The diagram shows the curve $y = x^3 \cos 2x$ for $0 \le x \le \frac{1}{4}\pi$. The curve has a maximum point at $M$, where $x = p$. (a) Show that $p$ satisfies the equation $p = \frac{1}{2}\tan^{-1}\left(\frac{3}{2p}\right)$. (b) Show by calculation that $0.5 < p < 0.7$. (c) Use an iterative formula based on the equation in part (a) to calculate $p$ correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

(a) p = 1/2 tan^-1(3/(2p)) · (b) 0.5 < p < 0.7 · (c) 0.596

9709/32 Feb/Mar 2025 — Mathematics Solved Past Paper

Q13MMedium-EasyLogarithmic and Exponential Functions

Solve the equation

\ln(1 - e^{-2x}) + 3 = 0

Give your final answer correct to 4 decimal places.

DifficultyMedium-Easy

Worked solution

Approach

Isolate the logarithm, then use the inverse relationship between $\ln$ and $e^x$ to remove it. Rearrange to isolate $e^{-2x}$ , take natural logarithms, and evaluate to 4 decimal places.

Working

\ln(1 - e^{-2x}) + 3 = 0

Subtract 3 from both sides:

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

Exponentiate both sides:

1 - e^{-2x} = e^{-3}

Rearrange:

e^{-2x} = 1 - e^{-3}

Take natural logarithms:

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

Divide by $-2$ :

x = -\frac{1}{2}\ln(1 - e^{-3})

Using $e^{-3} = 0.0497870684$ :

x = -\frac{1}{2}\ln(0.9502129316) = 0.025534...

Answer

x = 0.0255 \quad (4 \text{ d.p.})

Final answer

x = 0.0255

Detailed explanation

Walkthrough

Start with the equation $\ln(1 - e^{-2x}) + 3 = 0$ . The first step is to isolate the logarithm by subtracting 3 from both sides:

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

This is helpful because $\ln$ and $e^x$ are inverse functions: if $\ln A = B$ , then $A = e^B$ . So exponentiating both sides removes the logarithm and gives:

1 - e^{-2x} = e^{-3}

Next, rearrange to isolate the exponential term. Add $e^{-2x}$ to both sides and subtract $e^{-3}$ from both sides, which gives:

e^{-2x} = 1 - e^{-3}

Since the unknown $x$ is in the exponent, take natural logarithms of both sides. Because $\ln(e^{-2x}) = -2x$ , this gives:

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

Finally divide by $-2$ :

x = -\frac{1}{2}\ln(1 - e^{-3})

Evaluating with a calculator, using $e^{-3} = 0.0497870684$ , gives $1 - e^{-3} = 0.9502129316$ and $\ln(0.9502129316) = -0.051068...$ . Therefore $x = 0.025534...$ , which rounds to $0.0255$ correct to 4 decimal places.

Key Takeaways

The question tests the inverse relationship between the natural logarithm and the exponential function. To solve an equation with a logarithm, isolate the logarithm first and then exponentiate. To solve an equation with the unknown in an exponent, take logarithms of both sides. Always give the final answer with the required precision and show enough working to justify the answer.

Common Mistakes

Omitting working: the mark scheme states that no working seen scores 0.
Writing $1 - e^{-2x} = -3$ instead of $1 - e^{-2x} = e^{-3}$ after removing the logarithm.
Forgetting the negative sign when dividing by $-2$ , giving $x = +\frac{1}{2}\ln(1 - e^{-3})$ .
Using base 10 logarithms instead of natural logarithms; this gives a different, incorrect answer.
Rounding too early or giving the answer to the wrong number of decimal places.

Things to Be Careful About

The expression $1 - e^{-2x}$ must be positive for the logarithm to be defined; the solution $x = 0.0255$ is positive, so this condition is satisfied. Use $\ln$ , not $\log_{10}$ , throughout. Keep the exact expression $x = -\frac{1}{2}\ln(1 - e^{-3})$ until the final evaluation, then round to 4 decimal places. The final answer must be exactly $0.0255$ to 4 d.p., and evidence of the method must be shown.

Techniques used

remove logarithm by exponentiating both sidesisolate the exponential termtake natural logarithms of both sidesevaluate and round to 4 decimal places

Similar questions

Q25MMediumDifferentiation

The equation of a curve is $xy^2 + \ln(x + 2y) = 1$ .

Find the gradient of the curve at the point where $x = 0$ .

DifficultyMedium

Worked solution

Approach

Differentiate the equation implicitly with respect to $x$ , using the product rule for $xy^2$ and the chain rule for $\ln(x+2y)$ . Then substitute the point where $x=0$ : first find $y$ from the original equation, then solve for $\frac{dy}{dx}$ .

Working

Differentiate term by term.

For $xy^2$ , use the product rule:

\frac{d}{dx}(xy^2) = y^2 + x \cdot 2y\frac{dy}{dx}

For $\ln(x+2y)$ , use the chain rule:

\frac{d}{dx}\ln(x+2y) = \frac{1+2\frac{dy}{dx}}{x+2y}

Therefore the differentiated equation is:

y^2 + 2xy\frac{dy}{dx} + \frac{1+2\frac{dy}{dx}}{x+2y} = 0

At $x=0$ , substitute into the original equation:

0 \cdot y^2 + \ln(0+2y) = 1

\ln(2y) = 1

2y = e

y = \frac{e}{2}

Now substitute $x=0$ and $y=\frac{e}{2}$ into the differentiated equation. Here $x+2y = e$ and $y^2 = \frac{e^2}{4}$ :

\frac{e^2}{4} + 0 + \frac{1+2\frac{dy}{dx}}{e} = 0

Multiply through by $e$ :

\frac{e^3}{4} + 1 + 2\frac{dy}{dx} = 0

2\frac{dy}{dx} = -1 - \frac{e^3}{4}

\frac{dy}{dx} = -\frac{1}{2} - \frac{e^3}{8} = -\frac{e^3+4}{8}

Answer

\frac{dy}{dx} = -\frac{e^3+4}{8}

Final answer

dy/dx = -(e^3 + 4)/8

Detailed explanation

Walkthrough

We need the gradient of the curve, which is $\frac{dy}{dx}$ , at the point where $x=0$ . Since the curve is given by an equation mixing $x$ and $y$ , we differentiate implicitly with respect to $x$ .

The term $xy^2$ is a product of $x$ and $y^2$ . Because $y$ is a function of $x$ , differentiating $y^2$ gives $2y\frac{dy}{dx}$ . Applying the product rule gives:

\frac{d}{dx}(xy^2) = y^2 + x \cdot 2y\frac{dy}{dx}

The term $\ln(x+2y)$ is a composite function. Its derivative is $\frac{1}{x+2y}$ multiplied by the derivative of the inside, $1+2\frac{dy}{dx}$ :

\frac{d}{dx}\ln(x+2y) = \frac{1+2\frac{dy}{dx}}{x+2y}

Putting these together gives the differentiated equation shown in the Working.

To evaluate the gradient at $x=0$ , we first need the corresponding $y$ -value. Substituting $x=0$ into the original equation removes the $xy^2$ term and leaves $\ln(2y)=1$ . Taking exponentials gives $2y=e$ , so $y=\frac{e}{2}$ .

Finally, substitute $x=0$ and $y=\frac{e}{2}$ into the differentiated equation. The middle term $2xy\frac{dy}{dx}$ vanishes because $x=0$ , and $x+2y=e$ . Solving the resulting linear equation for $\frac{dy}{dx}$ gives the gradient.

Key Takeaways

This question tests implicit differentiation, the product rule, and differentiation of a logarithmic composite. It also tests the ability to find the missing coordinate before evaluating a derivative at a point. The key idea is that every occurrence of $y$ must be differentiated as a function of $x$ , producing a factor of $\frac{dy}{dx}$ .

Common Mistakes

Forgetting the $\frac{dy}{dx}$ factor when differentiating $y^2$ : the derivative of $y^2$ is $2y\frac{dy}{dx}$ , not just $2y$ .
Differentiating $\ln(x+2y)$ as $\frac{1}{x+2y}$ without also multiplying by the derivative of $x+2y$ , which is $1+2\frac{dy}{dx}$ .
Forgetting the product rule on $xy^2$ and writing only $2xy\frac{dy}{dx}$ , omitting the $y^2$ term.
Substituting $x=0$ without first finding $y$ , or substituting $y=0$ incorrectly.
Making a sign error when moving terms across the equation while solving for $\frac{dy}{dx}$ .

Things to Be Careful About

The logarithm requires $x+2y>0$ . At the point $(0,\frac{e}{2})$ , $x+2y=e>0$ , so the point is valid.
The mark scheme awards separate marks for the derivative of $xy^2$ , the derivative of $\ln(x+2y)$ , forming the differentiated equation, finding $y$ , and obtaining the final gradient. Show each stage clearly.
The final answer can be written as $-\frac{1}{2}-\frac{e^3}{8}$ or as $-\frac{e^3+4}{8}$ ; both are equivalent. Its numerical value is approximately $-3.01$ .
When substituting $x=0$ , the term $x \cdot 2y\frac{dy}{dx}$ disappears, but the $y^2$ term does not, because it came from differentiating the product.

Techniques used

differentiate implicitly with respect to xapply the product rule to xy^2apply the chain rule to ln(x+2y)substitute the given x-coordinate to find ysolve for dy/dx

Similar questions

Q3MediumComplex Numbers

Fig. 1

The shaded region on the Argand diagram shows points representing complex numbers $z$ defined by two inequalities. The shaded region is bounded by a circle and a line parallel to the real axis. The boundaries of the region are included in the shaded region.

(a)

Find two inequalities in terms of $z$ that define the shaded region.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The shaded region is bounded by two curves: a horizontal line and a circle. We read the inequalities directly from the diagram by identifying the position of the region relative to each boundary.

Working

Horizontal boundary:
The horizontal line passes through $\text{Im}(z) = -1$ . The shaded region lies below this line (including the line itself), so:

\text{Im}(z) \leq -1

Circular boundary:
The circle is centred at $-2 + i$ , which corresponds to the point $(-2, 1)$ on the Argand diagram. The circle passes through $(-5, 1)$ , $(1, 1)$ , $(-2, 4)$ and $(-2, -2)$ , so the radius is:

r = 1 - (-2) = 3

The shaded region lies inside the circle (including the boundary). The locus of points at a distance $r$ from a point $z_0$ is $|z - z_0| = r$ , so the interior is $|z - z_0| \leq r$ . Substituting $z_0 = -2 + i$ and $r = 3$ :

|z - (-2 + i)| \leq 3

which simplifies to:

|z + 2 - i| \leq 3

Answer

The two inequalities defining the shaded region are:

\text{Im}(z) \leq -1 \quad \text{and} \quad |z + 2 - i| \leq 3

Final answer

Im(z) ≤ -1 and |z + 2 - i| ≤ 3

Detailed explanation

Walkthrough

The shaded region is the intersection of two simpler regions, each defined by one inequality.

Step 1: The horizontal line. The line is drawn at $\text{Im}(z) = -1$ , which is the horizontal line $y = -1$ on the Argand diagram. The shaded region is below this line, and the boundary is included. This gives $\text{Im}(z) \leq -1$ .

Step 2: The circle. The circle is centred at $-2 + i$ , i.e. the point $(-2, 1)$ . By reading off the diagram, the circle passes through $(-5, 1)$ and $(1, 1)$ , which are 3 units left and right of the centre. So the radius is 3. The shaded region is inside the circle, with the boundary included. The locus $|z - z_0| = r$ is a circle of radius $r$ centred at $z_0$ , so the interior is $|z - z_0| \leq r$ . Substituting $z_0 = -2 + i$ and $r = 3$ gives $|z - (-2 + i)| \leq 3$ , or equivalently $|z + 2 - i| \leq 3$ .

Key Takeaways

On an Argand diagram, $\text{Im}(z) = c$ is a horizontal line, and $\text{Im}(z) \leq c$ is the half-plane below it.
The locus $|z - z_0| = r$ is a circle of radius $r$ centred at $z_0$ . The inequality $|z - z_0| \leq r$ represents the closed disk.
A region defined by multiple conditions is the intersection of the individual regions.

Common Mistakes

Writing $\text{Im}(z) \geq -1$ instead of $\leq -1$ by misreading which side of the line is shaded.
Getting the centre of the circle wrong, e.g. writing $|z + 2 + i| \leq 3$ instead of $|z + 2 - i| \leq 3$ . Remember that $z_0 = -2 + i$ , so $z - z_0 = z - (-2 + i) = z + 2 - i$ .
Using $<$ instead of $\leq$ when the boundary is included.

Things to Be Careful About

The mark scheme accepts strict inequalities throughout part (a), but the question states the boundaries are included, so $\leq$ is correct.
When writing $|z - (-2 + i)|$ , simplify to $|z + 2 - i|$ for the final answer.
The two inequalities together define the region; neither alone is sufficient.

Techniques used

read inequalities from Argand diagramidentify circle centre and radiuswrite modulus inequality for circular region

(b)

Find the greatest value of $|z|$ for points in this region.

3M Similar

DifficultyMedium

Worked solution

Approach

The greatest value of $|z|$ in the shaded region occurs at the point furthest from the origin. Since the shaded region is a circular segment below $y = -1$ , the maximum modulus will occur at one of the intersection points of the circle and the line $\text{Im}(z) = -1$ .

Working

Step 1: Find the intersection points.

The circle has equation $(x + 2)^2 + (y - 1)^2 = 9$ and the line is $y = -1$ . Substitute $y = -1$ into the circle equation:

(x + 2)^2 + (-1 - 1)^2 = 9

(x + 2)^2 + 4 = 9

(x + 2)^2 = 5

x + 2 = \pm\sqrt{5}

x = -2 \pm \sqrt{5}

The two intersection points are $(-2 - \sqrt{5}, -1)$ and $(-2 + \sqrt{5}, -1)$ , corresponding to the complex numbers $z_1 = -2 - \sqrt{5} - i$ and $z_2 = -2 + \sqrt{5} - i$ .

Step 2: Calculate $|z|$ at each intersection point.

For $z_1 = (-2 - \sqrt{5}) - i$ :

|z_1| = \sqrt{(-2 - \sqrt{5})^2 + (-1)^2} = \sqrt{4 + 4\sqrt{5} + 5 + 1} = \sqrt{10 + 4\sqrt{5}}

For $z_2 = (-2 + \sqrt{5}) - i$ :

|z_2| = \sqrt{(-2 + \sqrt{5})^2 + (-1)^2} = \sqrt{4 - 4\sqrt{5} + 5 + 1} = \sqrt{10 - 4\sqrt{5}}

Step 3: Compare and verify.

Since $\sqrt{10 + 4\sqrt{5}} > \sqrt{10 - 4\sqrt{5}}$ , the greater value is at $z_1$ .

We can also check the bottom of the circle at $(-2, -2)$ :

|z| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} \approx 2.83

This is less than $\sqrt{10 + 4\sqrt{5}} \approx 4.35$ , confirming the maximum is at the intersection point.

Step 4: Compute the numerical value.

\sqrt{10 + 4\sqrt{5}} = \sqrt{10 + 4(2.2361\ldots)} = \sqrt{10 + 8.9443\ldots} = \sqrt{18.9443\ldots} \approx 4.35

Answer

4.35 \quad \text{(or } \sqrt{10 + 4\sqrt{5}}\text{)}

Final answer

4.35

Detailed explanation

Walkthrough

Step 1: Locate where the maximum $|z|$ occurs.

The quantity $|z|$ is the distance from the origin to the point $z$ on the Argand diagram. We need the point in the shaded region (the circular segment below $y = -1$ ) that is furthest from the origin.

The shaded region is bounded by the arc of the circle below $y = -1$ and the chord along $y = -1$ . The furthest point from the origin on this closed region must lie on the boundary. By inspection, the leftmost intersection point $(-2 - \sqrt{5}, -1)$ is furthest from the origin.

Step 2: Find the intersection points.

The circle $(x+2)^2 + (y-1)^2 = 9$ meets the line $y = -1$ . Substituting gives $(x+2)^2 = 5$ , so $x = -2 \pm \sqrt{5}$ .

Step 3: Compute moduli.

At $(-2 - \sqrt{5}, -1)$ : $|z| = \sqrt{(-2-\sqrt{5})^2 + 1} = \sqrt{10 + 4\sqrt{5}} \approx 4.35$ .

At $(-2 + \sqrt{5}, -1)$ : $|z| = \sqrt{(-2+\sqrt{5})^2 + 1} = \sqrt{10 - 4\sqrt{5}} \approx 1.00$ .

The bottom of the circle $(-2, -2)$ gives $|z| = \sqrt{8} \approx 2.83$ , which is smaller.

So the greatest value is $\sqrt{10 + 4\sqrt{5}} \approx 4.35$ .

Key Takeaways

The maximum of $|z|$ over a closed bounded region occurs on the boundary.
For a circular segment cut by a line, the extreme values of $|z|$ often occur at the intersection points of the circle and the line.
Always check multiple candidate points to confirm the maximum.

Common Mistakes

Assuming the maximum $|z|$ is at the bottom of the circle $(-2, -2)$ without checking the intersection points.
Making sign errors when expanding $(-2 \pm \sqrt{5})^2$ .
Forgetting to include the $+1$ from the $y$ -coordinate when computing $|z| = \sqrt{x^2 + y^2}$ .

Things to Be Careful About

The question asks for the greatest value of $|z|$ , not the point $z$ itself.
AWRT (Answer Written To Required Total) 4.35 is accepted; give at least 3 significant figures.
The mark scheme accepts $\sqrt{10 + 4\sqrt{5}}$ as an exact answer.

Techniques used

find intersection of circle and linecalculate modulus from Cartesian coordinatescompare values to find maximum

Q46MMediumTrigonometry

By first expressing the equation $\tan(x - 60^\circ) = 2\cot x$ as a quadratic equation in $\tan x$ , solve the equation for $0^\circ \le x \le 180^\circ$ .

DifficultyMedium

Worked solution

Approach

Use the compound angle formula for $\tan(A-B)$ to expand $\tan(x - 60^\circ)$ , write $\cot x$ as $\frac{1}{\tan x}$ , then rearrange into a quadratic in $\tan x$ . Solve the quadratic and use the period of $\tan$ to find all solutions in $0^\circ \le x \le 180^\circ$ .

Working

Let $t = \tan x$ . Since $\tan 60^\circ = \sqrt{3}$ ,

\tan(x - 60^\circ) = \frac{\tan x - \tan 60^\circ}{1 + \tan x \tan 60^\circ} = \frac{t - \sqrt{3}}{1 + \sqrt{3}t}

Also,

2\cot x = \frac{2}{\tan x} = \frac{2}{t}

So the equation becomes

\frac{t - \sqrt{3}}{1 + \sqrt{3}t} = \frac{2}{t}

Cross-multiplying:

t(t - \sqrt{3}) = 2(1 + \sqrt{3}t)

t^2 - \sqrt{3}t = 2 + 2\sqrt{3}t

t^2 - 3\sqrt{3}t - 2 = 0

Solve using the quadratic formula:

t = \frac{3\sqrt{3} \pm \sqrt{(3\sqrt{3})^2 + 8}}{2} = \frac{3\sqrt{3} \pm \sqrt{35}}{2}

Thus

t_1 = \frac{3\sqrt{3} + \sqrt{35}}{2} \approx 5.556

t_2 = \frac{3\sqrt{3} - \sqrt{35}}{2} \approx -0.360

For $t_1 = 5.556$ :

x = \tan^{-1}(5.556) \approx 79.8^\circ

For $t_2 = -0.360$ , the principal value is negative:

\tan^{-1}(-0.360) \approx -19.8^\circ

Adding $180^\circ$ gives the solution in the required interval:

x = 180^\circ - 19.8^\circ = 160.2^\circ

The next addition of $180^\circ$ gives $340.2^\circ$ , outside the interval.

Answer

x = 79.8^\circ \quad \text{or} \quad x = 160.2^\circ

Final answer

x = 79.8° or x = 160.2°

Detailed explanation

Walkthrough

The equation is

\tan(x - 60^\circ) = 2\cot x

We need to turn this into an equation involving only $\tan x$ . Start with the left-hand side. The compound angle formula for tangent says

\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

Here $A = x$ and $B = 60^\circ$ , and $\tan 60^\circ = \sqrt{3}$ . So

\tan(x - 60^\circ) = \frac{\tan x - \sqrt{3}}{1 + \sqrt{3}\tan x}

This matches the first mark in the mark scheme.

Next, rewrite the right-hand side. Since $\cot x = \frac{1}{\tan x}$ ,

2\cot x = \frac{2}{\tan x}

Now the equation is

\frac{\tan x - \sqrt{3}}{1 + \sqrt{3}\tan x} = \frac{2}{\tan x}

Cross-multiply to clear the fractions:

\tan x(\tan x - \sqrt{3}) = 2(1 + \sqrt{3}\tan x)

Expand both sides:

\tan^2 x - \sqrt{3}\tan x = 2 + 2\sqrt{3}\tan x

Bring all terms to one side:

\tan^2 x - 3\sqrt{3}\tan x - 2 = 0

This is the quadratic equation in $\tan x$ required by the question.

Let $t = \tan x$ and solve:

t = \frac{3\sqrt{3} \pm \sqrt{(3\sqrt{3})^2 + 8}}{2} = \frac{3\sqrt{3} \pm \sqrt{35}}{2}

So the two values are approximately $5.556$ and $-0.360$ .

For $t \approx 5.556$ , the angle is

x = \tan^{-1}(5.556) \approx 79.8^\circ

For $t \approx -0.360$ , the calculator gives $\tan^{-1}(-0.360) \approx -19.8^\circ$ . Since we need $0^\circ \le x \le 180^\circ$ , add $180^\circ$ :

x = -19.8^\circ + 180^\circ = 160.2^\circ

Adding $180^\circ$ again gives $340.2^\circ$ , which is outside the interval, so there are no more solutions.

Key Takeaways

This question tests the compound angle formula for $\tan(A-B)$ , the reciprocal identity $\cot x = \frac{1}{\tan x}$ , and solving a quadratic equation in $\tan x$ . It also tests the periodic nature of the tangent function: $\tan$ has period $180^\circ$ , so once you have one angle for a given value of $\tan x$ , adding or subtracting $180^\circ$ gives another possible angle. You must check which angles lie in the required interval.

Common Mistakes

Forgetting the plus sign in the denominator of the $\tan(A-B)$ formula: the formula is $\frac{\tan A - \tan B}{1 + \tan A \tan B}$ , not $1 - \tan A \tan B$ .
Cross-multiplying incorrectly, especially missing the factor $2$ on the right-hand side.
Stopping after finding only one angle. The negative value of $\tan x$ gives a second solution in the second quadrant.
Using radians instead of degrees. The mark scheme treats answers in radians as a misread.
Giving an unsupported final answer without showing the quadratic equation or the solving method.

Things to Be Careful About

In the quadratic formula, $(3\sqrt{3})^2 = 27$ , so the discriminant is $27 + 8 = 35$ .
When $\tan x$ is negative, the principal value from a calculator is negative and outside $[0^\circ, 180^\circ]$ ; add $180^\circ$ to obtain the angle in the second quadrant.
The interval is $0^\circ \le x \le 180^\circ$ , so ignore any angles outside it, such as $340.2^\circ$ .
The mark scheme allows decimals throughout, so approximate answers such as $79.8^\circ$ and $160.2^\circ$ are acceptable.

Techniques used

expand tan(A - B) using the compound angle formularewrite cot x as the reciprocal of tan xform and solve a quadratic equation in tan xuse the periodicity of tan to find all solutions in the interval

Similar questions

Q55MMediumComplex Numbers

The square roots of $-4 + 6\sqrt{5}i$ can be expressed in the Cartesian form $x + iy$ , where $x$ and $y$ are real and exact.

By first forming a quartic equation in $x$ or $y$ , find the square roots of $-4 + 6\sqrt{5}i$ in exact Cartesian form.

DifficultyMedium

Worked solution

Approach

Let the square root be $x + iy$ . Squaring it and equating the real and imaginary parts to the target complex number gives two equations. The condition $xy = 3\sqrt{5}$ lets us eliminate one variable and obtain a quartic equation in the other. Solving the quartic gives the possible values of $x$ and $y$ , and the sign of $xy$ fixes the correct pairing.

Working

Let $(x + iy)^2 = -4 + 6\sqrt{5}\,i$ . Expanding:

(x + iy)^2 = x^2 - y^2 + 2xy\,i

Equating real and imaginary parts:

x^2 - y^2 = -4

2xy = 6\sqrt{5} \quad \Rightarrow \quad xy = 3\sqrt{5}

From $xy = 3\sqrt{5}$ :

x^2 y^2 = 45

Also $x^2 - y^2 = -4$ gives $y^2 = x^2 + 4$ . Substitute into $x^2 y^2 = 45$ :

x^2(x^2 + 4) = 45

x^4 + 4x^2 - 45 = 0

Let $u = x^2$ . Then:

u^2 + 4u - 45 = 0

(u + 9)(u - 5) = 0

Since $u = x^2 \geq 0$ , $u = 5$ , so:

x = \pm \sqrt{5}

Then:

y^2 = x^2 + 4 = 5 + 4 = 9

so $y = \pm 3$ . The condition $xy = 3\sqrt{5} > 0$ requires $x$ and $y$ to have the same sign. Hence the two square roots are:

\sqrt{5} + 3i \quad \text{and} \quad -\sqrt{5} - 3i

Answer

\pm(\sqrt{5} + 3i)

Final answer

±(√5 + 3i)

Detailed explanation

Walkthrough

We start by writing an unknown square root as $x + iy$ . Squaring it must give the original number, so we expand and compare real and imaginary parts. This is the key first step: it turns one complex equation into two real equations.

The two equations are $x^2 - y^2 = -4$ and $2xy = 6\sqrt{5}$ . The second simplifies to $xy = 3\sqrt{5}$ . To follow the instruction to form a quartic, we use $xy = 3\sqrt{5}$ to write $x^2 y^2 = 45$ , and use the first equation to replace $y^2$ by $x^2 + 4$ . Substituting gives $x^2(x^2+4)=45$ , i.e. $x^4 + 4x^2 - 45 = 0$ .

This is a quadratic in $x^2$ . Letting $u=x^2$ gives $(u+9)(u-5)=0$ . Since $u$ is a square, $u=-9$ is impossible, so $u=5$ and $x = \pm \sqrt{5}$ . Then $y^2 = x^2+4 = 9$ , so $y = \pm 3$ .

Finally, the sign condition $xy = 3\sqrt{5}$ is positive, so $x$ and $y$ must have the same sign. Therefore the valid pairs are $(\sqrt{5}, 3)$ and $(-\sqrt{5}, -3)$ , giving the two square roots.

Key Takeaways

To find square roots of a complex number, square $x+iy$ and equate real and imaginary parts.
The product condition $xy$ fixes the relative signs of $x$ and $y$ .
A quartic in $x$ or $y$ can often be reduced to a quadratic by substituting $u=x^2$ or $u=y^2$ .
Exact answers require exact surds and careful sign pairing.

Common Mistakes

Forgetting to equate both real and imaginary parts.
Losing the factor $2$ in $2xy$ .
Using $x^2 y^2 = 45$ but substituting the wrong expression for $y^2$ ; note $x^2 - y^2 = -4$ gives $y^2 = x^2 + 4$ , not $y^2 = x^2 - 4$ .
Accepting $u=-9$ as a possible value of $x^2$ ; since $x$ is real, $x^2$ cannot be negative.
Giving $x = \pm\sqrt{5}$ and $y = \pm 3$ without pairing them according to the sign of $xy$ . The mark scheme requires the pairs to be clearly stated.

Things to Be Careful About

The final answer must be written as two complex numbers, e.g. $\pm(\sqrt{5}+3i)$ , not just as four separate values of $x$ and $y$ .
The question asks for exact Cartesian form, so leave answers as surds, not decimals.
The sign condition is essential: $xy = 3\sqrt{5} > 0$ rules out $(\sqrt{5}, -3)$ and $(-\sqrt{5}, 3)$ .
If using the modulus method, remember $x^2 + y^2 = 14$ ; combining it with $x^2 - y^2 = -4$ gives the same values.

Techniques used

square a complex number and equate real and imaginary partseliminate one variable to form a quartic equationreduce the quartic to a quadratic by substitutionpair signs using the product condition

Similar questions

Q67MMediumDifferential EquationsTrigonometry

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

\frac{dx}{d\theta} = \left(\frac{1}{5}x + 1\right)\sin^2 2\theta

and $x = 5$ when $\theta = 0$ .

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

DifficultyMedium

Worked solution

Approach

Use the double-angle identity to rewrite $\sin^2 2\theta$ , then separate the variables and integrate both sides. Apply the initial condition $x=5$ when $\theta=0$ to find the constant, and finally rearrange to make $x$ the subject.

Working

Rewrite the trigonometric factor:

\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)

Separate the variables:

\frac{dx}{\frac{1}{5}x + 1} = \sin^2 2\theta \, d\theta

Integrate both sides. The left-hand side is

\int \frac{dx}{\frac{1}{5}x + 1} = 5\ln\left(\frac{1}{5}x + 1\right)

which is equivalent to $5\ln(x+5) - 5\ln 5$ . The right-hand side is

\int \sin^2 2\theta \, d\theta = \int \frac{1}{2}(1 - \cos 4\theta)\, d\theta = \frac{1}{2}\left(\theta - \frac{1}{4}\sin 4\theta\right)

Therefore,

5\ln\left(\frac{1}{5}x + 1\right) = \frac{1}{2}\left(\theta - \frac{1}{4}\sin 4\theta\right) + C

Use $x=5$ when $\theta=0$ . Since $\frac{1}{5}(5)+1=2$ , this gives

5\ln 2 = C

So

5\ln\left(\frac{1}{5}x + 1\right) = \frac{1}{2}\left(\theta - \frac{1}{4}\sin 4\theta\right) + 5\ln 2

Divide by 5:

\ln\left(\frac{1}{5}x + 1\right) = \frac{1}{10}\left(\theta - \frac{1}{4}\sin 4\theta\right) + \ln 2

Exponentiate:

\frac{1}{5}x + 1 = 2\exp\left[\frac{1}{10}\left(\theta - \frac{1}{4}\sin 4\theta\right)\right]

Hence

x = 10\exp\left[\frac{1}{10}\left(\theta - \frac{1}{4}\sin 4\theta\right)\right] - 5

Answer

x = 10\exp\left[\frac{1}{10}\left(\theta - \frac{1}{4}\sin 4\theta\right)\right] - 5

Final answer

x = 10 exp[(1/10)(θ - (1/4) sin 4θ)] - 5

Detailed explanation

Walkthrough

This is a first-order separable differential equation. The first difficulty is that the right-hand side contains $\sin^2 2\theta$ , which is not directly one of the standard integrable forms. Use the double-angle identity $\sin^2 A = \frac{1}{2}(1 - \cos 2A)$ with $A = 2\theta$ , giving $\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)$ . This converts the trigonometric factor into a sum of a constant and a cosine term, both easy to integrate.

Next, separate the variables: divide by $\frac{1}{5}x + 1$ and multiply by $d\theta$ , so all $x$ terms are on one side and all $\theta$ terms on the other. Then integrate each side separately.

For the left side, $\int \frac{dx}{\frac{1}{5}x + 1}$ is a reciprocal-linear integral. Since the derivative of $\ln\left(\frac{1}{5}x + 1\right)$ is $\frac{1/5}{\frac{1}{5}x + 1}$ , multiplying by 5 gives the correct integrand. Thus the integral is $5\ln\left(\frac{1}{5}x + 1\right)$ , or equivalently $5\ln(x+5) - 5\ln 5$ .

For the right side, integrate $\frac{1}{2}(1 - \cos 4\theta)$ . The integral of 1 is $\theta$ , and the integral of $\cos 4\theta$ is $\frac{1}{4}\sin 4\theta$ , so the result is $\frac{1}{2}\left(\theta - \frac{1}{4}\sin 4\theta\right)$ .

Add the constant of integration $C$ . Use the condition $x=5$ when $\theta=0$ : at that point $\frac{1}{5}x + 1 = 2$ , so the left side is $5\ln 2$ , while the right side is $0 + C$ . Hence $C = 5\ln 2$ .

Finally, divide by 5, write the constant as $\ln 2$ , exponentiate both sides, and rearrange to make $x$ the subject. The factor 2 from $e^{\ln 2}$ combines with the 5 from $x = 5\left(\frac{1}{5}x + 1\right) - 5$ to produce the factor 10 in the final answer.

Key Takeaways

Separable differential equations are solved by collecting each variable with its own differential and integrating both sides.
Powers of sine and cosine can often be integrated after applying a double-angle identity.
The constant of integration is determined by the initial condition.
When solving for the dependent variable, exponentiate carefully and simplify any constants inside the exponential.

Common Mistakes

Forgetting the factor 5 when integrating $\frac{1}{\frac{1}{5}x + 1}$ ; always check by differentiating the proposed antiderivative.
Omitting the $\frac{1}{4}$ when integrating $\cos 4\theta$ .
Dropping the constant of integration before using the initial condition.
Failing to show the separation step, which is required for the method mark.
Leaving the answer with a logarithm still present when the question asks for an explicit expression for $x$ .

Things to Be Careful About

At $\theta = 0$ , both $\theta$ and $\sin 4\theta$ are zero, so the right-hand side of the integrated equation is just the constant.
The argument $\frac{1}{5}x + 1$ is positive for the given initial value, so absolute-value signs are not needed here.
When exponentiating, remember that $e^{\ln 2} = 2$ , and that the 5 outside must be distributed correctly: $x = 5\left(2\exp[\dots]\right) - 5 = 10\exp[\dots] - 5$ .
The mark scheme requires the final answer to have the logarithm removed; an unsupported or implicit answer may not receive full credit.

Techniques used

separate variables and integrate both sidesapply a double-angle identity to rewrite a squared trigonometric termintegrate a reciprocal linear expressionuse an initial condition to determine the constant of integrationexponentiate and rearrange to solve for the dependent variable

Similar questions

Q7Medium-EasyDifferentiationNumerical Solution of Equations

Fig. 2

The diagram shows the curve $y = x^3 \cos 2x$ for $0 \le x \le \frac{1}{4}\pi$ . The curve has a maximum point at $M$ , where $x = p$ .

(a)

Show that $p$ satisfies the equation $p = \frac{1}{2}\tan^{-1}\left(\frac{3}{2p}\right)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

To find the equation satisfied by the x-coordinate $p$ of the maximum point $M$ , we differentiate $y = x^3 \cos 2x$ using the product rule, set the derivative equal to zero (since the gradient is zero at a maximum), and rearrange the resulting equation to match the required form.

Working

Let $y = x^3 \cos 2x$ . Differentiate with respect to $x$ using the product rule:

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

\frac{dy}{dx} = x^3(-2 \sin 2x) + \cos 2x(3x^2)

\frac{dy}{dx} = 3x^2 \cos 2x - 2x^3 \sin 2x

At the maximum point $M$ , $x = p$ and $\frac{dy}{dx} = 0$ :

3p^2 \cos 2p - 2p^3 \sin 2p = 0

Since $p > 0$ (as the maximum lies in $0 < x < \frac{1}{4}\pi$ ), divide through by $p^2$ :

3 \cos 2p - 2p \sin 2p = 0

Rearrange to separate the sine and cosine terms:

3 \cos 2p = 2p \sin 2p

\frac{\sin 2p}{\cos 2p} = \frac{3}{2p}

\tan 2p = \frac{3}{2p}

Take the inverse tangent of both sides:

2p = \tan^{-1}\left(\frac{3}{2p}\right)

p = \frac{1}{2} \tan^{-1}\left(\frac{3}{2p}\right)

Answer

p = \frac{1}{2} \tan^{-1}\left(\frac{3}{2p}\right)

Final answer

p = 1/2 tan^-1(3/(2p))

Detailed explanation

Walkthrough

First, we apply the product rule to differentiate $y = x^3 \cos 2x$ . The derivative of $x^3$ is $3x^2$ and the derivative of $\cos 2x$ is $-2\sin 2x$ . Combining these gives $\frac{dy}{dx} = 3x^2 \cos 2x - 2x^3 \sin 2x$ . At a maximum point, the gradient is zero, so we substitute $x = p$ and set the derivative to zero. Factoring out $p^2$ (which is non-zero) leaves $3 \cos 2p = 2p \sin 2p$ . Dividing both sides by $2p \cos 2p$ yields $\tan 2p = \frac{3}{2p}$ . Finally, applying the inverse tangent function and dividing by 2 gives the required equation.

Key Takeaways

The product rule is essential for differentiating products of functions.
Stationary points occur where the first derivative is zero.
Trigonometric identities like $\frac{\sin \theta}{\cos \theta} = \tan \theta$ are useful for rearranging equations involving multiple trig functions.

Common Mistakes

Forgetting the chain rule when differentiating $\cos 2x$ , resulting in $-\sin 2x$ instead of $-2\sin 2x$ .
Failing to divide by $p^2$ correctly or not justifying that $p \neq 0$ .
Incorrectly rearranging the trigonometric equation, such as writing $\tan 2p = \frac{2p}{3}$ instead of $\frac{3}{2p}$ .

Things to Be Careful About

Ensure all angles are in radians when using inverse trigonometric functions in later parts.
The division by $p^2$ is valid because the diagram shows $p > 0$ .
The mark scheme requires showing the intermediate step $\tan 2p = \frac{3}{2p}$ before arriving at the final answer.

Techniques used

apply product rule to differentiateset derivative to zero for stationary pointrearrange equation using inverse trigonometric function

(b)

Show by calculation that $0.5 < p < 0.7$ .

2M Similar

DifficultyEasy

Worked solution

Approach

To show that $p$ lies between 0.5 and 0.7, we define a function $f(x) = x - \frac{1}{2}\tan^{-1}\left(\frac{3}{2x}\right)$ based on the equation from part (a). Since $p$ is a root of $f(x) = 0$ , we evaluate $f(x)$ at $x = 0.5$ and $x = 0.7$ . If the signs are opposite, the Intermediate Value Theorem guarantees a root in the interval $(0.5, 0.7)$ .

Working

Let $f(x) = x - \frac{1}{2}\tan^{-1}\left(\frac{3}{2x}\right)$ .

Evaluate at $x = 0.5$ :

f(0.5) = 0.5 - \frac{1}{2}\tan^{-1}\left(\frac{3}{2(0.5)}\right) = 0.5 - \frac{1}{2}\tan^{-1}(3)

f(0.5) \approx 0.5 - 0.62452 = -0.12452

Evaluate at $x = 0.7$ :

f(0.7) = 0.7 - \frac{1}{2}\tan^{-1}\left(\frac{3}{2(0.7)}\right) = 0.7 - \frac{1}{2}\tan^{-1}\left(\frac{3}{1.4}\right)

f(0.7) \approx 0.7 - 0.56781 = 0.13219

Since $f(0.5) < 0$ and $f(0.7) > 0$ , there is a sign change in the interval $[0.5, 0.7]$ . Because $f(x)$ is continuous for $x > 0$ , there must be a root $p$ in the interval $0.5 < p < 0.7$ .

Answer

0.5 < p < 0.7

Final answer

0.5 < p < 0.7

Detailed explanation

Walkthrough

We rearrange the equation from part (a) into the form $f(x) = 0$ , where $f(x) = x - \frac{1}{2}\tan^{-1}\left(\frac{3}{2x}\right)$ . To locate the root $p$ , we calculate $f(0.5)$ and $f(0.7)$ . Using a calculator in radian mode, $f(0.5) \approx -0.12452$ and $f(0.7) \approx 0.13219$ . Since the function changes sign between these two values and is continuous, the Intermediate Value Theorem confirms a root exists in the interval $(0.5, 0.7)$ .

Key Takeaways

Setting up a function $f(x) = 0$ from an equation is a standard method for locating roots.
The Intermediate Value Theorem states that if a continuous function changes sign over an interval, it must have at least one root in that interval.

Common Mistakes

Using the wrong angle mode on the calculator (degrees instead of radians).
Failing to clearly state that there is a sign change and that the function is continuous.
Not showing the calculated values to sufficient decimal places.

Things to Be Careful About

Always ensure the calculator is in radian mode when dealing with inverse trigonometric functions in this context.
The mark scheme accepts comparing $x$ and $\frac{1}{2}\tan^{-1}\left(\frac{3}{2x}\right)$ directly, but using $f(x) = 0$ is a more rigorous approach.

Techniques used

define function from equationevaluate function at boundary valuesapply intermediate value theorem

(c)

Use an iterative formula based on the equation in part (a) to calculate $p$ correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

From part (a), we have the equation $p = \frac{1}{2}\tan^{-1}\left(\frac{3}{2p}\right)$ . We can use this to form an iterative formula: $x_{n+1} = \frac{1}{2}\tan^{-1}\left(\frac{3}{2x_n}\right)$ . We start with an initial value $x_1$ in the interval $(0.5, 0.7)$ and iterate until the result is correct to 3 decimal places.

Working

Use the iterative formula $x_{n+1} = \frac{1}{2}\tan^{-1}\left(\frac{3}{2x_n}\right)$ with $x_1 = 0.6$ :

x_1 = 0.60000

x_2 = \frac{1}{2}\tan^{-1}\left(\frac{3}{1.2}\right) = 0.59514

x_3 = \frac{1}{2}\tan^{-1}\left(\frac{3}{1.19028}\right) = 0.59654

x_4 = \frac{1}{2}\tan^{-1}\left(\frac{3}{1.19308}\right) = 0.59614

x_5 = \frac{1}{2}\tan^{-1}\left(\frac{3}{1.19228}\right) = 0.59626

x_6 = \frac{1}{2}\tan^{-1}\left(\frac{3}{1.19252}\right) = 0.59623

The values are converging to $0.596$ correct to 3 decimal places. To confirm, we check the bounds:

x_5 = 0.59626 > 0.5955

x_6 = 0.59623 < 0.5965

Since the iteration values lie within $(0.5955, 0.5965)$ , the root is $0.596$ correct to 3 decimal places.

Answer

p = 0.596

Final answer

0.596

Detailed explanation

Walkthrough

We rearrange the equation from part (a) into the form $x_{n+1} = F(x_n)$ , giving $x_{n+1} = \frac{1}{2}\tan^{-1}\left(\frac{3}{2x_n}\right)$ . Starting with $x_1 = 0.6$ , we substitute this value into the right-hand side to get $x_2$ , then substitute $x_2$ to get $x_3$ , and so on. We continue until two consecutive iterations agree to 3 decimal places. We then verify the result by checking that the value lies within the rounding interval $(0.5955, 0.5965)$ .

Key Takeaways

Iterative formulas allow us to approximate roots of equations numerically.
Convergence is confirmed when successive iterations yield the same value to the required number of decimal places.
Always check the rounding bounds to ensure the answer is correct to the specified precision.

Common Mistakes

Using the wrong iterative formula (e.g., forgetting the $\frac{1}{2}$ factor).
Not carrying enough decimal places during intermediate iterations, leading to rounding errors.
Failing to verify the answer using the rounding interval $(0.5955, 0.5965)$ .

Things to Be Careful About

Keep all intermediate values to at least 5 decimal places as required by the mark scheme.
Ensure the calculator is in radian mode.
The mark scheme accepts starting with any initial value in the interval $(0.5, 0.7)$ , such as $0.5$ or $0.7$ .

Techniques used

formulate iterative equationiterate to convergencecheck rounding bounds

Q8MediumVectors

Two lines have equations

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

(a)

Show that the lines are skew.

5M Similar

DifficultyMedium

Worked solution

Approach

To show that two lines are skew, we must demonstrate two things:

The lines do not intersect (the system of equations obtained by equating components has no solution).
The lines are not parallel (their direction vectors are not scalar multiples of each other).

We will express each line in component form, equate the components, solve the resulting system, and check that it is inconsistent. We will then verify the direction vectors are not parallel.

Working

Express the general point of each line in component form.

For line 1 with parameter $\lambda$ :

\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 + 2\lambda \\ 3 + 3\lambda \\ -4 - \lambda \end{pmatrix}

For line 2 with parameter $\mu$ :

\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 - \mu \\ -3 - 2\mu \\ -1 + \mu \end{pmatrix}

Equate the components to find any intersection point:

\begin{aligned} -1 + 2\lambda &= 2 - \mu \\ 3 + 3\lambda &= -3 - 2\mu \\ -4 - \lambda &= -1 + \mu \end{aligned}

Rearrange each equation into standard form:

\begin{aligned} 2\lambda + \mu &= 3 \quad &\text{...(1)}\\ 3\lambda + 2\mu &= -6 \quad &\text{...(2)}\\ \lambda + \mu &= -3 \quad &\text{...(3)} \end{aligned}

Subtract equation (3) from equation (1):

\lambda = 6

Substitute $\lambda = 6$ into equation (3):

6 + \mu = -3 \quad \Rightarrow \quad \mu = -9

Check these values in equation (2):

3(6) + 2(-9) = 18 - 18 = 0 \neq -6

Since equation (2) is not satisfied, the three component equations are inconsistent. The lines do not intersect.

Now check whether the lines are parallel. The direction vectors are:

\mathbf{d_1} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}, \quad \mathbf{d_2} = \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}

If they were parallel, there would exist a constant $k$ such that $\mathbf{d_1} = k\mathbf{d_2}$ . From the first component, $k = -2$ . But checking with the second component, $k = -3/2$ . Since $-2 \neq -3/2$ , the direction vectors are not scalar multiples of each other, so the lines are not parallel.

Because the lines are neither parallel nor intersecting, they are skew.

Answer

The lines are skew.

Final answer

The lines are skew.

Detailed explanation

Walkthrough

To prove two lines in 3D are skew, we need to rule out two simpler possibilities: that they intersect, and that they are parallel. A line given in vector form $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ sweeps out every point by adding scalar multiples of the direction vector $\mathbf{d}$ to a fixed point $\mathbf{a}$ .

First, we wrote each line's general point in component form by adding the parameter times each component of the direction vector to the corresponding component of the fixed point. This is a direct application of the line equation.

Next, we asked: can we choose $\lambda$ and $\mu$ so that the two lines pass through the same point? This is equivalent to solving the system of three equations obtained by equating the $x$ , $y$ , and $z$ components. Each line is parameterised by one variable, so we have three equations in two unknowns - the system is overdetermined, and a solution exists only if all three equations are consistent.

We picked the simplest pair of equations (the $x$ and $z$ equations) to solve. Subtracting them eliminated $\mu$ and gave us $\lambda = 6$ directly. Substituting back gave $\mu = -9$ . We then had to verify whether these values also satisfied the third equation ( $y$ component). Plugging in, the left side gave $0$ , but the right side required $-6$ . This contradiction proves the system is inconsistent, so the lines do not meet.

Finally, we ruled out parallelism. Two lines are parallel exactly when their direction vectors are scalar multiples. We tested whether $\mathbf{d_1} = k\mathbf{d_2}$ for some constant $k$ . The first component gave $k = -2$ , but the second gave $k = -3/2$ . These conflict, so the direction vectors are not proportional and the lines are not parallel.

A line that is neither intersecting nor parallel to another must be skew, so the proof is complete.

Key Takeaways

A pair of lines in 3D can be skew, intersecting, or parallel (or coincident). To prove they are skew, both non-intersection and non-parallelism must be established.
The non-intersection test uses the fact that equating components gives an overdetermined system (3 equations in 2 unknowns) that may be inconsistent.
Non-parallelism is shown by demonstrating that the direction vectors are not scalar multiples of one another.

Common Mistakes

Showing the system is inconsistent alone is not enough: parallel lines also have inconsistent component systems (because they can never meet, even if they lie in the same direction). The non-parallel check is essential.
A bare statement that "the direction vectors are not scalar multiples" without showing the calculation is not sufficient - mark scheme requires demonstrating this with at least two components.
Forgetting to check all three components of consistency - solving two equations and not verifying the third.

Things to Be Careful About

When subtracting equations, keep careful track of signs to avoid sign errors.
The parallel check should be done with at least two components, not just one (since two non-zero direction vectors always satisfy the first component relation for some $k$ - it is the second that usually fails).
The lines are skew in 3D, but in 2D the same definitions lead to only two possibilities (intersecting or parallel) because the third dimension is unavailable.

Techniques used

express general point in component formequate components to test for intersectionsolve simultaneous equationscheck for consistency across all three componentstest direction vectors for parallelism

(b)

Find the obtuse angle between the directions of the two lines.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The angle between two lines is determined entirely by their direction vectors. We will compute the scalar product and the magnitudes of the two direction vectors, then use the formula $\cos\theta = \dfrac{\mathbf{d_1} \cdot \mathbf{d_2}}{|\mathbf{d_1}||\mathbf{d_2}|}$ to find the angle. Because the scalar product is negative, the angle is obtuse.

Working

The direction vectors are:

\mathbf{d_1} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}, \quad \mathbf{d_2} = \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}

Compute the scalar product:

\mathbf{d_1} \cdot \mathbf{d_2} = (2)(-1) + (3)(-2) + (-1)(1) = -2 - 6 - 1 = -9

Compute the magnitudes:

|\mathbf{d_1}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}

|\mathbf{d_2}| = \sqrt{(-1)^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}

Apply the cosine formula:

\cos\theta = \frac{\mathbf{d_1} \cdot \mathbf{d_2}}{|\mathbf{d_1}||\mathbf{d_2}|} = \frac{-9}{\sqrt{14} \cdot \sqrt{6}} = \frac{-9}{\sqrt{84}}

Simplify $\sqrt{84} = 2\sqrt{21}$ , so:

\cos\theta = \frac{-9}{2\sqrt{21}}

Take the inverse cosine:

\theta = \cos^{-1}\left(\frac{-9}{2\sqrt{21}}\right)

Evaluating numerically: $\dfrac{9}{2\sqrt{21}} \approx 0.98198$ , so $\theta = \cos^{-1}(-0.98198) \approx 169.1°$ .

Answer

The obtuse angle between the directions of the two lines is approximately $169.1°$ .

Final answer

169.1° (AWRT)

Detailed explanation

Walkthrough

The angle between two lines in 3D (or any dimension) is defined as the angle between their direction vectors. Importantly, this angle is always taken to be between $0°$ and $180°$ , so it can be obtuse.

We begin with the direction vectors identified in part (a). The angle $\theta$ between two vectors is given by:

\cos\theta = \frac{\mathbf{d_1} \cdot \mathbf{d_2}}{|\mathbf{d_1}||\mathbf{d_2}|}

The scalar product is computed by multiplying corresponding components and summing. Here we obtained $-9$ , which is negative. This sign immediately tells us the angle is obtuse (between $90°$ and $180°$ ), since the cosine of an obtuse angle is negative.

The magnitude of each vector is computed as the square root of the sum of squared components. We obtained $\sqrt{14}$ and $\sqrt{6}$ , whose product is $\sqrt{84} = 2\sqrt{21}$ .

We then divided the scalar product by the product of magnitudes to obtain $\cos\theta$ , and used the inverse cosine function to find $\theta$ . Because $\cos\theta$ is negative, the inverse cosine returns an obtuse angle directly - we do not need to take $180°$ minus the acute result (the calculator's principal value already gives us the obtuse angle when the argument is negative).

Key Takeaways

The angle between two lines is the angle between their direction vectors, given by $\cos\theta = \dfrac{\mathbf{d_1} \cdot \mathbf{d_2}}{|\mathbf{d_1}||\mathbf{d_2}|}$ .
The result of $\cos^{-1}$ is always in $[0°, 180°]$ , so a negative cosine automatically gives the obtuse angle.
A negative scalar product means the angle is obtuse; a positive one means it is acute; zero means perpendicular.

Common Mistakes

Forgetting to take the modulus of vectors, or computing $|\mathbf{d}|^2$ instead of $|\mathbf{d}|$ .
Taking the acute angle by writing $\theta = \cos^{-1}(|\text{value}|)$ and reporting it when the question asks for the obtuse angle.
Switching direction vectors mid-calculation (using $\mathbf{d_1} \cdot \mathbf{d_2}$ for one part and $|\mathbf{d_2}||\mathbf{d_1}|$ for another).

Things to Be Careful About

The angle is taken between the directions, not necessarily the line segments - it is the same whether you walk along the direction vector or its negative.
Numerical answers should be quoted to a sensible accuracy; here AWRT $169.1°$ or $169°$ is acceptable.
The formula gives a unique angle in $[0°, 180°]$ , so no second solution is required.

Techniques used

compute scalar product of direction vectorscompute magnitudes of direction vectorsapply formula cos θ = (d₁·d₂)/(|d₁||d₂|)take inverse cosine to find obtuse angle

Q9MediumAlgebra

The polynomial $6x^3 + ax^2 + bx + 9$ is denoted by $p(x)$ , where $a$ and $b$ are constants. It is given that $(x - 3)$ is a factor of $p(x)$ , and when the first derivative $p'(x)$ is divided by $(x - 3)$ the remainder is 72.

(a)

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

5M Similar

DifficultyMedium

Worked solution

Approach

Since $(x-3)$ is a factor of $p(x)$ , the factor theorem gives $p(3)=0$ . Since $p'(x)$ leaves remainder $72$ when divided by $(x-3)$ , the remainder theorem gives $p'(3)=72$ . Differentiate $p(x)$ , substitute $x=3$ into both $p(x)$ and $p'(x)$ , and solve the resulting simultaneous equations.

Working

The polynomial is

p(x)=6x^3+ax^2+bx+9

and its derivative is

p'(x)=18x^2+2ax+b.

Using $p(3)=0$ :

6(3)^3+a(3)^2+b(3)+9=0

162+9a+3b+9=0

9a+3b=-171

3a+b=-57.

Using $p'(3)=72$ :

18(3)^2+2a(3)+b=72

162+6a+b=72

6a+b=-90.

Subtract the first simplified equation from the second:

3a=-33

a=-11.

Substitute into $3a+b=-57$ :

-33+b=-57

b=-24.

Answer

a=-11,\quad b=-24

Final answer

a = -11, b = -24

Detailed explanation

Walkthrough

The factor theorem says that if $(x-3)$ is a factor of $p(x)$ , then substituting $x=3$ into $p(x)$ must give $0$ . This gives the first equation in $a$ and $b$ . The remainder theorem says that when a polynomial is divided by $(x-3)$ , the remainder is the value of that polynomial at $x=3$ . Here the polynomial being divided is $p'(x)$ , so the remainder is $p'(3)$ , which is given as $72$ . First differentiate $p(x)$ to obtain $p'(x)=18x^2+2ax+b$ . Then substitute $x=3$ into both $p(x)$ and $p'(x)$ to form two linear equations. Solve them simultaneously, for example by subtracting one equation from the other to eliminate $b$ , then substitute back to find the other constant.

Key Takeaways

This question combines the factor theorem, the remainder theorem, and differentiation. It shows that the remainder theorem can be applied to any polynomial expression, including a derivative. The two conditions translate into a system of linear equations in the unknown coefficients.

Common Mistakes

Forgetting to differentiate before substituting into $p'(x)$ .
Using $p'(3)=0$ instead of $72$ .
Making sign errors when simplifying $162+9a+3b+9=0$ and $162+6a+b=72$ .
Solving the simultaneous equations incorrectly or not showing the substitution step.

Things to Be Careful About

The factor theorem gives $p(3)=0$ , not $p(-3)=0$ .
The remainder theorem for division by $x-3$ requires evaluating at $x=3$ .
When simplifying $171+9a+3b=0$ , divide by $3$ carefully to get $57+3a+b=0$ .
Check the final values by substituting them back into both equations.

Techniques used

apply the factor theoremapply the remainder theorem to the derivativesolve simultaneous linear equations

(b)

When $a$ and $b$ have the values found in part (a), factorise $p(x)$ completely.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the values of $a$ and $b$ from part (a), then write $p(x)$ as $(x-3)$ times a quadratic. Determine the quadratic by comparing coefficients, then factorise it completely.

Working

With $a=-11$ and $b=-24$ ,

p(x)=6x^3-11x^2-24x+9.

Let

p(x)=(x-3)(6x^2+Ax+B).

Expanding:

(x-3)(6x^2+Ax+B)=6x^3+(A-18)x^2+(B-3A)x-3B.

Comparing coefficients with $6x^3-11x^2-24x+9$ :

A-18=-11 \Rightarrow A=7,

-3B=9 \Rightarrow B=-3.

Check the middle coefficient:

B-3A=-3-21=-24.

So

p(x)=(x-3)(6x^2+7x-3).

Factorise the quadratic:

6x^2+7x-3=(2x+3)(3x-1).

Therefore

p(x)=(x-3)(2x+3)(3x-1).

Answer

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

Final answer

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

Detailed explanation

Walkthrough

After substituting $a=-11$ and $b=-24$ , write $p(x)$ as $6x^3-11x^2-24x+9$ . Since $(x-3)$ is already known to be a factor, the remaining factor must be a quadratic. Write it as $6x^2+Ax+B$ and expand $(x-3)(6x^2+Ax+B)$ . Comparing coefficients with the cubic gives equations for $A$ and $B$ : the $x^2$ coefficient gives $A-18=-11$ , and the constant term gives $-3B=9$ . Solving gives $A=7$ and $B=-3$ . The middle coefficient can be checked: $B-3A=-3-21=-24$ . Then factor the quadratic $6x^2+7x-3$ into $(2x+3)(3x-1)$ .

Key Takeaways

Once one factor of a cubic is known, the remaining factor is quadratic, found by division or coefficient comparison. Factorising the quadratic completely gives the full factorisation of the cubic.

Common Mistakes

Using the original $a$ and $b$ instead of the values found in part (a).
Sign errors in coefficient comparison, especially in $A-18$ and $-3B$ .
Failing to factor the quadratic completely.
Forgetting to include the $(x-3)$ factor in the final answer.

Things to Be Careful About

Check the middle coefficient: $B-3A=-3-21=-24$ , matching the cubic.
The final factorisation of a cubic must have three linear factors.
The order of the factors does not matter.

Techniques used

write polynomial as product with unknown quadraticcompare coefficientsfactorise quadratic

(c)

Hence solve the inequality $p(x) < 0$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the complete factorisation from part (b) to identify the roots. Then test the sign of $p(x)$ on the intervals between these roots to solve $p(x)<0$ .

Working

From part (b),

p(x)=(x-3)(2x+3)(3x-1).

The roots are

x=3,\quad x=-\frac{3}{2},\quad x=\frac{1}{3}.

Test one value in each interval:

For $x<-\frac{3}{2}$ , take $x=-2$ :

(-5)(-1)(-7)=-35<0.

For $-\frac{3}{2}<x<\frac{1}{3}$ , take $x=0$ :

(-3)(3)(-1)=9>0.

For $\frac{1}{3}<x<3$ , take $x=1$ :

(-2)(5)(2)=-20<0.

For $x>3$ , take $x=4$ :

(1)(11)(11)=121>0.

Thus $p(x)<0$ when

x<-\frac{3}{2}\quad \text{or}\quad \frac{1}{3}<x<3.

Final answer

x < -3/2 or 1/3 < x < 3

Detailed explanation

Walkthrough

Use the factorised form $p(x)=(x-3)(2x+3)(3x-1)$ . The roots are found by setting each factor equal to zero: $x=3$ , $x=-\frac{3}{2}$ , and $x=\frac{1}{3}$ . These roots split the real line into four intervals. Choose a convenient test value in each interval and evaluate the sign of the product. The product is negative when an odd number of the factors is negative. The intervals where the test value gives a negative product are the solution. Since the inequality is strict, the roots themselves are not included.

Key Takeaways

Solving a polynomial inequality after factorisation requires locating the roots and testing the sign on each interval. The sign of a product can be determined from the signs of its individual factors.

Common Mistakes

Including equality at the roots when the inequality is strict.
Missing one of the two solution intervals.
Testing only one interval and assuming the sign alternates without checking.
Writing the answer as a single combined interval.

Things to Be Careful About

The roots are $x=-\frac{3}{2}$ , $x=\frac{1}{3}$ , and $x=3$ .
The intervals are $x<-\frac{3}{2}$ , $-\frac{3}{2}<x<\frac{1}{3}$ , $\frac{1}{3}<x<3$ , and $x>3$ .
The answer must be written as two separate intervals: $x<-\frac{3}{2}$ or $\frac{1}{3}<x<3$ .
If the mark scheme uses and between the intervals, it means the union of the two regions.

Techniques used

find roots of factorised cubictest sign on each intervalwrite solution as union of intervals

Q10MediumAlgebraIntegration

Let $f(x) = \frac{-7x^2 + 2x - 6}{(1 + x)(4 + x^2)}$ .

(a)

Express $f(x)$ in partial fractions.

5M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since the denominator has a linear factor $(1+x)$ and an irreducible quadratic factor $(4+x^2)$ , use the partial fraction form

\frac{-7x^2+2x-6}{(1+x)(4+x^2)} = \frac{A}{1+x} + \frac{Bx+C}{4+x^2}.

Working

Multiply both sides by $(1+x)(4+x^2)$ :

-7x^2+2x-6 = A(4+x^2) + (Bx+C)(1+x).

Expand the right-hand side:

-7x^2+2x-6 = (A+B)x^2 + (B+C)x + (4A+C).

Compare coefficients:

A+B = -7, \qquad B+C = 2, \qquad 4A+C = -6.

From $A+B=-7$ , $B=-7-A$ . From $B+C=2$ ,

-7-A+C=2 \quad \Rightarrow \quad C=A+9.

Substitute into $4A+C=-6$ :

4A + (A+9) = -6 \quad \Rightarrow \quad 5A = -15 \quad \Rightarrow \quad A=-3.

Then

B = -7-(-3) = -4, \qquad C = -3+9 = 6.

Answer

f(x) = -\frac{3}{1+x} + \frac{6-4x}{4+x^2}.

Equivalently,

f(x) = -\frac{3}{1+x} - \frac{4x}{4+x^2} + \frac{6}{4+x^2}.

Final answer

f(x) = -3/(1+x) + (6-4x)/(4+x^2)

Detailed explanation

Walkthrough

We need to split the given rational function into partial fractions. Because the denominator is the product of a linear factor and a quadratic factor that has no real roots, the numerator over the quadratic must be allowed to be linear, $Bx+C$ .

Start by writing the proposed form and multiplying through by the denominator. This removes the fractions and gives a polynomial identity. Expanding and comparing coefficients of $x^2$ , $x$ and the constant term produces three equations in $A$ , $B$ and $C$ .

Solving the equations: from $A+B=-7$ we express $B$ in terms of $A$ ; from $B+C=2$ we express $C$ in terms of $A$ ; substituting into $4A+C=-6$ gives $A=-3$ . Back-substitution gives $B=-4$ and $C=6$ .

Key Takeaways

An irreducible quadratic denominator such as $4+x^2$ requires a linear numerator $Bx+C$ in the partial fraction.
Multiplying through by the denominator and comparing coefficients is a reliable method for finding the constants.
The final decomposition can be written with the linear term split into two separate fractions to make integration easier.

Common Mistakes

Using only a constant numerator $C$ over $4+x^2$ ; this is incorrect and loses the $Bx$ term.
Sign errors when expanding $(Bx+C)(1+x)$ .
Forgetting to compare the constant term as well as the $x^2$ and $x$ terms.

Things to Be Careful About

The quadratic $4+x^2$ has no real factors, so it must be treated as an irreducible quadratic.
Check the final decomposition by recombining the fractions or by substituting a convenient value of $x$ .
The mark scheme requires the form to be stated before the constants are found.

Techniques used

state the partial fraction form with a linear and an irreducible quadratic denominatormultiply through by the common denominatorcompare coefficients of powers of xsolve the resulting simultaneous linear equations

(b)

Hence find the exact value of $\int_0^2 f(x)\,dx$ . Give your answer in the form $a\pi - \ln b$ , where $a$ and $b$ are constants.

6M Similar

DifficultyMedium

Worked solution

Approach

Use the partial fraction decomposition from part (a) and integrate each term separately. The first term gives a logarithm, the second is of the form $k\frac{f'(x)}{f(x)}$ , and the third uses $\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a}$ .

Working

From part (a),

f(x) = -\frac{3}{1+x} - \frac{4x}{4+x^2} + \frac{6}{4+x^2}.

Integrate term by term:

\int f(x)\,dx = -3\ln(1+x) - 2\ln(4+x^2) + 3\tan^{-1}\left(\frac{x}{2}\right) + C.

Here $\int \frac{-4x}{4+x^2}\,dx = -2\ln(4+x^2)$ , since $\frac{d}{dx}(4+x^2)=2x$ , and $\int \frac{6}{4+x^2}\,dx = 3\tan^{-1}\left(\frac{x}{2}\right)$ .

Evaluate from $0$ to $2$ . At $x=2$ :

-3\ln 3 - 2\ln 8 + 3\tan^{-1}(1) = -3\ln 3 - 2\ln 8 + \frac{3\pi}{4}.

At $x=0$ :

-3\ln 1 - 2\ln 4 + 3\tan^{-1}(0) = -2\ln 4.

Subtract the lower-limit value:

\left(-3\ln 3 - 2\ln 8 + \frac{3\pi}{4}\right) - (-2\ln 4) = \frac{3\pi}{4} - 3\ln 3 - 2\ln 8 + 2\ln 4.

Simplify the logarithms:

-2\ln 8 + 2\ln 4 = -2(\ln 8 - \ln 4) = -2\ln 2.

Therefore

\frac{3\pi}{4} - 3\ln 3 - 2\ln 2 = \frac{3\pi}{4} - \ln(3^3 \cdot 2^2) = \frac{3\pi}{4} - \ln 108.

Answer

\int_0^2 f(x)\,dx = \frac{3\pi}{4} - \ln 108.

Final answer

3π/4 - ln 108

Detailed explanation

Walkthrough

After part (a), the integrand is a sum of three simpler fractions. Integrate each separately.

The first term, $-\frac{3}{1+x}$ , is of the form $\frac{k}{ax+b}$ and integrates to $-3\ln(1+x)$ .

The second term, $-\frac{4x}{4+x^2}$ , has numerator proportional to the derivative of $4+x^2$ . Since $\frac{d}{dx}(4+x^2)=2x$ , we write $-4x = -2(2x)$ , so its integral is $-2\ln(4+x^2)$ .

The third term, $\frac{6}{4+x^2}$ , is a standard arctan integral: $\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a}$ . With $a=2$ , this gives $3\tan^{-1}\frac{x}{2}$ .

Then substitute $x=2$ and $x=0$ and subtract. At $x=2$ , $\tan^{-1}(1)=\frac{\pi}{4}$ ; at $x=0$ , $\tan^{-1}(0)=0$ and $\ln 1=0$ . Finally combine the logarithms using $\ln p - \ln q = \ln\frac{p}{q}$ and $p\ln q = \ln q^p$ to obtain $-\ln 108$ .

Key Takeaways

Partial fractions turn a complicated rational integrand into terms that can be integrated using standard rules.
Recognising $k\frac{f'(x)}{f(x)}$ gives immediate logarithmic integrals.
The form $\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a}$ is essential for quadratic denominators without real roots.
Definite integrals require careful substitution of both limits and simplification of logarithms.

Common Mistakes

Forgetting the factor $\frac{1}{a}$ when integrating $\frac{1}{x^2+a^2}$ .
Missing the factor $2$ in $-2\ln(4+x^2)$ .
Incorrectly combining $\ln 8$ and $\ln 4$ ; remember $\ln 8 - \ln 4 = \ln 2$ .
Not subtracting the value at the lower limit $x=0$ .
Not simplifying the final answer to the required form $a\pi - \ln b$ .

Things to Be Careful About

The mark scheme allows follow-through from part (a), but the final answer must be exactly $\frac{3\pi}{4} - \ln 108$ .
At $x=0$ , $-3\ln(1)=0$ and $3\tan^{-1}(0)=0$ , but $-2\ln(4+x^2)$ contributes $-2\ln 4$ ; this must be subtracted.
Use $\ln 108 = \ln(27 \cdot 4) = 3\ln 3 + 2\ln 2$ to check the simplification.
The answer must be exact; do not give a decimal approximation.

Techniques used

integrate each partial fraction term separatelyuse the kf'(x)/f(x) rule for the term with a quadratic denominatorintegrate the reciprocal quadratic term using the arctan formevaluate the definite integral at the limits and simplify logarithms

Q116MMediumIntegration

Find the exact value of $\int_0^\pi x^2 \cos \frac{1}{3}x\,dx$ .

DifficultyMedium

Worked solution

Approach

Use integration by parts twice. Differentiating $x^2$ lowers its degree, and integrating $\cos\frac{x}{3}$ gives a sine term. A second integration by parts removes the polynomial factor completely, leaving only sine and cosine terms. Then substitute the limits and use the exact values of $\sin\frac{\pi}{3}$ and $\cos\frac{\pi}{3}$ .

Working

First integration by parts:

u = x^2, \quad \frac{du}{dx} = 2x

\frac{dv}{dx} = \cos\frac{x}{3}, \quad v = 3\sin\frac{x}{3}

Therefore

\int_0^\pi x^2 \cos\frac{x}{3}\,dx = [3x^2\sin\frac{x}{3}]_0^\pi - \int_0^\pi 6x\sin\frac{x}{3}\,dx.

Now integrate $6x\sin\frac{x}{3}$ by parts:

u = 6x, \quad \frac{du}{dx} = 6

\frac{dv}{dx} = \sin\frac{x}{3}, \quad v = -3\cos\frac{x}{3}

So

\int 6x\sin\frac{x}{3}\,dx = -18x\cos\frac{x}{3} + \int 18\cos\frac{x}{3}\,dx = -18x\cos\frac{x}{3} + 54\sin\frac{x}{3}.

Hence the antiderivative is

\int x^2 \cos\frac{x}{3}\,dx = 3x^2\sin\frac{x}{3} + 18x\cos\frac{x}{3} - 54\sin\frac{x}{3}.

At $x = \pi$ :

3\pi^2\cdot\frac{\sqrt{3}}{2} + 18\pi\cdot\frac{1}{2} - 54\cdot\frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}\pi^2 + 9\pi - 27\sqrt{3}.

At $x = 0$ , every term is zero, so only the upper limit contributes.

Answer

\frac{3\sqrt{3}}{2}\pi^2 + 9\pi - 27\sqrt{3}

Final answer

3√3/2 π^2 + 9π - 27√3

Detailed explanation

Walkthrough

The integrand is a product of a polynomial, $x^2$ , and a trigonometric function, $\cos\frac{x}{3}$ . Integration by parts is the natural tool because differentiating $x^2$ reduces it to $2x$ , lowering the polynomial degree.

In the first application, take $u = x^2$ and $\frac{dv}{dx} = \cos\frac{x}{3}$ . The antiderivative of $\cos\frac{x}{3}$ is $3\sin\frac{x}{3}$ , because differentiating $3\sin\frac{x}{3}$ gives $\cos\frac{x}{3}$ . This is where the factor $3$ comes from.

The integration-by-parts formula gives

\int x^2 \cos\frac{x}{3}\,dx = 3x^2\sin\frac{x}{3} - \int 2x \cdot 3\sin\frac{x}{3}\,dx = 3x^2\sin\frac{x}{3} - \int 6x\sin\frac{x}{3}\,dx.

This new integral still contains $x$ , so apply integration by parts again. Now choose $u = 6x$ and $\frac{dv}{dx} = \sin\frac{x}{3}$ . The antiderivative of $\sin\frac{x}{3}$ is $-3\cos\frac{x}{3}$ . The formula gives

\int 6x\sin\frac{x}{3}\,dx = -18x\cos\frac{x}{3} + \int 18\cos\frac{x}{3}\,dx.

Then $\int 18\cos\frac{x}{3}\,dx = 54\sin\frac{x}{3}$ , because the antiderivative of $\cos\frac{x}{3}$ is $3\sin\frac{x}{3}$ .

Substitute this back into the first expression. The negative sign in front of the second integral must be handled carefully:

3x^2\sin\frac{x}{3} - \left(-18x\cos\frac{x}{3} + 54\sin\frac{x}{3}\right) = 3x^2\sin\frac{x}{3} + 18x\cos\frac{x}{3} - 54\sin\frac{x}{3}.

For the definite integral, evaluate this antiderivative at $x = \pi$ and $x = 0$ . At $x = 0$ , all three terms contain either $x$ or $\sin 0$ , so the value is $0$ . At $x = \pi$ , use $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos\frac{\pi}{3} = \frac{1}{2}$ . This gives the exact final value.

Key Takeaways

When the integrand is a polynomial times a trigonometric or exponential function, try integration by parts.
Repeated integration by parts reduces the degree of the polynomial until only an integrable trigonometric or exponential term remains.
The factor produced by integrating $\cos(ax)$ or $\sin(ax)$ is $\frac{1}{a}$ , so here integrating $\cos\frac{x}{3}$ gives $3\sin\frac{x}{3}$ .
Exact trigonometric values allow the final definite integral to be written exactly in terms of $\pi$ and $\sqrt{3}$ .

Common Mistakes

Forgetting that $\int \cos\frac{x}{3}\,dx = 3\sin\frac{x}{3}$ , not $\sin\frac{x}{3}$ .
Forgetting the factor $3$ when integrating $\sin\frac{x}{3}$ , so writing the second integration incorrectly.
Sign errors: $\int 6x\sin\frac{x}{3}\,dx$ is $\left(-18x\cos\frac{x}{3} + 54\sin\frac{x}{3}\right)$ , and this whole quantity is subtracted in the first integration.
Substituting only the upper limit and forgetting that $x=0$ must be substituted into every term.
Using decimal approximations instead of the exact values of $\sin\frac{\pi}{3}$ and $\cos\frac{\pi}{3}$ when the question asks for an exact value.

Things to Be Careful About

The derivative of $v = 3\sin\frac{x}{3}$ is $\cos\frac{x}{3}$ , so it is consistent with choosing $\frac{dv}{dx} = \cos\frac{x}{3}$ .
When integrating $\sin\frac{x}{3}$ , the antiderivative is $-3\cos\frac{x}{3}$ ; the negative sign is essential.
In the second integration by parts, the integral of $18\cos\frac{x}{3}$ is $54\sin\frac{x}{3}$ , not $18\sin\frac{x}{3}$ .
The final answer must remain exact: keep $\pi$ and $\sqrt{3}$ rather than giving a decimal.

Techniques used

apply integration by partsdifferentiate the polynomial factor and integrate the trigonometric factorapply integration by parts a second timesubstitute the limits of integrationevaluate exact trigonometric values

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