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 point exists on the curve with y-coordinate e^-1. · (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°

9709/25 May/June 2025 — Mathematics Solved Past Paper

Q13MMedium-EasyIntegrationLogarithmic and Exponential Functions

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

Integrate $\frac{8}{4x+1}$ with respect to $x$ using the standard result $\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln(ax+b) + C$ . Then evaluate the definite integral between the given limits and simplify using logarithm laws.

Working

The integral is

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

Since $\int \frac{1}{4x+1}\,dx = \frac{1}{4}\ln(4x+1)$ ,

\int \frac{8}{4x+1}\,dx = 8 \cdot \frac{1}{4}\ln(4x+1) = 2\ln(4x+1)

Apply the limits:

\left[2\ln(4x+1)\right]_2^{11} = 2\ln(4(11)+1) - 2\ln(4(2)+1)

= 2\ln 45 - 2\ln 9

Use $\ln A - \ln B = \ln \frac{A}{B}$ :

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

Finally, use $k\ln x = \ln x^k$ :

2\ln 5 = \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 definite integral asks for the exact value of $\int_2^{11} \frac{8}{4x+1}\,dx$ . Because the denominator is a linear expression $4x+1$ , this is an integral of the form $\int \frac{1}{ax+b}\,dx$ , whose antiderivative is $\frac{1}{a}\ln(ax+b)$ . Here $a=4$ , so $\int \frac{1}{4x+1}\,dx = \frac{1}{4}\ln(4x+1)$ . Multiplying by 8 gives $2\ln(4x+1)$ .

Once the antiderivative is found, substitute the upper limit 11 and lower limit 2:

$2\ln(4(11)+1) - 2\ln(4(2)+1) = 2\ln 45 - 2\ln 9$ .

The two logarithms have the same coefficient 2, so factor it out and use the subtraction law $\ln A - \ln B = \ln(A/B)$ :

$2(\ln 45 - \ln 9) = 2\ln(45/9) = 2\ln 5$ .

Finally, use the power law $k\ln x = \ln x^k$ to write $2\ln 5 = \ln 25$ . Therefore the integral equals $\ln 25$ , so $a = 25$ .

Key Takeaways

The integral of $\frac{1}{ax+b}$ is $\frac{1}{a}\ln(ax+b)$ , not just $\ln(ax+b)$ .
Definite integration uses the antiderivative evaluated at the upper limit minus the lower limit.
Logarithm laws allow the difference of logarithms to be combined into a single logarithm.
The final form $\ln 25$ shows the value as a single logarithm with integer argument.

Common Mistakes

Forgetting the factor $\frac{1}{4}$ when integrating $\frac{1}{4x+1}$ , which would give $8\ln(4x+1)$ instead of $2\ln(4x+1)$ .
Substituting the limits incorrectly, especially evaluating $4(11)+1 = 45$ and $4(2)+1 = 9$ .
Stopping at $2\ln 5$ instead of converting it to $\ln 25$ ; the question asks for the form $\ln a$ with integer $a$ .
Using $\ln 45 - \ln 9 = \ln(45-9)$ rather than $\ln(45/9)$ .

Things to Be Careful About

The interval from 2 to 11 makes $4x+1$ positive, so no absolute-value signs are needed in the antiderivative.
The mark scheme awards B1 for the antiderivative $2\ln(4x+1)$ , M1 for applying limits and using at least one logarithm law, and A1 for the simplified result $\ln 25$ .
Keep the coefficient 2 outside the logarithms until the subtraction law has been applied; combining too early can lead to arithmetic errors.

Techniques used

integrate a reciprocal linear functionapply limits of integrationsimplify using logarithm laws

Approach

The graph of $y = |2x - 9|$ is V-shaped: it has a vertex where $2x - 9 = 0$ , and two straight arms of slopes $2$ and $-2$ . The graph of $y = 4x - 5$ is a straight line with slope $4$ and $y$ -intercept $-5$ . Draw both on the same axes, marking the vertex and the intersection of the line with the left arm of the modulus graph.

Working

For the modulus graph:

2x - 9 = 0 \quad \Rightarrow \quad x = \frac{9}{2}

So the vertex is $(\frac{9}{2}, 0)$ .

For $x \geq \frac{9}{2}$ , $y = 2x - 9$ (slope $2$ ).
For $x \leq \frac{9}{2}$ , $y = -2x + 9$ (slope $-2$ ).

The line $y = 4x - 5$ has slope $4$ , so it is steeper than either arm. It meets the left arm where:

-2x + 9 = 4x - 5

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

y = 4(\frac{7}{3}) - 5 = \frac{13}{3}

So the line crosses the left arm at $(\frac{7}{3}, \frac{13}{3})$ .

Answer

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

Final answer

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

Detailed explanation

Walkthrough

The modulus function $y = |2x - 9|$ can be thought of as two half-lines. The expression inside the modulus is zero at $x = \frac{9}{2}$ , so the vertex is $(\frac{9}{2}, 0)$ . To the right of this point, $2x - 9$ is positive, so the graph is the line $y = 2x - 9$ with slope $2$ . To the left, $2x - 9$ is negative, so the graph is $y = -(2x - 9) = -2x + 9$ , with slope $-2$ . This gives the V shape.

The line $y = 4x - 5$ is a straight line with slope $4$ and $y$ -intercept $-5$ . It is steeper than both arms of the V. To place it accurately, find where it meets the left arm of the V by setting $-2x + 9 = 4x - 5$ . This gives $x = \frac{7}{3}$ and $y = \frac{13}{3}$ , so the line crosses the V at that point.

Key Takeaways

A modulus graph $y = |ax + b|$ has a vertex where $ax + b = 0$ , and its two arms have slopes $a$ and $-a$ . A straight line can be sketched accurately by locating its intercept and any intersection with the modulus graph.

Common Mistakes

Drawing the vertex at $x = 9$ instead of $x = \frac{9}{2}$ .
Drawing the modulus graph as a smooth curve rather than a V with two straight arms.
Drawing the line $y = 4x - 5$ with the wrong slope or intercept.

Things to Be Careful About

The mark scheme awards one mark for the V shape with vertex on the positive $x$ -axis and one mark for the line $y = 4x - 5$ correctly placed relative to the modulus graph. The line must be steeper than the arms of the V, so its slope $4$ should be visibly greater than $2$ .

Techniques used

sketch a modulus graphidentify vertex and slopes of armslocate intersection with a straight line

(b)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the reflected branch of the modulus (where $2x - 9 < 0$ ) to find the intersection of the line with the V-shaped graph. The inequality holds where the line $y = 4x - 5$ lies above the modulus graph. Solve the linear equation with opposite signs, then give the interval.

Working

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

|2x - 9| = -2x + 9

The boundary of the inequality is when the line meets this branch:

-2x + 9 = 4x - 5

14 = 6x

x = \frac{7}{3}

For $x > \frac{7}{3}$ , the line is above the reflected branch, so:

|2x - 9| < 4x - 5

For $x \geq \frac{9}{2}$ , the modulus is $2x - 9$ , and:

2x - 9 < 4x - 5 \quad \Rightarrow \quad -4 < 2x \quad \Rightarrow \quad x > -2

so every $x \geq \frac{9}{2}$ also satisfies the inequality. Combining the two regions gives:

x > \frac{7}{3}

Answer

x > \frac{7}{3}

Final answer

x > 7/3

Detailed explanation

Walkthrough

The inequality asks for the values of $x$ for which the line $y = 4x - 5$ is above the V-shaped graph $y = |2x - 9|$ . The boundary occurs where the two graphs meet. Since the line meets the left arm of the V, use the reflected branch $|2x - 9| = -2x + 9$ for $x < \frac{9}{2}$ . Set $-2x + 9 = 4x - 5$ to find $x = \frac{7}{3}$ . To the right of this intersection the line is above the V, so the inequality is true. Check the right branch separately: for $x \geq \frac{9}{2}$ , the inequality becomes $2x - 9 < 4x - 5$ , which simplifies to $x > -2$ , so all values in that region also work. The union is $x > \frac{7}{3}$ .

Key Takeaways

Solving a modulus inequality can be done by considering the reflected branch where the modulus expression is negative, finding the boundary intersection, and then deciding which side satisfies the inequality. The graph from part (a) makes the direction clear.

Common Mistakes

Forgetting to consider the right branch $x \geq \frac{9}{2}$ .
Writing $x < \frac{7}{3}$ instead of $x > \frac{7}{3}$ .
If squaring both sides, forgetting that $4x - 5$ must be positive, which can introduce the extra region $x < -2$ .

Things to Be Careful About

The mark scheme accepts the answer as $x > \frac{7}{3}$ , or interval notation $(\frac{7}{3}, \infty)$ . If using the squaring method, solve $(2x - 9)^2 < (4x - 5)^2$ and then impose $4x - 5 > 0$ to remove the spurious region $x < -2$ . Also note the inequality is strict, so $x = \frac{7}{3}$ is not included.

Techniques used

use the reflected modulus branchsolve a linear equationcombine intervals to state the inequality solution

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 $\frac{8x}{2x+3}$ using the quotient rule, then differentiate the remaining linear terms. Set $\frac{dy}{dx}=0$ and solve the resulting quadratic equation for $x$ . Finally substitute each $x$ -value into the original equation to obtain the corresponding $y$ -coordinates.

Working

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

Using the quotient rule on $\frac{8x}{2x+3}$ :

\frac{d}{dx}\left(\frac{8x}{2x+3}\right) = \frac{(2x+3)(8) - 8x(2)}{(2x+3)^2} = \frac{16x+24-16x}{(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

2x+3 = \pm 2

So

2x+3 = 2 \quad \text{or} \quad 2x+3 = -2

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

Now find the corresponding $y$ -coordinates.

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

Final answer

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

Detailed explanation

Walkthrough

A stationary point of a curve is a point where the gradient is zero, so the first step is to differentiate the given equation.

The function has three terms: $\frac{8x}{2x+3}$ , $-6x$ and $5$ . The first term is a quotient, so use the quotient rule. For $u=8x$ and $v=2x+3$ , we have $u'=8$ and $v'=2$ . The quotient rule gives

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

so

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

The derivative of $-6x+5$ is simply $-6$ , so

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

To find stationary points, set this derivative equal to zero:

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

Rearrange to get $(2x+3)^2=4$ . Taking square roots gives $2x+3=\pm 2$ , so $2x+3=2$ or $2x+3=-2$ . These give $x=-\frac{1}{2}$ and $x=-\frac{5}{2}$ .

Finally, substitute each $x$ -value back into the original equation to find the $y$ -coordinate. For $x=-\frac{1}{2}$ , the value is $6$ ; for $x=-\frac{5}{2}$ , the value is $30$ . Therefore the stationary points are $(-\frac{1}{2},6)$ and $(-\frac{5}{2},30)$ .

Key Takeaways

Stationary points occur where $\frac{dy}{dx}=0$ .
The quotient rule is needed when differentiating a fraction such as $\frac{8x}{2x+3}$ .
Solving $(2x+3)^2=4$ requires taking both positive and negative square roots.
After finding the $x$ -coordinates, always substitute into the original equation to obtain the $y$ -coordinates.

Common Mistakes

Forgetting to differentiate the $-6x+5$ part; this would give an incomplete derivative.
Making a sign error in the quotient rule numerator. The $16x$ terms cancel, leaving $24$ .
Taking only the positive square root and missing $x=-\frac{5}{2}$ .
Substituting the $x$ -values into the derivative instead of the original equation when finding $y$ .
Swapping the $x$ and $y$ values in the final coordinates.

Things to Be Careful About

The denominator $2x+3$ cannot be zero, so $x=-\frac{3}{2}$ is not in the domain. Neither stationary point has this value.
The mark scheme allows an unsimplified derivative, but the final coordinates must be exact.
When solving $\frac{24}{(2x+3)^2}=6$ , multiply both sides by $(2x+3)^2$ before dividing; this avoids sign errors.
The stationary points must be written as coordinate pairs, not just the $x$ -values.

Techniques used

apply the quotient rule to differentiate a rational functiondifferentiate a linear expressionset the derivative equal to zerosolve the resulting quadratic equationsubstitute x-values back into the original equation

Approach

At the point of intersection $P$ , the $y$ -values of both curves are equal. Equate the two expressions for $y$ , then rearrange the equation to isolate $x$ using logarithms.

Working

Set the equations equal 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)

This confirms the required equation.

Answer

The equation is shown as required.

Final answer

Shown as required

Detailed explanation

Walkthrough

First, we recognize that at any point of intersection between two curves, their $y$ -coordinates must be equal. We set $4e^{-2x} = 1 + 0.5\sin 3x$ . To isolate $x$ , we divide by 4 to get $e^{-2x} = 0.25 + 0.125\sin 3x$ . Taking the natural logarithm removes the exponential, giving $-2x = \ln(0.25 + 0.125\sin 3x)$ . Finally, dividing by $-2$ yields the desired form $x = -0.5\ln(0.25 + 0.125\sin 3x)$ .

Key Takeaways

When finding intersections of curves involving exponentials and trigonometric functions, equating them and using logarithms is a standard technique to isolate the variable.

Common Mistakes

Forgetting to divide by 4 before taking the logarithm.
Sign errors when dividing by $-2$ .

Things to Be Careful About

Ensure all algebraic steps are shown clearly as the mark scheme requires 'necessary detail' for the mark. The final expression must match exactly.

Techniques used

equate curve equations at intersectionisolate exponential termapply natural logarithm to solve 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-Easy

Worked solution

Approach

Use the iterative formula $x_{n+1} = -0.5\ln(0.25 + 0.125\sin 3x_n)$ starting with $x_1 = 0.5$ . Calculate successive values to 6 significant figures until the answer stabilizes to 4 significant figures.

Working

Given $x_1 = 0.500000$ :

x_2 = -0.5\ln(0.25 + 0.125\sin(3 \times 0.5)) = -0.5\ln(0.25 + 0.125\sin(1.5))

x_2 \approx 0.490813

x_3 = -0.5\ln(0.25 + 0.125\sin(3 \times 0.490813)) = -0.5\ln(0.25 + 0.125\sin(1.47244))

x_3 \approx 0.491241

x_4 = -0.5\ln(0.25 + 0.125\sin(3 \times 0.491241)) = -0.5\ln(0.25 + 0.125\sin(1.47372))

x_4 \approx 0.491203

x_5 = -0.5\ln(0.25 + 0.125\sin(3 \times 0.491203)) = -0.5\ln(0.25 + 0.125\sin(1.47361))

x_5 \approx 0.491209

x_6 = -0.5\ln(0.25 + 0.125\sin(3 \times 0.491209)) = -0.5\ln(0.25 + 0.125\sin(1.47363))

x_6 \approx 0.491208

The values $x_5$ and $x_6$ both round to $0.4912$ to 4 significant figures.

Answer

x = 0.4912

Final answer

0.4912

Detailed explanation

Walkthrough

We use the formula from part (a) as an iterative process: $x_{n+1} = -0.5\ln(0.25 + 0.125\sin 3x_n)$ . Starting with $x_1 = 0.5$ , we plug in the value to get $x_2$ , then plug $x_2$ in to get $x_3$ , and so on. We continue until two consecutive values agree to the required precision (4 significant figures). Here, $x_5 = 0.491209$ and $x_6 = 0.491208$ both round to $0.4912$ .

Key Takeaways

Iterative formulas converge to a root when the function is well-behaved. Always show enough iterations to justify the final answer to the required number of significant figures.

Common Mistakes

Using radians instead of degrees for the sine function (calculator mode error).
Rounding intermediate values too early, which can lead to an incorrect final answer.

Things to Be Careful About

The mark scheme requires showing iterations to 6 significant figures. Ensure your calculator is in radian mode since the argument is $3x$ (not $3x^\circ$ ).

Techniques used

apply iterative formulacheck convergence to required 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

The shaded region is bounded above by $y = 4e^{-2x}$ and below by $y = 1 + 0.5\sin 3x$ , from $x = 0$ to $x = 0.4912$ (the $x$ -coordinate of $P$ from part (b)). The area is given by the definite integral of the upper curve minus the lower curve.

Working

\text{Area} = \int_0^{0.4912} \left( 4e^{-2x} - (1 + 0.5\sin 3x) \right) \, dx

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

Integrate each term:

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

\int -1 \, dx = -x

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

So the antiderivative is:

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

Apply the limits from $0$ to $0.4912$ :

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

Evaluate at upper limit $x = 0.4912$ :

-2e^{-2(0.4912)} - 0.4912 + \frac{1}{6}\cos(3 \times 0.4912)

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

\approx -2(0.37434) - 0.4912 + \frac{1}{6}(0.09733)

\approx -0.74868 - 0.4912 + 0.01622 = -1.22366

Evaluate at lower limit $x = 0$ :

-2e^0 - 0 + \frac{1}{6}\cos(0) = -2(1) + \frac{1}{6}(1) = -2 + 0.16667 = -1.83333

Subtract lower limit value from upper limit value:

\text{Area} = -1.22366 - (-1.83333) = 0.60967

Rounding to 2 significant figures:

\text{Area} \approx 0.61

Answer

0.61

Final answer

0.61

Detailed explanation

Walkthrough

The area between two curves $f(x)$ (upper) and $g(x)$ (lower) from $x = a$ to $x = b$ is $\int_a^b (f(x) - g(x)) \, dx$ . Here, $f(x) = 4e^{-2x}$ , $g(x) = 1 + 0.5\sin 3x$ , $a = 0$ , and $b = 0.4912$ . We integrate term by term: the integral of $4e^{-2x}$ is $-2e^{-2x}$ , the integral of $-1$ is $-x$ , and the integral of $-0.5\sin 3x$ is $\frac{1}{6}\cos 3x$ . We then substitute the upper and lower limits and subtract.

Key Takeaways

When finding the area between curves, always subtract the lower curve from the upper curve. Be careful with the signs when integrating trigonometric functions like $\sin(ax)$ , which gives $-\frac{1}{a}\cos(ax)$ .

Common Mistakes

Integrating $\sin 3x$ as $\frac{1}{3}\cos 3x$ instead of $-\frac{1}{3}\cos 3x$ .
Forgetting to evaluate the lower limit and subtract it.
Using the wrong value for the upper limit (must use the answer from part (b)).

Things to Be Careful About

Ensure you use the unrounded value from part (b) if possible, or at least the 4 significant figure value $0.4912$ . The final answer must be given to 2 significant figures as stated in the question.

Techniques used

set up definite integral for area between curvesintegrate exponential and trigonometric termsapply limits from part (b)

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

Use the factor theorem: if $(2x-1)$ is a factor of $p(x)$ , then $p\left(\frac{1}{2}\right)=0$ ; if $(x-3)$ is a factor, then $p(3)=0$ . This gives two linear equations in $a$ and $b$ , which can be solved simultaneously.

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

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

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

Multiplying by 16:

a + 2b + 12 = 0 \quad \Rightarrow \quad a + 2b = -12

Substitute $x=3$ :

p(3) = 81a + 27b + 13(9) - 35(3) + 15

81a + 27b + 117 - 105 + 15 = 0

81a + 27b + 27 = 0

Dividing by 27:

3a + b + 1 = 0 \quad \Rightarrow \quad 3a + b = -1

Solve simultaneously. From $3a+b=-1$ , $b=-1-3a$ . Substitute into $a+2b=-12$ :

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

The factor theorem states that if $(2x-1)$ is a factor of $p(x)$ , then substituting the root $x=\frac{1}{2}$ must make $p(x)=0$ . Similarly, if $(x-3)$ is a factor, then $p(3)=0$ . We substitute both values into $p(x)$ to create two equations involving $a$ and $b$ .

For $x=\frac{1}{2}$ , evaluate each power and combine the constant terms. The constants $\frac{13}{4}-\frac{35}{2}+15$ simplify to $\frac{3}{4}$ . Multiplying the whole equation by 16 clears the denominators and gives $a+2b+12=0$ .

For $x=3$ , the constant terms $117-105+15$ combine to 27, so $81a+27b+27=0$ . Dividing by 27 gives $3a+b+1=0$ .

Now solve the simultaneous equations. From $3a+b=-1$ , write $b=-1-3a$ . Substitute into $a+2b=-12$ to get $-5a=-10$ , so $a=2$ . Then $b=-1-3(2)=-7$ .

Key Takeaways

The factor theorem turns a known factor into a root condition. Substituting that root gives an equation linking the unknown coefficients. With two factors, we get two equations, which can be solved simultaneously.

Common Mistakes

A common error is using the wrong root for $(2x-1)$ : the root is $x=\frac{1}{2}$ , not $x=2$ or $x=-\frac{1}{2}$ . Another common error is making arithmetic mistakes when combining fractions; multiplying by 16 avoids this. Also, do not forget to divide the second equation by 27 before solving.

Things to Be Careful About

When multiplying by 16, multiply every term. Equivalent forms such as $a+2b+12=0$ and $3a+b+1=0$ are acceptable. The final values must satisfy both original equations; checking by substitution is a good safeguard.

Techniques used

apply the factor theoremsubstitute factor roots into the polynomialsolve simultaneous linear equations

(b)

Hence factorise $p(x)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

With $a=2$ and $b=-7$ , multiply the known linear factors to obtain the quadratic divisor $(2x-1)(x-3)=2x^2-7x+3$ , then divide $p(x)$ by this quadratic to find the remaining factor.

Working

The product of the given factors is:

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

Divide $p(x)=2x^4-7x^3+13x^2-35x+15$ by $2x^2-7x+3$ :

\frac{2x^4 - 7x^3 + 13x^2 - 35x + 15}{2x^2 - 7x + 3} = x^2 + 5

Check by expanding:

(2x^2 - 7x + 3)(x^2 + 5) = 2x^4 - 7x^3 + 13x^2 - 35x + 15

Hence:

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 is $p(x)=2x^4-7x^3+13x^2-35x+15$ . The two known factors multiply to $(2x-1)(x-3)=2x^2-7x+3$ . Since this quadratic is a factor of $p(x)$ , divide $p(x)$ by it to find the remaining factor.

The division gives the quotient $x^2+5$ . To check, expand $(2x^2-7x+3)(x^2+5)$ :

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

This matches $p(x)$ , so the factorisation is correct.

Key Takeaways

Multiplying known linear factors gives a quadratic divisor. Polynomial division or inspection then reveals the remaining factor. Always verify by expanding the factorised form.

Common Mistakes

A common mistake is dividing by only one linear factor and stopping with a cubic factor. Another is losing the quadratic factor $x^2+5$ in the final answer. Sign errors can occur if the divisor is written incorrectly.

Things to Be Careful About

The quotient $x^2+5$ cannot be factorised further over the real numbers. The fully factorised form must show all three factors: $(2x-1)(x-3)(x^2+5)$ .

Techniques used

multiply linear factors to form a quadratic divisordivide the polynomial by the quadratic factorverify the factorisation by expansion

(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 the factorisation, the real zeros of $p$ are $x=\frac{1}{2}$ and $x=3$ , since $x^2+5$ has no real zeros. Therefore $p(\cot 2\theta)=0$ requires $\cot 2\theta=\frac{1}{2}$ or $\cot 2\theta=3$ . Solve each equation and choose the least positive value of $\theta$ .

Working

For $\cot 2\theta = 3$ :

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

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

where $k$ is an integer.

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

The least positive value is obtained with $k=0$ :

\theta = \frac{1}{2}\arctan\left(\frac{1}{3}\right) \approx 0.1609

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

\tan 2\theta = 2

\theta = \frac{1}{2}\arctan(2) + \frac{k\pi}{2}

The least positive value is:

\theta = \frac{1}{2}\arctan(2) \approx 0.5536

The smaller value comes from $\cot 2\theta = 3$ .

Answer

\theta \approx 0.161

That is, $\theta \approx 0.161$ radians.

Final answer

theta = 0.161 radians (0.1609...)

Detailed explanation

Walkthrough

From the factorised form, $p(x)=0$ when $x=\frac{1}{2}$ , $x=3$ , or $x^2+5=0$ . Since $x^2+5=0$ has no real solutions, the only real roots are $\frac{1}{2}$ and $3$ .

For $p(\cot 2\theta)=0$ , the value of $\cot 2\theta$ must be one of these roots. So solve $\cot 2\theta=3$ and $\cot 2\theta=\frac{1}{2}$ .

For $\cot 2\theta=3$ , use $\tan 2\theta=\frac{1}{3}$ . The principal solution is $2\theta=\arctan\left(\frac{1}{3}\right)$ , so $\theta=\frac{1}{2}\arctan\left(\frac{1}{3}\right)\approx0.1609$ radians. Adding multiples of $\pi$ to $2\theta$ gives larger positive values.

For $\cot 2\theta=\frac{1}{2}$ , $\tan 2\theta=2$ , so the least positive value is $\theta=\frac{1}{2}\arctan(2)\approx0.5536$ radians.

The smaller of these is $0.1609$ , so the least positive value is $\theta\approx0.161$ radians.

Key Takeaways

$p(\cot 2\theta)=0$ means that $\cot 2\theta$ must equal a real root of $p$ . A cotangent equation can be converted to a tangent equation by taking reciprocals. The tangent function has period $\pi$ , so the general solution includes $+k\pi$ ; choose the smallest positive value.

Common Mistakes

A common mistake is forgetting that $\cot 2\theta=\frac{1}{2}$ is also possible. Another is solving for $2\theta$ and forgetting to halve it to get $\theta$ . Also, giving the answer in degrees instead of radians loses the required form.

Things to Be Careful About

The question asks for radians, so the answer is approximately $0.161$ radians, not $0.322$ (which would be $2\theta$ ). Compare both possible roots to justify that $0.161$ is the least positive value. The general solution is $2\theta=\arctan\left(\frac{1}{3}\right)+k\pi$ ; with $k=0$ the smallest positive $\theta$ is obtained.

Techniques used

identify real roots from the factorised polynomialconvert a cotangent equation to a tangent equationsolve an inverse trigonometric equationcompare periodic solutions to select the least positive value

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 reduces to a quadratic in $x$ . Show that this quadratic has no real roots by evaluating its discriminant.

Working

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

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

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

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

Simplify:

-x^2 + 3 + 6x = 14

-x^2 + 6x - 11 = 0

The discriminant is:

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

Since $\Delta < 0$ , the quadratic has no real solutions for $x$ . Therefore no point on the curve has $y$ -coordinate $e^{-1}$ .

Answer

There is no real $x$ satisfying the equation when $y = e^{-1}$ , so no such point exists on the curve.

Final answer

No point exists on the curve with y-coordinate e^-1.

Detailed explanation

Walkthrough

The curve is given implicitly. To test whether a point with $y$ -coordinate $e^{-1}$ exists, substitute $y = e^{-1}$ into the equation. Because $\ln$ and $e$ are inverse functions, $\ln(e^{-1}) = -1$ . This turns the equation into a quadratic in $x$ . For a point to exist, this quadratic must have at least one real solution for $x$ . We check the discriminant: if it is negative, no real $x$ satisfies the equation, so no such point lies on the curve.

After substitution, the equation becomes $-x^2 + 6x - 11 = 0$ . Its discriminant is $\Delta = 36 - 44 = -8$ , which is negative. Therefore the quadratic has no real roots, and the required conclusion follows.

Key Takeaways

$\ln(e^a) = a$ for any real $a$ .
A point on a curve exists only if the corresponding $x$ -value is real.
A negative discriminant means a quadratic equation has no real solutions.

Common Mistakes

Forgetting that $\ln(e^{-1}) = -1$ , or incorrectly writing it as $e^{-1}$ .
Making a sign error when expanding $-(x^2 - 3)$ ; the correct expansion is $-x^2 + 3$ .
Not showing the discriminant calculation. This part is an AG (answer given) result, so necessary detail must be shown.

Things to Be Careful About

The quadratic can be written as $-x^2 + 6x - 11 = 0$ , so $a = -1$ , $b = 6$ , $c = -11$ .
Alternatively, completing the square gives $(x - 3)^2 + 2 = 0$ , which also shows there are no real roots.
State clearly that a negative discriminant means no real $x$ exists.

Techniques used

substitute a given y-value into an implicit equationuse the inverse relationship between ln and eform a quadratic equationevaluate 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 $x = 2$ and $y = e^2$ to find $\frac{dy}{dx}$ , then use the point-gradient form of a line to write the tangent.

Working

Differentiate both sides with respect to $x$ :

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

Using the product rule on $(x^2 - 3)\ln y$ :

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

At the point $(2, e^2)$ , $x = 2$ , $y = e^2$ and $\ln y = 2$ . Substituting:

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

So the gradient of the tangent is $-14e^2$ . The tangent passes through $(2, e^2)$ :

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

Differentiate both sides of the curve equation with respect to $x$ . The term $(x^2 - 3)\ln y$ is a product of $x^2 - 3$ and $\ln y$ , so it requires the product rule. The derivative of $x^2 - 3$ is $2x$ , and the derivative of $\ln y$ is $\frac{1}{y}\frac{dy}{dx}$ because $y$ depends on $x$ . The derivative of $6x$ is $6$ , and the derivative of the constant $14$ is $0$ .

This gives $2x\ln y + \frac{x^2 - 3}{y}\frac{dy}{dx} + 6 = 0$ . To find the gradient at the given point, substitute $x = 2$ and $y = e^2$ . Since $\ln(e^2) = 2$ , the equation becomes $8 + \frac{1}{e^2}\frac{dy}{dx} + 6 = 0$ , so $\frac{dy}{dx} = -14e^2$ .

Finally, use the point-gradient form $y - y_1 = m(x - x_1)$ with $m = -14e^2$ and $(x_1, y_1) = (2, e^2)$ . Expanding gives the tangent in the required form $y = -14e^2 x + 29e^2$ .

Key Takeaways

In implicit differentiation, every term containing $y$ is differentiated with an extra factor $\frac{dy}{dx}$ .
The product rule is needed when a term is a product of a function of $x$ and a function of $y$ .
The gradient of a tangent at a point is the value of $\frac{dy}{dx}$ at that point.
The equation of a line can be written using $y - y_1 = m(x - x_1)$ .

Common Mistakes

Forgetting to include $\frac{dy}{dx}$ when differentiating $\ln y$ .
Omitting the $+6$ term when differentiating $6x$ .
Making a sign error when solving for $\frac{dy}{dx}$ .
Not substituting both $x = 2$ and $y = e^2$ ; the mark scheme requires an attempt at substitution.
Using an approximate value such as $-103$ instead of the exact value $-14e^2$ .

Things to Be Careful About

At $(2, e^2)$ , $\ln y = 2$ and $x^2 - 3 = 1$ .
The final answer must be in the form $y = mx + c$ with exact constants.
When expanding $-14e^2(x - 2)$ , the constant term is $+28e^2$ ; adding $e^2$ gives $29e^2$ .

Techniques used

differentiate implicitly using the product rulesolve for dy/dxsubstitute a point to find the gradientwrite 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, then apply the double angle formulae to rewrite the expression in terms of $\sin 2\theta$ and $\cos 2\theta$ . Finally, express $a\sin 2\theta + b\cos 2\theta$ in the form $R\cos(2\theta - \alpha)$ by comparing coefficients.

Working

Expand $\sin(\theta + 30°)$ using the compound angle formula:

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

Substitute into the given expression:

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 formulae $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$ :

2\sqrt{3}\cos\theta\sin\theta + 2\cos^2\theta = \sqrt{3}\sin 2\theta + (\cos 2\theta + 1)

= \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 + R\sin 2\theta\sin\alpha

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

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

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

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

Therefore:

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

This problem asks us to transform $4\cos\theta\sin(\theta + 30°)$ into the form $R\cos(2\theta - \alpha) + k$ . The target form involves $2\theta$ , but we start with $\theta$ and $\theta + 30°$ . This tells us we need to use the double angle formulae at some point.

Step 1 — Expand the compound angle. The expression contains $\sin(\theta + 30°)$ . We use the compound angle formula $\sin(A + B) = \sin A\cos B + \cos A\sin B$ with $A = \theta$ and $B = 30°$ . This gives $\sin(\theta + 30°) = \frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta$ . Multiplying by $4\cos\theta$ distributes the factor and gives $2\sqrt{3}\cos\theta\sin\theta + 2\cos^2\theta$ .

Step 2 — Convert to double angles. Now we have terms involving $\sin\theta\cos\theta$ and $\cos^2\theta$ . The double angle formulae tell us $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$ . So $\sin\theta\cos\theta = \frac{1}{2}\sin 2\theta$ and $\cos^2\theta = \frac{1}{2}(\cos 2\theta + 1)$ . Substituting: $2\sqrt{3} \cdot \frac{1}{2}\sin 2\theta + 2 \cdot \frac{1}{2}(\cos 2\theta + 1) = \sqrt{3}\sin 2\theta + \cos 2\theta + 1$ .

Step 3 — R-form. We now have $\sqrt{3}\sin 2\theta + \cos 2\theta + 1$ . The constant $+1$ is our $k$ . We need to write $\sqrt{3}\sin 2\theta + \cos 2\theta$ as $R\cos(2\theta - \alpha)$ . Expanding: $R\cos(2\theta - \alpha) = R(\cos 2\theta\cos\alpha + \sin 2\theta\sin\alpha) = R\cos\alpha\cos 2\theta + R\sin\alpha\sin 2\theta$ . Comparing with $\sqrt{3}\sin 2\theta + \cos 2\theta$ , we get $R\sin\alpha = \sqrt{3}$ and $R\cos\alpha = 1$ . Squaring and adding: $R^2 = 3 + 1 = 4$ , so $R = 2$ . Dividing: $\tan\alpha = \sqrt{3}$ , so $\alpha = 60°$ (within the given range $0° < \alpha < 90°$ ).

Key Takeaways

The compound angle formula $\sin(A + B) = \sin A\cos B + \cos A\sin B$ allows us to expand expressions like $\sin(\theta + 30°)$ .
The double angle formulae $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$ convert products of $\sin\theta$ and $\cos\theta$ into functions of $2\theta$ .
The R-form $R\cos(2\theta - \alpha)$ is found by expanding and comparing coefficients with $a\sin 2\theta + b\cos 2\theta$ , giving $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \frac{a}{b}$ (where $a$ is the coefficient of $\sin 2\theta$ and $b$ is the coefficient of $\cos 2\theta$ ).

Common Mistakes

Incorrectly applying the double angle formula: $\cos^2\theta = \frac{1}{2}(\cos 2\theta + 1)$ , not $\frac{1}{2}\cos 2\theta$ .
Mixing up which coefficient is $R\sin\alpha$ and which is $R\cos\alpha$ — this leads to the wrong value of $\alpha$ .
Forgetting the constant $+1$ in the final answer.
Taking the negative square root for $R$ (the problem states $R > 0$ ).

Things to Be Careful About

The domain $0° < \alpha < 90°$ ensures $\alpha = 60°$ is the correct choice (not $240°$ or $-300°$ ).
The mark scheme requires exact coefficients at each stage — do not round $\sqrt{3}$ to a decimal.
The "FT" (follow through) marks in the mark scheme mean that even if you make an earlier error, you can still earn later marks if your method is correct.

Techniques used

expand compound angle sineapply double angle formulae to convert to 2θ termsexpress a sin 2θ + b cos 2θ in R cos(2θ − α) form by comparing coefficients

(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 coefficient 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

Multiply both sides by 3:

12\cos(2\phi)\sin(2\phi + 30°) = 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 $0° < \phi < 90°$ gives $0° < 4\phi < 360°$ , so $-60° < 4\phi - 60° < 300°$ .

\cos^{-1}\left(\frac{1}{3}\right) \approx 70.5288°

First solution:

4\phi - 60° = 70.5288°

4\phi = 130.5288°

\phi = 32.6322° \approx 32.6°

Second solution, using the symmetry of cosine about $0°$ :

4\phi - 60° = 360° - 70.5288° = 289.4712°

4\phi = 349.4712°

\phi = 87.3678° \approx 87.4°

Both values lie within $0° < \phi < 90°$ . Since $4\phi - 60°$ spans an interval of width $360°$ (exactly one full period of cosine), there are no other solutions.

Answer

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

Final answer

φ = 32.6° or φ = 87.4°

Detailed explanation

Walkthrough

This part asks us to use the result from part (a) to solve a trigonometric equation.

Step 1 — Apply the part (a) result. The expression $12\cos 2\phi\sin(2\phi + 30°)$ is exactly 3 times the expression from part (a) with $\theta$ replaced by $2\phi$ . From part (a), $4\cos\theta\sin(\theta + 30°) = 2\cos(2\theta - 60°) + 1$ . Substituting $\theta = 2\phi$ : $4\cos(2\phi)\sin(2\phi + 30°) = 2\cos(4\phi - 60°) + 1$ .

Step 2 — Scale and solve. Multiplying by 3: $12\cos(2\phi)\sin(2\phi + 30°) = 6\cos(4\phi - 60°) + 3$ . Setting this equal to 5: $6\cos(4\phi - 60°) + 3 = 5$ , so $\cos(4\phi - 60°) = \frac{1}{3}$ .

Step 3 — Find solutions. The domain $0° < \phi < 90°$ means $0° < 4\phi < 360°$ , so $-60° < 4\phi - 60° < 300°$ . We need $\cos x = \frac{1}{3}$ where $x = 4\phi - 60°$ . The principal value is $\cos^{-1}(\frac{1}{3}) \approx 70.5288°$ . The two solutions in our range are $x = 70.5288°$ (first quadrant) and $x = 360° - 70.5288° = 289.4712°$ (fourth quadrant, where cosine is also positive). Converting back: $\phi = \frac{70.5288° + 60°}{4} = 32.6322°$ and $\phi = \frac{289.4712° + 60°}{4} = 87.3678°$ .

Step 4 — Verify the domain. Both $32.6°$ and $87.4°$ lie within $0° < \phi < 90°$ . Since the range of $4\phi - 60°$ spans exactly one full period of cosine (360°), there are exactly two solutions and no others.

Key Takeaways

The result of part (a) can be reused by substitution — this is a common "hence" question pattern.
Solving $\cos x = k$ requires finding ALL solutions in the relevant range, using the symmetry $\cos x = \cos(360° - x)$ .
The domain of the original variable must be translated into a domain for the transformed variable.

Common Mistakes

Forgetting to multiply by 3 when applying the part (a) result.
Only finding the principal value solution and missing the second one.
Not checking whether the solutions fall within the given domain.
Using degrees vs radians inconsistently (here everything is in degrees).

Things to Be Careful About

The mark scheme states "no others between 0 and 90" — you must verify there are exactly two solutions in the domain.
The range of $4\phi - 60°$ is $(-60°, 300°)$ — this is exactly one full period, so there are exactly two solutions of $\cos x = \frac{1}{3}$ in this range.
The final answers should be given to an appropriate degree of accuracy (the mark scheme accepts 32.6 and 87.4, or greater accuracy).

Techniques used

substitute the part (a) identity with θ = 2φsolve trigonometric equation for the transformed anglecheck all solutions against the given domain

Mathematics 9709/25 — May/June 2025

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