(a) Express $4\cos\theta\sin(\theta + 30°)$ in the form $R\cos(2\theta - \alpha) + k$, where $R > 0$, $0° < \alpha < 90°$ and $k$ is a constant. (b) Hence solve the equation $$ 12\cos 2\phi\sin(2\phi + 30°) = 5 $$ for $0° < \phi < 90°$.

(a) 2 cos(2θ - 60°) + 1, with R = 2, α = 60°, k = 1 · (b) φ = 32.6° or φ = 87.4°

9709/22 May/June 2025 — Mathematics Solved Past Paper

Q13MMedium-EasyIntegration

Show that $\int_2^{11} \frac{8}{4x+1}\,dx = \ln a$ , where $a$ is an integer to be found.

DifficultyMedium-Easy

Worked solution

Approach

Recognise that $\frac{8}{4x+1}$ is a constant multiple of $\frac{1}{4x+1}$ , so integrate to a natural logarithm of the linear denominator. Then substitute the limits and use the logarithm property $\ln A - \ln B = \ln\left(\frac{A}{B}\right)$ to express the answer as a single logarithm.

Working

Integrate:

\int_2^{11} \frac{8}{4x+1}\,dx = 2\ln(4x+1) \Big|_2^{11}

Apply the limits:

= 2\ln(4(11)+1) - 2\ln(4(2)+1)

= 2\ln 45 - 2\ln 9

Use the logarithm law $\ln A - \ln B = \ln\left(\frac{A}{B}\right)$ :

= 2\ln\left(\frac{45}{9}\right)

= 2\ln 5

Rewrite in the form $\ln a$ :

= \ln\left(5^2\right) = \ln 25

Answer

a = 25

Final answer

a = 25

Detailed explanation

Walkthrough

We need to evaluate the definite integral $\int_2^{11} \frac{8}{4x+1}\,dx$ . The integrand is a constant multiple of the reciprocal of a linear function, so the antiderivative is $2\ln(4x+1)$ . The factor $2$ comes from dividing $8$ by the derivative of $4x+1$ , which is $4$ : $\frac{8}{4} = 2$ .

Next, substitute the upper limit $x=11$ and the lower limit $x=2$ into $2\ln(4x+1)$ and subtract. This gives $2\ln 45 - 2\ln 9$ . Since both terms have the same coefficient $2$ , factor it out and use the logarithm law $\ln A - \ln B = \ln\left(\frac{A}{B}\right)$ to combine them: $2\ln\left(\frac{45}{9}\right) = 2\ln 5$ .

Finally, write $2\ln 5$ as $\ln(5^2) = \ln 25$ so the answer has the required form $\ln a$ . Therefore $a = 25$ .

Key Takeaways

This question tests the standard integral $\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + c$ and the laws of logarithms. It also shows how to combine logarithms after applying limits. The key skill is recognising the constant factor needed to integrate a reciprocal linear function.

Common Mistakes

Forgetting the factor $\frac{1}{a}$ when integrating $\frac{1}{ax+b}$ ; here that factor is $\frac{1}{4}$ , so $8$ becomes $2$ .
Omitting the logarithm law and leaving the answer as $2\ln 45 - 2\ln 9$ instead of simplifying to $\ln 25$ .
Incorrectly applying the limits by subtracting in the wrong order, which would give the wrong sign.
Writing $\ln 45 - \ln 9 = \ln 36$ instead of $\ln 5$ .

Things to Be Careful About

The mark scheme requires the antiderivative $2\ln(4x+1)$ to be shown explicitly.
At least one relevant logarithm property must be used when applying the limits.
The final answer must be a single logarithm $\ln a$ with integer $a$ ; here $a = 25$ .
Since $4x+1$ is positive on the interval $[2, 11]$ , no absolute value signs are needed.

Techniques used

integrate reciprocal linear expressionapply limits of integrationsimplify using logarithm laws

Approach

Sketch the V-shaped modulus graph and the straight line on the same axes, marking the vertex and the intersection point.

Working

For $y = |2x - 9|$ , the vertex occurs when $2x - 9 = 0$ , so $x = \frac{9}{2}$ and $y = 0$ .

For $x < \frac{9}{2}$ , $|2x - 9| = -2x + 9$ ; for $x \ge \frac{9}{2}$ , $|2x - 9| = 2x - 9$ .

The line $y = 4x - 5$ has gradient $4$ and $y$ -intercept $-5$ .

The intersection is on the left branch, so solve

-2x + 9 = 4x - 5

14 = 6x \quad \Rightarrow \quad x = \frac{7}{3}

Then

y = 4\left(\frac{7}{3}\right) - 5 = \frac{28}{3} - \frac{15}{3} = \frac{13}{3}

So the graphs meet at $\left(\frac{7}{3}, \frac{13}{3}\right)$ .

Answer

Draw the V-shaped graph with vertex at $\left(\frac{9}{2}, 0\right)$ , and the line $y = 4x - 5$ , with intersection at $\left(\frac{7}{3}, \frac{13}{3}\right)$ .

Final answer

V-shaped graph with vertex at (9/2, 0) and line y = 4x - 5, intersecting at (7/3, 13/3)

Detailed explanation

Walkthrough

We need to sketch two graphs on the same diagram. The graph of $y = |2x - 9|$ is V-shaped because the modulus changes the sign of $2x - 9$ when it is negative. The vertex is found by setting $2x - 9 = 0$ , giving $x = \frac{9}{2}$ and $y = 0$ . For $x < \frac{9}{2}$ , the graph is the line $y = -2x + 9$ ; for $x \ge \frac{9}{2}$ , it is the line $y = 2x - 9$ . The graph of $y = 4x - 5$ is a straight line with gradient $4$ and $y$ -intercept $-5$ . To mark the intersection accurately, solve the equation of the left branch with the line; the right branch does not meet the line in its region. The intersection is at $\left(\frac{7}{3}, \frac{13}{3}\right)$ .

Key Takeaways

The modulus graph $y = |ax + b|$ is V-shaped with vertex where $ax + b = 0$ .
The left branch has gradient $-a$ and the right branch has gradient $a$ .
A straight line is determined by its gradient and intercept.

Common Mistakes

Drawing the modulus graph as a single straight line instead of a V-shape.
Putting the vertex at the wrong point, such as $x = 0$ .
Using the wrong gradient for one of the branches.
Forgetting to show the intersection point on the sketch.

Things to Be Careful About

The vertex is at $x = \frac{9}{2}$ , not at $x = 0$ .
The left branch has gradient $-2$ and the right branch has gradient $2$ .
The line $y = 4x - 5$ is steeper than either branch of the modulus graph.
Only the left-branch intersection is valid; the right-branch equation gives $x = -2$ , which is not in the region $x \ge \frac{9}{2}$ .

Techniques used

sketch a modulus graphsketch a linear graphfind the intersection of two graphs

(b)

Solve the inequality $|2x - 9| < 4x - 5$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Split the modulus into its two cases according to the sign of $2x - 9$ , solve the resulting linear inequality in each case, then combine the valid intervals.

Working

The critical value is $x = \frac{9}{2}$ .

Case 1: $x < \frac{9}{2}$ . Then $|2x - 9| = -2x + 9$ , so

-2x + 9 < 4x - 5

14 < 6x

x > \frac{7}{3}

With $x < \frac{9}{2}$ , this gives $\frac{7}{3} < x < \frac{9}{2}$ .

Case 2: $x \ge \frac{9}{2}$ . Then $|2x - 9| = 2x - 9$ , so

2x - 9 < 4x - 5

-4 < 2x \quad \Rightarrow \quad x > -2

With $x \ge \frac{9}{2}$ , this gives $x \ge \frac{9}{2}$ .

Combining both cases:

\frac{7}{3} < x < \frac{9}{2} \quad \text{or} \quad x \ge \frac{9}{2}

which is simply

x > \frac{7}{3}

Answer

x > \frac{7}{3}

Final answer

x > 7/3

Detailed explanation

Walkthrough

The inequality $|2x - 9| < 4x - 5$ contains a modulus, so the expression inside the modulus can be positive or negative. The sign changes at $x = \frac{9}{2}$ . Split into two cases.

For $x < \frac{9}{2}$ , the modulus is $-2x + 9$ . Substitute this into the inequality and solve:

-2x + 9 < 4x - 5

Add $2x$ and $5$ to both sides:

14 < 6x

so $x > \frac{7}{3}$ . Since this case already requires $x < \frac{9}{2}$ , the solutions in this case are $\frac{7}{3} < x < \frac{9}{2}$ .

For $x \ge \frac{9}{2}$ , the modulus is $2x - 9$ . Then

2x - 9 < 4x - 5

Subtract $2x$ from both sides:

-9 < 2x - 5

Add $5$ : $-4 < 2x$ , so $x > -2$ . Since this case requires $x \ge \frac{9}{2}$ , every value in this case satisfies the inequality. Therefore the second case contributes $x \ge \frac{9}{2}$ .

Taking the union of the two intervals gives $x > \frac{7}{3}$ . This matches the graph from part (a): the line is above the modulus graph for all $x > \frac{7}{3}$ .

Key Takeaways

To solve a modulus inequality, split at the point where the expression inside the modulus is zero.
Solve each linear inequality separately and intersect with the case condition.
The final answer is the union of the valid intervals.

Common Mistakes

Solving only one case and ignoring the other.
Forgetting to intersect the solution of each case with its case condition.
Writing the answer as $x = \frac{7}{3}$ instead of an inequality.
Reversing the inequality sign when no multiplication or division by a negative number has occurred.

Things to Be Careful About

The critical value is $x = \frac{9}{2}$ , not $x = \frac{7}{3}$ ; $\frac{7}{3}$ is the boundary of the solution.
The solution $x > \frac{7}{3}$ includes both the interval $\frac{7}{3} < x < \frac{9}{2}$ and the interval $x \ge \frac{9}{2}$ .
The mark scheme also accepts interval notation such as $\left(\frac{7}{3}, \infty\right)$ .

Techniques used

split the modulus into casessolve a linear inequalitycombine solution intervals

Q35MMediumDifferentiation

Find the coordinates of the stationary points of the curve with equation $y = \frac{8x}{2x+3} - 6x + 5$ .

DifficultyMedium

Worked solution

Approach

Differentiate the rational term using the quotient rule, then differentiate the linear terms, set the derivative equal to zero, solve the resulting quadratic, and substitute back to find the $y$ -coordinates.

Working

Let

u = 8x, \quad v = 2x+3

Then

\frac{du}{dx} = 8, \quad \frac{dv}{dx} = 2

Using the quotient rule,

\frac{d}{dx}\left(\frac{8x}{2x+3}\right) = \frac{(2x+3)(8) - 8x(2)}{(2x+3)^2} = \frac{24}{(2x+3)^2}

Therefore

\frac{dy}{dx} = \frac{24}{(2x+3)^2} - 6

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

\frac{24}{(2x+3)^2} - 6 = 0

Multiply through by $(2x+3)^2$ :

24 - 6(2x+3)^2 = 0

Divide by 6 and expand:

4 - (2x+3)^2 = 0

4 - (4x^2 + 12x + 9) = 0

4x^2 + 12x + 5 = 0

Factorise:

(2x+1)(2x+5) = 0

So

x = -\frac12 \quad \text{or} \quad x = -\frac52

For $x = -\frac12$ :

y = \frac{8(-\frac12)}{2(-\frac12)+3} - 6(-\frac12) + 5 = \frac{-4}{2} + 3 + 5 = 6

For $x = -\frac52$ :

y = \frac{8(-\frac52)}{2(-\frac52)+3} - 6(-\frac52) + 5 = \frac{-20}{-2} + 15 + 5 = 30

Answer

The stationary points are

\left(-\frac52, 30\right) \quad \text{and} \quad \left(-\frac12, 6\right)

Final answer

(-5/2, 30) and (-1/2, 6)

Detailed explanation

Walkthrough

We need the points where the curve has zero gradient. A stationary point occurs where $\frac{dy}{dx}=0$ , so the first task is to differentiate $y = \frac{8x}{2x+3} - 6x + 5$ .

The term $\frac{8x}{2x+3}$ is a quotient, so we use the quotient rule:

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

Here $u = 8x$ and $v = 2x+3$ , so $u' = 8$ and $v' = 2$ . Substituting gives the numerator $8(2x+3) - 8x(2) = 24$ , so the derivative of the quotient is $\frac{24}{(2x+3)^2}$ . The derivative of $-6x+5$ is just $-6$ , because the derivative of a constant is zero and the derivative of $-6x$ is $-6$ . Hence

\frac{dy}{dx} = \frac{24}{(2x+3)^2} - 6

Next, set this equal to zero. Multiplying through by $(2x+3)^2$ removes the denominator and gives a quadratic equation. Expanding and simplifying leads to $4x^2 + 12x + 5 = 0$ , which factorises as $(2x+1)(2x+5)=0$ . Therefore the two $x$ -coordinates are $x = -\frac12$ and $x = -\frac52$ .

Finally, substitute each $x$ -value back into the original equation to find the corresponding $y$ -coordinate. This gives the two stationary points $(-\frac12, 6)$ and $(-\frac52, 30)$ .

Key Takeaways

Stationary points are found by solving $\frac{dy}{dx}=0$ .
The quotient rule is the correct tool when differentiating a fraction such as $\frac{8x}{2x+3}$ .
After differentiating, solving the resulting quadratic may require expansion and factorisation.
Always substitute the $x$ -values back into the original curve to get the $y$ -coordinates.

Common Mistakes

Forgetting to differentiate the $-6x+5$ part of the expression.
Making a sign error in the quotient rule: the formula is $\frac{u'v - uv'}{v^2}$ , not $\frac{uv' - u'v}{v^2}$ .
Losing one root when solving $(2x+3)^2 = 4$ ; remember that $2x+3$ can be $2$ or $-2$ .
Stopping after finding the $x$ -values and not finding the corresponding $y$ -coordinates.
Not showing the quotient rule working; the mark scheme requires a visible method.

Things to Be Careful About

The denominator $(2x+3)^2$ is never negative and is zero only at $x = -\frac32$ , which is a vertical asymptote and not a stationary point. Multiplying by $(2x+3)^2$ is valid for all other $x$ .
When substituting $x = -\frac52$ , note that $2x+3 = -2$ , so the fraction $\frac{8x}{2x+3}$ becomes $\frac{-20}{-2}=10$ . Watch the signs carefully.
The mark scheme awards marks for the method: an unsupported answer is not sufficient. Show the quotient rule, the derivative, the equation set to zero, and the substitution back.
State the final answers as coordinate pairs.

Techniques used

apply the quotient ruledifferentiate a linear expressionset the derivative to zerosolve the resulting quadratic equationsubstitute back to find y-coordinates

Approach

Since P lies on both curves, the y-coordinates are equal. Equate the two expressions and use the inverse relationship between the exponential and natural logarithm to obtain the required form.

Working

At point P:

4e^{-2x} = 1 + 0.5\sin 3x

Divide both sides by 4:

e^{-2x} = 0.25 + 0.125\sin 3x

Take the natural logarithm of both sides:

-2x = \ln(0.25 + 0.125\sin 3x)

Divide by $-2$ :

x = -0.5\ln(0.25 + 0.125\sin 3x)

Answer

x = -0.5\ln(0.25 + 0.125\sin 3x)

Final answer

x = -0.5ln(0.25 + 0.125 sin 3x)

Detailed explanation

Walkthrough

We need to show that the x-coordinate of P satisfies the given equation. The key fact is that P lies on both curves, so its coordinates satisfy both equations. This means the y-coordinates are equal: $4e^{-2x} = 1 + 0.5\sin 3x$ .

To get from this equation to the required form, we want to isolate x. The challenge is that the exponential $e^{-2x}$ contains x, and $\sin 3x$ also contains x. So we have a transcendental equation.

The trick is to isolate the exponential on one side. We can do this by dividing both sides by 4:
$e^{-2x} = 0.25 + 0.125\sin 3x$

Now, the natural logarithm is the inverse of the exponential function. Taking $\ln$ of both sides eliminates the exponential:
$\ln(e^{-2x}) = \ln(0.25 + 0.125\sin 3x)$
$-2x = \ln(0.25 + 0.125\sin 3x)$

Finally, divide by $-2$ to get $x$ alone:
$x = -0.5\ln(0.25 + 0.125\sin 3x)$

This matches the required form.

Key Takeaways

When solving equations involving both exponential and trigonometric terms, isolate the exponential first.
The natural logarithm is the inverse of the exponential function: $\ln(e^y) = y$ .
Watch out for the negative sign when dividing by $-2$ .

Common Mistakes

Forgetting to divide by 4, leading to a coefficient other than 0.25.
Not taking the logarithm of both sides correctly, or missing the negative sign.
Errors in the coefficients (0.25 vs 0.5, 0.125 vs 0.25) when dividing.

Things to Be Careful About

The required form must match exactly. Pay attention to coefficients (0.25, 0.125, -0.5).
Make sure the logarithm is the natural log ( $\ln$ ), not log base 10.

Techniques used

set two curves equal at intersection pointtake natural logarithm of both sidesdivide to isolate the unknown

(b)

Use an iterative formula, based on the equation in part (a), to find the $x$ -coordinate of $P$ correct to 4 significant figures. Use an initial value of 0.5 and give the result of each iteration to 6 significant figures.

3M Similar

DifficultyMedium

Worked solution

Approach

Rearrange the equation from part (a) into the form $x_{n+1} = F(x_n)$ and iterate from $x_0 = 0.5$ until the value stabilises to 4 significant figures.

Working

The iterative formula is:

x_{n+1} = -0.5\ln(0.25 + 0.125\sin 3x_n)

Starting with $x_0 = 0.5$ :

\begin{aligned} x_1 &= -0.5\ln(0.25 + 0.125\sin(3 \times 0.5)) \\ &= -0.5\ln(0.25 + 0.125\sin(1.5)) \\ &= -0.5\ln(0.25 + 0.125 \times 0.997495) \\ &= -0.5\ln(0.374687) \\ &= -0.5 \times (-0.981665) \\ &= 0.490832 \end{aligned}

\begin{aligned} x_2 &= -0.5\ln(0.25 + 0.125\sin(3 \times 0.490832)) \\ &= -0.5\ln(0.25 + 0.125\sin(1.472497)) \\ &= -0.5\ln(0.25 + 0.125 \times 0.995173) \\ &= -0.5\ln(0.374397) \\ &= -0.5 \times (-0.982440) \\ &= 0.491220 \end{aligned}

\begin{aligned} x_3 &= -0.5\ln(0.25 + 0.125\sin(3 \times 0.491220)) \\ &= -0.5\ln(0.25 + 0.125\sin(1.473660)) \\ &= -0.5\ln(0.25 + 0.125 \times 0.995286) \\ &= -0.5\ln(0.374411) \\ &= -0.5 \times (-0.982402) \\ &= 0.491201 \end{aligned}

Since $x_2 = 0.491220$ and $x_3 = 0.491201$ both lie in the interval $[0.49115, 0.49125]$ , the value has stabilised to 4 significant figures.

Answer

x = 0.4912

Final answer

x = 0.4912

Detailed explanation

Walkthrough

The equation from part (a) gives us an iterative formula directly: $x_{n+1} = -0.5\ln(0.25 + 0.125\sin 3x_n)$ . This formula is already in the standard $x_{n+1} = F(x_n)$ form, where $F(x) = -0.5\ln(0.25 + 0.125\sin 3x)$ .

Starting from $x_0 = 0.5$ , we apply the formula repeatedly. At each step, we:

Compute $3x_n$ (the input to sine).
Take the sine of that value.
Multiply by 0.125 and add 0.25.
Take the natural log.
Multiply by $-0.5$ .

The iterations:

$x_1$ : $3 \times 0.5 = 1.5$ , $\sin(1.5) = 0.997495$ , $0.25 + 0.125 \times 0.997495 = 0.374687$ , $\ln(0.374687) = -0.981665$ , so $x_1 = 0.490832$ .
$x_2$ : $3 \times 0.490832 = 1.472497$ , $\sin(1.472497) = 0.995173$ , $0.25 + 0.125 \times 0.995173 = 0.374397$ , $\ln(0.374397) = -0.982440$ , so $x_2 = 0.491220$ .
$x_3$ : $3 \times 0.491220 = 1.473660$ , $\sin(1.473660) = 0.995286$ , $0.25 + 0.125 \times 0.995286 = 0.374411$ , $\ln(0.374411) = -0.982402$ , so $x_3 = 0.491201$ .

The values are converging rapidly. After 3 iterations, the value has stabilised to 4 significant figures. The values $x_2$ and $x_3$ both lie in the required interval $[0.49115, 0.49125]$ , confirming that 0.4912 is the correct answer.

Key Takeaways

An equation of the form $x = F(x)$ always gives an iterative formula $x_{n+1} = F(x_n)$ directly.
Iterate until consecutive values agree to the required accuracy.
Showing that the value lies in a specific interval (e.g., $[0.49115, 0.49125]$ ) provides a check on the answer.

Common Mistakes

Using too few iterations, failing to show convergence.
Rounding intermediate values too aggressively, which can slow convergence or change the final answer.
Using the wrong angle mode (degrees vs radians) - the question uses radians.
Forgetting to use natural log (not log base 10).

Things to Be Careful About

The argument of $\sin$ is $3x$ in radians, not degrees.
Show all intermediate values to enough precision (6 sig figs here) to justify the final answer.
Final answer must be to 4 significant figures exactly: 0.4912 (not 0.491 or 0.49120).

Techniques used

apply iterative formula in form x_{n+1} = F(x_n)iterate to convergenceverify value lies in required tolerance interval

(c)

Hence find the area of the shaded region. Give your answer correct to 2 significant figures.

4M Similar

DifficultyMedium-Hard

Worked solution

Approach

The shaded region lies between the y-axis and the two curves, with $y = 4e^{-2x}$ on top and $y = 1 + 0.5\sin 3x$ below, from $x = 0$ to $x = x_P = 0.4912$ . Compute the area as the definite integral of (top − bottom) over this interval.

Working

The area is:

A = \int_0^{0.4912} \left[4e^{-2x} - (1 + 0.5\sin 3x)\right] dx

Integrating each term separately:

\int 4e^{-2x}\, dx = -2e^{-2x}

\int 1\, dx = x

\int 0.5\sin 3x\, dx = -\frac{1}{6}\cos 3x

So the antiderivative of the integrand is:

-2e^{-2x} - x + \frac{1}{6}\cos 3x

Therefore:

A = \left[-2e^{-2x} - x + \frac{1}{6}\cos 3x\right]_0^{0.4912}

At the upper limit $x = 0.4912$ :

-2e^{-2(0.4912)} = -2e^{-0.9824} = -2(0.37441) = -0.74882

-x = -0.4912

\frac{1}{6}\cos(3 \times 0.4912) = \frac{1}{6}\cos(1.4736) = \frac{1}{6}(0.09705) = 0.01617

Sum at upper limit: $-0.74882 - 0.4912 + 0.01617 = -1.22385$

At the lower limit $x = 0$ :

-2e^0 = -2

-0 = 0

\frac{1}{6}\cos(0) = \frac{1}{6} \approx 0.16667

Sum at lower limit: $-2 + 0 + 0.16667 = -1.83333$

Subtracting:

A = -1.22385 - (-1.83333) = 0.60948

Answer

A \approx 0.61

Final answer

A ≈ 0.61

Detailed explanation

Walkthrough

The shaded region is bounded on the left by the y-axis ( $x = 0$ ) and on the right by the intersection point P. Between these vertical lines, the upper boundary is $y = 4e^{-2x}$ (the decreasing exponential, which starts at 4) and the lower boundary is $y = 1 + 0.5\sin 3x$ (the oscillating curve, which starts at 1). At $x = 0$ , the exponential is at 4 and the sine curve is at 1, so the exponential is above the sine curve - which matches the diagram.

The area of such a region is the integral of (top − bottom) over the interval:
$A = \int_0^{x_P} [4e^{-2x} - (1 + 0.5\sin 3x)]\, dx$

We integrate each term:

$\int 4e^{-2x}\, dx$ : use the rule $\int e^{ax}\, dx = \frac{1}{a}e^{ax}$ , so $4e^{-2x}$ integrates to $4 \times \frac{1}{-2} e^{-2x} = -2e^{-2x}$ .
$\int 1\, dx = x$ (a constant integrates to itself times x).
$\int 0.5\sin 3x\, dx$ : use $\int \sin(ax)\, dx = -\frac{1}{a}\cos(ax)$ , so $0.5\sin 3x$ integrates to $0.5 \times (-\frac{1}{3})\cos 3x = -\frac{1}{6}\cos 3x$ .

Combining (the integrand contains $-0.5\sin 3x$ , so its antiderivative is $+\frac{1}{6}\cos 3x$ ):
$A = [-2e^{-2x} - x + \frac{1}{6}\cos 3x]_0^{0.4912}$

At $x = 0.4912$ :

$-2e^{-0.9824} = -2 \times 0.37441 = -0.74882$
$-x = -0.4912$
$\frac{1}{6}\cos(1.4736) = \frac{1}{6} \times 0.09705 = 0.01617$
Total: $-1.22385$

At $x = 0$ :

$-2e^0 = -2$
$0$
$\frac{1}{6}\cos 0 = \frac{1}{6} = 0.16667$
Total: $-1.83333$

Subtracting (upper minus lower): $A = -1.22385 - (-1.83333) = 0.60948$ .

To 2 significant figures, $A \approx 0.61$ .

Key Takeaways

The area between two curves $y_1 \geq y_2$ from $x = a$ to $x = b$ is $\int_a^b (y_1 - y_2)\, dx$ .
Standard integrals to remember: $\int e^{ax}\, dx = \frac{1}{a}e^{ax}$ , $\int \sin(ax)\, dx = -\frac{1}{a}\cos(ax)$ , $\int \cos(ax)\, dx = \frac{1}{a}\sin(ax)$ .
Always subtract in the right order: (upper limit value) − (lower limit value).

Common Mistakes

Integrating $0.5\sin 3x$ as $0.5\cos 3x$ (forgetting the negative and the $1/3$ factor).
Confusing which curve is on top - the curve with the larger y-value at each x is the upper boundary. Check at $x = 0$ : $4 > 1$ , so the exponential is on top.
Using the wrong limits (e.g., using 1 instead of $x_P = 0.4912$ from part (b)).
Sign errors when subtracting the lower-limit value from the upper-limit value.
Forgetting to use radians in the trig functions.

Things to Be Careful About

The area is positive, so the final answer should be positive after subtraction.
Use the value from part (b) for the upper limit, since this question is sequential (the answer from (b) is $x_P = 0.4912$ ).
The $\cos 3x$ term contributes a small but nonzero value to the area - don't forget it.
The final answer must be to 2 significant figures, so $0.60948$ rounds to $0.61$ .

Techniques used

set up area integral as difference between curvesintegrate exponential, constant, and sine functionsevaluate at limits and subtract

Q5Medium-EasyAlgebraTrigonometry

The polynomial $p(x)$ is defined by

p(x) = ax^4 + bx^3 + 13x^2 - 35x + 15,

where $a$ and $b$ are constants. It is given that $(2x - 1)$ and $(x - 3)$ are factors of $p(x)$ .

(a)

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

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $(2x - 1)$ and $(x - 3)$ are factors of $p(x)$ , the factor theorem gives $p\left(\frac{1}{2}\right) = 0$ and $p(3) = 0$ . Substitute these values into the polynomial to obtain two equations in $a$ and $b$ , then solve the simultaneous equations.

Working

Substitute $x = \frac{1}{2}$ :

p\left(\frac{1}{2}\right) = a\left(\frac{1}{2}\right)^4 + b\left(\frac{1}{2}\right)^3 + 13\left(\frac{1}{2}\right)^2 - 35\left(\frac{1}{2}\right) + 15 = 0

\frac{a}{16} + \frac{b}{8} + \frac{13}{4} - \frac{35}{2} + 15 = 0

\frac{a}{16} + \frac{b}{8} + \frac{3}{4} = 0

\frac{a}{16} + \frac{b}{8} = -\frac{3}{4}

Multiply through by 16:

a + 2b = -12 \tag{1}

Substitute $x = 3$ :

p(3) = 81a + 27b + 117 - 105 + 15 = 0

81a + 27b + 27 = 0

81a + 27b = -27

Divide through by 27:

3a + b = -1 \tag{2}

Solve (1) and (2). From (2):

b = -1 - 3a

Substitute into (1):

a + 2(-1 - 3a) = -12

a - 2 - 6a = -12

-5a = -10

a = 2

Then:

b = -1 - 3(2) = -7

Answer

a = 2, \quad b = -7

Final answer

a = 2, b = -7

Detailed explanation

Walkthrough

We are told that $(2x - 1)$ and $(x - 3)$ are factors of $p(x)$ . By the factor theorem, if a linear factor $(cx - d)$ divides a polynomial $p(x)$ , then $p\left(\frac{d}{c}\right) = 0$ . Here, $(2x - 1)$ being a factor means $x = \frac{1}{2}$ is a root, and $(x - 3)$ being a factor means $x = 3$ is a root.

We substitute $x = \frac{1}{2}$ into $p(x)$ and set the result to zero. This converts the statement "this is a factor" into a concrete equation involving the unknown constants $a$ and $b$ . When substituting, we must compute the fractional powers carefully: $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$ , $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$ , and $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ . The constant terms combine: $\frac{13}{4} - \frac{35}{2} + 15 = \frac{3}{4}$ . So we get $\frac{a}{16} + \frac{b}{8} + \frac{3}{4} = 0$ , i.e., $\frac{a}{16} + \frac{b}{8} = -\frac{3}{4}$ . Multiplying through by 16 clears the denominators and gives $a + 2b = -12$ .

Similarly, substituting $x = 3$ gives $81a + 27b + 117 - 105 + 15 = 0$ , which simplifies to $81a + 27b = -27$ , and dividing by 27 gives $3a + b = -1$ .

We now have two linear equations in $a$ and $b$ :

$a + 2b = -12$
$3a + b = -1$

We solve by substitution: from equation (2), $b = -1 - 3a$ . Substituting into equation (1) gives $a + 2(-1 - 3a) = -12$ , which simplifies to $-5a = -10$ , so $a = 2$ . Then $b = -1 - 3(2) = -7$ .

Key Takeaways

The factor theorem is the bridge between "has a factor" and "evaluates to zero at a specific point".
For a factor like $(2x - 1)$ , the root is found by solving $2x - 1 = 0$ , giving $x = \frac{1}{2}$ .
Substituting roots into a polynomial with unknown coefficients produces equations that can be solved simultaneously.

Common Mistakes

Using the wrong root: for $(2x - 1)$ , some students substitute $x = -1$ or $x = 1$ instead of $x = \frac{1}{2}$ . Always solve the factor equal to zero.
Arithmetic slips with fractional powers, e.g., writing $\left(\frac{1}{2}\right)^4 = \frac{1}{8}$ .
Sign errors when simplifying $-35 \times \frac{1}{2} = -\frac{35}{2}$ .
The mark scheme gives "SC B1 for a correct equation, if M0 otherwise" — so even if the method is not fully shown, a correct equation earns a special-case mark. Show all substitutions clearly.

Things to Be Careful About

When clearing fractions, multiply every term by the common denominator.
After finding $a$ and $b$ , verify by checking $p\left(\frac{1}{2}\right) = 0$ and $p(3) = 0$ .
Keep the equations labelled; it helps when solving the simultaneous system.

Techniques used

apply the factor theorem to substitute known rootsform simultaneous equations in the unknown coefficientssolve the simultaneous linear equations

(b)

Hence factorise $p(x)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

With $a = 2$ and $b = -7$ , write out $p(x)$ . Since $(2x - 1)$ and $(x - 3)$ are factors, their product $(2x - 1)(x - 3) = 2x^2 - 7x + 3$ is a factor of $p(x)$ . Divide $p(x)$ by this quadratic to find the remaining factor.

Working

With $a = 2$ and $b = -7$ :

p(x) = 2x^4 - 7x^3 + 13x^2 - 35x + 15

The product of the known factors:

(2x - 1)(x - 3) = 2x^2 - 7x + 3

Divide $p(x)$ by $2x^2 - 7x + 3$ . First term of the quotient is $x^2$ , since $2x^2 \times x^2 = 2x^4$ :

x^2(2x^2 - 7x + 3) = 2x^4 - 7x^3 + 3x^2

Subtracting from $p(x)$ :

(2x^4 - 7x^3 + 13x^2 - 35x + 15) - (2x^4 - 7x^3 + 3x^2) = 10x^2 - 35x + 15

Next term of the quotient is $5$ , since $2x^2 \times 5 = 10x^2$ :

5(2x^2 - 7x + 3) = 10x^2 - 35x + 15

Subtracting gives remainder $0$ . The quotient is $x^2 + 5$ .

Therefore:

p(x) = (2x - 1)(x - 3)(x^2 + 5)

Answer

p(x) = (2x - 1)(x - 3)(x^2 + 5)

Final answer

p(x) = (2x - 1)(x - 3)(x^2 + 5)

Detailed explanation

Walkthrough

With $a = 2$ and $b = -7$ , the polynomial becomes $p(x) = 2x^4 - 7x^3 + 13x^2 - 35x + 15$ .

Since $(2x - 1)$ and $(x - 3)$ are both factors, their product $(2x - 1)(x - 3) = 2x^2 - 7x + 3$ is also a factor of $p(x)$ . So we divide $p(x)$ by $2x^2 - 7x + 3$ to find the remaining quadratic factor.

Polynomial long division: the leading term $2x^4$ divided by the leading term of the divisor $2x^2$ gives $x^2$ . Multiply the divisor by $x^2$ to get $2x^4 - 7x^3 + 3x^2$ , and subtract from $p(x)$ . This eliminates the $x^4$ and $x^3$ terms, leaving $10x^2 - 35x + 15$ . Now divide $10x^2$ by $2x^2$ to get $5$ . Multiply the divisor by $5$ to get $10x^2 - 35x + 15$ , which subtracts exactly to zero. The quotient is $x^2 + 5$ .

Therefore $p(x) = (2x - 1)(x - 3)(x^2 + 5)$ .

Key Takeaways

When two linear factors are known, their product is a quadratic factor of the polynomial.
Polynomial long division by a quadratic factor reveals the remaining factor.
The fully factorised form is the product of all known and found factors.

Common Mistakes

Sign errors during subtraction in long division — always subtract the entire product, not just some terms.
Forgetting that the quotient is the remaining factor and must be included in the final factorisation.
The mark scheme notes that synthetic division may give $2(x^2 + 5)$ ; this is condoned as an intermediate, but the final answer should be $(2x - 1)(x - 3)(x^2 + 5)$ .

Things to Be Careful About

Check the factorisation by expanding: $(2x - 1)(x - 3)(x^2 + 5)$ should equal $2x^4 - 7x^3 + 13x^2 - 35x + 15$ .
All coefficients should be integers; if a fractional coefficient appears, recheck the division.

Techniques used

multiply known linear factors to form the divisorperform polynomial long divisionwrite the fully factorised form

(c)

Find the least positive value of $\theta$ in radians such that $p(\cot 2\theta) = 0$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

From part (b), $p(x) = (2x - 1)(x - 3)(x^2 + 5)$ . Setting $p(\cot 2\theta) = 0$ means $\cot 2\theta$ must equal one of the real roots of $p$ . The real roots are $\frac{1}{2}$ and $3$ . Since $\cot$ is decreasing on $(0, \pi)$ , the larger value $\cot 2\theta = 3$ gives the smaller angle, hence the least positive $\theta$ . Solve $\cot 2\theta = 3$ .

Working

From part (b):

p(x) = (2x - 1)(x - 3)(x^2 + 5)

Setting $p(\cot 2\theta) = 0$ :

(2\cot 2\theta - 1)(\cot 2\theta - 3)(\cot^2 2\theta + 5) = 0

The factor $\cot^2 2\theta + 5 \geq 5 > 0$ , so it gives no real solutions. Hence:

\cot 2\theta = \frac{1}{2} \quad \text{or} \quad \cot 2\theta = 3

Since $\cot$ is decreasing on $(0, \pi)$ , the larger value $\cot 2\theta = 3$ corresponds to the smaller $2\theta$ , giving the least positive $\theta$ . Solve:

\cot 2\theta = 3

\tan 2\theta = \frac{1}{3}

2\theta = \arctan\left(\frac{1}{3}\right) + k\pi, \quad k \in \mathbb{Z}

For the least positive $\theta$ , take $k = 0$ :

\theta = \frac{1}{2}\arctan\left(\frac{1}{3}\right)

\theta \approx 0.16087

Answer

\theta = 0.161 \text{ radians}

Final answer

θ = 0.161 radians

Detailed explanation

Walkthrough

We need $\theta$ such that $p(\cot 2\theta) = 0$ . From part (b), $p(x) = (2x - 1)(x - 3)(x^2 + 5)$ . So $p(\cot 2\theta) = 0$ means at least one of the three factors is zero:

$2\cot 2\theta - 1 = 0$ , i.e., $\cot 2\theta = \frac{1}{2}$
$\cot 2\theta - 3 = 0$ , i.e., $\cot 2\theta = 3$
$\cot^2 2\theta + 5 = 0$

The third factor $\cot^2 2\theta + 5$ is always positive (since $\cot^2 \geq 0$ , so $\cot^2 + 5 \geq 5$ ), so it has no real solutions. Thus $\cot 2\theta = \frac{1}{2}$ or $\cot 2\theta = 3$ .

Both equations have solutions. To find the least positive $\theta$ , we use the fact that $\cot$ is decreasing on $(0, \pi)$ : as the angle increases, the cotangent decreases. So a larger value of $\cot 2\theta$ corresponds to a smaller angle $2\theta$ . Since $3 > \frac{1}{2}$ , the equation $\cot 2\theta = 3$ gives the smaller $2\theta$ , hence the least positive $\theta$ .

Solve $\cot 2\theta = 3$ . Since $\cot x = \frac{1}{\tan x}$ , we have $\tan 2\theta = \frac{1}{3}$ . The principal solution is $2\theta = \arctan\left(\frac{1}{3}\right)$ . The general solution adds multiples of $\pi$ : $2\theta = \arctan\left(\frac{1}{3}\right) + k\pi$ . For the least positive $\theta$ , take $k = 0$ :

\theta = \frac{1}{2}\arctan\left(\frac{1}{3}\right) \approx 0.16087 \text{ radians}

Rounded to 3 significant figures: $\theta = 0.161$ radians.

Key Takeaways

Using the factorised form of a polynomial to solve $p(\text{expression}) = 0$ by setting each factor to zero.
The identity $\cot x = \frac{1}{\tan x}$ for converting cotangent equations to tangent equations.
The monotonicity of $\cot$ on $(0, \pi)$ helps identify which root gives the least positive angle.
The general solution of $\tan x = c$ is $x = \arctan(c) + k\pi$ .

Common Mistakes

Forgetting to check the factor $\cot^2 2\theta + 5 = 0$ — it has no real solutions, but a complete solution should note this.
Converting $\cot 2\theta = 3$ incorrectly to $\tan 2\theta = 3$ instead of $\tan 2\theta = \frac{1}{3}$ .
Not recognising that both $\cot 2\theta = \frac{1}{2}$ and $\cot 2\theta = 3$ give solutions, and that the least positive comes from the one with the larger cotangent value.
Giving the answer in degrees instead of radians.

Things to Be Careful About

The question specifies radians; the answer must be in radians.
$\arctan\left(\frac{1}{3}\right)$ is in radians when computed on a calculator in radian mode.
The mark scheme requires solving $\cot 2\theta = 3$ (M1) and obtaining $\theta = 0.161$ (A1).
The period of $\tan 2\theta$ is $\frac{\pi}{2}$ in terms of $\theta$ , so other solutions exist, but the least positive is the principal one.

Techniques used

set each factor of the polynomial to zeroconvert cotangent to tangentsolve the trigonometric equationselect the least positive solution

Q6MediumLogarithmic and Exponential FunctionsDifferentiation

A curve has equation $(x^2 - 3)\ln y + 6x = 14$ .

(a)

Show that there is no point on the curve at which the $y$ -coordinate is $e^{-1}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute $y = e^{-1}$ into the curve equation. Since $\ln(e^{-1}) = -1$ , the equation becomes a quadratic in $x$ . If this quadratic has no real roots, then no real $x$ -coordinate exists, so no point on the curve has $y = e^{-1}$ .

Working

Substitute $y = e^{-1}$ :

(x^2 - 3)\ln(e^{-1}) + 6x = 14

Since $\ln(e^{-1}) = -1$ :

-(x^2 - 3) + 6x = 14

-x^2 + 3 + 6x = 14

-x^2 + 6x - 11 = 0

Equivalently, multiplying by $-1$ :

x^2 - 6x + 11 = 0

The discriminant is

\Delta = (-6)^2 - 4(1)(11) = 36 - 44 = -8

Since $\Delta = -8 < 0$ , the quadratic has no real solutions. Therefore there is no point on the curve with $y = e^{-1}$ .

Answer

No real $x$ exists because the discriminant is $-8 < 0$ , so no point on the curve has $y = e^{-1}$ .

Final answer

No real x exists; discriminant is -8, so no point on the curve has y = e^-1

Detailed explanation

Walkthrough

The curve is defined by $(x^2 - 3)\ln y + 6x = 14$ . To test whether $y = e^{-1}$ can occur, substitute this value into the equation. The key simplification is $\ln(e^{-1}) = -1$ , because $\ln$ and $e$ are inverse functions. This turns the equation into a quadratic in $x$ : $-(x^2 - 3) + 6x = 14$ , which rearranges to $-x^2 + 6x - 11 = 0$ , or equivalently $x^2 - 6x + 11 = 0$ . A real point on the curve would require a real $x$ satisfying this quadratic. The discriminant $\Delta = b^2 - 4ac$ tells whether real roots exist. Here $\Delta = (-6)^2 - 4(1)(11) = -8 < 0$ , so there are no real solutions. Hence no point on the curve has $y = e^{-1}$ .

Key Takeaways

$\ln(e^{-1}) = -1$ follows from the inverse relationship between $\ln$ and $e$ .
A curve point requires a real $x$ -coordinate; if the resulting equation has no real solutions, the point cannot exist.
The discriminant of a quadratic determines whether real roots exist.

Common Mistakes

Forgetting that $\ln(e^{-1}) = -1$ , not $-e$ or $1/e$ .
Sign errors when rearranging: from $(x^2 - 3)(-1) + 6x = 14$ to $-x^2 + 3 + 6x = 14$ .
Using the discriminant incorrectly on $-x^2 + 6x - 11 = 0$ ; it still works, but must compute with $a = -1$ carefully. Multiplying by $-1$ first avoids sign errors.
Not showing the discriminant explicitly. The mark scheme requires necessary detail for an answer-given proof.

Things to Be Careful About

The discriminant is $b^2 - 4ac$ , not $b^2 + 4ac$ .
A negative discriminant means no real roots, so no real point.
State the conclusion clearly: no real $x$ exists, hence no point on the curve has $y = e^{-1}$ .

Techniques used

substitute a given y-value into an implicit equationuse the inverse property of ln and eevaluate a discriminant to show no real roots

(b)

Find the equation of the tangent to the curve at the point $(2, e^2)$ . Give your answer in the form $y = mx + c$ , where $m$ and $c$ are exact constants.

6M Similar

DifficultyMedium

Worked solution

Approach

Differentiate the curve equation implicitly with respect to $x$ , using the product rule on $(x^2 - 3)\ln y$ . Substitute the point $(2, e^2)$ to find the gradient, then form the equation of the tangent.

Working

Differentiate both sides with respect to $x$ :

\frac{d}{dx}\left((x^2 - 3)\ln y\right) + \frac{d}{dx}(6x) = \frac{d}{dx}(14)

Using the product rule:

2x\ln y + (x^2 - 3)\cdot \frac{1}{y}\frac{dy}{dx} + 6 = 0

At $(2, e^2)$ , we have $x = 2$ , $y = e^2$ and $\ln y = 2$ :

2(2)(2) + \frac{2^2 - 3}{e^2}\frac{dy}{dx} + 6 = 0

8 + \frac{1}{e^2}\frac{dy}{dx} + 6 = 0

\frac{1}{e^2}\frac{dy}{dx} = -14

\frac{dy}{dx} = -14e^2

The tangent line through $(2, e^2)$ with gradient $-14e^2$ is

y - e^2 = -14e^2(x - 2)

y = -14e^2 x + 28e^2 + e^2

y = -14e^2 x + 29e^2

Answer

y = -14e^2 x + 29e^2

Final answer

y = -14e^2 x + 29e^2

Detailed explanation

Walkthrough

We need the gradient of the tangent at $(2, e^2)$ . The curve is defined implicitly, so differentiate both sides with respect to $x$ . The term $(x^2 - 3)\ln y$ is a product of $x^2 - 3$ and $\ln y$ ; use the product rule. The derivative of $x^2 - 3$ is $2x$ , and the derivative of $\ln y$ with respect to $x$ is $\frac{1}{y}\frac{dy}{dx}$ by the chain rule. This gives $2x\ln y + \frac{x^2 - 3}{y}\frac{dy}{dx}$ . The derivative of $6x$ is $6$ , and the derivative of the constant $14$ is $0$ . Substitute $x = 2$ , $y = e^2$ ; then $\ln y = 2$ and $\frac{x^2 - 3}{y} = \frac{1}{e^2}$ . Solving for $\frac{dy}{dx}$ gives $-14e^2$ . Finally, the tangent line through $(2, e^2)$ with gradient $m$ is $y - e^2 = m(x - 2)$ ; simplifying gives $y = -14e^2 x + 29e^2$ .

Key Takeaways

Implicit differentiation: differentiate every term with respect to $x$ , treating $y$ as a function of $x$ .
The product rule is needed for products involving both $x$ and $y$ .
The chain rule gives $\frac{d}{dx}(\ln y) = \frac{1}{y}\frac{dy}{dx}$ .
The tangent line uses the point-slope form with the gradient found at the point.

Common Mistakes

Forgetting the $\frac{dy}{dx}$ factor when differentiating $\ln y$ .
Applying the product rule incorrectly, e.g. writing $(x^2 - 3)\frac{1}{y}$ without the $2x\ln y$ term.
Forgetting the $+6$ term from differentiating $6x$ .
Substituting incorrectly: at $y = e^2$ , $\ln y = 2$ , not $e^2$ or $1$ .
Not simplifying the tangent equation to the required form $y = mx + c$ with exact constants.

Things to Be Careful About

The derivative of $\ln y$ is $\frac{1}{y}\frac{dy}{dx}$ , not $\frac{1}{y}$ .
When solving $\frac{1}{e^2}\frac{dy}{dx} = -14$ , multiply both sides by $e^2$ .
The constant term in the tangent is $28e^2 + e^2 = 29e^2$ , not $28e^2$ .

Techniques used

differentiate implicitly with respect to xapply the product rule to a logarithmic termsubstitute coordinates to find the gradientform the equation of a tangent line

Q7MediumTrigonometry

(a)

Express $4\cos\theta\sin(\theta + 30°)$ in the form $R\cos(2\theta - \alpha) + k$ , where $R > 0$ , $0° < \alpha < 90°$ and $k$ is a constant.

6M Similar

DifficultyMedium

Worked solution

Approach

Expand $\sin(\theta + 30°)$ using the compound angle formula, simplify using double angle formulae, then express the resulting $a\sin 2\theta + b\cos 2\theta$ in the form $R\cos(2\theta - \alpha)$ .

Working

Using the compound angle formula for sine:

\sin(\theta + 30°) = \sin\theta\cos 30° + \cos\theta\sin 30° = \frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta

Therefore:

4\cos\theta\sin(\theta + 30°) = 4\cos\theta\left(\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta\right) = 2\sqrt{3}\cos\theta\sin\theta + 2\cos^2\theta

Now apply the double angle formulae $2\sin\theta\cos\theta = \sin 2\theta$ and $2\cos^2\theta = 1 + \cos 2\theta$ :

4\cos\theta\sin(\theta + 30°) = \sqrt{3}\sin 2\theta + \cos 2\theta + 1

Now express $\sqrt{3}\sin 2\theta + \cos 2\theta$ in the form $R\cos(2\theta - \alpha)$ . Expanding:

R\cos(2\theta - \alpha) = R(\cos 2\theta\cos\alpha + \sin 2\theta\sin\alpha)

Comparing coefficients of $\cos 2\theta$ and $\sin 2\theta$ :

R\cos\alpha = 1, \qquad R\sin\alpha = \sqrt{3}

Therefore:

R = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2

And:

\tan\alpha = \frac{R\sin\alpha}{R\cos\alpha} = \frac{\sqrt{3}}{1} = \sqrt{3}

Since $0° < \alpha < 90°$ , we have $\alpha = 60°$ .

Answer

4\cos\theta\sin(\theta + 30°) = 2\cos(2\theta - 60°) + 1

with $R = 2$ , $\alpha = 60°$ , $k = 1$ .

Final answer

2 cos(2θ - 60°) + 1, with R = 2, α = 60°, k = 1

Detailed explanation

Walkthrough

The goal is to rewrite $4\cos\theta\sin(\theta + 30°)$ in the form $R\cos(2\theta - \alpha) + k$ . The presence of $\sin(\theta + 30°)$ suggests expanding it using the compound angle formula $\sin(A + B) = \sin A\cos B + \cos A\sin B$ . This gives $\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta$ , using the standard values $\cos 30° = \frac{\sqrt{3}}{2}$ and $\sin 30° = \frac{1}{2}$ .

Multiplying by $4\cos\theta$ distributes the cosine: $2\sqrt{3}\cos\theta\sin\theta + 2\cos^2\theta$ . The form $R\cos(2\theta - \alpha)$ involves $2\theta$ , so we need to convert the products of $\sin\theta$ and $\cos\theta$ into double angles. The double angle formulae give $2\sin\theta\cos\theta = \sin 2\theta$ and $2\cos^2\theta = 1 + \cos 2\theta$ . Applying these converts the expression to $\sqrt{3}\sin 2\theta + \cos 2\theta + 1$ .

Now we have $a\sin 2\theta + b\cos 2\theta + 1$ with $a = \sqrt{3}$ and $b = 1$ . To express $a\sin 2\theta + b\cos 2\theta$ in the form $R\cos(2\theta - \alpha)$ , we expand $R\cos(2\theta - \alpha) = R\cos 2\theta\cos\alpha + R\sin 2\theta\sin\alpha$ and match coefficients. This gives $R\cos\alpha = 1$ and $R\sin\alpha = \sqrt{3}$ . Squaring and adding gives $R^2 = 1 + 3 = 4$ , so $R = 2$ . Dividing gives $\tan\alpha = \sqrt{3}$ , and since $\alpha$ must be between $0°$ and $90°$ , $\alpha = 60°$ .

Key Takeaways

The compound angle formula for sine allows expansion of $\sin(\theta + 30°)$ .
Double angle formulae convert products of $\sin\theta$ and $\cos\theta$ into terms of $2\theta$ .
The R-form conversion $a\sin x + b\cos x = R\cos(x - \alpha)$ is performed by comparing coefficients.

Common Mistakes

Forgetting the constant term $+1$ when converting $2\cos^2\theta$ to $1 + \cos 2\theta$ .
Mixing up the coefficients in the R-form comparison (e.g., putting $\sqrt{3}$ with $\cos\alpha$ instead of $\sin\alpha$ ).
Not checking that $\alpha$ lies in the required range $0° < \alpha < 90°$ .

Things to Be Careful About

When comparing coefficients, ensure $R\cos\alpha$ matches the coefficient of $\cos 2\theta$ and $R\sin\alpha$ matches the coefficient of $\sin 2\theta$ .
The mark scheme requires exact coefficients ( $2\sqrt{3}$ and $2$ ) at the intermediate step.
The final answer must be in the exact form requested: $R\cos(2\theta - \alpha) + k$ .

Techniques used

expand compound angle sineapply double angle identitiesexpress a sin θ + b cos θ in R cos form

(b)

Hence solve the equation

12\cos 2\phi\sin(2\phi + 30°) = 5

for $0° < \phi < 90°$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Substitute $\theta = 2\phi$ into the result from part (a), multiply by 3, then solve the resulting equation $\cos(4\phi - 60°) = \frac{1}{3}$ for $0° < \phi < 90°$ .

Working

From part (a), with $\theta = 2\phi$ :

4\cos(2\phi)\sin(2\phi + 30°) = 2\cos(4\phi - 60°) + 1

Multiplying both sides by 3:

12\cos(2\phi)\sin(2\phi + 30°) = 6\cos(4\phi - 60°) + 3

The equation $12\cos 2\phi\sin(2\phi + 30°) = 5$ becomes:

6\cos(4\phi - 60°) + 3 = 5

\cos(4\phi - 60°) = \frac{1}{3}

Let $X = 4\phi - 60°$ . Since $0° < \phi < 90°$ , we have $0° < 4\phi < 360°$ , so:

-60° < X < 300°

The principal solution is:

X = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.53°

Since $\cos$ is positive in the first and fourth quadrants, the solutions in the range are:

X = 70.53° \quad \text{and} \quad X = 360° - 70.53° = 289.47°

(The value $X = -70.53°$ is outside the range since $-70.53° < -60°$ .)

Now $\phi = \frac{X + 60°}{4}$ :

\phi = \frac{70.53° + 60°}{4} = \frac{130.53°}{4} \approx 32.6°

\phi = \frac{289.47° + 60°}{4} = \frac{349.47°}{4} \approx 87.4°

Both values lie in the range $0° < \phi < 90°$ .

Answer

\phi = 32.6° \quad \text{or} \quad \phi = 87.4°

(to 1 decimal place)

Final answer

φ = 32.6° or φ = 87.4°

Detailed explanation

Walkthrough

The word "Hence" tells us to use the result from part (a). Substituting $\theta = 2\phi$ into $4\cos\theta\sin(\theta + 30°) = 2\cos(2\theta - 60°) + 1$ gives $4\cos(2\phi)\sin(2\phi + 30°) = 2\cos(4\phi - 60°) + 1$ . The given equation has coefficient 12 on the left, so we multiply both sides by 3 to get $12\cos(2\phi)\sin(2\phi + 30°) = 6\cos(4\phi - 60°) + 3$ .

Setting this equal to 5 gives $6\cos(4\phi - 60°) + 3 = 5$ , so $\cos(4\phi - 60°) = \frac{1}{3}$ .

Now we solve $\cos X = \frac{1}{3}$ where $X = 4\phi - 60°$ . The range of $X$ follows from $0° < \phi < 90°$ : multiplying by 4 gives $0° < 4\phi < 360°$ , and subtracting $60°$ gives $-60° < X < 300°$ .

The principal value is $X = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.53°$ . Since cosine is positive in quadrants I and IV, the other solution in the range is $X = 360° - 70.53° = 289.47°$ . We must check that $X = -70.53°$ is not in the range (it is less than $-60°$ ), and that $X = 430.53°$ is too large.

Finally, converting back: $\phi = \frac{X + 60°}{4}$ gives $\phi = 32.6°$ and $\phi = 87.4°$ , both within $0° < \phi < 90°$ .

Key Takeaways

The R-form result can be used directly to solve equations of the form $a\cos x + b\sin x = c$ .
When solving $\cos X = k$ , all solutions in the given range must be found, not just the principal value.
The range of the substituted variable $X$ must be computed carefully from the original range of $\phi$ .

Common Mistakes

Forgetting to multiply by 3 when substituting into the part (a) result (the coefficient must match 12).
Only finding the principal value of $\cos^{-1}\left(\frac{1}{3}\right)$ and missing the second solution.
Not checking whether solutions lie within the required range for $\phi$ .

Things to Be Careful About

The range $-60° < X < 300°$ excludes $X = -70.53°$ but includes both $70.53°$ and $289.47°$ .
The mark scheme requires showing the method for finding both values of $\phi$ .
Answers should be given to a sensible degree of accuracy (1 decimal place is standard here).

Techniques used

substitute into R-form resultsolve trigonometric equationidentify all solutions in restricted range

Mathematics 9709/22 — May/June 2025

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