Let $f(x) = 8x^3 + 54x^2 - 17x - 21$. (a) Show that $x + 7$ is a factor of $f(x)$. (b) Find the quotient when $f(x)$ is divided by $x + 7$. (c) Hence solve the equation $$ 8\cos^3\theta + 54\cos^2\theta - 17\cos\theta - 21 = 0 $$ for $0^\circ \leq \theta \leq 360^\circ$.

(a) x + 7 is a factor of f(x) · (b) 8x^2 - 2x - 3 · (c) θ = 41.4°, 120°, 240°, 318.6° (accept 318.6° or 319°)

(a) Express $3\cos 2x - \sqrt{3}\sin 2x$ in the form $R\cos(2x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$. Give the exact values of $R$ and $\alpha$. (b) Hence find the exact value of $$ \int_0^{\frac{1}{12}\pi} \frac{3}{(3\cos 2x - \sqrt{3}\sin 2x)^2}\,dx $$ simplifying your answer.

(a) R = 2√3, α = π/6 · (b) √3/12

9709/33 May/June 2024 — Mathematics Solved Past Paper

Q14MMedium-EasyLogarithmic and Exponential Functions

Solve the equation $8^{3-6x} = 4 \times 5^{-2x}$ . Give your answer correct to 3 decimal places.

DifficultyMedium-Easy

Worked solution

Approach

Take natural logarithms of both sides to bring the variable exponents down, then use the laws of logarithms to obtain a linear equation in $x$ and solve it.

Working

Take logs of both sides:

\ln\left(8^{3-6x}\right) = \ln\left(4 \times 5^{-2x}\right)

Using the power law on the left and the product law on the right:

(3-6x)\ln 8 = \ln 4 + \ln\left(5^{-2x}\right)

(3-6x)\ln 8 = \ln 4 - 2x \ln 5

Expand the left-hand side:

3\ln 8 - 6x\ln 8 = \ln 4 - 2x\ln 5

Collect the $x$ -terms on one side:

-6x\ln 8 + 2x\ln 5 = \ln 4 - 3\ln 8

x\left(2\ln 5 - 6\ln 8\right) = \ln 4 - 3\ln 8

Solve for $x$ :

x = \frac{\ln 4 - 3\ln 8}{2\ln 5 - 6\ln 8}

x = 0.524096\ldots

Answer

x = 0.524 \quad (3\text{ d.p.})

Final answer

x = 0.524

Detailed explanation

Walkthrough

We start with the equation $8^{3-6x} = 4 \times 5^{-2x}$ . The unknown $x$ appears in the exponents, so taking logarithms of both sides is the natural first step: it converts powers into products and brings the exponents down.

Taking natural logs gives

\ln\left(8^{3-6x}\right) = \ln\left(4 \times 5^{-2x}\right).

On the left, the power law $\ln(a^b) = b\ln a$ gives $(3-6x)\ln 8$ . On the right, the product law $\ln(ab) = \ln a + \ln b$ gives $\ln 4 + \ln\left(5^{-2x}\right)$ , and then the power law gives $\ln 4 - 2x\ln 5$ .

The equation is now linear in $x$ :

(3-6x)\ln 8 = \ln 4 - 2x\ln 5.

Expanding the left-hand side and collecting all $x$ -terms on one side gives

x\left(2\ln 5 - 6\ln 8\right) = \ln 4 - 3\ln 8.

Finally, dividing by the coefficient of $x$ gives $x = 0.524096\ldots$ , which rounds to $0.524$ to 3 decimal places.

Key Takeaways

This question tests the ability to solve an equation with the unknown in the exponent. The key idea is to take logarithms of both sides and then apply the laws of logarithms: the power law to bring exponents down, and the product law to split a product inside a logarithm. Once the equation is linear in $x$ , standard algebraic rearrangement completes the solution.

Common Mistakes

Forgetting to take logs of the whole right-hand side, e.g. writing $\ln\left(4 \times 5^{-2x}\right)$ incorrectly as $\ln 4 \times \ln\left(5^{-2x}\right)$ instead of $\ln 4 + \ln\left(5^{-2x}\right)$ .
Failing to apply the power law correctly when the exponent contains $x$ , such as writing $\ln\left(5^{-2x}\right) = 5^{-2x}\ln$ or missing the minus sign.
Not showing working: the mark scheme states that no working scores 0 marks, and an unsupported final answer cannot receive full credit.
Rounding too early, which can change the final decimal answer.

Things to Be Careful About

The answer must be given correct to 3 decimal places, so keep sufficient precision during the calculation and round only at the end.
Any base of logarithm is acceptable, but natural logs are usually the most convenient. If decimals are used, small errors in the second and third decimal places may be tolerated in intermediate working, but the final answer must still be $0.524$ .
Be careful with signs when collecting $x$ -terms: the term $-2x\ln 5$ moves across the equation as $+2x\ln 5$ .

Techniques used

take logarithms of both sidesapply the power law for logarithmsapply the product law for logarithmssolve the resulting linear equation

Similar questions

Q25MMediumDifferentiation

Find the exact coordinates of the stationary point of the curve $y = e^{2x} \sin 2x$ for $0 \leq x \leq \frac{1}{2}\pi$ .

DifficultyMedium

Worked solution

Approach

Use the product rule to differentiate $y = e^{2x}\sin 2x$ . Since $e^{2x}$ is never zero, set the derivative equal to zero and solve the resulting trigonometric equation for $x$ in the given interval. Then substitute back into $y$ to find the exact $y$ -coordinate.

Working

\frac{dy}{dx} = 2e^{2x}\sin 2x + e^{2x}(2\cos 2x)

\frac{dy}{dx} = 2e^{2x}(\sin 2x + \cos 2x)

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

2e^{2x}(\sin 2x + \cos 2x) =0

Since $e^{2x} >0$ for all $x$ , this gives

\sin 2x + \cos 2x =0

\tan 2x = -1

For $0 \leq x \leq \frac{1}{2}\pi$ , we have $0 \leq 2x \leq \pi$ . In this interval, $\tan \theta = -1$ when $\theta = \frac{3\pi}{4}$ . Thus

2x = \frac{3\pi}{4} \quad\Rightarrow\quad x = \frac{3\pi}{8}

Now substitute into $y$ :

y = e^{2\cdot \frac{3\pi}{8}}\sin\left(2\cdot \frac{3\pi}{8}\right)

y = e^{\frac{3\pi}{4}}\sin\left(\frac{3\pi}{4}\right) = e^{\frac{3\pi}{4}}\cdot \frac{\sqrt{2}}{2}

Answer

\left(\frac{3\pi}{8},\frac{\sqrt{2}}{2}e^{\frac{3\pi}{4}}\right)

Final answer

x = 3π/8, y = (√2/2)e^(3π/4)

Detailed explanation

Walkthrough

Start by differentiating $y = e^{2x}\sin 2x$ . The product rule says that $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$ . Here $u = e^{2x}$ and $v = \sin 2x$ . The derivative of $e^{2x}$ is $2e^{2x}$ by the chain rule, and the derivative of $\sin 2x$ is $2\cos 2x$ .

At a stationary point the gradient is zero, so set $\frac{dy}{dx} =0$ . Because $e^{2x}$ is never zero, divide both sides by $2e^{2x}$ . This gives $\sin 2x + \cos 2x =0$ , or $\tan 2x = -1$ .

Now solve $\tan 2x = -1$ on the interval $0 \leq x \leq \frac{\pi}{2}$ . Multiplying by 2 gives $0 \leq 2x \leq \pi$ . On this interval tangent is negative only in the second quadrant, so $2x = \frac{3\pi}{4}$ , hence $x = \frac{3\pi}{8}$ .

Finally substitute this $x$ into the original equation. Since $2x = \frac{3\pi}{4}$ , we get $y = e^{\frac{3\pi}{4}}\sin\frac{3\pi}{4}$ . The exact value $\sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}$ , so $y = \frac{\sqrt{2}}{2}e^{\frac{3\pi}{4}}$ .

Key Takeaways

The product rule and chain rule are both needed when differentiating a product of composite functions. A stationary point is found by solving $\frac{dy}{dx} =0$ ; an exponential factor is never zero and can be divided out. Exact stationary-point coordinates require substituting back into the original function and simplifying using exact trig values.

Common Mistakes

Forgetting the factor 2 in the derivative of $e^{2x}$ or $\sin 2x$ ; this loses the A1 for the derivative. Solving $\tan 2x = -1$ incorrectly:the angle must be $\frac{3\pi}{4}$ , not $\frac{\pi}{4}$ or $-\frac{\pi}{4}$ , because the interval for $2x$ is $[0,\pi]$ . Using degrees for the final exact answer: $67.5^\circ$ is not accepted;the A1 requires the exact radian value. Substituting the wrong $x$ or ignoring the interval can produce extra values;answers outside $0 \leq x \leq \frac{\pi}{2}$ should be ignored. Not simplifying $\sin\frac{3\pi}{4}$ to $\frac{\sqrt{2}}{2}$ loses the final exact form.

Things to Be Careful About

The domain restriction is essential: $2x$ ranges from $0$ to $\pi$ , so there is exactly one solution to $\tan 2x = -1$ . If the interval were different, there could be more solutions. The derivative factor $2e^{2x}$ is positive for all real $x$ , so it cannot cause the derivative to be zero. Mark scheme guidance: CWO (correct working only) and ISW (ignore subsequent working) apply;unsupported answers are not enough. The exact coordinate must be in radians and in simplified form.

Techniques used

apply the product ruleuse the chain rule for composite functionsset the derivative equal to zero and solve a trigonometric equationevaluate the original function at the stationary point

Similar questions

Q35MMedium-HardComplex Numbers

The square roots of $24 - 7i$ can be expressed in the Cartesian form $x + iy$ , where $x$ and $y$ are real and exact.

By first forming a quartic equation in $x$ or $y$ , find the square roots of $24 - 7i$ in exact Cartesian form.

DifficultyMedium-Hard

Worked solution

Approach

Let a square root be $x + iy$ . Square it and equate the real and imaginary parts to $24$ and $-7$ . This gives two equations. Eliminate $y$ to obtain a quartic in $x$ , then reduce it to a quadratic by substituting $u = x^2$ . Finally use $2xy = -7$ to determine the correct signs.

Working

Let $z = x + iy$ . Then

z^2 = (x+iy)^2 = x^2 - y^2 + 2xyi.

Since $z^2 = 24 - 7i$ , equating real and imaginary parts gives

x^2 - y^2 = 24, \qquad 2xy = -7.

From $2xy = -7$ , $y = -\frac{7}{2x}$ . Substituting into $x^2 - y^2 = 24$ :

x^2 - \frac{49}{4x^2} = 24.

Multiply by $4x^2$ :

4x^4 - 49 = 96x^2,

so

4x^4 - 96x^2 - 49 = 0.

Let $u = x^2$ . Then

4u^2 - 96u - 49 = 0.

Using the quadratic formula,

u = \frac{96 \pm \sqrt{96^2 + 4(4)(49)}}{8} = \frac{96 \pm 100}{8}.

Thus $u = \frac{49}{2}$ or $u = -\frac{1}{2}$ . Since $u = x^2 \ge 0$ ,

x^2 = \frac{49}{2}, \qquad x = \pm \frac{7\sqrt{2}}{2}.

Now use $2xy = -7$ . If $x = \frac{7\sqrt{2}}{2}$ , then

y = -\frac{7}{2x} = -\frac{7}{7\sqrt{2}} = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}.

If $x = -\frac{7\sqrt{2}}{2}$ , then $y = \frac{\sqrt{2}}{2}$ .

Answer

\frac{7\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \quad \text{and} \quad -\frac{7\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i.

Final answer

x = 7√2/2, y = -√2/2 and x = -7√2/2, y = √2/2

Detailed explanation

Walkthrough

We want numbers $z$ such that $z^2 = 24 - 7i$ . Write $z = x + iy$ . Squaring gives $x^2 - y^2 + 2xyi$ . The real part must be $24$ and the imaginary part must be $-7$ , so we get $x^2 - y^2 = 24$ and $2xy = -7$ . The second equation tells us $x$ and $y$ have opposite signs, because their product is negative.

To remove $y$ , solve $y = -\frac{7}{2x}$ and substitute. This gives $4x^4 - 96x^2 - 49 = 0$ . Although this is a quartic, it is quadratic in $x^2$ , so set $u = x^2$ . Solving gives $u = \frac{49}{2}$ or $u = -\frac{1}{2}$ . Since $u$ is a square, only $u = \frac{49}{2}$ is possible, so $x = \pm \frac{7\sqrt{2}}{2}$ .

Finally, use $2xy = -7$ to find which $y$ goes with each $x$ . For positive $x$ , $y$ is negative; for negative $x$ , $y$ is positive. This produces exactly two square roots.

Key Takeaways

To find square roots of a complex number, write the unknown as $x+iy$ and equate real and imaginary parts after squaring.
The equation $2xy = -7$ not only links $x$ and $y$ but also fixes the sign pairing.
A quartic that contains only even powers can be solved as a quadratic in $x^2$ .
Only real values of $x$ and $y$ are allowed, so discard any squared value that is negative.

Common Mistakes

Forgetting that $i^2 = -1$ ; this would give the wrong real part.
Equating only the real parts or only the imaginary parts.
Multiplying by $x^2$ without noting that $x \neq 0$ ; here $2xy = -7$ guarantees this.
Taking all four combinations of $x = \pm a$ and $y = \pm b$ ; the condition $2xy = -7$ allows only two.
Writing $\pm\left(\frac{7\sqrt{2}}{2} \mp \frac{\sqrt{2}}{2}i\right)$ in a way that hides the correct pairing, or failing to simplify $\sqrt{49}$ .

Things to Be Careful About

The signs of $x$ and $y$ must be opposite because $2xy = -7$ .
When solving the quadratic in $u = x^2$ , discard the negative root.
The final answer should be in exact Cartesian form with simplified surds.
If you use the quartic in $y$ instead, $4y^4 + 96y^2 - 49 = 0$ , the same sign-pairing condition must be applied.
Check that the modulus of each square root is $5$ , since $|24 - 7i| = 25$ .

Techniques used

square a complex number in Cartesian formequate real and imaginary partseliminate one variable to form a quarticsubstitute u = x^2 to solve a quadraticpair signs using the product condition

Similar questions

Q44MMedium-EasyLogarithmic and Exponential Functions

Fig. 1

The variables $x$ and $y$ satisfy the equation $ky = e^{cx}$ , where $k$ and $c$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $(2.80, 0.372)$ and $(5.10, 2.21)$ , as shown in the diagram.

Find the values of $k$ and $c$ . Give each value correct to 2 significant figures.

DifficultyMedium-Easy

Worked solution

Approach

Take the natural logarithm of both sides of $ky = e^{cx}$ to linearise the relationship, giving $\ln y = cx - \ln k$ . This is a straight line with gradient $c$ and y-intercept $-\ln k$ . Use the two given points to find the gradient $c$ , then substitute back to find $k$ .

Working

Take natural logarithms of both sides:

\ln(ky) = \ln(e^{cx})

\ln k + \ln y = cx

Rearrange into straight-line form:

\ln y = cx - \ln k

This is a straight line equation $Y = mX + b$ with $Y = \ln y$ , $X = x$ , gradient $m = c$ , and y-intercept $b = -\ln k$ .

The gradient $c$ is found from the two points $(2.80, 0.372)$ and $(5.10, 2.21)$ :

c = \frac{2.21 - 0.372}{5.10 - 2.80} = \frac{1.838}{2.30} = 0.79913\ldots

Rounding to 2 significant figures:

c = 0.80

Substitute $c = 0.80$ and the point $(2.80, 0.372)$ into $\ln y = cx - \ln k$ :

0.372 = 0.80 \times 2.80 - \ln k

0.372 = 2.24 - \ln k

\ln k = 2.24 - 0.372 = 1.868

Therefore:

k = e^{1.868} = 6.475\ldots

Rounding to 2 significant figures:

k = 6.5

Answer

c = 0.80, \quad k = 6.5

Final answer

c = 0.80, k = 6.5

Detailed explanation

Walkthrough

First, we linearise the equation $ky = e^{cx}$ by taking natural logarithms of both sides. This gives $\ln k + \ln y = cx$ , which rearranges to $\ln y = cx - \ln k$ . This is now in the form of a straight line $Y = mX + b$ where the vertical axis is $\ln y$ , the horizontal axis is $x$ , the gradient is $c$ , and the y-intercept is $-\ln k$ .

Since the graph of $\ln y$ against $x$ is a straight line passing through $(2.80, 0.372)$ and $(5.10, 2.21)$ , we can calculate the gradient directly using the standard gradient formula:

c = \frac{2.21 - 0.372}{5.10 - 2.80} = \frac{1.838}{2.30} \approx 0.80

With $c$ known, we substitute one of the points (say $(2.80, 0.372)$ ) and the gradient into the straight-line equation to find the y-intercept:

0.372 = 0.80 \times 2.80 - \ln k

\ln k = 2.24 - 0.372 = 1.868

Finally, we exponentiate to find $k$ :

k = e^{1.868} \approx 6.5

Key Takeaways

Taking logarithms of both sides of an exponential equation can transform it into a linear form suitable for straight-line graph analysis.
The gradient and y-intercept of the linearised graph directly correspond to the constants in the original equation.
Always round final answers to the specified number of significant figures.

Common Mistakes

Forgetting to take the logarithm of both sides, or only taking the logarithm of one side.
Incorrectly rearranging $\ln k + \ln y = cx$ ; the y-intercept is $-\ln k$ , not $\ln k$ .
Using the wrong points or swapping the numerator and denominator when calculating the gradient.
Rounding intermediate values too early, which can cause the final answer for $k$ to be slightly off.
Giving $k$ or $c$ as fractions instead of decimal values to the required significant figures.

Things to Be Careful About

The mark scheme accepts $c = 0.80$ or $0.8$ , and $k = 6.5$ ; these must be given as decimals to 2 significant figures, not as fractions.
When using the gradient to find $k$ , using a rounded value of $c = 0.80$ is acceptable (AWRT), but using the exact gradient $1.838/2.30$ gives a more precise $\ln k \approx 1.870$ , still yielding $k \approx 6.5$ .
The y-intercept of the line $\ln y = cx - \ln k$ is $-\ln k$ , so $\ln k = -b$ where $b$ is the y-intercept. A sign error here is a common mistake.
Always verify the answer by substituting the second point $(5.10, 2.21)$ : $0.80 \times 5.10 - \ln 6.5 \approx 4.08 - 1.872 = 2.208 \approx 2.21$ , confirming consistency.

Techniques used

take natural logarithm of both sidesrearrange into straight line formcalculate gradient from two pointssubstitute to find intercept and hence k

Similar questions

Q55MMediumAlgebra

Express $\frac{6x^2 - 2x + 2}{(x - 1)(2x + 1)}$ in partial fractions.

DifficultyMedium

Worked solution

Approach

Since the numerator and denominator have the same degree, the partial fraction form must include a constant term:

\frac{6x^2 - 2x + 2}{(x - 1)(2x + 1)} = A + \frac{B}{x - 1} + \frac{C}{2x + 1}.

Find $A$ by division or by comparing coefficients, then find $B$ and $C$ by substituting the roots of the denominator.

Working

Multiply both sides by $(x - 1)(2x + 1)$ :

6x^2 - 2x + 2 = A(x - 1)(2x + 1) + B(2x + 1) + C(x - 1).

Comparing the $x^2$ coefficients gives $6 = 2A$ , so $A = 3$ . With $A = 3$ :

6x^2 - 2x + 2 = 3(x - 1)(2x + 1) + B(2x + 1) + C(x - 1).

Subtracting $3(x - 1)(2x + 1) = 6x^2 - 3x - 3$ from both sides:

x + 5 = B(2x + 1) + C(x - 1).

Substitute $x = 1$ : $6 = 3B$ , so $B = 2$ .

Substitute $x = -\frac{1}{2}$ : $\frac{9}{2} = -\frac{3}{2}C$ , so $C = -3$ .

Answer

\frac{6x^2 - 2x + 2}{(x - 1)(2x + 1)} = 3 + \frac{2}{x - 1} - \frac{3}{2x + 1}.

Final answer

3 + 2/(x - 1) - 3/(2x + 1)

Detailed explanation

Walkthrough

First notice that the numerator and denominator have the same degree. In partial fractions, when the numerator degree is at least the denominator degree, there must be a polynomial part. Here that polynomial part is a constant, so we write

\frac{6x^2 - 2x + 2}{(x - 1)(2x + 1)} = A + \frac{B}{x - 1} + \frac{C}{2x + 1}.

Multiplying by the denominator clears the fractions. To find $A$ , compare the coefficients of $x^2$ : the left side has $6x^2$ , and on the right only the $A$ term contributes $2Ax^2$ . Hence $A = 3$ .

Next, subtract $3(x - 1)(2x + 1)$ from the numerator. This leaves $x + 5$ , so we only need to decompose

\frac{x + 5}{(x - 1)(2x + 1)} = \frac{B}{x - 1} + \frac{C}{2x + 1}.

Substitute $x = 1$ to make the $C$ term vanish; this gives $B = 2$ . Substitute $x = -\frac{1}{2}$ to make the $B$ term vanish; this gives $C = -3$ . Finally, combine the constant and the two fractions to write the answer.

Key Takeaways

When the numerator and denominator have equal degree, the partial fraction decomposition must include a constant term.
Substituting the roots of the denominator is an efficient way to find the constants one at a time.
The final answer can be checked by combining the fractions back over the common denominator.

Common Mistakes

Omitting the constant term $A$ and trying to write only $\frac{B}{x-1} + \frac{C}{2x+1}$ .
Making a sign error when substituting $x = -\frac{1}{2}$ , especially with $C(x - 1)$ .
Not showing the initial form; the mark scheme awards a method mark for stating or implying the correct form.
Forgetting to multiply the whole right-hand side by the denominator when clearing fractions.

Things to Be Careful About

At $x = -\frac{1}{2}$ , the factor $x - 1$ equals $-\frac{3}{2}$ , so the resulting equation gives a negative value for $C$ .
If using polynomial division instead, the remainder must be $x + 5$ ; any other remainder indicates an arithmetic error.
The two linear factors are distinct, so no repeated-factor terms such as $\frac{D}{(x-1)^2}$ are needed.

Techniques used

recognise improper fraction and include a constant termdivide numerator by denominator or compare coefficientssubstitute denominator roots to find partial fraction constants

Similar questions

Q6Medium-HardComplex Numbers

(a)

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

5M Similar

DifficultyMedium-Easy

Worked solution

Approach

Interpret each inequality geometrically on the Argand diagram. The modulus inequality defines a circular region, and the argument inequality defines a half-plane bounded by a half-line. Shade the intersection of these two regions.

Working

The first inequality is $|z - (4 + 3i)| \leq 2$ . This represents all points $z$ whose distance from the point $4 + 3i$ (coordinates $(4, 3)$ ) is less than or equal to 2. This is a closed disk with centre $(4, 3)$ and radius 2.

The second inequality is $\arg(z - (2 + i)) \geq \frac{\pi}{3}$ . The boundary is a half-line starting from the point $2 + i$ (coordinates $(2, 1)$ ) making an angle of $\frac{\pi}{3}$ with the positive real axis. The condition $\geq$ means the region lies to the left (or above) this half-line.

The required shaded region is the intersection of the interior of the circle and the half-plane above the half-line.

Answer

The shaded region is bounded by the circle with centre $(4, 3)$ and radius 2, and the half-line from $(2, 1)$ at angle $\frac{\pi}{3}$ , as shown in the diagram above.

Final answer

Shaded region inside the circle |z - (4 + 3i)| <= 2 and above the half-line from (2, 1) at angle pi/3.

Detailed explanation

Walkthrough

First, analyze the modulus inequality $|z - (4 + 3i)| \leq 2$ . By definition, $|z - z_0|$ is the distance between the point $z$ and the fixed point $z_0$ on the Argand diagram. Here, $z_0 = 4 + 3i$ , which corresponds to the coordinates $(4, 3)$ . The inequality states that this distance is at most 2, which describes a closed disk (a circle and its interior) with centre $(4, 3)$ and radius 2.

Next, analyze the argument inequality $\arg(z - (2 + i)) \geq \frac{\pi}{3}$ . The expression $z - (2 + i)$ represents the vector from the point $2 + i$ (coordinates $(2, 1)$ ) to $z$ . The argument is the angle this vector makes with the positive real axis. The boundary is a half-line originating at $(2, 1)$ and extending at an angle of $\frac{\pi}{3}$ (or $60^\circ$ ) to the positive real axis. The condition $\geq$ indicates that we want all points $z$ that make an angle greater than or equal to $\frac{\pi}{3}$ with the positive real axis relative to $(2, 1)$ , which is the region to the left (or above) this half-line.

Finally, shade the region that satisfies both conditions simultaneously. This is the part of the circular disk that lies above the half-line.

Key Takeaways

The locus $|z - z_0| = r$ is a circle with centre $z_0$ and radius $r$ . The inequality $\leq$ includes the interior.
The locus $\arg(z - z_0) = \theta$ is a half-line starting at $z_0$ at angle $\theta$ . The inequality $\geq$ gives the region to the left of this half-line.
Intersecting these loci gives the required shaded region.

Common Mistakes

Drawing a full line instead of a half-line for the argument boundary, or extending it in the wrong direction.
Shading the region below the half-line instead of above it.
Forgetting to include the boundary of the circle (the inequality is $\leq$ , not $<$ ).

Things to Be Careful About

Ensure the half-line starts exactly at $(2, 1)$ and not at the origin.
The angle $\frac{\pi}{3}$ is measured from the positive real axis, not from the vertical axis.
The mark scheme notes that a full circle is not strictly required, but the centre and relevant arc must be shown clearly.

Techniques used

interpret modulus inequality as a circular regioninterpret argument inequality as a half-plane bounded by a half-lineshade the intersection of the regions

(b)

Calculate the greatest value of $\arg z$ for points in this region.

2M Similar

DifficultyMedium-Hard

Worked solution

Approach

The argument $\arg z$ is the angle the vector from the origin to $z$ makes with the positive real axis. To maximize $\arg z$ for points in the shaded region, we find the line from the origin that is tangent to the circle and has the steepest gradient, then verify this tangent point lies within the shaded region.

Working

The centre of the circle is $C(4, 3)$ and its radius is $r = 2$ . The distance from the origin $O(0, 0)$ to $C$ is:

OC = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5

Let $\alpha$ be the angle of $OC$ with the positive real axis:

\tan \alpha = \frac{3}{4} \implies \alpha = \tan^{-1}\left(\frac{3}{4}\right)

Let $\beta$ be the angle between $OC$ and the tangent from the origin to the circle. From the right-angled triangle formed by the origin, the centre, and the point of tangency:

\sin \beta = \frac{r}{OC} = \frac{2}{5} \implies \beta = \sin^{-1}\left(\frac{2}{5}\right)

The greatest value of $\arg z$ is $\alpha + \beta$ :

\arg z_{\text{max}} = \tan^{-1}\left(\frac{3}{4}\right) + \sin^{-1}\left(\frac{2}{5}\right)

Calculating this value:

\alpha \approx 0.6435 \text{ rad}, \quad \beta \approx 0.4115 \text{ rad}

\arg z_{\text{max}} \approx 0.6435 + 0.4115 = 1.055 \text{ rad} \approx 60.5^\circ

We must verify that this tangent point lies in the shaded region. The half-line boundary is at angle $\frac{\pi}{3} \approx 60.0^\circ$ . Since $60.5^\circ > 60.0^\circ$ , the tangent point is above the half-line and thus lies within the shaded region.

Alternatively, using the discriminant method, let $y = kx$ . Substituting into $(x-4)^2 + (y-3)^2 = 4$ and setting the discriminant to 0 gives $\tan^{-1}\left(\frac{6 + \sqrt{21}}{6}\right) \approx 1.056$ rad.

Answer

1.06 \text{ rad} \text{ (or } 60.5^\circ\text{)}

Final answer

1.06 rad (or 60.5 degrees)

Detailed explanation

Walkthrough

To find the greatest value of $\arg z$ , we need to find the point in the shaded region that makes the largest angle with the positive real axis. Geometrically, this occurs at the point on the boundary of the region where a line from the origin is tangent to the circle, provided this tangent point is within the shaded area.

First, calculate the distance from the origin to the centre of the circle $C(4, 3)$ : $OC = \sqrt{4^2 + 3^2} = 5$ . The angle of $OC$ with the positive real axis is $\alpha = \tan^{-1}(3/4)$ .

The angle between $OC$ and the tangent line from the origin to the circle is $\beta$ , where $\sin \beta = \frac{\text{radius}}{OC} = \frac{2}{5}$ . Thus, $\beta = \sin^{-1}(2/5)$ .

The maximum argument is $\alpha + \beta = \tan^{-1}(3/4) + \sin^{-1}(2/5) \approx 1.055$ rad or $60.5^\circ$ .

Finally, check that this point is in the shaded region. The half-line boundary is at $\pi/3 \approx 60.0^\circ$ . Since $60.5^\circ > 60.0^\circ$ , the tangent point is above the half-line and inside the shaded region. Thus, the maximum argument is indeed $1.06$ rad.

Key Takeaways

The maximum argument of points on a circle from the origin is found using the tangent from the origin to the circle.
The angle can be found using right-angled triangle trigonometry: $\sin \beta = r/d$ where $d$ is the distance from the origin to the centre.
Always verify that the optimal point lies within any additional constraints (like the half-line boundary).

Common Mistakes

Assuming the maximum argument is simply the angle to the centre plus the angle to the top of the circle.
Forgetting to check whether the tangent point satisfies the argument inequality $\arg(z - 2 - i) \geq \pi/3$ .
Using the wrong trigonometric ratio (e.g., $\cos \beta = 2/5$ instead of $\sin \beta = 2/5$ ).

Things to Be Careful About

The answer must be in radians or degrees as appropriate; the mark scheme accepts 1.06, 1.055, or $60.5^\circ$ .
If the tangent point did not lie in the shaded region (e.g., if it were below the half-line), the maximum argument would occur at the intersection of the half-line and the circle, not at the tangent point.

Techniques used

find the tangent from the origin to the circleuse trigonometry to calculate the maximum argumentverify the tangent point lies in the shaded region

Q7MediumAlgebra

Let $f(x) = 8x^3 + 54x^2 - 17x - 21$ .

(a)

Show that $x + 7$ is a factor of $f(x)$ .

1M Similar

DifficultyEasy

Worked solution

Approach

Use the factor theorem: $x + 7$ is a factor of $f(x)$ if and only if $f(-7) = 0$ .

Working

Substitute $x = -7$ :

f(-7) = 8(-7)^3 + 54(-7)^2 - 17(-7) - 21 = -2744 + 2646 + 119 - 21 = 0

Since $f(-7) = 0$ , $x + 7$ is a factor of $f(x)$ .

Answer

$x + 7$ is a factor of $f(x)$ .

Final answer

x + 7 is a factor of f(x)

Detailed explanation

Walkthrough

The factor theorem says that for a polynomial $f(x)$ , if $f(a) = 0$ then $(x - a)$ is a factor. Here we want $x + 7$ as a factor, which corresponds to $a = -7$ because $x + 7 = x - (-7)$ . So substitute $x = -7$ into $f(x)$ and evaluate. If the result is $0$ , the factor theorem confirms that $x + 7$ is a factor. We compute each term: $8(-7)^3 = -2744$ , $54(-7)^2 = 2646$ , $-17(-7) = 119$ , and $-21$ . Their sum is $0$ . Since there are no arithmetic errors, $x + 7$ is a factor. Alternatively, long division would produce the quotient and a zero remainder.

Key Takeaways

The factor theorem connects roots to factors. To show that a linear factor $x - a$ exists, evaluate $f(a)$ and show that it is zero. This avoids doing a full polynomial division.

Common Mistakes

Arithmetic errors with negatives are common: $(-7)^3 = -343$ , so $8(-7)^3 = -2744$ ; $(-7)^2 = 49$ , so $54(-7)^2 = 2646$ ; and $-17(-7) = +119$ . Sign errors here are easy to make. The mark scheme says no errors are allowed in this substitution. If division is used instead, the quotient and remainder must be correct.

Things to Be Careful About

Remember that $x + 7$ corresponds to $a = -7$ , not $+7$ . Show $f(-7) = 0$ explicitly. If using division, the remainder must be $0$ .

Techniques used

apply the factor theoremevaluate polynomial at a root

(b)

Find the quotient when $f(x)$ is divided by $x + 7$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use polynomial division (or synthetic division) to divide $f(x)$ by $x + 7$ .

Working

Using synthetic division with $-7$ :

\begin{array}{c|cccc} -7 & 8 & 54 & -17 & -21 \\ & -56 & 14 & 21 \\ \hline & 8 & -2 & -3 & 0 \end{array}

The quotient is $8x^2 - 2x - 3$ and the remainder is $0$ .

Answer

8x^2 - 2x - 3

Final answer

8x^2 - 2x - 3

Detailed explanation

Walkthrough

We need the quotient when $f(x)$ is divided by $x + 7$ . Synthetic division with $-7$ is efficient. Bring down the leading coefficient $8$ . Multiply $8$ by $-7$ to get $-56$ , then add to $54$ to get $-2$ . Multiply $-2$ by $-7$ to get $14$ , then add to $-17$ to get $-3$ . Multiply $-3$ by $-7$ to get $21$ , then add to $-21$ to get $0$ . The bottom row gives coefficients $8$ , $-2$ , $-3$ and a remainder of $0$ , so the quotient is $8x^2 - 2x - 3$ . This agrees with the factorisation $f(x) = (x + 7)(8x^2 - 2x - 3)$ .

Key Takeaways

Synthetic division is a compact method for dividing by a linear factor $x - a$ . The quotient has degree one less than the original polynomial. A zero remainder confirms that the divisor is a factor.

Common Mistakes

Sign errors in the synthetic multiplication step are common, as is forgetting to bring down the first coefficient. Misplacing coefficients can also occur. The mark scheme allows partial credit for reaching $8x^2 \pm 2x$ or for finding $A = 8$ , $B = \pm 2$ , or $C = -3$ in the form $(x + 7)(Ax^2 + Bx + C)$ .

Things to Be Careful About

The quotient must be stated in part (b) even if the division was shown in part (a). If using coefficient comparison, make sure every coefficient of the cubic is matched correctly.

Techniques used

divide polynomial by a linear factoruse synthetic division

(c)

Hence solve the equation

8\cos^3\theta + 54\cos^2\theta - 17\cos\theta - 21 = 0

for $0^\circ \leq \theta \leq 360^\circ$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the factorisation from part (b): $f(x) = (x + 7)(8x^2 - 2x - 3)$ . Let $x = \cos\theta$ . Since $\cos\theta$ is always between $-1$ and $1$ , the factor $x + 7 = 0$ gives no valid solution. Solve the quadratic factor for $\cos\theta$ , then find all angles in the interval.

Working

Factor the quadratic:

8x^2 - 2x - 3 = (4x - 3)(2x + 1)

So

\cos\theta = \frac{3}{4} \quad \text{or} \quad \cos\theta = -\frac{1}{2}

For $\cos\theta = \frac{3}{4}$ :

\theta = \cos^{-1}\left(\frac{3}{4}\right) \approx 41.4^\circ \quad \text{or} \quad \theta = 360^\circ - 41.4^\circ = 318.6^\circ

For $\cos\theta = -\frac{1}{2}$ :

\theta = 120^\circ \quad \text{or} \quad \theta = 240^\circ

Answer

\theta = 41.4^\circ,\ 120^\circ,\ 240^\circ,\ 318.6^\circ

(accept $318.6^\circ$ or $319^\circ$ ; no other solutions in the interval).

Final answer

θ = 41.4°, 120°, 240°, 318.6° (accept 318.6° or 319°)

Detailed explanation

Walkthrough

We have the equation $8\cos^3\theta + 54\cos^2\theta - 17\cos\theta - 21 = 0$ . Let $x = \cos\theta$ . Then the equation becomes $f(x) = 0$ . From parts (a) and (b), $f(x) = (x + 7)(8x^2 - 2x - 3)$ . Factor the quadratic further: $8x^2 - 2x - 3 = (4x - 3)(2x + 1)$ . So the equation is $(x + 7)(4x - 3)(2x + 1) = 0$ . The possible values are $x = -7$ , $x = \frac{3}{4}$ , and $x = -\frac{1}{2}$ . But $x = \cos\theta$ must lie between $-1$ and $1$ , so $x = -7$ is impossible and is discarded. Now solve $\cos\theta = \frac{3}{4}$ and $\cos\theta = -\frac{1}{2}$ for $0^\circ \leq \theta \leq 360^\circ$ . For $\cos\theta = \frac{3}{4}$ , the principal value is $\cos^{-1}\left(\frac{3}{4}\right) \approx 41.4^\circ$ . Cosine is positive in the first and fourth quadrants, so the second solution is $360^\circ - 41.4^\circ = 318.6^\circ$ . For $\cos\theta = -\frac{1}{2}$ , cosine is negative in the second and third quadrants. The reference angle is $60^\circ$ , so the solutions are $180^\circ - 60^\circ = 120^\circ$ and $180^\circ + 60^\circ = 240^\circ$ . Thus there are four solutions in the interval.

Key Takeaways

A trigonometric equation can often be reduced to a polynomial equation by substitution such as $x = \cos\theta$ . Factorisation using the factor theorem then makes solving straightforward. Always check the range of the trigonometric function to discard impossible roots. For $\cos\theta = k$ , use the symmetry of the cosine graph to find all solutions in the required interval.

Common Mistakes

A common mistake is including $\cos\theta = -7$ as a solution, but this is impossible because $-1 \leq \cos\theta \leq 1$ . Another mistake is giving only the principal value for $\cos\theta = \frac{3}{4}$ and forgetting the fourth-quadrant solution $318.6^\circ$ . For $\cos\theta = -\frac{1}{2}$ , students sometimes give only $120^\circ$ and forget $240^\circ$ . Giving answers in radians when the interval is in degrees loses marks; the mark scheme allows a maximum of $2/3$ for radian answers. Including solutions outside $0^\circ$ to $360^\circ$ is also penalised.

Things to Be Careful About

Use degree mode on the calculator because the interval is given in degrees. The angles $120^\circ$ and $240^\circ$ are exact. The angles for $\cos\theta = \frac{3}{4}$ are approximate, and $318.6^\circ$ is often accepted as $319^\circ$ . Make sure no extra solutions are included from the discarded factor $x + 7$ .

Techniques used

factorise a cubic using known factorsubstitute trigonometric variablesolve for all angles in an interval

Q8MediumTrigonometryIntegration

(a)

Express $3\cos 2x - \sqrt{3}\sin 2x$ in the form $R\cos(2x + \alpha)$ , where $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$ . Give the exact values of $R$ and $\alpha$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write the expression as $R\cos(2x+\alpha)$ by expanding the right-hand side and comparing coefficients.

Working

Expand:

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

Comparing with $3\cos 2x - \sqrt{3}\sin 2x$ :

R\cos\alpha=3, \quad R\sin\alpha=\sqrt{3}

Then

R^2=(R\cos\alpha)^2+(R\sin\alpha)^2=3^2+(\sqrt{3})^2=12

so

R=\sqrt{12}=2\sqrt{3}

and

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

Since $0<\alpha<\frac{\pi}{2}$ , $\alpha=\frac{\pi}{6}$ .

Answer

R=2\sqrt{3}, \quad \alpha=\frac{\pi}{6}

Final answer

R = 2√3, α = π/6

Detailed explanation

Walkthrough

We want to write $3\cos 2x - \sqrt{3}\sin 2x$ as $R\cos(2x+\alpha)$ . Expanding the right-hand side gives $R\cos 2x\cos\alpha - R\sin 2x\sin\alpha$ . Comparing coefficients with the target expression gives $R\cos\alpha=3$ and $R\sin\alpha=\sqrt{3}$ . Squaring and adding eliminates $\alpha$ and gives $R^2=12$ , so $R=2\sqrt{3}$ . Dividing the second equation by the first gives $\tan\alpha=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$ . Because $0<\alpha<\frac{\pi}{2}$ , the angle is $\frac{\pi}{6}$ .

Key Takeaways

The expansion of $R\cos(2x+\alpha)$ has coefficients $R\cos\alpha$ and $R\sin\alpha$ .
$R$ can be found by $R^2=a^2+b^2$ .
The quadrant of $\alpha$ is determined by the signs of $\cos\alpha$ and $\sin\alpha$ ; here both are positive.

Common Mistakes

Treating $R\cos\alpha=3$ and $R\sin\alpha=\sqrt{3}$ as $\cos\alpha=3$ and $\sin\alpha=\sqrt{3}$ ; this is not valid and the mark scheme awards M0 A0 if this is seen.
Using $\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)$ would give a negative angle; since the answer is CWO, this would lose the final A1.
Forgetting the condition $0<\alpha<\frac{\pi}{2}$ , which selects the positive angle.

Things to Be Careful About

The original expression has a minus sign before $\sqrt{3}\sin 2x$ ; after comparison this becomes $R\sin\alpha=+\sqrt{3}$ , not negative.
$\tan\alpha=\frac{1}{\sqrt{3}}$ corresponds to $\frac{\pi}{6}$ , not $\frac{\pi}{3}$ ; check exact trig values carefully.

Techniques used

expand R cos(2x+alpha)compare coefficientsfind R using Pythagorasfind alpha using tan ratio

(b)

Hence find the exact value of

\int_0^{\frac{1}{12}\pi} \frac{3}{(3\cos 2x - \sqrt{3}\sin 2x)^2}\,dx

simplifying your answer.

5M Similar

DifficultyMedium

Worked solution

Approach

Use part (a) to rewrite the denominator, convert the integrand to a $\sec^2$ form, integrate, and evaluate at the limits.

Working

From part (a),

3\cos 2x - \sqrt{3}\sin 2x = 2\sqrt{3}\cos\left(2x+\frac{\pi}{6}\right)

Therefore

(3\cos 2x - \sqrt{3}\sin 2x)^2 = 12\cos^2\left(2x+\frac{\pi}{6}\right)

So the integrand is

\frac{3}{(3\cos 2x - \sqrt{3}\sin 2x)^2} = \frac{3}{12\cos^2\left(2x+\frac{\pi}{6}\right)} = \frac{1}{4}\sec^2\left(2x+\frac{\pi}{6}\right)

Thus

I = \frac{1}{4}\int_0^{\frac{\pi}{12}} \sec^2\left(2x+\frac{\pi}{6}\right)\,dx

Integrating,

I = \frac{1}{4}\cdot\frac{1}{2}\left[\tan\left(2x+\frac{\pi}{6}\right)\right]_0^{\frac{\pi}{12}} = \frac{1}{8}\left[\tan\left(2x+\frac{\pi}{6}\right)\right]_0^{\frac{\pi}{12}}

At $x=\frac{\pi}{12}$ , $2x+\frac{\pi}{6}=\frac{\pi}{3}$ , so $\tan\left(\frac{\pi}{3}\right)=\sqrt{3}$ .

At $x=0$ , $2x+\frac{\pi}{6}=\frac{\pi}{6}$ , so $\tan\left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}$ .

Therefore

I = \frac{1}{8}\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = \frac{1}{8}\left(\frac{3-1}{\sqrt{3}}\right) = \frac{1}{8}\cdot\frac{2}{\sqrt{3}} = \frac{1}{4\sqrt{3}} = \frac{\sqrt{3}}{12}

Answer

\frac{\sqrt{3}}{12}

Final answer

√3/12

Detailed explanation

Walkthrough

Use the result from part (a): $3\cos 2x - \sqrt{3}\sin 2x = 2\sqrt{3}\cos\left(2x+\frac{\pi}{6}\right)$ . Squaring gives $12\cos^2\left(2x+\frac{\pi}{6}\right)$ . Therefore the integrand becomes $\frac{3}{12\cos^2\left(2x+\frac{\pi}{6}\right)}=\frac{1}{4}\sec^2\left(2x+\frac{\pi}{6}\right)$ . Integrate $\sec^2(2x+\frac{\pi}{6})$ ; since the derivative of $2x+\frac{\pi}{6}$ is 2, the integral is $\frac{1}{2}\tan\left(2x+\frac{\pi}{6}\right)$ . Multiplying by $\frac{1}{4}$ gives $\frac{1}{8}\tan\left(2x+\frac{\pi}{6}\right)$ . Substitute limits: at $x=\frac{\pi}{12}$ , the angle is $\frac{\pi}{3}$ , with $\tan=\sqrt{3}$ ; at $x=0$ , the angle is $\frac{\pi}{6}$ , with $\tan=\frac{1}{\sqrt{3}}$ . Subtract lower from upper and simplify: $\frac{1}{8}\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=\frac{1}{8}\cdot\frac{2}{\sqrt{3}}=\frac{\sqrt{3}}{12}$ .

Key Takeaways

Squaring a $R\cos$ form gives $R^2\cos^2$ , which can be converted to $\sec^2$ for integration.
$\int \sec^2(ax+b)\,dx = \frac{1}{a}\tan(ax+b)+C$ .
Exact values $\tan\frac{\pi}{3}=\sqrt{3}$ and $\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}$ are needed.

Common Mistakes

Forgetting the $\frac{1}{2}$ factor when integrating $\sec^2(2x+\frac{\pi}{6})$ .
Forgetting the $\frac{1}{4}$ factor from $\frac{3}{12}$ ; the mark scheme requires the coefficient $\frac{1}{8}$ to be correct.
Not simplifying $\frac{1}{8}\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)$ ; the mark scheme requires a single term exact equivalent.
Substituting limits in the wrong order; always upper limit minus lower limit.

Things to Be Careful About

The mark scheme allows follow-through on $\alpha$ from part (a), but the coefficient $\frac{1}{8}$ must still be correct.
The final answer may be written as $\frac{\sqrt{3}}{12}$ , $\frac{1}{4\sqrt{3}}$ or $\frac{1}{\sqrt{48}}$ ; all are accepted.
Keep the angle in radians throughout, since the limits are in radians.

Techniques used

rewrite denominator using part (a) resultconvert to sec^2 formintegrate sec^2(2x+alpha)evaluate definite integral at limitssimplify exact surd result

Q9Medium-HardDifferential Equations

Fig. 2

A container in the shape of a cuboid has a square base of side $x$ and a height of $(10 - x)$ . It is given that $x$ varies with time, $t$ , where $t > 0$ . The container decreases in volume at a rate which is inversely proportional to $t$ .

When $t = \frac{1}{10}$ , $x = \frac{1}{2}$ and the rate of decrease of $x$ is $\frac{20}{37}$ .

(a)

Show that $x$ and $t$ satisfy the differential equation

\frac{dx}{dt} = \frac{-1}{2t(20x - 3x^2)}

5M Similar

DifficultyMedium

Worked solution

Approach

Express the volume in terms of $x$ , write the rate-of-change statement using a constant of proportionality, then use the chain rule to convert $\frac{dV}{dt}$ into $\frac{dx}{dt}$ . Finally substitute the three given data values to find the constant and reach the stated differential equation.

Working

The cuboid has a square base of side $x$ and height $10 - x$ , so its volume is

V = x^2(10 - x) = 10x^2 - x^3.

Differentiating with respect to $x$ :

\frac{dV}{dx} = 20x - 3x^2.

Since the volume decreases at a rate inversely proportional to $t$ , with $k$ the constant of proportionality,

\frac{dV}{dt} = -\frac{k}{t}.

Apply the chain rule $\frac{dx}{dt} = \dfrac{dV}{dt} \times \dfrac{dx}{dV} = \dfrac{dV/dt}{dV/dx}$ :

\frac{dx}{dt} = \frac{-k/t}{20x - 3x^2} = \frac{-k}{t(20x - 3x^2)}.

Now use $t = \tfrac{1}{10}$ , $x = \tfrac{1}{2}$ and $\dfrac{dx}{dt} = -\dfrac{20}{37}$ . First compute $20x - 3x^2$ at $x = \tfrac{1}{2}$ :

20\left(\tfrac{1}{2}\right) - 3\left(\tfrac{1}{2}\right)^2 = 10 - \tfrac{3}{4} = \tfrac{37}{4}.

Substitute:

-\frac{20}{37} = \frac{-k}{\tfrac{1}{10} \cdot \tfrac{37}{4}} = \frac{-k}{\tfrac{37}{40}} = -\frac{40k}{37}.

Hence $\tfrac{20}{37} = \tfrac{40k}{37}$ , giving $k = \tfrac{1}{2}$ . Substituting $k = \tfrac{1}{2}$ back:

\frac{dx}{dt} = \frac{-\tfrac{1}{2}}{t(20x - 3x^2)} = \frac{-1}{2t(20x - 3x^2)}.

Answer

\frac{dx}{dt} = \frac{-1}{2t(20x - 3x^2)}

Final answer

dx/dt = -1/(2t(20x - 3x^2))

Detailed explanation

Walkthrough

The cuboid has volume $V = x \times x \times (10 - x) = 10x^2 - x^3$ , so $\frac{dV}{dx} = 20x - 3x^2$ . The problem says the volume is decreasing at a rate inversely proportional to $t$ , so we write $\frac{dV}{dt} = -\frac{k}{t}$ for some positive constant $k$ (the minus sign is what makes it a decrease). Because $V$ depends on $x$ which depends on $t$ , the chain rule links the two derivatives: $\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}$ , hence $\frac{dx}{dt} = \frac{dV/dt}{dV/dx} = \frac{-k/t}{20x - 3x^2} = \frac{-k}{t(20x - 3x^2)}$ . This general form still has an unknown $k$ , so we use the three given data values. At $x = \tfrac{1}{2}$ we have $20x - 3x^2 = \tfrac{37}{4}$ , and substituting $t = \tfrac{1}{10}$ and $\frac{dx}{dt} = -\tfrac{20}{37}$ gives $-\tfrac{20}{37} = -\tfrac{40k}{37}$ , so $k = \tfrac{1}{2}$ . Putting this back produces the required differential equation.

Key Takeaways

Translate "decreases at a rate inversely proportional to $t$ " into $\frac{dV}{dt} = -k/t$ (the sign is what makes it a decrease).
Use the chain rule to convert between rates with respect to different variables, here $\frac{dx}{dt} = \frac{dV/dt}{dV/dx}$ .
Substitute known values of $t$ , $x$ and $\frac{dx}{dt}$ to determine the constant of proportionality.

Common Mistakes

Forgetting the minus sign in $\frac{dV}{dt} = -k/t$ when the volume is decreasing.
Using $\frac{dV}{dx} = 2x(10 - x) = 20x - 2x^2$ instead of correctly differentiating $10x^2 - x^3$ , giving $20x - 3x^2$ .
Inverting the chain rule and writing $\frac{dx}{dt} = \frac{dV/dx}{dV/dt}$ .
Writing $\frac{dx}{dt} = \tfrac{20}{37}$ (positive) when the problem states the rate of decrease of $x$ is $\tfrac{20}{37}$ , so $\frac{dx}{dt} = -\tfrac{20}{37}$ .

Things to Be Careful About

The mark scheme requires the value $\frac{dx}{dt} = -\frac{20}{37}$ to appear (not $+\frac{20}{37}$ ) for the final A1.
When you compute $20x - 3x^2$ at $x = \tfrac{1}{2}$ , remember it is $10 - \tfrac{3}{4} = \tfrac{37}{4}$ , not $10 - \tfrac{1}{4}$ .
Keep $k$ symbolic until the substitution step; only then determine $k = \tfrac{1}{2}$ .

Techniques used

express volume as a function of xapply the chain rule to related ratesuse initial conditions to determine the proportionality constant

(b)

Solve the differential equation, obtaining an expression for $t$ in terms of $x$ .

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Separate the variables so that all $x$ -terms are on the left and all $t$ -terms are on the right, integrate both sides, and use the given initial condition $x = \tfrac{1}{2}$ when $t = \tfrac{1}{10}$ to fix the constant of integration. Then isolate $t$ by collecting the logarithms and exponentiating.

Working

Starting from

\frac{dx}{dt} = \frac{-1}{2t(20x - 3x^2)},

separate the variables:

(20x - 3x^2)\,dx = -\frac{1}{2t}\,dt.

Integrate both sides:

\int (20x - 3x^2)\,dx = \int -\frac{1}{2t}\,dt.

The left side gives

10x^2 - x^3,

and the right side gives

-\tfrac{1}{2}\ln t + C.

So

10x^2 - x^3 = -\tfrac{1}{2}\ln t + C.

Apply the initial condition $x = \tfrac{1}{2}$ , $t = \tfrac{1}{10}$ :

10\left(\tfrac{1}{2}\right)^2 - \left(\tfrac{1}{2}\right)^3 = -\tfrac{1}{2}\ln\left(\tfrac{1}{10}\right) + C.

Evaluate the left side: $10 \cdot \tfrac{1}{4} - \tfrac{1}{8} = \tfrac{5}{2} - \tfrac{1}{8} = \tfrac{19}{8}$ .

And $-\tfrac{1}{2}\ln(\tfrac{1}{10}) = -\tfrac{1}{2}(-\ln 10) = \tfrac{1}{2}\ln 10$ . Therefore

\tfrac{19}{8} = \tfrac{1}{2}\ln 10 + C \quad\Rightarrow\quad C = \tfrac{19}{8} - \tfrac{1}{2}\ln 10.

So the implicit solution is

10x^2 - x^3 = -\tfrac{1}{2}\ln t + \tfrac{19}{8} - \tfrac{1}{2}\ln 10.

Rearrange to isolate $\ln t$ :

\tfrac{1}{2}\ln t = -10x^2 + x^3 + \tfrac{19}{8} - \tfrac{1}{2}\ln 10,

\ln t = -20x^2 + 2x^3 + \tfrac{19}{4} - \ln 10.

Combine the logarithms:

\ln t + \ln 10 = 2x^3 - 20x^2 + \tfrac{19}{4},

\ln(10t) = 2x^3 - 20x^2 + \tfrac{19}{4}.

Exponentiate both sides and divide by $10$ :

10t = e^{2x^3 - 20x^2 + \tfrac{19}{4}} \quad\Rightarrow\quad t = \tfrac{1}{10}e^{2x^3 - 20x^2 + \tfrac{19}{4}}.

Answer

t = \frac{1}{10}e^{2x^3 - 20x^2 + \frac{19}{4}}

Final answer

t = (1/10) e^(2x^3 - 20x^2 + 19/4)

Detailed explanation

Walkthrough

The differential equation is separable: every $x$ lives on the right-hand side inside the bracket, so we move it to the left and move $t$ to the right. This produces $(20x - 3x^2)\,dx = -\tfrac{1}{2t}\,dt$ . Integrating the left side with respect to $x$ uses the power rule: $\int 20x\,dx = 10x^2$ and $\int 3x^2\,dx = x^3$ . Integrating $-\tfrac{1}{2t}\,dt$ gives $-\tfrac{1}{2}\ln|t|$ , and since $t > 0$ we may drop the absolute value, leaving $-\tfrac{1}{2}\ln t$ . Adding a constant $C$ on the right gives the implicit general solution $10x^2 - x^3 = -\tfrac{1}{2}\ln t + C$ . The initial condition $x = \tfrac{1}{2}$ at $t = \tfrac{1}{10}$ then determines $C$ : the left side is $\tfrac{5}{2} - \tfrac{1}{8} = \tfrac{19}{8}$ , and the right side is $-\tfrac{1}{2}\ln(\tfrac{1}{10}) + C = \tfrac{1}{2}\ln 10 + C$ , so $C = \tfrac{19}{8} - \tfrac{1}{2}\ln 10$ . To solve explicitly for $t$ , multiply the implicit relation by $2$ to clear the fraction on the $\ln t$ term, then collect all logarithms on the left to form $\ln(10t) = 2x^3 - 20x^2 + \tfrac{19}{4}$ , and finally exponentiate to obtain $t = \tfrac{1}{10}e^{2x^3 - 20x^2 + \tfrac{19}{4}}$ .

Key Takeaways

A separable ODE of the form $\frac{dx}{dt} = f(t)g(x)$ is solved by writing $\frac{dx}{g(x)} = f(t)\,dt$ and integrating both sides.
After integration, an implicit relation between $x$ and $t$ is produced; algebraic manipulation (combining logs, exponentiating) is needed to make $t$ the subject.
The constant of integration is found by substituting the initial data into the general (with $C$ ) relation.

Common Mistakes

Forgetting to put brackets around the $t$ side: writing $\int \tfrac{-1}{2t}\,dt = -\tfrac{1}{2}\ln t$ is correct, but $\int -\tfrac{1}{2}t\,dt = -\tfrac{1}{4}t^2$ would mean misreading the differential equation.
Integrating the right side as $\ln t$ instead of $-\tfrac{1}{2}\ln t$ (dropping the factor $-\tfrac{1}{2}$ ).
Forgetting the negative sign when separating, producing the wrong sign throughout.
Failing to combine $\ln t - \ln 10$ into $\ln(10t)$ (or $\ln(\tfrac{t}{10})$ ) and getting stuck with logs on both sides.
Writing $t = \tfrac{1}{10} \cdot e^{\ln(\tfrac{1}{10})} \cdot e^{2x^3 - 20x^2 + \tfrac{19}{4}}$ (or similar): the mark scheme forbids an $e^{\ln(1/10)}$ term sitting in the final answer.

Things to Be Careful About

Domain: $t > 0$ is given, so $\ln t$ is defined without absolute value bars.
Sign of $C$ : many equivalent final forms are accepted, but the working must be self-consistent.
The mark scheme allows several equivalent forms of the constant: $1.14 \le \tfrac{1}{2}\ln 10 \le 1.16$ and $2.44 \le \tfrac{19}{4} \le 2.46$ are both permitted as decimal approximations.

Techniques used

separate variables in the differential equationintegrate both sides with respect to the appropriate variableuse initial conditions to find the integration constantapply logarithmic identities to isolate texponentiate to obtain an explicit expression for t

Q10MediumVectors

The equations of two straight lines are

\mathbf{r} = \mathbf{i} + \mathbf{j} + 2a\mathbf{k} + \lambda(3\mathbf{i} + 4\mathbf{j} + a\mathbf{k}) \quad \text{and} \quad \mathbf{r} = -3\mathbf{i} - \mathbf{j} + 4\mathbf{k} + \mu(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})

where $a$ is a constant.

(a)

Given that the acute angle between the directions of these lines is $\frac{1}{4}\pi$ , find the possible values of $a$ .

6M Similar

DifficultyMedium

Worked solution

Approach

The angle between two lines is found from their direction vectors using

\cos \theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1| |\mathbf{d}_2|}.

Since the acute angle is $\frac{\pi}{4}$ , set this equal to $\pm \frac{\sqrt{2}}{2}$ and solve for $a$ .

Working

The direction vectors are

\mathbf{d}_1 = 3\mathbf{i} + 4\mathbf{j} + a\mathbf{k}, \quad \mathbf{d}_2 = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}.

Their scalar product is

\mathbf{d}_1 \cdot \mathbf{d}_2 = 3(-1) + 4(2) + a(2) = -3 + 8 + 2a = 5 + 2a.

The magnitudes are

|\mathbf{d}_1| = \sqrt{9 + 16 + a^2} = \sqrt{25 + a^2}, \quad |\mathbf{d}_2| = \sqrt{1 + 4 + 4} = 3.

Therefore

\frac{5 + 2a}{3\sqrt{25 + a^2}} = \pm \frac{\sqrt{2}}{2}.

Squaring both sides:

(5 + 2a)^2 = \frac{9}{2}(25 + a^2).

Expand and simplify:

25 + 20a + 4a^2 = \frac{225}{2} + \frac{9}{2}a^2

Multiply by 2:

50 + 40a + 8a^2 = 225 + 9a^2

a^2 - 40a + 175 = 0.

Factorising:

(a - 5)(a - 35) = 0.

So $a = 5$ or $a = 35$ .

Answer

a = 5 \quad \text{or} \quad a = 35.

Final answer

a = 5 or a = 35

Detailed explanation

Walkthrough

The two lines are given in vector form, so their directions are the vectors multiplying the parameters: $\mathbf{d}_1 = 3\mathbf{i} + 4\mathbf{j} + a\mathbf{k}$ and $\mathbf{d}_2 = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ . The angle between the lines is the angle between these direction vectors, so we use the scalar product formula.

First compute the dot product: $3(-1) + 4(2) + a(2) = 5 + 2a$ . Then compute the magnitudes: $|\mathbf{d}_1| = \sqrt{9 + 16 + a^2} = \sqrt{25 + a^2}$ and $|\mathbf{d}_2| = \sqrt{1 + 4 + 4} = 3$ .

The cosine of the angle is $\frac{5 + 2a}{3\sqrt{25 + a^2}}$ . Since the acute angle is $\frac{\pi}{4}$ , this cosine should be $\frac{\sqrt{2}}{2}$ . Because the acute angle between two lines can correspond to either direction vector being reversed, the mark scheme allows $\pm\frac{\sqrt{2}}{2}$ .

Square both sides to remove the square root. This gives $(5 + 2a)^2 = \frac{9}{2}(25 + a^2)$ . Expanding and simplifying gives $a^2 - 40a + 175 = 0$ , which factors as $(a - 5)(a - 35) = 0$ . Hence $a = 5$ or $a = 35$ . Both values give a positive dot product and satisfy the acute angle condition.

Key Takeaways

The scalar product formula $\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$ connects vector directions to angles. Magnitudes of vectors are found by taking the square root of the sum of squares of components. When an angle condition leads to a square root, squaring both sides often produces a quadratic equation. The $\pm$ sign accounts for the acute angle between lines.

Common Mistakes

Forgetting to use both direction vectors consistently; the same two vectors must be used throughout.
Forgetting the $\pm$ when setting the cosine equal to $\frac{\sqrt{2}}{2}$ , or using only the positive value without justification.
Squaring only one side, or not squaring the whole product of moduli, leading to an incorrect quadratic.
Sign errors in the dot product, especially $3(-1) + 4(2) + 2a = 5 + 2a$ .

Things to Be Careful About

The mark scheme requires a correct equation such as $\frac{5 + 2a}{3\sqrt{25 + a^2}} = \pm\frac{\sqrt{2}}{2}$ before squaring.
When squaring, you must square $(5 + 2a)$ to get three terms and remove the square root from the other side.
Both $a = 5$ and $a = 35$ are valid; do not discard one because it seems large.

Techniques used

compute scalar product of direction vectorscompute magnitudes of direction vectorsset cosine equal to ±√2/2 and square to form a quadraticsolve the resulting quadratic equation

(b)

Given instead that the lines intersect, find the value of $a$ and the position vector of the point of intersection.

5M Similar

DifficultyMedium

Worked solution

Approach

For two lines to intersect, their position vectors must be equal for some parameters $\lambda$ and $\mu$ . Write each line in component form, equate corresponding components, solve for $\lambda$ and $\mu$ , then use the remaining component to find $a$ .

Working

Line 1:

(1 + 3\lambda, \, 1 + 4\lambda, \, 2a + a\lambda)

Line 2:

(-3 - \mu, \, -1 + 2\mu, \, 4 + 2\mu)

Equating components:

1 + 3\lambda = -3 - \mu \quad \Rightarrow \quad 3\lambda + \mu = -4

1 + 4\lambda = -1 + 2\mu \quad \Rightarrow \quad 4\lambda - 2\mu = -2 \quad \Rightarrow \quad 2\lambda - \mu = -1.

Add the two equations:

5\lambda = -5 \quad \Rightarrow \quad \lambda = -1.

Substitute $\lambda = -1$ into $2\lambda - \mu = -1$ :

-2 - \mu = -1 \quad \Rightarrow \quad \mu = -1.

Use the third component:

a(2 + \lambda) = 4 + 2\mu

a(2 - 1) = 4 - 2 \quad \Rightarrow \quad a = 2.

The point of intersection is obtained using $\lambda = -1$ in line 1:

(1 - 3, \, 1 - 4, \, 4 - 2) = (-2, \, -3, \, 2).

So the position vector is

-2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}.

Answer

a = 2, \quad \text{position vector} = -2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}.

Final answer

a = 2; position vector = -2i - 3j + 2k

Detailed explanation

Walkthrough

For the lines to intersect, there must be values of $\lambda$ and $\mu$ such that the position vectors are equal. Write each line in component form: line 1 is $(1 + 3\lambda, 1 + 4\lambda, 2a + a\lambda)$ and line 2 is $(-3 - \mu, -1 + 2\mu, 4 + 2\mu)$ .

Equate the $x$ -components: $1 + 3\lambda = -3 - \mu$ , so $3\lambda + \mu = -4$ . Equate the $y$ -components: $1 + 4\lambda = -1 + 2\mu$ , so $4\lambda - 2\mu = -2$ , or $2\lambda - \mu = -1$ . These two equations involve only $\lambda$ and $\mu$ , so solve them simultaneously. Adding gives $5\lambda = -5$ , so $\lambda = -1$ . Substituting back gives $\mu = -1$ .

Now use the $z$ -component: $2a + a\lambda = 4 + 2\mu$ . With $\lambda = -1$ and $\mu = -1$ , this becomes $a = 2$ . Finally substitute $\lambda = -1$ into line 1 to get $(-2, -3, 2)$ , so the position vector is $-2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}$ .

Key Takeaways

Two lines intersect when their vector equations can be made equal for some parameters. Equating components produces a system of equations. Solving for the parameters first is often easiest because the third component then determines any unknown constant. The intersection point is found by substituting the parameter back into either line.

Common Mistakes

Only equating two components and solving for $\lambda$ and $\mu$ without using the third component to find $a$ .
Using different parameter values for the two lines, e.g. using $\lambda$ in both equations.
Sign errors when equating components, especially with the negative signs in line 2.
Substituting $\lambda$ or $\mu$ into the wrong line or using the wrong sign, giving an incorrect point.

Things to Be Careful About

The mark scheme expects the general point of at least one line in component form before equating.
When solving, $\lambda = -1$ and $\mu = -1$ must both be obtained.
The final position vector must be written as a single vector, not as separate components with $\mathbf{i}, \mathbf{j}, \mathbf{k}$ in a tuple such as $(-2\mathbf{i}, -3\mathbf{j}, 2\mathbf{k})$ .
If two different points are given, the answer is marked wrong even if one is correct.

Techniques used

write lines in component formequate corresponding componentssolve simultaneous equations for parameterssubstitute parameters to find intersection point

Q119MMedium-HardIntegration

Use the substitution $2x = \tan\theta$ to find the exact value of

\int_0^{\frac{1}{2}} \frac{12}{(1 + 4x^2)^2}\,dx

Give your answer in the form $a + b\pi$ , where $a$ and $b$ are rational numbers.

DifficultyMedium-Hard

Worked solution

Approach

We are told to use the substitution $2x = \tan\theta$ . This is a natural choice because it converts $1+4x^2$ into $1+\tan^2\theta = \sec^2\theta$ . We will differentiate the substitution to replace $dx$ , change the limits of integration, simplify the integrand using the Pythagorean identity, and then integrate.

Working

Let $2x = \tan\theta$ . Then

\frac{dx}{d\theta} = \frac{1}{2}\sec^2\theta \quad \Rightarrow \quad dx = \frac{1}{2}\sec^2\theta \, d\theta.

Also,

1+4x^2 = 1+\tan^2\theta = \sec^2\theta.

Change the limits. When $x=0$ , $\tan\theta = 0$ , so $\theta=0$ . When $x=\frac{1}{2}$ , $\tan\theta = 1$ , so $\theta = \frac{\pi}{4}$ . Hence

\begin{aligned} I &= \int_0^{\pi/4} \frac{12}{(\sec^2\theta)^2}\cdot \frac{1}{2}\sec^2\theta \, d\theta \\[3pt] &= \int_0^{\pi/4} \frac{6}{\sec^2\theta}\,d\theta \\[3pt] &= 6\int_0^{\pi/4} \cos^2\theta \, d\theta. \end{aligned}

Use $\cos^2\theta = \frac{1+\cos 2\theta}{2}$ :

\begin{aligned} I &= 6\int_0^{\pi/4} \frac{1+\cos 2\theta}{2}\,d\theta \\[3pt] &= 3\int_0^{\pi/4} (1+\cos 2\theta)\,d\theta \\[3pt] &= 3\left[\theta + \frac{1}{2}\sin 2\theta\right]_0^{\pi/4}. \end{aligned}

Evaluate:

I = 3\left[\left(\frac{\pi}{4} + \frac{1}{2}\sin\frac{\pi}{2}\right) - \left(0+0\right)\right] = 3\left(\frac{\pi}{4} + \frac{1}{2}\right) = \frac{3\pi}{4} + \frac{3}{2}.

Answer

\frac{3}{2} + \frac{3\pi}{4}

so $a = \frac{3}{2}$ and $b = \frac{3}{4}$ .

Final answer

3/2 + (3/4)π

Detailed explanation

Walkthrough

We are given the substitution $2x = \tan\theta$ . The key reason this substitution is effective is the identity $1+\tan^2\theta = \sec^2\theta$ , which exactly matches the denominator $1+4x^2$ . First differentiate the substitution to find $dx$ in terms of $d\theta$ . Then replace all $x$ 's and $dx$ in the integral. Because this is a definite integral, change the limits: substitute the original limits into $2x=\tan\theta$ . Then simplify the integrand: $(1+4x^2)^2 = \sec^4\theta$ , and the $dx$ brings a factor $\frac12\sec^2\theta\,d\theta$ , so the secant powers cancel, leaving $\cos^2\theta$ . Then use the double-angle identity to integrate $\cos^2\theta$ . Finally evaluate at the limits and simplify to the form $a+b\pi$ .

Key Takeaways

Recognising when a trigonometric substitution is appropriate: when the integrand contains $1+x^2$ or $1+4x^2$ , the identity $1+\tan^2\theta=\sec^2\theta$ is the key.
The importance of changing the limits of integration when using substitution in a definite integral.
Using trigonometric identities to reduce powers of cosine before integrating.
Simplifying the final answer into the requested linear form in $\pi$ .

Common Mistakes

Forgetting to change the limits of integration when substituting.
Incorrectly differentiating $2x=\tan\theta$ ; remember $dx = \frac12\sec^2\theta\,d\theta$ .
Failing to simplify $(1+4x^2)^2$ correctly; note $1+4x^2 = \sec^2\theta$ .
Forgetting the factor $\frac12$ that appears from $dx$ .
Incorrectly integrating $\cos^2\theta$ without using the double-angle identity.

Things to Be Careful About

The substitution gives $dx = \frac{1}{2}\sec^2\theta\,d\theta$ ; do not omit the $\frac12$ .
When simplifying, $\frac{12}{\sec^4\theta}\cdot\frac12\sec^2\theta = 6\cos^2\theta$ , not $12\cos^2\theta$ .
The limits: $x=0\Rightarrow\theta=0$ and $x=\frac12\Rightarrow\theta=\frac{\pi}{4}$ .
The final answer must be in the form $a+b\pi$ ; here $a=\frac32$ and $b=\frac34$ .
Since no mark scheme was provided, ensure every algebraic step is shown clearly to gain method marks.

Techniques used

perform a trigonometric substitutionchange the limits of integrationsimplify using the Pythagorean identityintegrate a power of cosine using a double-angle identity

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