin 2x = 2\sin x\cos x$ , then apply the substitution $u = \cos x$ as instructed. Convert both the integrand and the limits of integration, evaluate the resulting polynomial integral, and simplify.

Working

The required area is:

A = \int_{0}^{\frac{1}{4}\pi} 5\sin 2x \cos^2 x \,dx

Use the double angle identity $\sin 2x = 2\sin x \cos x$ :

A = \int_{0}^{\frac{1}{4}\pi} 5 \cdot 2\sin x \cos x \cdot \cos^2 x \,dx = 10\int_{0}^{\frac{1}{4}\pi} \sin x \cos^3 x \,dx

Substitute $u = \cos x$ , so $\dfrac{du}{dx} = -\sin x$ , i.e. $\sin x \,dx = -du$ .

Change the limits:

\begin{aligned} x = 0 &: \quad u = \cos 0 = 1 \\ x = \tfrac{1}{4}\pi &: \quad u = \cos\tfrac{1}{4}\pi = \tfrac{\sqrt{2}}{2} \end{aligned}

Substitute into the integral:

\begin{aligned} A &= 10\int_{1}^{\frac{\sqrt{2}}{2}} u^3 \cdot (-du) \\ &= -10\int_{1}^{\frac{\sqrt{2}}{2}} u^3 \,du \\ &= 10\int_{\frac{\sqrt{2}}{2}}^{1} u^3 \,du \\ &= 10\left[\frac{u^4}{4}\right]_{\frac{\sqrt{2}}{2}}^{1} \\ &= \frac{10}{4}\left[1 - \left(\frac{\sqrt{2}}{2}\right)^{4}\right] \end{aligned}

Since $\left(\dfrac{\sqrt{2}}{2}\right)^{4} = \dfrac{4}{16} = \dfrac{1}{4}$ :

\begin{aligned} A &= \frac{5}{2}\left(1 - \frac{1}{4}\right) \\ &= \frac{5}{2} \cdot \frac{3}{4} \\ &= \frac{15}{8} \end{aligned}

Answer

A = \frac{15}{8}

Final answer

15/8

Detailed explanation

Walkthrough

We need the area enclosed by the curve $y = 5\sin 2x \cos^2 x$ , the $x$ -axis, the line $x = 0$ , and the line $x = \frac{1}{4}\pi$ . Since the curve lies above the $x$ -axis on $[0, \frac{1}{4}\pi]$ , this area is just the definite integral of the curve between these limits.

Step 1 — Use the double angle identity. $\sin 2x = 2\sin x \cos x$ , so the integrand becomes $5 \cdot 2\sin x \cos x \cdot \cos^2 x = 10\sin x \cos^3 x$ . This is necessary because the substitution $u = \cos x$ requires the integrand to involve $\cos x$ (and a single $\sin x$ to match the differential $du$ ).

Step 2 — Apply the substitution. Let $u = \cos x$ , then $du = -\sin x \,dx$ , so $\sin x \,dx = -du$ . The integrand $\sin x \cos^3 x \,dx$ becomes $u^3 \cdot (-du) = -u^3\,du$ , giving the constant factor $10 \cdot (-1) = -10$ in front.

Step 3 — Convert the limits. When $x = 0$ , $u = \cos 0 = 1$ . When $x = \frac{1}{4}\pi$ , $u = \cos \frac{1}{4}\pi = \frac{\sqrt{2}}{2}$ . Note that $\frac{\sqrt{2}}{2} < 1$ , so the integral goes from a larger to a smaller value of $u$ . The minus sign in front of the integral flips the limits back, giving the integral from $\frac{\sqrt{2}}{2}$ to $1$ — alternatively, one may keep the limits as written and carefully apply the minus sign when evaluating.

Step 4 — Evaluate. The antiderivative of $u^3$ is $\frac{u^4}{4}$ , so the value is $\frac{10}{4}\bigl[1 - (\frac{\sqrt{2}}{2})^4\bigr] = \frac{5}{2}\bigl[1 - \frac{1}{4}\bigr] = \frac{5}{2} \cdot \frac{3}{4} = \frac{15}{8}$ .

Key Takeaways

The double angle identity $\sin 2x = 2\sin x \cos x$ is often the key step that prepares a trig integrand for a $u = \cos x$ (or $u = \sin x$ ) substitution.
Always convert the limits of integration when using a definite integral with a substitution — this avoids the need to convert the antiderivative back to $x$ .
A minus sign appears whenever the substitution has a negative derivative (here $du = -\sin x\,dx$ ); this sign flips the order of the limits.

Common Mistakes

Forgetting the factor of 2 from the double angle identity, leaving the integrand as $5\sin 2x \cos^2 x$ and being unable to make progress.
Failing to convert the limits and instead substituting $u = \cos x$ into the antiderivative while keeping the $x$ -limits — this mixes up the variable of integration.
Losing the minus sign from $du = -\sin x\,dx$ and getting a negative final answer.
Miscalculating $\left(\frac{\sqrt{2}}{2}\right)^4$ as $\frac{\sqrt{2}}{2}$ or $\frac{1}{2}$ instead of $\frac{1}{4}$ .

Things to Be Careful About

The integrand $5\sin 2x \cos^2 x$ is non-negative on $[0, \frac{1}{4}\pi]$ , so the geometric area equals the definite integral with no absolute value needed.
The mark scheme accepts either keeping the original limits with the minus sign explicit, or swapping the limits so the minus sign is absorbed. Both are correct.
The final answer is $\frac{15}{8}$ exactly, which is approximately $1.875$ .

Techniques used

use the double angle identity to express the integrand in terms of sin x and cos xapply the substitution u = cos xconvert the limits of integration to u-valuesevaluate the resulting polynomial definite integral

Mathematics 9709/32 — May/June 2025

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