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 $e^{-1}$. (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.

(a) No real x exists; discriminant is -8. · (b) y = -14e^2 x + 29e^2

(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) 2cos(2θ - 60°) + 1 · (b) φ = 32.6° or φ = 87.4°

9709/23 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

Use the standard integral $\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + c$ to integrate $\frac{8}{4x+1}$ . Then substitute the upper and lower limits and simplify the resulting logarithms using the laws of logarithms.

Working

\int_2^{11} \frac{8}{4x+1}\,dx = \left[2\ln(4x+1)\right]_2^{11}

Substitute the limits:

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

= 2\ln 45 - 2\ln 9

Use the logarithm law $p\ln x - p\ln y = p\ln\left(\frac{x}{y}\right)$ :

= 2\ln\left(\frac{45}{9}\right) = 2\ln 5

Convert the coefficient to a power:

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

Answer

\int_2^{11} \frac{8}{4x+1}\,dx = \ln 25

So $a = 25$ .

Final answer

a = 25

Detailed explanation

Walkthrough

The integral is of the form $\int \frac{1}{ax+b}\,dx$ , whose antiderivative is $\frac{1}{a}\ln|ax+b| + c$ . Here $a = 4$ and the numerator is $8$ , so the integral becomes $2\ln(4x+1)$ .

After integrating, we evaluate the antiderivative at the upper limit $x = 11$ and the lower limit $x = 2$ , subtracting the lower value from the upper value. This gives $2\ln 45 - 2\ln 9$ .

To combine these logarithms, we use the law $p\ln x - p\ln y = p\ln\left(\frac{x}{y}\right)$ . This gives $2\ln\left(\frac{45}{9}\right) = 2\ln 5$ . Finally, we use $p\ln x = \ln(x^p)$ to write $2\ln 5 = \ln(5^2) = \ln 25$ . Therefore $a = 25$ .

Key Takeaways

This question tests the integration of $\frac{1}{ax+b}$ , which produces a natural logarithm. It also tests the laws of logarithms, especially combining terms like $p\ln x - p\ln y$ and moving a coefficient into the argument as a power.

Common Mistakes

Forgetting the factor $\frac{1}{a}$ when integrating $\frac{1}{ax+b}$ . Here the correct antiderivative is $2\ln(4x+1)$ , not $\frac{8}{4}\ln(4x+1)$ incorrectly simplified.
Applying the limits incorrectly, such as evaluating at only one limit or subtracting in the wrong order.
Stopping at $2\ln 5$ instead of simplifying to $\ln 25$ when the question asks for the answer in the form $\ln a$ .
Using an incorrect logarithm law, such as writing $\ln 45 - \ln 9 = \ln(45-9)$ .

Things to Be Careful About

On the interval $[2, 11]$ , the expression $4x+1$ is always positive, so no absolute value sign is needed in the final answer.
The mark scheme awards B1 for integrating to $2\ln(4x+1)$ , M1 for applying limits and using at least one relevant logarithm property, and A1 for the final simplified answer $\ln 25$ .
Since the question asks for $\ln a$ with $a$ an integer, the final answer must be written as $\ln 25$ , not just $2\ln 5$ .

Techniques used

integrate a reciprocal linear functionapply limits of integrationsimplify using logarithm laws

Approach

Locate the vertex of the modulus graph, then sketch its two linear branches. Sketch the straight line using its gradient and intercept.

Working

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

For $x \geq \frac{9}{2}$ :

y = 2x - 9

For $x < \frac{9}{2}$ :

y = -2x + 9

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

Answer

A V-shaped graph with vertex at $\left(\frac{9}{2}, 0\right)$ and a straight line of gradient $4$ crossing the $y$ -axis at $-5$ .

Final answer

A V-shaped graph with vertex at (4.5, 0) and the line y = 4x - 5.

Detailed explanation

Walkthrough

The modulus function $y = |2x - 9|$ is V-shaped. It changes behaviour at the value of $x$ that makes $2x - 9 = 0$ , namely $x = \frac{9}{2}$ . To the right of this point the expression inside the modulus is positive, so the graph is the line $y = 2x - 9$ ; to the left it is negative, so the graph is the reflected line $y = -2x + 9$ . The vertex is at $\left(\frac{9}{2}, 0\right)$ .

The line $y = 4x - 5$ is a straight line with gradient $4$ and $y$ -intercept $-5$ . It is steeper than either branch of the modulus graph. Drawing both on the same axes lets you see where the modulus graph lies below the line, which is useful for part (b).

Key Takeaways

A modulus graph of the form $y = |ax+b|$ is V-shaped, with its vertex at the root of $ax+b=0$ . The two branches have gradients $a$ and $-a$ . A linear graph is determined by its gradient and intercept.

Common Mistakes

Drawing the modulus graph as a smooth curve instead of two straight lines meeting at a vertex.
Placing the vertex at the wrong point; the vertex is at $x = \frac{9}{2}$ , not at $x = 9$ .
Drawing the line $y = 4x - 5$ with the wrong gradient or intercept.

Things to Be Careful About

The vertex must be on the positive $x$ -axis. The line $y = 4x - 5$ has gradient $4$ , so it must be steeper than the modulus branches, which have gradients $\pm 2$ .

Techniques used

locate the vertex of a modulus graphsketch the two linear branches of a modulus graphdraw a straight line using its gradient and intercept

(b)

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

3M Similar

DifficultyMedium-Easy

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 intervals.

Working

The modulus changes sign at $x = \frac{9}{2}$ .

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

-2x + 9 < 4x - 5.

Solving:

14 < 6x \quad \Rightarrow \quad x > \frac{7}{3}.

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

\frac{7}{3} < x < \frac{9}{2}.

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

2x - 9 < 4x - 5.

Solving:

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

Since this is automatically true for $x \geq \frac{9}{2}$ , the whole interval $x \geq \frac{9}{2}$ is included.

Combining the two cases:

x > \frac{7}{3}.

Answer

x > \frac{7}{3}

Final answer

x > 7/3

Detailed explanation

Walkthrough

The inequality $|2x - 9| < 4x - 5$ cannot be solved by simply removing the modulus sign, because the sign of $2x - 9$ changes at $x = \frac{9}{2}$ . Split the problem into two cases.

For $x < \frac{9}{2}$ , $2x - 9$ is negative, so $|2x - 9| = -(2x - 9) = -2x + 9$ . The inequality becomes $-2x + 9 < 4x - 5$ . Rearranging gives $14 < 6x$ , so $x > \frac{7}{3}$ . Since we are in the case $x < \frac{9}{2}$ , this gives $\frac{7}{3} < x < \frac{9}{2}$ .

For $x \geq \frac{9}{2}$ , $2x - 9$ is non-negative, so $|2x - 9| = 2x - 9$ . The inequality becomes $2x - 9 < 4x - 5$ , which simplifies to $x > -2$ . Every $x \geq \frac{9}{2}$ already satisfies this, so all of $x \geq \frac{9}{2}$ is included.

Taking the union of the two intervals gives $x > \frac{7}{3}$ .

Key Takeaways

To solve a modulus inequality, identify the critical point where the expression inside the modulus changes sign, solve the inequality separately on each side, and then combine the solution sets. The graph from part (a) confirms the answer: the modulus graph is below the line exactly when $x > \frac{7}{3}$ .

Common Mistakes

Forgetting to split into cases and simply writing $2x - 9 < 4x - 5$ .
Losing the restriction $x < \frac{9}{2}$ in the first case, which would give the wrong final interval.
Choosing the wrong direction of the inequality after rearranging, e.g. writing $x < \frac{7}{3}$ .

Things to Be Careful About

The final answer must be strict, since the original inequality is strict. The boundary point $x = \frac{7}{3}$ is not included. Also, when combining cases, remember that the second case contributes the whole interval $x \geq \frac{9}{2}$ , not just a small part of it.

Techniques used

split the modulus into its two casessolve a linear inequality in each casecombine the 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 equation term by term, using the quotient rule for the rational term. Set the derivative equal to zero and solve for $x$ , then substitute back to find the corresponding $y$ -coordinates.

Working

Let

y = \frac{8x}{2x+3} - 6x + 5.

Differentiate the first term using the quotient rule:

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

Therefore

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

At a stationary point, $\frac{dy}{dx} = 0$ :

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

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

(2x+3)^2 = 4.

Expanding and rearranging gives $4x^2 + 12x + 5 = 0$ , so

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

Hence

x = -\frac{1}{2} \quad \text{or} \quad x = -\frac{5}{2}.

Substitute into $y$ :

For $x = -\frac{1}{2}$ :

y = \frac{8(-1/2)}{2(-1/2)+3} - 6(-1/2) + 5 = \frac{-4}{2} + 3 + 5 = 6.

For $x = -\frac{5}{2}$ :

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

Answer

The stationary points are

\left(-\frac{5}{2},\,30\right) \quad \text{and} \quad \left(-\frac{1}{2},\,6\right).

Final answer

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

Detailed explanation

Walkthrough

We want the points where the curve is stationary, i.e. where $\frac{dy}{dx} = 0$ . The expression has two parts: a rational term $\frac{8x}{2x+3}$ and a polynomial term $-6x+5$ . Differentiate the rational term with the quotient rule. With $u = 8x$ and $v = 2x+3$ , we have $u' = 8$ and $v' = 2$ , so

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

Then the derivative of $-6x+5$ is just $-6$ , so

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

Set this equal to zero and rearrange:

\frac{24}{(2x+3)^2} = 6 \quad \Rightarrow \quad (2x+3)^2 = 4.

Taking square roots gives $2x+3 = \pm 2$ , so $x = -\frac{1}{2}$ or $x = -\frac{5}{2}$ . Finally, substitute these values back into the original equation for $y$ . For $x = -\frac{1}{2}$ , $y = 6$ ; for $x = -\frac{5}{2}$ , $y = 30$ . Thus the stationary points are $\left(-\frac{1}{2}, 6\right)$ and $\left(-\frac{5}{2}, 30\right)$ .

Key Takeaways

This question tests the quotient rule for differentiating a rational function, the fact that stationary points occur where $\frac{dy}{dx} = 0$ , and solving the resulting equation. It also reminds you to substitute back into the original function to find the $y$ -coordinates.

Common Mistakes

Forgetting the derivative of $-6x+5$ is $-6$ , not zero.
Making a sign error in the quotient rule numerator: the correct numerator is $8(2x+3) - 8x(2)$ , not $8x(2) - 8(2x+3)$ .
Solving $(2x+3)^2 = 4$ by only taking the positive square root, losing $x = -\frac{5}{2}$ .
Substituting the $x$ -values into $\frac{dy}{dx}$ instead of the original equation when finding the $y$ -coordinates.
Not showing the quadratic equation or equivalent rearrangement, which is needed for the method mark.

Things to Be Careful About

The derivative may be left unsimplified and still earns the first accuracy mark, but simplifying to $\frac{24}{(2x+3)^2} - 6$ makes the next step clearer.
The denominator $(2x+3)^2$ is never zero at the stationary points found, since $x = -\frac{3}{2}$ is not among them; however, the original function is undefined at $x = -\frac{3}{2}$ , so that value must be excluded.
When giving final coordinates, use exact fractions, not decimals, unless the question asks otherwise.
Both $x$ -values must be found; finding only one stationary point loses the final accuracy mark.

Techniques used

apply quotient rule to differentiate a rational termdifferentiate linear polynomial termsset derivative equal to zero and solve the resulting quadraticsubstitute x-values back into the original equation

Approach

At point $P$ the two curves share the same $y$ -value. Equate the two expressions, isolate the exponential term, then take the natural logarithm of both sides to remove the exponential.

Working

At $P$ :

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)

Solve for $x$ :

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

Answer

The $x$ -coordinate of $P$ satisfies the equation $x = -0.5\ln(0.25 + 0.125\sin 3x)$ , as required.

Final answer

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

Detailed explanation

Walkthrough

At a point of intersection, both curves have the same $y$ -coordinate at the same $x$ -coordinate. So the first step is to set the two given expressions for $y$ equal to each other.

Once we have $4e^{-2x} = 1 + 0.5\sin 3x$ , the next step is to get the exponential on its own on the left side. Dividing through by 4 gives $e^{-2x} = 0.25 + 0.125\sin 3x$ .

To remove the exponential we take the natural logarithm: $\ln(e^{u}) = u$ , so $\ln(e^{-2x}) = -2x$ . This produces a linear equation in $x$ which we solve by dividing by $-2$ .

Key Takeaways

The first step in any intersection problem is equating the two expressions in $y$ .
The natural logarithm is the inverse of the exponential function $e^u$ , so $\ln(e^u) = u$ .
This is an 'AG' (Answer Given) question, so the working must end at the stated form.

Common Mistakes

Forgetting to divide both sides of the equation by 4.
Dropping the minus sign when solving $-2x = \ln(\ldots)$ for $x$ .
Trying to take the logarithm before isolating the exponential.

Things to Be Careful About

All algebraic steps must be shown because this is an 'AG' question.
The final form is $x = \ldots$ , not $\ln(\ldots) = \ldots$ .

Techniques used

equate curves at intersectionrearrange to isolate exponentialapply natural logarithmsolve linear equation for x

(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

The equation in part (a) has the form $x = g(x)$ . Apply the iteration $x_{n+1} = g(x_n) = -0.5\ln(0.25 + 0.125\sin 3x_n)$ , starting from $x_0 = 0.5$ , and continue until the iterates have stabilised to 6 significant figures.

Working

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

$x_0 = 0.500000$

x_1 = -0.5\ln(0.25 + 0.125\sin(1.50000)) = -0.5\ln(0.374687) = 0.490832

x_2 = -0.5\ln(0.25 + 0.125\sin(1.47250)) = -0.5\ln(0.374397) = 0.491220

x_3 = -0.5\ln(0.25 + 0.125\sin(1.47366)) = -0.5\ln(0.374411) = 0.491201

x_4 = -0.5\ln(0.25 + 0.125\sin(1.47360)) = -0.5\ln(0.374410) = 0.491202

The iterates have stabilised to 6 significant figures. The final value lies in the interval $[0.49115, 0.49125]$ , confirming the result.

Answer

x = 0.4912 \text{ (to 4 s.f.)}

Final answer

x = 0.4912

Detailed explanation

Walkthrough

The equation from part (a) is in the form $x = g(x)$ , where $g(x) = -0.5\ln(0.25 + 0.125\sin 3x)$ . Such an equation is solved numerically using a fixed-point iteration $x_{n+1} = g(x_n)$ .

For the first iterate, $x_1 = g(0.5)$ : compute $3 \times 0.5 = 1.5$ , then $\sin(1.5) \approx 0.99749$ , then $0.25 + 0.125 \times 0.99749 = 0.37469$ , then $\ln(0.37469) \approx -0.98166$ , and finally $x_1 = -0.5 \times -0.98166 = 0.49083$ . Rounded to 6 significant figures, $x_1 = 0.490832$ .

For $x_2$ , repeat with $x_1$ in place of $x_0$ : $3 \times 0.490832 = 1.47250$ , $\sin(1.47250) \approx 0.99517$ , $0.25 + 0.125 \times 0.99517 = 0.37440$ , $\ln(0.37440) \approx -0.98244$ , giving $x_2 = 0.491220$ to 6 s.f.

Continue with $x_3$ and $x_4$ similarly. By $x_3 = 0.491201$ and $x_4 = 0.491202$ the iterates have essentially stabilised. Rounding to 4 significant figures gives $x = 0.4912$ .

Key Takeaways

A fixed-point iteration solves $x = g(x)$ by repeatedly applying $x_{n+1} = g(x_n)$ .
Iterates should be recorded to one more significant figure than the final answer requires, so that the convergence can be verified.
The iteration must be applied using the previous value, never the value two steps back.

Common Mistakes

Using too few iterations, so the answer is not actually stable to 6 s.f.
Rounding intermediate values too early, which can shift the final answer.
Confusing the iteration $x = g(x)$ with $f(x) = 0$ and using the wrong rearrangement.

Things to Be Careful About

The angle in $\sin 3x$ is in radians (no degree symbol is given).
Each iteration must use the previous iterate, not the value two steps back.
The mark scheme also accepts a sign-change check in $[0.49115, 0.49125]$ as alternative justification.

Techniques used

apply fixed-point iterationcompute successive iteratesround to 6 significant figuresconclude to 4 significant figures

(c)

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

4M Similar

DifficultyMedium

Worked solution

Approach

From the diagram, the upper curve is $y = 4e^{-2x}$ and the lower curve is $y = 1 + 0.5\sin 3x$ on the interval $[0, x_P]$ . The shaded area is therefore the integral of (upper) $-$ (lower) from $0$ to $x_P = 0.4912$ .

Working

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

Find the antiderivative of each term:

$\int 4e^{-2x}\,dx = -2e^{-2x}$
$\int 1\,dx = x$ , so $\int(-1)\,dx = -x$
$\int \sin 3x\,dx = -\dfrac{1}{3}\cos 3x$ , so $\int(-0.5\sin 3x)\,dx = \dfrac{1}{6}\cos 3x$

Combining the three pieces:

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

Substitute the upper limit $x = 0.4912$ :

-2e^{-0.9824} - 0.4912 + \dfrac{1}{6}\cos(1.4736)

Using $e^{-0.9824} \approx 0.37441$ and $\cos(1.4736) \approx 0.09705$ :

= -2(0.37441) - 0.4912 + \dfrac{1}{6}(0.09705) = -0.74882 - 0.4912 + 0.01617 = -1.22385

Substitute the lower limit $x = 0$ :

-2e^{0} - 0 + \dfrac{1}{6}\cos 0 = -2 + \dfrac{1}{6} = -1.83333

Subtract (upper $-$ lower):

A = -1.22385 - (-1.83333) = 0.60948

Answer

A \approx 0.61 \text{ (to 2 s.f.)}

Final answer

0.61

Detailed explanation

Walkthrough

The shaded region in Fig. 1 is bounded above by $y = 4e^{-2x}$ and below by $y = 1 + 0.5\sin 3x$ , on the interval $0 \leq x \leq x_P$ . The area is therefore the integral of (upper $-$ lower) across this interval.

Each term integrates using a standard rule:

$\int e^{ax}\,dx = \dfrac{1}{a}e^{ax}$ , so $\int 4e^{-2x}\,dx = \dfrac{4}{-2}e^{-2x} = -2e^{-2x}$ .
$\int 1\,dx = x$ .
$\int \sin(ax)\,dx = -\dfrac{1}{a}\cos(ax)$ , so $\int \sin 3x\,dx = -\dfrac{1}{3}\cos 3x$ . The $-0.5$ factor in the integrand then turns this into $+\dfrac{1}{6}\cos 3x$ in the antiderivative.

After combining the three antiderivatives, substitute the upper and lower limits. At the upper limit $x = 0.4912$ , we need the numerical values of $e^{-0.9824}$ and $\cos(1.4736)$ , which come from a calculator. At the lower limit $x = 0$ , $e^{0} = 1$ and $\cos 0 = 1$ are exact.

Subtracting the two substituted values gives $0.60948$ , which rounds to $0.61$ to 2 significant figures.

Key Takeaways

The area between two curves on $[a, b]$ is $\int_a^b (y_{\text{upper}} - y_{\text{lower}})\,dx$ .
Each term of an integrand is integrated separately, using standard rules for exponentials and trigonometric functions.
The question's 'hence' links part (b) to part (c), so the answer from part (b) is used as the upper limit.

Common Mistakes

Forgetting the sign when integrating $\sin 3x$ , or confusing the sign of the $\cos 3x$ term because of the $-0.5$ in the integrand.
Mixing up which curve is on top — the diagram is the reference, not the equations alone.
Using the wrong limits, e.g. not using the answer from part (b) as the upper limit.
Forgetting to subtract the two substituted values; some students only evaluate at the upper limit.

Things to Be Careful About

The value from part (b) is approximate, but we use it as the upper limit because that is what 'hence' implies.
Keep enough decimal places in intermediate steps so the final answer is reliable to 2 s.f.
The question asks for 2 s.f. — round accordingly.

Techniques used

set up area between curves as a definite integralintegrate an exponential functionintegrate a sine functionapply limits and evaluateround to required accuracy

Q5MediumAlgebraTrigonometry

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

Worked solution

Approach

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

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

Multiply by $16$ :

a + 2b + 12 = 0

Substitute $x=3$ :

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

81a + 27b + 27 = 0

Divide by $27$ :

3a + b + 1 = 0

Solve the simultaneous equations:

a + 2b = -12, \qquad 3a + b = -1

From the second, $b = -1 - 3a$ . Substitute into the first:

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

-5a - 2 = -12 \quad \Rightarrow \quad a = 2

Then $b = -1 - 3(2) = -7$ .

Answer

a = 2, \qquad b = -7

Final answer

a = 2, b = -7

Detailed explanation

Walkthrough

Because $(2x-1)$ is a factor, $x=\frac{1}{2}$ is a root, so $p\left(\frac{1}{2}\right)=0$ . Similarly $x=3$ is a root, so $p(3)=0$ . Substitute these values into the polynomial. For $x=\frac{1}{2}$ , the powers are $\left(\frac{1}{2}\right)^4=\frac{1}{16}$ , $\left(\frac{1}{2}\right)^3=\frac{1}{8}$ , and $\left(\frac{1}{2}\right)^2=\frac{1}{4}$ . Combine the constant terms: $\frac{13}{4}-\frac{35}{2}+15=\frac{3}{4}$ . So $\frac{a}{16}+\frac{b}{8}+\frac{3}{4}=0$ ; multiplying by $16$ gives $a+2b+12=0$ . For $x=3$ : $81a+27b+117-105+15=81a+27b+27=0$ ; dividing by $27$ gives $3a+b+1=0$ . Solve: from $3a+b=-1$ , $b=-1-3a$ . Substitute into $a+2b=-12$ : $a+2(-1-3a)=-12$ , so $-5a-2=-12$ , giving $a=2$ , then $b=-7$ .

Key Takeaways

The factor theorem turns a known factor into a root condition. Substituting roots produces equations in unknown coefficients. Clearing fractions and dividing common factors simplifies solving.

Common Mistakes

Forgetting that the factor $(2x-1)$ gives the root $x=\frac{1}{2}$ , not $x=-\frac{1}{2}$ or $x=2$ .
Arithmetic errors when combining constants at $x=\frac{1}{2}$ and $x=3$ .
Not simplifying the equations before solving.
Solving simultaneous equations incorrectly; check by substituting back.

Things to Be Careful About

The factor theorem says $p(k)=0$ if $(x-k)$ is a factor; for $(2x-1)$ , set $2x-1=0$ , so $x=\frac{1}{2}$ . When simplifying, divide both sides by common factors correctly. Keep exact fractions rather than decimals.

Techniques used

substitute roots into the polynomial using the factor theoremform and solve simultaneous equations

(b)

Hence factorise $p(x)$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Multiply the two known linear factors to form the quadratic divisor $2x^2-7x+3$ , then divide $p(x)$ by this quadratic to find the remaining quadratic factor.

Working

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

Divide $2x^4 - 7x^3 + 13x^2 - 35x + 15$ by $2x^2 - 7x + 3$ :

\begin{aligned} &2x^4 - 7x^3 + 13x^2 - 35x + 15 \\ &= (2x^2 - 7x + 3)(x^2 + 5) \end{aligned}

So

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

Since $x^2+5$ has no real roots, this is the fully factorised form.

Answer

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

Final answer

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

Detailed explanation

Walkthrough

We already know two factors. Multiplying them gives $2x^2-7x+3$ . Since $p(x)$ has degree 4, the remaining factor must be quadratic. Divide $p(x)$ by $2x^2-7x+3$ . The leading term $2x^4$ divided by $2x^2$ gives $x^2$ . Multiplying the divisor by $x^2$ gives $2x^4-7x^3+3x^2$ ; subtracting leaves $10x^2-35x+15$ . Then $10x^2$ divided by $2x^2$ gives $5$ ; multiplying the divisor by $5$ gives $10x^2-35x+15$ , which subtracts to zero. So the quotient is $x^2+5$ . Therefore $p(x)=(2x-1)(x-3)(x^2+5)$ . Since $x^2+5$ cannot be factored into real linear factors, this is the fully factorised form.

Key Takeaways

Polynomial division by a quadratic factor is efficient when two linear factors are known. Checking coefficients can also be used. A quadratic with no real roots remains as an irreducible factor over the reals.

Common Mistakes

Dividing incorrectly; keep track of subtraction signs.
Stopping at the quotient without writing the full factorisation.
Trying to factor $x^2+5$ over the reals; it has no real roots.
If using synthetic division by the two linear factors, forgetting the leading coefficient and writing $2(x^2+5)$ ; the final factorised form must be $(2x-1)(x-3)(x^2+5)$ .

Things to Be Careful About

The divisor should be the product of the two known factors. Ensure quotient times divisor equals the original polynomial. "Fully factorise" over the reals means leaving $x^2+5$ as it is.

Techniques used

multiply known linear factors to form a quadratic divisordivide the quartic by the quadratic factor

(c)

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

2M Similar

DifficultyMedium

Worked solution

Approach

Use the factorised form of $p(x)$ . Since $p(\cot 2\theta)=0$ , one of the factors must be zero. The quadratic factor gives no real solution, so solve $\cot 2\theta = \frac{1}{2}$ and $\cot 2\theta = 3$ , then choose the least positive $\theta$ .

Working

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

So $p(\cot 2\theta)=0$ requires:

2\cot 2\theta - 1 = 0 \quad \text{or} \quad \cot 2\theta - 3 = 0

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

For $\cot 2\theta = 3$ :

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

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

For $\cot 2\theta = \frac{1}{2}$ :

\tan 2\theta = 2

\theta = \frac{1}{2} \arctan(2) \approx 0.55357

The smaller positive value is $0.16087$ .

Answer

\theta \approx 0.161 \text{ radians}

Final answer

θ ≈ 0.161 radians

Detailed explanation

Walkthrough

Since $p$ is factorised, $p(\cot 2\theta)=0$ when any factor is zero. The factor $x^2+5$ cannot be zero for real $x$ , so ignore it. The linear factors give $2\cot 2\theta-1=0$ or $\cot 2\theta-3=0$ , i.e. $\cot 2\theta=\frac{1}{2}$ or $\cot 2\theta=3$ . Because $\cot$ and $\tan$ are reciprocals, these become $\tan 2\theta=2$ and $\tan 2\theta=\frac{1}{3}$ . For the least positive $\theta$ , take the smallest positive angle $2\theta$ : $2\theta=\arctan 2$ or $2\theta=\arctan\frac{1}{3}$ . Then $\theta$ is half of that. The values are approximately $0.5536$ and $0.1609$ ; the least is $0.1609$ , so $\theta \approx 0.161$ radians.

Key Takeaways

Factorised polynomials let us solve composite equations by setting each factor to zero. Cotangent is the reciprocal of tangent. The least positive solution comes from the smallest positive principal angle.

Common Mistakes

Forgetting the root $x=\frac{1}{2}$ gives another possible value of $\cot 2\theta$ ; but the least positive angle still comes from $\cot 2\theta=3$ .
Writing $\tan 2\theta=3$ instead of $\tan 2\theta=\frac{1}{3}$ for $\cot 2\theta=3$ .
Solving for $2\theta$ and forgetting to halve to get $\theta$ .
Not checking which of the two solutions is smaller.

Things to Be Careful About

$x^2+5=0$ has no real solutions, so it contributes no values of $\theta$ . Answer in radians. The final answer is $0.161$ radians (3 s.f.), or $0.16087\ldots$ if more accuracy is required. General solutions add multiples of $\pi$ to $2\theta$ , which increase $\theta$ , so they are not needed for the least positive value.

Techniques used

use the factorised form to set factors equal to zeroconvert cotangent equations to tangent equationsfind and compare least positive solutions

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 the given $y$ -coordinate into the curve equation, simplify using $\ln(e^{-1}) = -1$ , and show that the resulting quadratic has no real roots by evaluating its discriminant.

Working

The curve is

(x^2 - 3)\ln y + 6x = 14.

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

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

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

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

Simplify:

-x^2 + 3 + 6x = 14

-x^2 + 6x - 11 = 0.

Equivalently,

x^2 - 6x + 11 = 0.

The discriminant is

b^2 - 4ac = (-6)^2 - 4(1)(11) = 36 - 44 = -8.

Since the discriminant is negative, the quadratic has no real solutions for $x$ . Therefore there is no point on the curve with $y$ -coordinate $e^{-1}$ .

Answer

Substitution gives $-x^2 + 6x - 11 = 0$ , with discriminant $-8 < 0$ , so no real $x$ exists.

Final answer

No real x exists; discriminant is -8.

Detailed explanation

Walkthrough

We want to know whether any point on the curve has $y$ -coordinate $e^{-1}$ . A point lies on the curve exactly when its coordinates satisfy the curve equation, so we substitute $y = e^{-1}$ and see whether the resulting equation can be solved for a real value of $x$ .

First use the logarithm fact $\ln(e^{-1}) = -1$ . This is because $\ln$ and $e^x$ are inverse functions: $\ln(e^k) = k$ for any real $k$ . This turns the equation into a quadratic in $x$ .

After simplifying, we get $-x^2 + 6x - 11 = 0$ , or equivalently $x^2 - 6x + 11 = 0$ . To check whether real solutions exist, calculate the discriminant $b^2 - 4ac$ . Here $a = 1$ , $b = -6$ , $c = 11$ , so

b^2 - 4ac = (-6)^2 - 4(1)(11) = -8.

A negative discriminant means the quadratic has no real roots. Therefore no real $x$ can satisfy the equation when $y = e^{-1}$ , so there is no such point on the curve.

Key Takeaways

The inverse relationship between $e^x$ and $\ln x$ gives $\ln(e^k) = k$ .
A quadratic equation has no real roots exactly when its discriminant is negative.
Substituting a candidate coordinate into an equation is a standard way to test whether a point lies on a curve.

Common Mistakes

Writing $\ln(e^{-1})$ incorrectly; it is $-1$ , not $1$ or $-e$ .
Making a sign error when expanding $-(x^2 - 3)$ ; it becomes $-x^2 + 3$ .
Stating the discriminant is negative without showing the calculation. Because the question says 'Show that', the working must be shown.

Things to Be Careful About

The discriminant must be computed from the simplified quadratic; using $-x^2 + 6x - 11$ gives $a = -1$ , $b = 6$ , $c = -11$ and the same discriminant $-8$ .
An alternative acceptable method is completing the square: $(x - 3)^2 + 2 = 0$ , which also has no real solution.
Since the question is 'Show that', a bare statement such as 'no real roots' without supporting detail may not earn full marks.

Techniques used

substitute a candidate coordinate into the curve equationuse the logarithm identity for e and lnform a quadratic equationevaluate the discriminant

(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 for $(x^2 - 3)\ln y$ . Substitute the coordinates of the given point to find the gradient, then write the equation of the tangent in the form $y = mx + c$ .

Working

Differentiate both sides of

(x^2 - 3)\ln y + 6x = 14

with respect to $x$ :

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 $\ln y = 2$ . Substituting:

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

So

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

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

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

The gradient of the tangent is $-14e^2$ . Using $y - y_1 = m(x - x_1)$ :

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

Expand:

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

The curve is defined by an equation involving both $x$ and $y$ , so we differentiate implicitly with respect to $x$ . This means every time we differentiate a term containing $y$ , we multiply by $\frac{dy}{dx}$ .

The term $(x^2 - 3)\ln y$ is a product of $x^2 - 3$ and $\ln y$ , so we use the product rule:

\frac{d}{dx}\left((x^2 - 3)\ln y\right) = 2x\ln y + (x^2 - 3)\cdot \frac{1}{y}\frac{dy}{dx}.

The derivative of $6x$ is $6$ , and the derivative of the constant $14$ is $0$ . This gives the differentiated equation.

Next, substitute the point $(2, e^2)$ . Here $x = 2$ , $y = e^2$ , and $\ln(e^2) = 2$ . Substituting gives an equation that can be solved for $\frac{dy}{dx}$ , which is the gradient of the curve at that point.

Finally, use the point-gradient form of a straight line:

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

Expanding and collecting the constant terms gives the tangent in the required form $y = mx + c$ .

Key Takeaways

Implicit differentiation: differentiate both sides with respect to $x$ and multiply any $y$ -derivative by $\frac{dy}{dx}$ .
The product rule is needed when a term is a product of a function of $x$ and a function of $y$ .
The gradient of the tangent at a point is the value of $\frac{dy}{dx}$ at that point.

Common Mistakes

Forgetting the $\frac{dy}{dx}$ factor when differentiating $\ln y$ .
Applying the product rule incorrectly; both terms $2x\ln y$ and $(x^2 - 3)\frac{1}{y}\frac{dy}{dx}$ must appear.
Substituting $\ln(e^2)$ as $e^2$ instead of $2$ .
Losing the constant $6$ when differentiating $6x$ .
Forgetting to add $e^2$ to $28e^2$ when forming the $c$ term.

Things to Be Careful About

The mark scheme allows follow-through on the derivative of $(x^2 - 3)\ln y$ , so a small error in the product rule may still gain later method marks if the substitution and tangent method are correct.
The final answer must be in the form $y = mx + c$ with exact constants; do not round $e^2$ .
Check the sign: the gradient is negative, so the tangent slopes downward.

Techniques used

differentiate implicitly with respect to xapply the product rulesubstitute the given point to find the gradientwrite the tangent equation in gradient-intercept form

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, rewrite the product using double-angle identities to obtain an expression in $\sin 2\theta$ and $\cos 2\theta$ , then express the result in the form $R\cos(2\theta - \alpha) + k$ by comparing coefficients.

Working

Expand $\sin(\theta + 30°)$ using $\sin(A + B) = \sin A\cos B + \cos A\sin B$ :

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

Multiply by $4\cos\theta$ :

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

Apply the double-angle identities $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$ :

2\sqrt{3}\cos\theta\sin\theta = \sqrt{3}\sin 2\theta

2\cos^2\theta = 1 + \cos 2\theta

Therefore:

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 $R\cos(2\theta - \alpha)$ :

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

Comparing coefficients with $\cos 2\theta + \sqrt{3}\sin 2\theta$ :

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

Compute $R$ :

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

Compute $\alpha$ :

\tan\alpha = \frac{\sqrt{3}}{1} = \sqrt{3} \Rightarrow \alpha = 60°

Hence:

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

Answer

2\cos(2\theta - 60°) + 1

Final answer

2cos(2θ - 60°) + 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 expression contains a product of $\cos\theta$ and $\sin(\theta + 30°)$ , and the target form involves $\cos(2\theta - \alpha)$ with a double angle. The strategy proceeds in four stages.

Stage 1 — Expand the compound angle. Use $\sin(A + B) = \sin A\cos B + \cos A\sin B$ with $A = \theta$ and $B = 30°$ . This gives $\sin\theta\cos 30° + \cos\theta\sin 30° = \frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta$ . This expands the compound angle into single-angle terms.

Stage 2 — Multiply by $4\cos\theta$ . Distributing $4\cos\theta$ gives $2\sqrt{3}\cos\theta\sin\theta + 2\cos^2\theta$ . This is the form $k_1\cos\theta\sin\theta + k_2\cos^2\theta$ that the mark scheme requires.

Stage 3 — Apply double-angle identities. The term $2\cos\theta\sin\theta$ equals $\sin 2\theta$ , so $2\sqrt{3}\cos\theta\sin\theta = \sqrt{3}\sin 2\theta$ . The term $2\cos^2\theta$ equals $1 + \cos 2\theta$ since $\cos 2\theta = 2\cos^2\theta - 1$ . This converts everything to double angles.

Stage 4 — Express in R-form. We now have $\sqrt{3}\sin 2\theta + \cos 2\theta + 1$ . The constant $+1$ is the $k$ in the target form. The remaining part $\sqrt{3}\sin 2\theta + \cos 2\theta$ needs to be written as $R\cos(2\theta - \alpha)$ . Using $R\cos(2\theta - \alpha) = R\cos 2\theta\cos\alpha + R\sin 2\theta\sin\alpha$ , we compare coefficients: $R\cos\alpha = 1$ (coefficient of $\cos 2\theta$ ) and $R\sin\alpha = \sqrt{3}$ (coefficient of $\sin 2\theta$ ). Then $R = \sqrt{1^2 + (\sqrt{3})^2} = 2$ and $\tan\alpha = \sqrt{3}$ , giving $\alpha = 60°$ since both $\cos\alpha$ and $\sin\alpha$ are positive.

Key Takeaways

The compound angle formula $\sin(A + B) = \sin A\cos B + \cos A\sin B$ allows us to expand expressions like $\sin(\theta + 30°)$ .
Double-angle identities $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$ convert single-angle products into double-angle terms.
The R-form $a\sin\theta + b\cos\theta = R\cos(\theta - \alpha)$ with $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \frac{a}{b}$ (careful with quadrant) is a powerful tool for simplifying trigonometric expressions.

Common Mistakes

Forgetting the constant term: $2\cos^2\theta = 1 + \cos 2\theta$ includes a $+1$ that must be carried through to the final answer.
Mixing up coefficients: when comparing $R\cos(2\theta - \alpha) = (R\cos\alpha)\cos 2\theta + (R\sin\alpha)\sin 2\theta$ with $\cos 2\theta + \sqrt{3}\sin 2\theta$ , make sure $R\cos\alpha$ matches the $\cos 2\theta$ coefficient and $R\sin\alpha$ matches the $\sin 2\theta$ coefficient.
Using the wrong quadrant for $\alpha$ : since both $\cos\alpha$ and $\sin\alpha$ are positive, $\alpha$ must be in the first quadrant.

Things to Be Careful About

The mark scheme requires exact coefficients ( $2\sqrt{3}$ and $2$ ) — do not use decimal approximations.
The angle $\alpha$ must satisfy $0° < \alpha < 90°$ , which confirms the first-quadrant choice.
When the mark scheme says "follow through", it means later marks can still be earned even if an earlier step had an error, as long as the method is correct.

Techniques used

expand compound angle formulaapply double-angle identitiesexpress in R cos formcompare coefficients to find R and alpha

(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), scale the equation to match the given coefficient of 12, then solve the resulting trigonometric equation $\cos(4\phi - 60°) = \frac{1}{3}$ within the domain $0° < \phi < 90°$ .

Working

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

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

The given equation is $12\cos 2\phi\sin(2\phi + 30°) = 5$ . Since $12 = 3 \times 4$ , multiply the result from part (a) by 3:

12\cos 2\phi\sin(2\phi + 30°) = 3\left(2\cos(4\phi - 60°) + 1\right) = 6\cos(4\phi - 60°) + 3

Set this equal to 5:

6\cos(4\phi - 60°) + 3 = 5

6\cos(4\phi - 60°) = 2

\cos(4\phi - 60°) = \frac{1}{3}

The domain is $0° < \phi < 90°$ , so $0° < 4\phi < 360°$ and hence $-60° < 4\phi - 60° < 300°$ .

The principal value is $\cos^{-1}\left(\frac{1}{3}\right) = 70.53°$ . The solutions of $\cos x = \frac{1}{3}$ in the range $(-60°, 300°)$ are:

x = 70.53° \quad \text{or} \quad x = 360° - 70.53° = 289.47°

First solution:

4\phi - 60° = 70.53° \Rightarrow 4\phi = 130.53° \Rightarrow \phi = 32.6°

Second solution:

4\phi - 60° = 289.47° \Rightarrow 4\phi = 349.47° \Rightarrow \phi = 87.4°

Both values lie in $0° < \phi < 90°$ , and there are no other solutions in this range.

Answer

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

Final answer

φ = 32.6° or φ = 87.4°

Detailed explanation

Walkthrough

The equation $12\cos 2\phi\sin(2\phi + 30°) = 5$ is a scaled version of the expression from part (a) with $\theta$ replaced by $2\phi$ .

Stage 1 — Substitute the result from part (a). From part (a), $4\cos\theta\sin(\theta + 30°) = 2\cos(2\theta - 60°) + 1$ . Setting $\theta = 2\phi$ gives $4\cos 2\phi\sin(2\phi + 30°) = 2\cos(4\phi - 60°) + 1$ .

Stage 2 — Scale to match the equation. The given equation has coefficient 12 on the left, which is $3 \times 4$ . So we multiply both sides of the substituted result by 3: $12\cos 2\phi\sin(2\phi + 30°) = 3(2\cos(4\phi - 60°) + 1) = 6\cos(4\phi - 60°) + 3$ .

Stage 3 — Solve for the cosine. Setting the right side equal to 5: $6\cos(4\phi - 60°) + 3 = 5$ , so $6\cos(4\phi - 60°) = 2$ , giving $\cos(4\phi - 60°) = \frac{1}{3}$ .

Stage 4 — Determine the domain. Since $0° < \phi < 90°$ , we have $0° < 4\phi < 360°$ , so $-60° < 4\phi - 60° < 300°$ . This is the range in which we must find solutions for $x = 4\phi - 60°$ .

Stage 5 — Find all solutions. The principal value is $\cos^{-1}\left(\frac{1}{3}\right) = 70.53°$ . The general solution of $\cos x = \frac{1}{3}$ is $x = \pm 70.53° + 360°n$ . In the range $(-60°, 300°)$ :

$x = 70.53°$ (from $n = 0$ )
$x = -70.53° + 360° = 289.47°$ (from $n = 1$ )

The value $x = -70.53°$ is below $-60°$ and $x = 430.53°$ is above $300°$ , so these are excluded.

Stage 6 — Solve for $\phi$ .

$4\phi - 60° = 70.53°$ gives $4\phi = 130.53°$ , so $\phi = 32.6°$ .
$4\phi - 60° = 289.47°$ gives $4\phi = 349.47°$ , so $\phi = 87.4°$ .

Both values lie in $0° < \phi < 90°$ , so both are valid solutions.

Key Takeaways

A result from one part of a question can be reused in a later part — this is a common exam pattern.
When solving $\cos x = k$ over a domain, the general solution is $x = \pm\cos^{-1}(k) + 360°n$ , and you must check which values fall in the domain.
The domain of a transformed variable (here $4\phi - 60°$ ) is different from the domain of the original variable ( $\phi$ ), and must be computed carefully.

Common Mistakes

Forgetting to multiply the constant term by 3: $3(2\cos(4\phi - 60°) + 1) = 6\cos(4\phi - 60°) + 3$ , not $6\cos(4\phi - 60°) + 1$ .
Only finding the principal value solution and missing the second solution in the fourth quadrant.
Not checking that both solutions are within $0° < \phi < 90°$ .

Things to Be Careful About

The mark scheme explicitly requires "no others between 0 and 90" — you must verify that no additional solutions exist in the domain.
The domain of $4\phi - 60°$ is $(-60°, 300°)$ , which is wider than $[0°, 360°)$ — be careful about solutions near the boundaries.
The mark scheme accepts $32.6$ and $87.4$ to 1 decimal place, or greater accuracy $32.6321\dots$ and $87.3678\dots$ .

Techniques used

substitute result from part (a)solve trigonometric equationcheck domain restrictions

Mathematics 9709/23 — May/June 2025

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working