(a) Show that $\cos 4x + 2\sin^2 x - 1 \equiv 8\sin^4 x - 6\sin^2 x$. (b) Hence solve the equation $\cos 4x + 2\sin^2 x - 1 = 0$ for $-180^\circ \leq x \leq 180^\circ$.

(a) 8 sin^4 x - 6 sin^2 x · (b) x = -180°, -120°, -60°, 0°, 60°, 120°, 180°

The curve with equation $y = e^{-5x} \ln 5x$ has a stationary point at $x = p$. (a) Show that $p$ satisfies the equation $\ln 5p = \frac{1}{5p}$. (b) By sketching a suitable pair of graphs, show that the equation in part (a) has only one root. (c) Show by calculation that $0.2 < p < 0.6$. (d) It is given that the equation in part (a) can be written in the form $p = \frac{1}{5} \exp\left(\frac{1}{5p}\right)$, where $\exp(x)$ denotes $e^x$. Use an iterative formula based on…

(a) ln(5p) = 1/(5p) · (b) The equation has only one root. · (c) 0.2 < p < 0.6 · (d) p = 0.35

The diagram shows the graph of $y = \sec^2 x \sqrt{3 + 2\tan x}$ for $-\frac{1}{4}\pi \leq x \leq \frac{1}{4}\pi$, and its minimum point $M$. (a) Find the $x$-coordinate of $M$. (b) Using the substitution $u = 3 + 2\tan x$, find the exact value of the area of the region bounded by the curve, the $x$-axis and the lines $x = -\frac{1}{4}\pi$ and $x = \frac{1}{4}\pi$.

(a) x ≈ -0.197 · (b) (5√5 - 1)/3

9709/33 Oct/Nov 2025 — Mathematics Solved Past Paper

Q14MMediumAlgebra

Solve the inequality $|3x + 2| < 3|2x - 1|$ .

DifficultyMedium

Worked solution

Approach

Since both sides of the inequality are non-negative, square both sides to remove the modulus signs. Then expand, collect terms on one side, factorise the quadratic, find its critical values, and use the sign of the quadratic to state the solution intervals.

Working

(3x + 2)^2 < 9(2x - 1)^2

Expanding both sides:

9x^2 + 12x + 4 < 36x^2 - 36x + 9

Subtract the left-hand side from both sides:

0 < 27x^2 - 48x + 5

Factorise:

27x^2 - 48x + 5 = (9x - 1)(3x - 5)

So the critical values are:

x = \frac{1}{9} \quad \text{or} \quad x = \frac{5}{3}

Since the quadratic has a positive $x^2$ coefficient, it is positive outside its roots. Checking $x = 0$ , $x = 1$ and $x = 2$ confirms this.

Answer

x < \frac{1}{9} \quad \text{or} \quad x > \frac{5}{3}

Final answer

x < 1/9 or x > 5/3

Detailed explanation

Walkthrough

We need to solve $|3x+2| < 3|2x-1|$ . The modulus signs make direct algebraic manipulation awkward, but because both $|3x+2|$ and $3|2x-1|$ are non-negative, squaring both sides preserves the inequality. This is the key step: it removes the absolute values without changing the solution set.

Squaring gives $(3x+2)^2 < 9(2x-1)^2$ . Notice the factor $3$ outside the second modulus is also squared, so the right-hand side becomes $9(2x-1)^2$ , not just $(2x-1)^2$ .

Expanding both sides gives $9x^2 + 12x + 4 < 36x^2 - 36x + 9$ . To compare the quadratic with zero, subtract the left-hand side from the right-hand side: $0 < 27x^2 - 48x + 5$ . This is equivalent to $27x^2 - 48x + 5 > 0$ .

Factorising gives $(9x-1)(3x-5) > 0$ . The roots, or critical values, are $x = \frac{1}{9}$ and $x = \frac{5}{3}$ . A quadratic with a positive coefficient of $x^2$ is positive outside the interval between its roots and negative between them. Testing $x=0$ , $x=1$ and $x=2$ confirms this: the expression is positive at $x=0$ and $x=2$ , but negative at $x=1$ .

Therefore the solution is $x < \frac{1}{9}$ or $x > \frac{5}{3}$ . The endpoints are not included because the original inequality is strict.

Key Takeaways

Squaring is a powerful method for modulus inequalities because both sides are non-negative.
Remember to square any coefficient outside a modulus: $3|2x-1|$ squared is $9(2x-1)^2$ .
A quadratic inequality $ax^2 + bx + c > 0$ with $a > 0$ is satisfied outside the interval between its roots.
The final answer must be written as two separate intervals joined by 'or', not by 'and'.

Common Mistakes

Forgetting to square the factor $3$ , giving $(2x-1)^2$ instead of $9(2x-1)^2$ .
Making a sign error when subtracting terms to collect the quadratic on one side.
Confusing the direction of the quadratic inequality: for $> 0$ with a positive leading coefficient, the solution is outside the roots, not between them.
Writing the final answer with 'and' instead of 'or', or using $\cap$ instead of $\cup$ .
Giving only an unsupported answer; the mark scheme states that no marks can be scored if no working is seen.

Things to Be Careful About

The inequality is strict, so the critical values $x = \frac{1}{9}$ and $x = \frac{5}{3}$ must not be included in the final answer.
When squaring, the inequality sign does not reverse because both sides are non-negative.
If using the alternative method of solving $3x+2 = \pm 3(2x-1)$ , the two linear equations give the critical values, but you must still decide the correct intervals by testing or by considering the sign of the quadratic.
The mark scheme allows 'OR' but not 'AND', and allows ' $\cup$ ' but not ' $\cap$ '.

Techniques used

square both sides to remove modulus signsexpand and factorise the resulting quadraticdetermine the solution intervals using the sign of the quadratic

Similar questions

Q23MMedium-EasyAlgebra

Find the quotient and the remainder when $3x^4 - 2x^2$ is divided by $x + 1$ .

DifficultyMedium-Easy

Worked solution

Approach

Use polynomial long division by writing the dividend with zero coefficients for the missing terms. Divide term by term, multiply back by $x+1$ , subtract, and continue until the remainder is a constant.

Working

Write the dividend with all powers of $x$ present:

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

First term: divide $3x^4$ by $x$ to get $3x^3$ .

(x+1)(3x^3) = 3x^4 + 3x^3

Subtracting:

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

Next term: divide $-3x^3$ by $x$ to get $-3x^2$ .

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

Subtracting:

(-3x^3 - 2x^2) - (-3x^3 - 3x^2) = x^2

Next term: divide $x^2$ by $x$ to get $x$ .

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

Subtracting:

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

Next term: divide $-x$ by $x$ to get $-1$ .

(x+1)(-1) = -x - 1

Subtracting:

-x - (-x - 1) = 1

The division stops with remainder $1$ . Therefore:

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

Answer

Quotient $3x^3 - 3x^2 + x - 1$ and remainder $1$ .

Final answer

Quotient = 3x^3 - 3x^2 + x - 1, remainder = 1

Detailed explanation

Walkthrough

The expression $3x^4 - 2x^2$ has no $x^3$ term and no $x$ term. In long division we must include these as $0x^3$ and $0x$ so that the subtraction aligns correctly by powers of $x$ .

Then we divide the leading term $3x^4$ by $x$ , giving $3x^3$ . We multiply this by the whole divisor $x+1$ and subtract. This removes the $3x^4$ term and gives a new remainder-like expression of lower degree. We repeat the same process with $-3x^3$ , then $x^2$ , then $-x$ .

The last subtraction leaves $1$ , which has degree lower than the divisor $x+1$ , so it is the remainder. Collecting the terms we divided by gives the quotient $3x^3 - 3x^2 + x - 1$ .

As an alternative check, the remainder theorem says the remainder when dividing $\text{f}(x)$ by $x+1$ is $\text{f}(-1)$ . Here $\text{f}(-1) = 3 - 2 = 1$ , confirming the remainder.

Key Takeaways

The key skill is setting up polynomial long division correctly, including zero coefficients for missing powers. It is also useful to know that division by $x+1$ can be checked using the remainder theorem: substitute $x = -1$ .

Common Mistakes

Forgetting to include $0x^3$ and $0x$ as placeholders, which leads to misaligned subtraction.
Making sign errors when subtracting, especially after multiplying by a negative term such as $-3x^2$ .
Confusing the quotient and the remainder. The quotient is the polynomial $3x^3 - 3x^2 + x - 1$ and the remainder is the constant $1$ , not $\dfrac{1}{x+1}$ .
Not giving both the quotient and the remainder when the question asks for both.

The mark scheme warns not to ignore subsequent working; if the quotient and remainder are stated with wrong labels, marks may be lost.

Things to Be Careful About

When writing the final division statement, keep the remainder outside the product:

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

It is acceptable to write the result as $3x^3 - 3x^2 + x - 1 + \dfrac{1}{x+1}$ , but it is not acceptable to state that the remainder is $\dfrac{1}{x+1}$ .
The divisor is $x+1$ , so the remainder theorem uses $x = -1$ , not $x = 1$ .
If using synthetic division, the value used is $-1$ , and the last number in the synthetic division row gives the remainder.

Techniques used

write the dividend with zero coefficients for missing termsperform polynomial long division by a linear divisoridentify the quotient and constant remainder from the division

Similar questions

Q34MMedium-EasyLogarithmic and Exponential Functions

Solve the equation $2^{3x-4} = \frac{3}{5^x}$ . Give your answer in the form $\frac{\ln m}{\ln n}$ , where $m$ and $n$ are integers.

DifficultyMedium-Easy

Worked solution

Approach

Take natural logarithms of both sides so the unknown index $x$ can be brought down using the power law. Apply the quotient and power laws of logarithms, then rearrange to make $x$ the subject and combine the logarithms into single terms.

Working

Take $\ln$ of both sides:

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

Use the quotient law on the right-hand side:

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

Use the power law on both sides:

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

Expand the left-hand side:

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

Collect the $x$ terms on one side:

3x\ln 2 + x\ln 5 = \ln 3 + 4\ln 2

Factor out $x$ :

x(3\ln 2 + \ln 5) = \ln 3 + 4\ln 2

Divide:

x = \frac{\ln 3 + 4\ln 2}{3\ln 2 + \ln 5}

Combine the logarithms:

\ln 3 + 4\ln 2 = \ln 3 + \ln 2^4 = \ln(3 \times 16) = \ln 48

3\ln 2 + \ln 5 = \ln 2^3 + \ln 5 = \ln(8 \times 5) = \ln 40

Therefore,

x = \frac{\ln 48}{\ln 40}

Answer

x = \frac{\ln 48}{\ln 40}

with $m = 48$ and $n = 40$ .

Final answer

x = ln(48)/ln(40)

Detailed explanation

Walkthrough

The equation has the unknown in the exponents, so the key move is to take logarithms of both sides. Taking $\ln$ is convenient because the required answer is written with $\ln$ . This turns the exponential equation into a linear equation in $x$ .

Start with

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

On the left, the power law gives $(3x-4)\ln 2$ . On the right, the quotient law gives $\ln 3 - \ln(5^x)$ , and then the power law gives $\ln 3 - x\ln 5$ . The equation is now linear in $x$ :

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

Expand the left side and bring all the $x$ -terms to one side and the constants to the other:

3x\ln 2 + x\ln 5 = \ln 3 + 4\ln 2

Factor out $x$ and divide:

x = \frac{\ln 3 + 4\ln 2}{3\ln 2 + \ln 5}

Finally, combine the logarithms using $a\ln b = \ln b^a$ and $\ln p + \ln q = \ln(pq)$ :

\ln 3 + 4\ln 2 = \ln 3 + \ln 2^4 = \ln(3 \times 16) = \ln 48

3\ln 2 + \ln 5 = \ln 2^3 + \ln 5 = \ln(8 \times 5) = \ln 40

So $x = \frac{\ln 48}{\ln 40}$ .

Key Takeaways

This question tests the ability to solve equations where the unknown appears in an index. The essential tools are the laws of logarithms: the power law, the quotient law, and the product law. It also tests careful algebraic rearrangement and the ability to combine logarithmic terms into a single logarithm. The final step shows how $a\ln b$ can be absorbed into a logarithm to produce a compact answer in the required form.

Common Mistakes

Forgetting to apply a logarithm to both sides of the equation.
Misapplying the quotient law: $\ln\left(\frac{3}{5^x}\right)$ is $\ln 3 - \ln(5^x)$ , not $\frac{\ln 3}{\ln(5^x)}$ .
Dropping the brackets when expanding $(3x-4)\ln 2$ ; this leads to sign errors unless corrected later.
Stopping at $\frac{\ln 3 + 4\ln 2}{3\ln 2 + \ln 5}$ without combining into $\frac{\ln 48}{\ln 40}$ .
Giving the final answer as $\log_{40} 48$ : the mark scheme awards only 3 marks for this, because the question asks for the form $\frac{\ln m}{\ln n}$ .
Writing a final answer without any working; the mark scheme awards 0 marks for an unsupported answer.

Things to Be Careful About

The logarithm may be taken to any base for the first three marks, but the final answer must be expressed with natural logarithms in the form $\frac{\ln m}{\ln n}$ .
Keep the bracket $(3x-4)$ intact until you expand it; if you write $3x-4\ln 2$ without brackets, you must recover the correct expansion later.
When combining logarithms, remember $4\ln 2 = \ln(2^4)$ , not $\ln(4\times 2)$ .
The integers are $m=48$ and $n=40$ ; check that both numerator and denominator are simplified fully.

Techniques used

take logarithms of both sidesapply the quotient law of logarithmsapply the power law of logarithmsrearrange to isolate xcombine logarithmic terms into single logarithms

Similar questions

Q45MMediumComplex Numbers

On an Argand diagram shade the region whose points represent complex numbers $z$ which satisfy both the inequalities $|z + 2i| \leq 3$ and $|z + 2i| \leq |z - 2 + 4i|$ .

DifficultyMedium

Worked solution

Approach

Interpret each modulus inequality geometrically on the Argand diagram. The first inequality represents a disk (circle and its interior). The second inequality represents a half-plane bounded by the perpendicular bisector of the segment joining two fixed points. The solution is the intersection of these two regions.

Working

Let $z = x + iy$ .

First inequality: $|z + 2i| \leq 3$

|x + i(y + 2)| \leq 3

\sqrt{x^2 + (y + 2)^2} \leq 3

x^2 + (y + 2)^2 \leq 9

This is the region on and inside a circle with centre $(0, -2)$ and radius $3$ .

Second inequality: $|z + 2i| \leq |z - 2 + 4i|$

Rewrite as $|z - (-2i)| \leq |z - (2 - 4i)|$ . This states that the distance from $z$ to $-2i$ (the point $(0, -2)$ ) is less than or equal to the distance from $z$ to $2 - 4i$ (the point $(2, -4)$ ). The boundary is the perpendicular bisector of the line segment joining $(0, -2)$ and $(2, -4)$ .

Midpoint:

\left(\frac{0+2}{2}, \frac{-2-4}{2}\right) = (1, -3)

Gradient of the line joining $(0, -2)$ and $(2, -4)$ :

m = \frac{-4 - (-2)}{2 - 0} = \frac{-2}{2} = -1

Gradient of the perpendicular bisector:

m_\perp = 1

Equation of the perpendicular bisector:

y - (-3) = 1(x - 1) \implies y = x - 4

The inequality $|z + 2i| \leq |z - 2 + 4i|$ selects the half-plane containing $(0, -2)$ . Testing $(0, -2)$ : $|-2i| = 2$ and $|2 - 6i| = \sqrt{40}$ . Since $2 < \sqrt{40}$ , the region is above and to the left of the line $y = x - 4$ .

Shading:

The correct region is the part of the disk $x^2 + (y + 2)^2 \leq 9$ that lies above and to the left of the line $y = x - 4$ .

Answer

The shaded region is the intersection of the disk $|z + 2i| \leq 3$ and the half-plane $|z + 2i| \leq |z - 2 + 4i|$ , bounded by the circle centred at $(0, -2)$ with radius $3$ and the line $y = x - 4$ .

Final answer

The shaded region is the part of the disk |z + 2i| <= 3 that lies above and to the left of the line y = x - 4.

Detailed explanation

Walkthrough

First, we translate the algebraic modulus inequalities into geometric conditions on the Argand diagram. Let $z = x + iy$ .

For the first inequality, $|z + 2i| \leq 3$ , we recognize this as $|z - (-2i)| \leq 3$ . The locus $|z - z_1| = r$ is a circle with centre $z_1$ and radius $r$ . Here, the centre is $-2i$ , which corresponds to the point $(0, -2)$ , and the radius is $3$ . The $\leq$ sign means we include the interior of the circle, forming a closed disk.

For the second inequality, $|z + 2i| \leq |z - 2 + 4i|$ , we rewrite it as $|z - (-2i)| \leq |z - (2 - 4i)|$ . This compares the distance from a variable point $z$ to two fixed points: $A = (0, -2)$ and $B = (2, -4)$ . The locus of points equidistant from $A$ and $B$ is the perpendicular bisector of the segment $AB$ . The inequality selects the side of this bisector that is closer to $A$ .

To find the equation of the perpendicular bisector, we first find the midpoint of $AB$ : $\left(\frac{0+2}{2}, \frac{-2-4}{2}\right) = (1, -3)$ . The gradient of $AB$ is $\frac{-4 - (-2)}{2 - 0} = -1$ . The gradient of the perpendicular bisector is the negative reciprocal, which is $1$ . Using the point-slope form with the midpoint $(1, -3)$ , the equation is $y + 3 = 1(x - 1)$ , simplifying to $y = x - 4$ .

To determine which side of the line to shade, we test the point $A(0, -2)$ . The distance from $A$ to $A$ is $0$ , and the distance from $A$ to $B$ is $\sqrt{20}$ . Since $0 < \sqrt{20}$ , the region satisfying the inequality is the half-plane containing $A$ , which is above and to the left of the line $y = x - 4$ .

Finally, we shade the region that satisfies both conditions: inside the circle and above/left of the bisector line. This is the intersection of the two regions.

Key Takeaways

The modulus $|z - z_1| = r$ represents a circle with centre $z_1$ and radius $r$ .
The modulus $|z - z_1| = |z - z_2|$ represents the perpendicular bisector of the line segment joining $z_1$ and $z_2$ .
Inequalities with $\leq$ or $\geq$ include the boundary and extend into a region (disk or half-plane).
When shading intersections of regions, always test a point to confirm which side of a boundary line satisfies the inequality.

Common Mistakes

Forgetting that $|z + 2i|$ corresponds to the point $(0, -2)$ , not $(0, 2)$ . The sign inside the modulus is reversed for the coordinate.
Drawing the perpendicular bisector with the wrong gradient (e.g., using $-1$ instead of $1$ ).
Shading the wrong half-plane (e.g., shading below the line $y = x - 4$ instead of above it).
Failing to include the boundary of the circle or the line in the shading (though typically shading implies the region, the boundary must be drawn as solid, not dashed, for $\leq$ ).

Things to Be Careful About

Ensure the scale on both axes is consistent, as this is an Argand diagram (complex plane) where the x-axis is the real part and the y-axis is the imaginary part.
The centre of the circle is at $(0, -2)$ , which is below the real axis. Check that the circle intersects the axes correctly: it crosses the y-axis at $y = 1$ and $y = -5$ , and the x-axis at $x = \pm\sqrt{5} \approx \pm 2.24$ .
The line $y = x - 4$ crosses the y-axis at $(0, -4)$ and the x-axis at $(4, 0)$ . Ensure the shading is clearly bounded and does not extend beyond the circle.
Condone dashed circles and dashed perpendicular bisectors for all marks according to the mark scheme, but solid lines are preferred for $\leq$ inequalities.

Techniques used

interpret modulus inequalities geometricallyidentify circle locus from |z - z1| = ridentify perpendicular bisector locus from |z - z1| = |z - z2|shade intersection of regions

Similar questions

Q5MediumTrigonometry

(a)

Show that $\cos 4x + 2\sin^2 x - 1 \equiv 8\sin^4 x - 6\sin^2 x$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Apply the double angle formula $\cos 4x = 1 - 2\sin^2 2x$ to remove $\cos 4x$ , then the double angle formula $\sin 2x = 2\sin x\cos x$ , then the Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ to express everything purely in $\sin x$ .

Working

Start with the left-hand side:

\cos 4x + 2\sin^2 x - 1

Apply $\cos 4x = 1 - 2\sin^2 2x$ :

= 1 - 2\sin^2 2x + 2\sin^2 x - 1

Simplify (the $1$ and $-1$ cancel):

= -2\sin^2 2x + 2\sin^2 x

Apply $\sin 2x = 2\sin x\cos x$ :

= -2(2\sin x\cos x)^2 + 2\sin^2 x

= -8\sin^2 x\cos^2 x + 2\sin^2 x

Apply $\cos^2 x = 1 - \sin^2 x$ :

= -8\sin^2 x(1 - \sin^2 x) + 2\sin^2 x

Expand the bracket:

= -8\sin^2 x + 8\sin^4 x + 2\sin^2 x

Collect like terms:

= 8\sin^4 x - 6\sin^2 x

Answer

\cos 4x + 2\sin^2 x - 1 \equiv 8\sin^4 x - 6\sin^2 x

Final answer

8 sin^4 x - 6 sin^2 x

Detailed explanation

Walkthrough

We need to show that the trigonometric expression $\cos 4x + 2\sin^2 x - 1$ is identically equal to $8\sin^4 x - 6\sin^2 x$ . The strategy is to remove $\cos 4x$ and $\cos^2 x$ step by step using standard double angle and Pythagorean identities, until only $\sin x$ remains.

First, we use the double angle identity $\cos 2\theta = 1 - 2\sin^2 \theta$ with $\theta = 2x$ to get $\cos 4x = 1 - 2\sin^2 2x$ . This is the most useful form because the $1$ cancels with the $-1$ in the original expression, leaving only even powers of $\sin$ in sight.

After simplifying $1 - 2\sin^2 2x + 2\sin^2 x - 1$ , we get $-2\sin^2 2x + 2\sin^2 x$ . Now we have $\sin^2 2x$ to deal with, and the double angle formula $\sin 2x = 2\sin x\cos x$ converts this into $4\sin^2 x\cos^2 x$ .

At this stage the expression contains $\sin^2 x$ and $\cos^2 x$ . Using the Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ allows us to eliminate $\cos^2 x$ entirely. Substituting this in gives $-8\sin^2 x(1 - \sin^2 x) + 2\sin^2 x$ , which when expanded is $-8\sin^2 x + 8\sin^4 x + 2\sin^2 x$ . Collecting like terms of $\sin^2 x$ yields the final $8\sin^4 x - 6\sin^2 x$ .

Key Takeaways

The identity $\cos 2\theta = 1 - 2\sin^2 \theta$ is the right tool when the expression has a $\cos$ of an even angle and you want to convert to a $\sin$ power.
For identities involving $\sin 2x$ or $\cos 2x$ , remember to substitute these and then use $\cos^2 x = 1 - \sin^2 x$ (or equivalently $\sin^2 x = 1 - \cos^2 x$ ) as the next step.
"Show that" questions require every intermediate step to be shown; the mark scheme expects to see $\sin^4 x$ explicitly before the final AG line.

Common Mistakes

Forgetting to expand $(2\sin x\cos x)^2$ correctly (the 2 is squared to give 4).
Sign errors when expanding $-8\sin^2 x(1 - \sin^2 x)$ — the second term is $+8\sin^4 x$ , not $-8\sin^4 x$ .
Collecting $-8\sin^2 x$ and $+2\sin^2 x$ incorrectly.
Writing $\cos 4x = 2\cos^2 2x - 1$ and getting stuck because the work then leads back to $\cos 2x$ instead of $\sin x$ .

Things to Be Careful About

Since this is an "AG" (answer given) question, every mark depends on showing the right method at the right time. The mark scheme requires a clear intermediate line containing $\sin^4 x$ before stating the AG.
The domain does not matter for an identity, but watch for sign slips when expanding negative brackets.
The double angle formula for $\cos 2\theta$ has three equivalent forms: $\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta$ . The third is the right one here.

Techniques used

apply double angle formula for cos 4xapply double angle formula for sin 2xsubstitute Pythagorean identity cos²x = 1 − sin²xexpand and collect like terms

(b)

Hence solve the equation $\cos 4x + 2\sin^2 x - 1 = 0$ for $-180^\circ \leq x \leq 180^\circ$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Using the result of part (a), the equation becomes $8\sin^4 x - 6\sin^2 x = 0$ . Factor out $2\sin^2 x$ , then solve the remaining quadratic $\sin^2 x = 3/4$ , considering both positive and negative square roots, and listing all solutions in $-180° \leq x \leq 180°$ .

Working

Substitute the identity proved in (a):

8\sin^4 x - 6\sin^2 x = 0

Factor out $2\sin^2 x$ :

2\sin^2 x(4\sin^2 x - 3) = 0

So either $\sin^2 x = 0$ or $\sin^2 x = \dfrac{3}{4}$ .

Case 1: $\sin^2 x = 0$ gives $\sin x = 0$ . In the interval $-180° \leq x \leq 180°$ this yields

x = 0°,\ 180°,\ -180°

Case 2: $\sin^2 x = \dfrac{3}{4}$ gives $\sin x = \pm\dfrac{\sqrt{3}}{2}$ .

For $\sin x = \dfrac{\sqrt{3}}{2}$ :

x = 60°\ \text{or}\ 120°

For $\sin x = -\dfrac{\sqrt{3}}{2}$ :

x = -60°\ \text{or}\ -120°

Answer

x = -180°,\ -120°,\ -60°,\ 0°,\ 60°,\ 120°,\ 180°

Final answer

x = -180°, -120°, -60°, 0°, 60°, 120°, 180°

Detailed explanation

Walkthrough

From part (a) we know that $\cos 4x + 2\sin^2 x - 1 \equiv 8\sin^4 x - 6\sin^2 x$ , so the equation $\cos 4x + 2\sin^2 x - 1 = 0$ becomes the much simpler polynomial equation $8\sin^4 x - 6\sin^2 x = 0$ in $\sin x$ .

The expression $8\sin^4 x - 6\sin^2 x$ is a quadratic in $\sin^2 x$ . Let $u = \sin^2 x$ to give $8u^2 - 6u = 0$ , or $2u(4u - 3) = 0$ . So $u = 0$ or $u = 3/4$ . Translating back: $\sin^2 x = 0$ or $\sin^2 x = 3/4$ .

The case $\sin^2 x = 0$ is the easiest: $\sin x = 0$ has solutions $x = 0°, 180°, -180°$ in the given closed interval.

The case $\sin^2 x = 3/4$ means $\sin x = +\sqrt{3}/2$ or $\sin x = -\sqrt{3}/2$ . The positive case gives first-quadrant and second-quadrant solutions $60°$ and $120°$ . The negative case gives third- and fourth-quadrant solutions $-60°$ and $-120°$ .

Combining all cases, the full solution set is $\{-180°, -120°, -60°, 0°, 60°, 120°, 180°\}$ , seven solutions in total.

Key Takeaways

Recognising a quartic in $\sin x$ as a quadratic in $\sin^2 x$ lets you factor it directly.
When solving $\sin x = k$ in a closed interval, remember to take both signs when $k$ came from $\sin^2 x = k^2$ , and to find solutions in all four quadrants.
The endpoints $-180°$ and $180°$ are valid solutions for $\sin x = 0$ — these are often forgotten.

Common Mistakes

Forgetting that $\sin^2 x = 3/4$ has $\sin x = +\sqrt{3}/2$ AND $\sin x = -\sqrt{3}/2$ (and missing four of the seven solutions).
Omitting the boundary values $x = -180°$ and $x = 180°$ .
Writing the solution set in degrees and forgetting the degree symbol.
Including solutions outside the given interval (e.g., $x = 240°$ or $x = -240°$ ).

Things to Be Careful About

The interval $-180° \leq x \leq 180°$ is closed, so both $-180°$ and $180°$ count.
The mark scheme awards the B1 only for the three solutions of $\sin^2 x = 0$ ( $0°, 180°, -180°$ ). The A1's require the four solutions of $\sin^2 x = 3/4$ .
The mark scheme notes: "Allow M1A1A1 if dividing by $\sin^2 x$ , but B1 is not scored." This means if you wrongly divide by $\sin^2 x$ and lose the $0°$ solutions, you forfeit the B1. Better to factor out $2\sin^2 x$ instead of dividing by it.

Techniques used

substitute the proven identity from part (a)factorise the polynomial in sin²xsolve sin x = 0 in the given intervalsolve sin x = ±√3/2 in the given interval

Q66MMediumIntegration

Find the exact value of $\int_0^{\frac{1}{6}\pi} x^2 \sin 2x \, dx$ .

DifficultyMedium

Worked solution

Approach

The integrand $x^2 \sin 2x$ is a product of a polynomial and a trigonometric function, so integrate by parts. Because differentiating $x^2$ gives a linear term $x$ , a second integration by parts is required to remove the polynomial factor completely. Then substitute the limits using exact trigonometric values.

Working

Let

I = \int_0^{\frac{\pi}{6}} x^2 \sin 2x \, dx.

First integration by parts with $u = x^2$ , $\mathrm{d}v = \sin 2x \, \mathrm{d}x$ , so $\mathrm{d}u = 2x\, \mathrm{d}x$ and $v = -\frac{1}{2}\cos 2x$ :

I = \left[-\frac{1}{2}x^2 \cos 2x\right]_0^{\frac{\pi}{6}} + \int_0^{\frac{\pi}{6}} x \cos 2x \, dx.

Now integrate $\int x \cos 2x \, dx$ by parts with $u = x$ , $\mathrm{d}v = \cos 2x \, \mathrm{d}x$ , so $\mathrm{d}u = \mathrm{d}x$ and $v = \frac{1}{2}\sin 2x$ :

\begin{aligned} \int x \cos 2x \, dx &= \frac{1}{2}x \sin 2x - \int \frac{1}{2}\sin 2x \, dx \\ &= \frac{1}{2}x \sin 2x + \frac{1}{4}\cos 2x. \end{aligned}

Thus an antiderivative is

F(x) = -\frac{1}{2}x^2 \cos 2x + \frac{1}{2}x \sin 2x + \frac{1}{4}\cos 2x.

At the upper limit $x = \frac{\pi}{6}$ , we have $2x = \frac{\pi}{3}$ , so

\cos \frac{\pi}{3} = \frac{1}{2}, \qquad \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}.

Then

\begin{aligned} F\left(\frac{\pi}{6}\right) &= -\frac{1}{2}\left(\frac{\pi}{6}\right)^2\left(\frac{1}{2}\right) + \frac{1}{2}\left(\frac{\pi}{6}\right)\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{4}\left(\frac{1}{2}\right) \\ &= -\frac{\pi^2}{144} + \frac{\sqrt{3}\pi}{24} + \frac{1}{8}. \end{aligned}

At the lower limit,

F(0) = \frac{1}{4}\cos 0 = \frac{1}{4}.

Therefore

\begin{aligned} I &= F\left(\frac{\pi}{6}\right) - F(0) \\ &= -\frac{\pi^2}{144} + \frac{\sqrt{3}\pi}{24} + \frac{1}{8} - \frac{1}{4} \\ &= -\frac{\pi^2}{144} + \frac{\sqrt{3}\pi}{24} - \frac{1}{8}. \end{aligned}

Answer

\boxed{-\frac{\pi^2}{144} + \frac{\sqrt{3}\pi}{24} - \frac{1}{8}}

Final answer

-(1/144)π² + (√3/24)π - 1/8

Detailed explanation

Walkthrough

The integrand is a product $x^2 \sin 2x$ , which signals that integration by parts is the appropriate method. In the first application we choose $u = x^2$ , because differentiating a polynomial lowers its degree and simplifies the integral. Integrating $\sin 2x$ gives $-\frac{1}{2}\cos 2x$ , so one integration by parts leaves

\int_0^{\pi/6} x \cos 2x \, dx

rather than the original $x^2\sin 2x$ . The polynomial factor is now only $x$ , so one more integration by parts is required. In the second application, differentiating $x$ gives $\mathrm{d}u = \mathrm{d}x$ and integrating $\cos 2x$ gives $v = \frac{1}{2}\sin 2x$ . This produces

\frac{1}{2}x\sin 2x - \int \frac{1}{2}\sin 2x\, dx = \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x.

Notice the sign: the minus sign in front of the integral combined with $\int \frac{1}{2}\sin 2x\, dx = -\frac{1}{4}\cos 2x$ gives the positive $+\frac{1}{4}\cos 2x$ . We now have the complete antiderivative

F(x) = -\frac{1}{2}x^2\cos 2x + \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x.

For a definite integral the value is $F(b)-F(a)$ . At $x = \pi/6$ , since $2x = \pi/3$ , the exact values are $\cos(\pi/3)=\frac12$ and $\sin(\pi/3)=\frac{\sqrt3}{2}$ . Substituting gives $F(\pi/6)=-\pi^2/144 + \sqrt3\pi/24 + 1/8$ . At $x=0$ , only the $\frac14\cos0$ term survives, so $F(0)=1/4$ . Subtracting $F(0)$ from $F(\pi/6)$ gives the final exact answer.

Key Takeaways

This question tests the repeated use of integration by parts on a product of a polynomial and a trigonometric function. The key skill is to choose the polynomial for differentiation each time, and to keep careful track of the factors of $\frac12$ that arise when integrating $\sin 2x$ and $\cos 2x$ . It also requires exact evaluation of trigonometric functions at $\pi/3$ and the correct treatment of the lower limit in a definite integral.

Common Mistakes

Sign errors in integration by parts. The formula is $\int u\,dv = uv - \int v\,du$ , and the minus sign is easy to lose, especially when $v$ is already negative.
Stopping after one application. After the first integration by parts the remaining integral contains $x\cos2x$ , so a second application is essential.
Mixing up exact trigonometric values. At $\pi/3$ , $\cos = \frac12$ and $\sin = \frac{\sqrt3}{2}$ ; swapping these would give a wrong final answer.
Forgetting the lower limit. $F(0)=1/4$ , not $0$ .
Using decimals. The mark scheme requires an exact value and disallows answers given only as decimals.

Things to Be Careful About

When integrating $\sin 2x$ or $\cos 2x$ , the factor $\frac12$ must be tracked in every term. The third term in the antiderivative comes from $\int \frac12\sin 2x\,dx$ with the sign carefully handled. In the final evaluation, simplify $1/8 - 1/4 = -1/8$ to combine the constants. No approximation should be used: leave the answer with $\pi^2$ , $\pi$ , and $\sqrt3$ terms exactly as requested.

Techniques used

apply integration by parts twicedifferentiate the polynomial factor and integrate the trigonometric factorevaluate the antiderivative at the given limitsuse exact trigonometric values at pi/3

Similar questions

Q76MMedium-HardComplex Numbers

Solve the equation $\frac{5z}{2 - i} - zz^* + 20 + 8i = 0$ . Give your answers in the form $x + iy$ , where $x$ and $y$ are real.

DifficultyMedium-Hard

Worked solution

Approach

Let $z = x + \mathrm{i}y$ and $z^* = x - \mathrm{i}y$ . Remove the fraction by multiplying the whole equation by $2 - \mathrm{i}$ , or by multiplying the fraction by $\frac{2+\mathrm{i}}{2+\mathrm{i}}$ . Then use $z z^* = x^2 + y^2$ and $\mathrm{i}^2 = -1$ , equate real and imaginary parts to zero, and solve the resulting pair of equations.

Working

Let $z = x + \mathrm{i}y$ . Then $z^* = x - \mathrm{i}y$ and $z z^* = x^2 + y^2$ .

Multiply the equation by $2 - \mathrm{i}$ :

5z - z z^*(2 - \mathrm{i}) + (20 + 8\mathrm{i})(2 - \mathrm{i}) = 0

Compute the product:

(20 + 8\mathrm{i})(2 - \mathrm{i}) = 40 - 20\mathrm{i} + 16\mathrm{i} - 8\mathrm{i}^2 = 48 - 4\mathrm{i}

Substitute $z = x + \mathrm{i}y$ and $z z^* = x^2 + y^2$ :

5(x + \mathrm{i}y) - (2 - \mathrm{i})(x^2 + y^2) + 48 - 4\mathrm{i} = 0

Expanding and using $\mathrm{i}^2 = -1$ :

5x + 5\mathrm{i}y - 2(x^2 + y^2) + \mathrm{i}(x^2 + y^2) + 48 - 4\mathrm{i} = 0

Equate real and imaginary parts to zero.

Real part:

5x - 2(x^2 + y^2) + 48 = 0

Imaginary part:

5y + x^2 + y^2 - 4 = 0

Let $r = x^2 + y^2$ . Then:

2r = 5x + 48, \qquad r = 4 - 5y

Eliminate $r$ :

\frac{5x + 48}{2} = 4 - 5y

5x + 48 = 8 - 10y

x + 2y + 8 = 0

So $x = -2y - 8$ . Substitute into $x^2 + y^2 = 4 - 5y$ :

(-2y - 8)^2 + y^2 = 4 - 5y

4y^2 + 32y + 64 + y^2 = 4 - 5y

5y^2 + 37y + 60 = 0

Factorise:

(5y + 12)(y + 5) = 0

Hence $y = -\frac{12}{5}$ or $y = -5$ .

Using $x = -2y - 8$ :

If $y = -5$ , then $x = -2(-5) - 8 = 2$ .

If $y = -\frac{12}{5}$ , then $x = -2\left(-\frac{12}{5}\right) - 8 = -\frac{16}{5}$ .

Answer

z = 2 - 5\mathrm{i} \quad \text{or} \quad z = -\frac{16}{5} - \frac{12}{5}\mathrm{i}

Final answer

z = 2 - 5i, z = -16/5 - 12/5 i

Detailed explanation

Walkthrough

The equation contains $z$ and $z^*$ , so the natural first step is to write $z = x + \mathrm{i}y$ and $z^* = x - \mathrm{i}y$ . This turns the complex equation into two real equations, one from the real parts and one from the imaginary parts.

Before substituting, remove the denominator $2 - \mathrm{i}$ . Multiplying the entire equation by $2 - \mathrm{i}$ clears the fraction. The product $(20 + 8\mathrm{i})(2 - \mathrm{i})$ must be expanded carefully, using $\mathrm{i}^2 = -1$ to replace $-8\mathrm{i}^2$ by $+8$ , giving $48 - 4\mathrm{i}$ .

After substituting $z = x + \mathrm{i}y$ and $z z^* = x^2 + y^2$ , expand the term $-(2 - \mathrm{i})(x^2 + y^2)$ as $-2(x^2 + y^2) + \mathrm{i}(x^2 + y^2)$ . Then separate the real and imaginary parts. Since the whole expression equals $0$ , both the real part and the imaginary part must be zero. This gives the two equations

5x - 2(x^2 + y^2) + 48 = 0

and

5y + x^2 + y^2 - 4 = 0.

Let $r = x^2 + y^2$ to see the structure: $2r = 5x + 48$ and $r = 4 - 5y$ . Eliminating $r$ gives a linear relation $x + 2y + 8 = 0$ , i.e. $x = -2y - 8$ .

Substitute this linear relation into $x^2 + y^2 = 4 - 5y$ . This produces a quadratic in $y$ , $5y^2 + 37y + 60 = 0$ , which factorises as $(5y + 12)(y + 5) = 0$ . The two values of $y$ give the two corresponding values of $x$ from $x = -2y - 8$ .

Key Takeaways

The conjugate product $z z^*$ equals the real number $x^2 + y^2$ , i.e. the squared modulus.
A single complex equation can be solved by equating real and imaginary parts after writing $z$ in Cartesian form.
Clearing a complex denominator by multiplying by its conjugate is a standard technique.
The resulting system often reduces to one linear and one quadratic equation, solvable by substitution.

Common Mistakes

Forgetting to multiply every term by $2 - \mathrm{i}$ ; only removing the fraction from the first term is not enough.
Forgetting that $\mathrm{i}^2 = -1$ when expanding, which changes signs in the real and imaginary parts.
Equating the real and imaginary parts incorrectly, or omitting the imaginary unit when collecting imaginary terms.
Losing one of the two solutions when solving the quadratic.
Giving only the values of $x$ and $y$ without writing the final complex answers in the form $x + \mathrm{i}y$ .

Things to Be Careful About

The product $(20 + 8\mathrm{i})(2 - \mathrm{i})$ must be computed with the correct sign: $-8\mathrm{i}^2 = +8$ , so the constant term is $48$ , not $32$ .
When equating imaginary parts, the imaginary unit $\mathrm{i}$ must be factored out correctly. For example, $5\mathrm{i}y + \mathrm{i}(x^2 + y^2) - 4\mathrm{i} = 0$ gives $5y + x^2 + y^2 - 4 = 0$ .
The linear relation $x + 2y + 8 = 0$ must be used consistently to substitute back; a sign error here gives the wrong pairs.
The final answers must be given as complex numbers $2 - 5\mathrm{i}$ and $-\frac{16}{5} - \frac{12}{5}\mathrm{i}$ , not just as coordinate pairs, unless the question explicitly accepts coordinates.

Techniques used

multiply by the conjugate to clear a complex denominatorsubstitute z and z* in Cartesian formequate real and imaginary parts to zerosolve simultaneous equations involving a quadratic

Similar questions

Q8MediumDifferentiationLogarithmic and Exponential FunctionsNumerical Solution of Equations

The curve with equation $y = e^{-5x} \ln 5x$ has a stationary point at $x = p$ .

(a)

Show that $p$ satisfies the equation $\ln 5p = \frac{1}{5p}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The curve has equation $y = e^{-5x} \ln 5x$ . At a stationary point, $\frac{dy}{dx} = 0$ . Differentiate using the product rule, then substitute $x = p$ and set the derivative to zero. Rearrange to obtain the required equation.

Working

Let $u = e^{-5x}$ and $v = \ln 5x$ . Then:

\frac{du}{dx} = -5e^{-5x}

\frac{dv}{dx} = \frac{1}{5x} \cdot 5 = \frac{1}{x}

Apply the product rule $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ :

\frac{dy}{dx} = e^{-5x} \cdot \frac{1}{x} + \ln 5x \cdot \left(-5e^{-5x}\right)

\frac{dy}{dx} = \frac{1}{x}e^{-5x} - 5e^{-5x}\ln 5x

At the stationary point $x = p$ , set $\frac{dy}{dx} = 0$ :

\frac{1}{p}e^{-5p} - 5e^{-5p}\ln 5p = 0

Factor out $e^{-5p}$ (which is never zero):

e^{-5p}\left(\frac{1}{p} - 5\ln 5p\right) = 0

Since $e^{-5p} \neq 0$ :

\frac{1}{p} - 5\ln 5p = 0

\frac{1}{p} = 5\ln 5p

Divide both sides by 5:

\ln 5p = \frac{1}{5p}

Answer

\ln 5p = \frac{1}{5p}

Final answer

ln(5p) = 1/(5p)

Detailed explanation

Walkthrough

The question asks us to show that the $x$ -coordinate $p$ of the stationary point satisfies a given equation. A stationary point occurs where the gradient of the curve is zero, so we need to differentiate $y = e^{-5x}\ln 5x$ and set it equal to zero.

Step 1: Differentiate using the product rule. The function is a product of two parts: $e^{-5x}$ and $\ln 5x$ . We let $u = e^{-5x}$ and $v = \ln 5x$ . The derivative of $u$ is $-5e^{-5x}$ (using the chain rule on the exponential). The derivative of $v$ is $\frac{1}{5x} \cdot 5 = \frac{1}{x}$ (using the chain rule on the logarithm). Applying the product rule $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ gives $\frac{1}{x}e^{-5x} - 5e^{-5x}\ln 5x$ .

Step 2: Set the derivative to zero at $x = p$ . Substituting $x = p$ and setting $\frac{dy}{dx} = 0$ yields $\frac{1}{p}e^{-5p} - 5e^{-5p}\ln 5p = 0$ .

Step 3: Rearrange. Factor out $e^{-5p}$ , which is never zero, leaving $\frac{1}{p} - 5\ln 5p = 0$ . Rearranging gives $\frac{1}{p} = 5\ln 5p$ , and dividing by 5 gives the required result $\ln 5p = \frac{1}{5p}$ .

Key Takeaways

The product rule is essential when differentiating a product of two functions of $x$ .
Stationary points are found by setting $\frac{dy}{dx} = 0$ .
Exponential functions like $e^{-5x}$ are never zero, so they can be safely divided out.

Common Mistakes

Forgetting the chain rule when differentiating $e^{-5x}$ or $\ln 5x$ (e.g., writing $\frac{d}{dx}(\ln 5x) = \frac{1}{5x}$ instead of $\frac{1}{x}$ ).
Sign errors when differentiating $e^{-5x}$ (the derivative is $-5e^{-5x}$ , not $5e^{-5x}$ ).
Not setting the derivative equal to zero before rearranging.

Things to Be Careful About

Always show the product rule explicitly; the mark scheme awards a method mark for this.
The factor $e^{-5p}$ must be noted as non-zero to justify dividing it out.
The final rearrangement must clearly lead to $\ln 5p = \frac{1}{5p}$ .

Techniques used

apply the product rule to differentiatedifferentiate composite exponential and logarithmic functionsset derivative equal to zero for stationary pointrearrange to obtain required form

(b)

By sketching a suitable pair of graphs, show that the equation in part (a) has only one root.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The equation from part (a) is $\ln 5p = \frac{1}{5p}$ . To show it has only one root, sketch $y = \ln 5x$ and $y = \frac{1}{5x}$ on the same axes and argue that they intersect exactly once for $x > 0$ .

Working

Sketch the two curves:

$y = \ln 5x$ : This is a logarithmic curve. It crosses the $x$ -axis when $\ln 5x = 0$ , i.e., $5x = 1$ , so $x = 0.2$ . It has a vertical asymptote at $x = 0$ and increases without bound as $x \to \infty$ .
$y = \frac{1}{5x}$ : This is a reciprocal curve (hyperbola) in the first quadrant. It has a vertical asymptote at $x = 0$ and a horizontal asymptote at $y = 0$ . It is strictly decreasing for $x > 0$ .

For $x > 0$ :

$y = \ln 5x$ is strictly increasing from $-\infty$ to $+\infty$ .
$y = \frac{1}{5x}$ is strictly decreasing from $+\infty$ to $0$ .

Since one curve is strictly increasing and the other is strictly decreasing, and both are continuous for $x > 0$ , they can intersect at most once. From part (c), we know a root exists in $(0.2, 0.6)$ , so there is exactly one intersection point.

Therefore, the equation $\ln 5p = \frac{1}{5p}$ has only one root.

Answer

The equation has only one root, as shown by the single intersection of the two graphs.

Final answer

The equation has only one root.

Detailed explanation

Walkthrough

The equation $\ln 5p = \frac{1}{5p}$ can be interpreted as finding the intersection of two curves: $y = \ln 5x$ and $y = \frac{1}{5x}$ . By sketching both curves on the same axes, we can visually and logically argue how many times they meet.

Step 1: Sketch $y = \ln 5x$ . This is a logarithmic curve shifted horizontally. It crosses the $x$ -axis at $x = 0.2$ (since $\ln(5 \times 0.2) = \ln 1 = 0$ ). As $x \to 0^+$ , $y \to -\infty$ (vertical asymptote at $x = 0$ ). As $x \to \infty$ , $y \to \infty$ slowly. The curve is strictly increasing.

Step 2: Sketch $y = \frac{1}{5x}$ . This is a reciprocal curve. As $x \to 0^+$ , $y \to \infty$ (vertical asymptote at $x = 0$ ). As $x \to \infty$ , $y \to 0^+$ (horizontal asymptote at $y = 0$ ). The curve is strictly decreasing for $x > 0$ and never touches either axis.

Step 3: Argue about intersections. Since $\ln 5x$ is strictly increasing and $\frac{1}{5x}$ is strictly decreasing on $(0, \infty)$ , they can intersect at most once. Part (c) confirms a root exists in $(0.2, 0.6)$ , so there is exactly one intersection.

Key Takeaways

Sketching graphs is a powerful way to determine the number of roots of an equation.
A strictly increasing function and a strictly decreasing function can intersect at most once.
Asymptotic behaviour is crucial for correct sketches.

Common Mistakes

Sketching $\ln 5x$ crossing the $x$ -axis at $x = 0$ instead of $x = 0.2$ .
Not showing the asymptotic behaviour of $y = \frac{1}{5x}$ (it must not touch the axes).
Failing to state why there is only one intersection (must mention increasing/decreasing or single intersection explicitly).

Things to Be Careful About

The sketch must show both curves in the correct quadrants. If either curve is shown in other quadrants, it must still be correct.
Allowance is made for unlabeled graphs or graphs labeled with $p$ instead of $x$ , but the key features (asymptotes, axis crossings) must be correct.
A statement or mark (dot/cross) indicating only one intersection is required for full marks.

Techniques used

sketch logarithmic curve y = ln(5x)sketch reciprocal curve y = 1/(5x)identify asymptotic behaviourreason about number of intersections

(c)

Show by calculation that $0.2 < p < 0.6$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Define $f(p) = \ln 5p - \frac{1}{5p}$ . If $f(0.2)$ and $f(0.6)$ have opposite signs, and $f$ is continuous on $[0.2, 0.6]$ , then by the sign change theorem there is a root in $(0.2, 0.6)$ .

Working

Let $f(p) = \ln 5p - \frac{1}{5p}$ .

Evaluate at $p = 0.2$ :

f(0.2) = \ln(5 \times 0.2) - \frac{1}{5 \times 0.2}

f(0.2) = \ln 1 - \frac{1}{1}

f(0.2) = 0 - 1 = -1

So $f(0.2) = -1 < 0$ .

Evaluate at $p = 0.6$ :

f(0.6) = \ln(5 \times 0.6) - \frac{1}{5 \times 0.6}

f(0.6) = \ln 3 - \frac{1}{3}

f(0.6) = 1.0986... - 0.3333... = 0.7653...

So $f(0.6) \approx 0.765 > 0$ .

Since $f(0.2) = -1 < 0$ and $f(0.6) \approx 0.765 > 0$ , and $f(p)$ is continuous for $p > 0$ , there is a sign change in the interval $(0.2, 0.6)$ .

By the sign change theorem, there is at least one root of $f(p) = 0$ in $(0.2, 0.6)$ , i.e., $0.2 < p < 0.6$ .

Answer

0.2 < p < 0.6

Final answer

0.2 < p < 0.6

Detailed explanation

Walkthrough

We need to show that the root $p$ lies between 0.2 and 0.6. The standard method is to define a function whose root we seek, evaluate it at the endpoints of the interval, and check for a sign change.

Step 1: Define the function. From part (a), the equation is $\ln 5p = \frac{1}{5p}$ , which we rewrite as $f(p) = \ln 5p - \frac{1}{5p} = 0$ .

Step 2: Evaluate at $p = 0.2$ . $f(0.2) = \ln(1) - \frac{1}{1} = 0 - 1 = -1$ . This is negative.

Step 3: Evaluate at $p = 0.6$ . $f(0.6) = \ln(3) - \frac{1}{3} \approx 1.0986 - 0.3333 = 0.7653$ . This is positive.

Step 4: Apply the sign change theorem. Since $f$ is continuous for $p > 0$ and $f(0.2) < 0$ while $f(0.6) > 0$ , there must be at least one value of $p$ in $(0.2, 0.6)$ where $f(p) = 0$ . From part (b), we know there is exactly one root, so $0.2 < p < 0.6$ .

Key Takeaways

The sign change theorem (a consequence of the Intermediate Value Theorem) is used to locate roots of equations.
You must show that the function is continuous on the interval and that the values at the endpoints have opposite signs.
Calculations should be given to at least 2 significant figures.

Common Mistakes

Not defining $f(p)$ explicitly.
Forgetting to state that $f$ is continuous.
Not stating that the signs are opposite (i.e., $f(0.2) < 0$ and $f(0.6) > 0$ ).
Calculating $\ln 3$ incorrectly.

Things to Be Careful About

The mark scheme accepts alternative intervals that work (e.g., using $p = 0.3$ and $p = 0.5$ ), but the endpoints 0.2 and 0.6 must be used to prove the specific statement $0.2 < p < 0.6$ .
At least 3 correct values to at least 2 significant figures are needed if using intermediate points.
The argument must be completed correctly with the calculated values to earn the final mark.

Techniques used

define a function f(p) = ln(5p) - 1/(5p)evaluate f at the boundary values 0.2 and 0.6apply the sign change theorem to locate the root

(d)

It is given that the equation in part (a) can be written in the form $p = \frac{1}{5} \exp\left(\frac{1}{5p}\right)$ , where $\exp(x)$ denotes $e^x$ .

Use an iterative formula based on this rearrangement to calculate $p$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

3M Similar

DifficultyMedium

Worked solution

Approach

The iterative formula is $p_{n+1} = \frac{1}{5}\exp\left(\frac{1}{5p_n}\right)$ . Choose an initial value $p_1$ in the interval $(0.2, 0.6)$ (e.g., $p_1 = 0.2$ ), and iterate, rounding each result to 4 decimal places, until the value has settled to 2 decimal places.

Working

Use the iterative formula $p_{n+1} = \frac{1}{5}e^{1/(5p_n)}$ with $p_1 = 0.2$ :

p_2 = \frac{1}{5}e^{1/(5 \times 0.2)} = 0.2 \times e^{1} = 0.2 \times 2.71828... = 0.5437

p_3 = \frac{1}{5}e^{1/(5 \times 0.5437)} = 0.2 \times e^{1/2.7185} = 0.2 \times e^{0.36790} = 0.2 \times 1.4447 = 0.2889

p_4 = \frac{1}{5}e^{1/(5 \times 0.2889)} = 0.2 \times e^{1/1.4445} = 0.2 \times e^{0.69228} = 0.2 \times 1.9983 = 0.3997

Continuing the iterations:

p_5 = \frac{1}{5}e^{1/(5 \times 0.3997)} = 0.2 \times e^{0.4996} = 0.3299

p_6 = \frac{1}{5}e^{1/(5 \times 0.3299)} = 0.2 \times e^{0.6060} = 0.3667

p_7 = \frac{1}{5}e^{1/(5 \times 0.3667)} = 0.2 \times e^{0.5454} = 0.3451

p_8 = \frac{1}{5}e^{1/(5 \times 0.3451)} = 0.2 \times e^{0.5795} = 0.3571

p_9 = \frac{1}{5}e^{1/(5 \times 0.3571)} = 0.2 \times e^{0.5599} = 0.3502

p_{10} = \frac{1}{5}e^{1/(5 \times 0.3502)} = 0.2 \times e^{0.5711} = 0.3541

The values are oscillating and converging:

p_9 = 0.3502, \quad p_{10} = 0.3541

Both round to $0.35$ to 2 decimal places. Further iterations confirm convergence:

p_{11} = 0.3507, \quad p_{12} = 0.3538, \quad p_{13} = 0.3520, \quad p_{14} = 0.3530

All values from $p_9$ onwards round to $0.35$ to 2 decimal places.

Therefore, $p = 0.35$ correct to 2 decimal places.

Answer

p = 0.35

Final answer

p = 0.35

Detailed explanation

Walkthrough

We are given the rearranged equation $p = \frac{1}{5}\exp\left(\frac{1}{5p}\right)$ , which gives us the iterative formula $p_{n+1} = \frac{1}{5}e^{1/(5p_n)}$ . We start with an initial guess in the interval $(0.2, 0.6)$ and iterate until the answer is stable to 2 decimal places.

Step 1: Choose $p_1 = 0.2$ . This is within the interval found in part (c).

Step 2: Iterate. At each step, substitute the current value into the right-hand side of the formula and round to 4 decimal places:

$p_2 = 0.2 \times e^1 = 0.5437$
$p_3 = 0.2 \times e^{1/2.7185} = 0.2889$
$p_4 = 0.2 \times e^{1/1.4445} = 0.3997$
Continue until consecutive values agree to 2 decimal places.

Step 3: Check convergence. The sequence oscillates but converges. Values $p_9 = 0.3502$ and $p_{10} = 0.3541$ both round to $0.35$ to 2 dp. Further iterations ( $p_{11} = 0.3507$ , $p_{12} = 0.3538$ , etc.) confirm this. The root is $p = 0.35$ to 2 decimal places.

Key Takeaways

Iterative formulas can be used to find roots of equations numerically.
The sequence may oscillate around the root before converging; you must continue until the answer is stable to the required number of decimal places.
Always show sufficient iterations (to 4 dp) to justify the final answer.

Common Mistakes

Using too few iterations (must show enough to justify 2 dp).
Not rounding intermediate results to 4 dp as instructed, which can lead to accumulated errors.
Starting the iteration with a value outside the interval (though convergence is generally robust here).
Stating the answer as $x = 0.35$ instead of $p = 0.35$ (the variable is $p$ ).
Writing $p_7 = 0.35...$ or $p_\infty = 0.35$ instead of simply stating $p = 0.35$ or answer $= 0.35$ .

Things to Be Careful About

The mark scheme requires showing sufficient iterations to 4 dp to justify the answer to 2 dp, or showing a sign change in $(0.345, 0.355)$ .
Starting from $p_1 = 0.3526$ or $0.3527$ with correct iterations can earn partial marks (special case), but starting from these values directly without showing the iteration process earns 0 marks.
The final answer must be stated as $p = 0.35$ or answer $= 0.35$ or simply $0.35$ .

Techniques used

use the given iterative formula p_(n+1) = (1/5) exp(1/(5p_n))iterate starting from an initial value in (0.2, 0.6)round each result to 4 decimal placesdetermine when the value has converged to 2 decimal places

Q9MediumVectors

The line $l_1$ passes through the point $(3, 1, -6)$ and is parallel to the vector $2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ .

The line $l_2$ passes through the point $(-1, 3, -6)$ and is perpendicular to the vector $3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ . The direction vector for $l_2$ has no component in the $x$ -direction.

(a)

Write down a vector equation for $l_1$ and find a vector equation for $l_2$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For $l_1$ , use the given point as position vector $\mathbf{a}$ and the given parallel vector as direction $\mathbf{b}$ , so $\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}$ . For $l_2$ , write its direction vector as $y\mathbf{j} + z\mathbf{k}$ because it has no $x$ -component. Since $l_2$ is perpendicular to $3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ , their scalar product is zero, which determines $y$ and $z$ .

Working

For $l_1$ :

\mathbf{r} = 3\mathbf{i} + \mathbf{j} - 6\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j} + 4\mathbf{k})

For $l_2$ , let the direction vector be $y\mathbf{j} + z\mathbf{k}$ . Perpendicular to $3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ gives:

(3\mathbf{i} - 2\mathbf{j} + \mathbf{k}) \cdot (y\mathbf{j} + z\mathbf{k}) = 0

-2y + z = 0

So $z = 2y$ . Taking $y = 1$ gives $z = 2$ , so a direction vector is $\mathbf{j} + 2\mathbf{k}$ . Hence:

\mathbf{r} = -\mathbf{i} + 3\mathbf{j} - 6\mathbf{k} + \mu(\mathbf{j} + 2\mathbf{k})

Answer

\mathbf{r} = 3\mathbf{i} + \mathbf{j} - 6\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j} + 4\mathbf{k})

\mathbf{r} = -\mathbf{i} + 3\mathbf{j} - 6\mathbf{k} + \mu(\mathbf{j} + 2\mathbf{k})

Final answer

l1: r = 3i + j - 6k + λ(2i + j + 4k); l2: r = -i + 3j - 6k + μ(j + 2k)

Detailed explanation

Walkthrough

Start with $l_1$ . A line through a point with position vector $\mathbf{a}$ and direction vector $\mathbf{b}$ has equation $\mathbf{r} = \mathbf{a} + t\mathbf{b}$ . Here $\mathbf{a} = 3\mathbf{i} + \mathbf{j} - 6\mathbf{k}$ and $\mathbf{b} = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ , so the equation can be written down directly.

For $l_2$ , the direction vector has no $x$ -component, so it must have the form $y\mathbf{j} + z\mathbf{k}$ . The line is perpendicular to $3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ , so the scalar product of these two vectors is zero:

(3\mathbf{i} - 2\mathbf{j} + \mathbf{k}) \cdot (y\mathbf{j} + z\mathbf{k}) = 0

This gives $-2y + z = 0$ , so $z = 2y$ . Choosing $y = 1$ gives $z = 2$ , so one valid direction vector is $\mathbf{j} + 2\mathbf{k}$ . Using the given point $(-1, 3, -6)$ gives the required equation for $l_2$ .

Key Takeaways

A line in vector form is determined by a point and a direction vector. Perpendicularity is expressed by a zero scalar product. A direction vector can be scaled by any non-zero constant, so choosing a convenient value such as $y = 1$ is acceptable.

Common Mistakes

Omitting $\mathbf{r} =$ ; the mark scheme penalises this only once in part (a).
Forgetting that $l_2$ has no $x$ -component, so including an $x$ -component in its direction vector.
Making a sign error when expanding $(3\mathbf{i} - 2\mathbf{j} + \mathbf{k}) \cdot (y\mathbf{j} + z\mathbf{k})$ .
Writing column vectors that still contain $\mathbf{i}$ , $\mathbf{j}$ and $\mathbf{k}$ ; this is not allowed in the mark scheme.

Things to Be Careful About

Use different parameters for the two lines, for example $\lambda$ and $\mu$ , so that the equations remain distinct. Any non-zero multiple of $\mathbf{j} + 2\mathbf{k}$ is also a correct direction vector, so answers such as $2\mathbf{j} + 4\mathbf{k}$ are equally valid. Always present the final line equation in the form $\mathbf{r} = \ldots$ .

Techniques used

write the vector equation of a line through a point with a given directionset the x-component of the l2 direction vector to zeroimpose perpendicularity using the scalar product

(b)

Calculate the acute angle between $l_1$ and $l_2$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The angle between two lines is found from their direction vectors. Use the direction vectors from part (a): $\mathbf{d}_1 = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ and $\mathbf{d}_2 = \mathbf{j} + 2\mathbf{k}$ . Apply

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

and take the acute angle.

Working

\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(0) + (1)(1) + (4)(2) = 9

|\mathbf{d}_1| = \sqrt{2^2 + 1^2 + 4^2} = \sqrt{21}

|\mathbf{d}_2| = \sqrt{0^2 + 1^2 + 2^2} = \sqrt{5}

So

\cos \theta = \frac{9}{\sqrt{21}\sqrt{5}} = \frac{9}{\sqrt{105}}

\theta = \cos^{-1}\left(\frac{9}{\sqrt{105}}\right) = 28.6^\circ \quad \text{or} \quad 0.498 \text{ radians}

Answer

\theta = 28.6^\circ \quad (0.498 \text{ radians})

Final answer

28.6° (or 0.498 radians)

Detailed explanation

Walkthrough

For the angle between two lines, only their direction vectors matter. From part (a), $l_1$ has direction vector $\mathbf{d}_1 = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ and $l_2$ has direction vector $\mathbf{d}_2 = \mathbf{j} + 2\mathbf{k}$ . The scalar product is

\mathbf{d}_1 \cdot \mathbf{d}_2 = 2(0) + 1(1) + 4(2) = 9

The magnitudes are $\sqrt{21}$ and $\sqrt{5}$ . Therefore

\cos \theta = \frac{9}{\sqrt{21}\sqrt{5}} = \frac{9}{\sqrt{105}}

Since the scalar product is positive, this already gives the acute angle. Evaluating gives $28.6^\circ$ or $0.498$ radians.

Key Takeaways

The cosine formula for the angle between two vectors is $\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$ . For lines, use their direction vectors, not their position vectors. If the cosine is negative, the angle found is obtuse, and the acute angle is obtained by taking the absolute value of the scalar product.

Common Mistakes

Using position vectors of the given points instead of direction vectors.
Forgetting to divide by the product of the magnitudes.
Using the wrong direction vector for $l_2$ from part (a).
Giving the obtuse angle instead of the acute angle.

Things to Be Careful About

The mark scheme allows using the direction vectors found in part (a), so a mistake in part (a) will affect this part. The final answer should be $28.6^\circ$ or $0.498$ radians; do not mix units or give an unreasonably rounded value. The symbol used for the product is not important, but the scalar product must be computed correctly.

Techniques used

compute the scalar product of the direction vectorscompute the magnitudes of the direction vectorsapply the cosine formula for the angle between two vectors

(c)

Find the position vector of the point of intersection of $l_1$ and $l_2$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Write a general point on each line using its own parameter. Equate the $x$ , $y$ and $z$ components, solve for one parameter, then substitute back to find the intersection point.

Working

A general point on $l_1$ is

(3 + 2\lambda, \; 1 + \lambda, \; -6 + 4\lambda)

A general point on $l_2$ is

(-1, \; 3 + \mu, \; -6 + 2\mu)

Equating $x$ -components:

3 + 2\lambda = -1

\lambda = -2

Equating $z$ -components:

-6 + 4\lambda = -6 + 2\mu

4\lambda = 2\mu

\mu = 2\lambda = -4

Check the $y$ -components: $1 + \lambda = -1$ and $3 + \mu = -1$ , so the components agree. Substitute $\lambda = -2$ into $l_1$ :

(3 - 4, \; 1 - 2, \; -6 - 8) = (-1, -1, -14)

So the position vector of the point of intersection is

-\mathbf{i} - \mathbf{j} - 14\mathbf{k}

Answer

-\mathbf{i} - \mathbf{j} - 14\mathbf{k}

Final answer

-i - j - 14k

Detailed explanation

Walkthrough

To find where two lines meet, use a different parameter for each line: $\lambda$ for $l_1$ and $\mu$ for $l_2$ . A general point on $l_1$ is $(3 + 2\lambda, 1 + \lambda, -6 + 4\lambda)$ and a general point on $l_2$ is $(-1, 3 + \mu, -6 + 2\mu)$ . At intersection these points are equal, so their components must match. The $x$ -equation gives $3 + 2\lambda = -1$ , so $\lambda = -2$ . The $z$ -equation then gives $-6 + 4\lambda = -6 + 2\mu$ , so $\mu = 2\lambda = -4$ . Substituting $\lambda = -2$ into $l_1$ gives $(-1, -1, -14)$ , whose position vector is $-\mathbf{i} - \mathbf{j} - 14\mathbf{k}$ . The $y$ -components agree, confirming the point lies on both lines.

Key Takeaways

Intersecting lines can be solved by equating component expressions. A different parameter must be used for each line when solving. Once one parameter is found, the other can be checked or found from another component.

Common Mistakes

Using the same parameter for both lines when solving; the mark scheme does not allow this if solving using two linear equations.
Stopping after finding $\lambda = -2$ without substituting it back to obtain the intersection point.
Giving the answer as coordinates $(-1, -1, -14)$ instead of the position vector; the mark scheme requires a position vector.
Making sign errors when substituting negative parameters.

Things to Be Careful About

The mark scheme allows the $x$ -equation alone to earn the method mark, even if the other equations are wrong. The final answer must be a position vector, not coordinates. It is worth checking the unused component to confirm the lines actually intersect.

Techniques used

write a general point on each line using its own parameterequate components of the two general pointssolve the resulting simultaneous equations for the parameterssubstitute back to find the position vector of the intersection

Q10Medium-HardAlgebraDifferential EquationsIntegration

(a)

Express $\frac{2}{1 - 9y^2}$ in partial fractions.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Recognise that $1-9y^2$ is a difference of two squares, so it factorises as $(1+3y)(1-3y)$ . Express the fraction as a sum of two partial fractions with unknown constants and determine those constants.

Working

Set

\frac{2}{1-9y^2} = \frac{A}{1+3y} + \frac{B}{1-3y}

Multiplying through by $(1+3y)(1-3y)$ :

2 = A(1-3y) + B(1+3y)

Put $y = \frac{1}{3}$ to eliminate $A$ :

2 = 2B \Rightarrow B = 1

Put $y = -\frac{1}{3}$ to eliminate $B$ :

2 = 2A \Rightarrow A = 1

Therefore

\frac{2}{1-9y^2} = \frac{1}{1+3y} + \frac{1}{1-3y}

Answer

\frac{2}{1-9y^2} = \frac{1}{1+3y} + \frac{1}{1-3y}

Final answer

1/(1+3y) + 1/(1-3y)

Detailed explanation

Walkthrough

Set

\frac{2}{1-9y^2} = \frac{A}{1+3y} + \frac{B}{1-3y}

sine the denominator is

1-9y^2 = (1+3y)(1-3y)

Multiplying the partial fraction identity by the common denominator leaves

2 = A(1-3y) + B(1+3y)

The useful values to substitute are the roots of the factors. When $y = \frac13$ , the factor $1-3y$ is zero, so the $A$ term disappears and $B$ can be read off immediately. When $y = -\frac13$ , the factor $1+3y$ is zero, so the $B$ term disappears and $A$ can be found. This 'cover-up' method is exactly the fastest way to find the constants.

Key Takeaways

The denominator $1-9y^2$ factorises as $(1+3y)(1-3y)$ because it is a difference of squares.
For distinct linear factors, the unknown constants are best found by substituting the roots of the factors.
The final decomposition should be checked by recombining the two fractions.

Common Mistakes

Forgetting the minus sign in $1-3y$ when forming the second factor.
Using the wrong sign in the identity $2 = A(1-3y) + B(1+3y)$ .
Swapping the values of $A$ and $B$ .
Not showing the method and simply writing the final partial fraction result; the mark scheme requires a relevant method for the M1 mark.

Things to Be Careful About

When substituting $y = \frac13$ , the factor $1-3y$ is zero, so $A$ is correctly eliminated.
When substituting $y = -\frac13$ , the factor $1+3y$ is zero, so $B$ is eliminated.
If the problem is approached with $x$ instead of $y$ , the mark scheme allows M1 but only awards A1 if $y$ appears on the right-hand side.

Techniques used

factorise the denominator as a difference of squaresset up partial fractions with unknown constantssubstitute the roots of the factors to find the constants

(b)

The variables $x$ and $y$ satisfy the differential equation

2\cos^2 3x \frac{dy}{dx} = 1 - 9y^2

and $y = 0$ when $x = \frac{1}{12}\pi$ .

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

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Separate the variables so that all $y$ terms are on the left and all $x$ terms are on the right. Use the partial fraction result from part (a) to integrate the $y$ side, use the standard integral of $\sec^2 3x$ , then apply the initial condition to find the constant. Finally rearrange to make $y$ the subject.

Working

Separate variables:

\frac{2}{1-9y^2}\,dy = \sec^2 3x\,dx

Using part (a), $\frac{2}{1-9y^2} = \frac{1}{1+3y} + \frac{1}{1-3y}$ , so integrate both sides:

\int \left(\frac{1}{1+3y} + \frac{1}{1-3y}\right)dy = \int \sec^2 3x\,dx

\frac{1}{3}\ln(1+3y) - \frac{1}{3}\ln(1-3y) = \frac{1}{3}\tan 3x + C

Equivalently,

\frac{1}{3}\ln\left(\frac{1+3y}{1-3y}\right) = \frac{1}{3}\tan 3x + C

Apply $y=0$ when $x=\frac{1}{12}\pi$ :

0 = \frac{1}{3}\tan\left(\frac{\pi}{4}\right) + C

Since $\tan \frac{\pi}{4} = 1$ ,

C = -\frac{1}{3}

Thus

\frac{1}{3}\ln\left(\frac{1+3y}{1-3y}\right) = \frac{1}{3}\tan 3x - \frac{1}{3}

Multiplying by 3:

\ln\left(\frac{1+3y}{1-3y}\right) = \tan 3x - 1

Exponentiating both sides:

\frac{1+3y}{1-3y} = e^{\tan 3x - 1}

Let $E = e^{\tan 3x - 1}$ . Then

1+3y = E(1-3y)

3y + 3Ey = E - 1

y = \frac{E - 1}{3(E + 1)}

or equivalently

y = \frac{1}{3} - \frac{2}{3(E + 1)}

Answer

y = \frac{e^{\tan 3x - 1} - 1}{3\left(e^{\tan 3x - 1} + 1\right)}

Final answer

y = (e^(tan3x - 1) - 1)/(3(e^(tan3x - 1) + 1))

Detailed explanation

Walkthrough

The differential equation is first-order and separable. Dividing by $(1-9y^2)$ and by $\cos^2 3x$ gives

\frac{2}{1-9y^2}\,dy = \sec^2 3x\,dx

The left-hand side is exactly the fraction from part (a), so its integral uses the partial fraction decomposition

\frac{2}{1-9y^2} = \frac{1}{1+3y} + \frac{1}{1-3y}

Integrating gives logarithms on the left and a tangent on the right. The $\frac13$ factors come from the coefficients of $3y$ inside the logarithms and from the $3x$ inside $\sec^2 3x$ . Including the constant of integration is essential.

The initial condition $y=0$ at $x=\frac{1}{12}\pi$ is used to find $C$ . At that value, $3x = \frac{\pi}{4}$ and $\tan\frac{\pi}{4}=1$ , so $C=-\frac13$ .

Finally, the equation is rewritten as a single logarithm, exponentiated, and solved algebraically for $y$ :

\frac{1+3y}{1-3y} = e^{\tan 3x - 1}

Letting $E = e^{\tan 3x - 1}$ and rearranging gives

y = \frac{E-1}{3(E+1)} = \frac13 - \frac{2}{3(E+1)}

Techniques used

separate the variablesintegrate using partial fractionsintegrate a secant squared termapply the initial condition to find the constantrearrange the resulting logarithmic equation to obtain y explicitly

Q11MediumDifferentiationTrigonometryIntegration

Fig. 1

The diagram shows the graph of $y = \sec^2 x \sqrt{3 + 2\tan x}$ for $-\frac{1}{4}\pi \leq x \leq \frac{1}{4}\pi$ , and its minimum point $M$ .

(a)

Find the $x$ -coordinate of $M$ .

6M Similar

DifficultyMedium

Worked solution

Approach

To find the x-coordinate of the minimum point M, differentiate $y = \sec^2 x \sqrt{3 + 2\tan x}$ with respect to $x$ using the product rule and chain rule, then set $\frac{dy}{dx} = 0$ and solve for $x$ .

Working

Let $y = \sec^2 x \cdot (3 + 2\tan x)^{1/2}$ .

Differentiate each factor:

\frac{d}{dx}(\sec^2 x) = 2\sec x \cdot \sec x \tan x = 2\sec^2 x \tan x

\frac{d}{dx}\left((3 + 2\tan x)^{1/2}\right) = \frac{1}{2}(3 + 2\tan x)^{-1/2} \cdot 2\sec^2 x = \sec^2 x (3 + 2\tan x)^{-1/2}

Apply the product rule $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ :

\frac{dy}{dx} = \sec^2 x \cdot \sec^2 x (3 + 2\tan x)^{-1/2} + (3 + 2\tan x)^{1/2} \cdot 2\sec^2 x \tan x

\frac{dy}{dx} = \sec^4 x (3 + 2\tan x)^{-1/2} + 2\sec^2 x \tan x (3 + 2\tan x)^{1/2}

Set $\frac{dy}{dx} = 0$ and multiply through by $(3 + 2\tan x)^{1/2}$ :

\sec^4 x + 2\sec^2 x \tan x (3 + 2\tan x) = 0

Divide by $\sec^2 x$ (which is never zero):

\sec^2 x + 2\tan x (3 + 2\tan x) = 0

Substitute $\sec^2 x = 1 + \tan^2 x$ :

1 + \tan^2 x + 6\tan x + 4\tan^2 x = 0

5\tan^2 x + 6\tan x + 1 = 0

Factorise:

(5\tan x + 1)(\tan x + 1) = 0

So $\tan x = -\frac{1}{5}$ or $\tan x = -1$ .

For $\tan x = -1$ , $x = -\frac{\pi}{4}$ , which is the left endpoint of the domain, not the minimum point M.

For $\tan x = -\frac{1}{5}$ :

x = \arctan\left(-\frac{1}{5}\right) \approx -0.197

Answer

x \approx -0.197

Final answer

x ≈ -0.197

Detailed explanation

Walkthrough

We need to find the x-coordinate of the minimum point M on the curve $y = \sec^2 x \sqrt{3 + 2\tan x}$ . The minimum occurs where the derivative equals zero.

First, we identify the two factors in the product: $u = \sec^2 x$ and $v = (3 + 2\tan x)^{1/2}$ . We differentiate each using the chain rule. For $\sec^2 x$ , we treat it as $(\sec x)^2$ and apply the chain rule to get $2\sec x \cdot \sec x \tan x = 2\sec^2 x \tan x$ . For $(3 + 2\tan x)^{1/2}$ , we bring down the power $1/2$ , reduce it to $-1/2$ , and multiply by the derivative of the inside, $2\sec^2 x$ , giving $\sec^2 x (3 + 2\tan x)^{-1/2}$ .

Next, we apply the product rule: $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$ . This gives us two terms involving $\sec^2 x$ and powers of $(3 + 2\tan x)$ .

Setting the derivative to zero, we multiply through by $(3 + 2\tan x)^{1/2}$ to eliminate the negative exponent. This is valid because $3 + 2\tan x > 0$ on the given domain. Dividing by $\sec^2 x$ (never zero) simplifies the equation. We then use the Pythagorean identity $\sec^2 x = 1 + \tan^2 x$ to convert everything into an equation in $\tan x$ alone, yielding the quadratic $5\tan^2 x + 6\tan x + 1 = 0$ .

Factorising gives $\tan x = -1/5$ or $\tan x = -1$ . The solution $\tan x = -1$ corresponds to $x = -\pi/4$ , which is the boundary of the domain, not the interior minimum. The valid solution is $\tan x = -1/5$ , giving $x = \arctan(-1/5) \approx -0.197$ .

Key Takeaways

The product rule combined with the chain rule is essential for differentiating products of composite trigonometric functions.
Setting the derivative to zero and using trigonometric identities to reduce to a single trig function is a standard technique for finding turning points.
Always check boundary values when solving trigonometric equations to distinguish between endpoints and actual turning points.

Common Mistakes

Forgetting to apply the chain rule when differentiating $\sec^2 x$ or $(3 + 2\tan x)^{1/2}$ .
Not using the product rule correctly and instead differentiating the whole expression as a single function.
Failing to use $\sec^2 x = 1 + \tan^2 x$ to convert the equation into a quadratic in $\tan x$ .
Accepting $\tan x = -1$ as the answer without checking that it corresponds to the domain boundary $x = -\pi/4$ .

Things to Be Careful About

When multiplying through by $(3 + 2\tan x)^{1/2}$ , ensure this factor is non-zero. On the given domain $-\pi/4 \leq x \leq \pi/4$ , we have $-1 \leq \tan x \leq 1$ , so $3 + 2\tan x \geq 1 > 0$ .
The mark scheme requires showing the derivative in a form that includes both $(\ldots)^{1/2}$ and $(\ldots)^{-1/2}$ terms before equating to zero.
The answer $x \approx -0.197$ is to 3 decimal places; more accuracy is acceptable but decimals alone without showing exact form $\arctan(-1/5)$ may not earn full marks.

Techniques used

apply product rule to composite functiondifferentiate sec^2 x using chain ruledifferentiate sqrt(3 + 2tan x) using chain ruleuse sec^2 x = 1 + tan^2 x identitysolve quadratic in tan x

(b)

Using the substitution $u = 3 + 2\tan x$ , find the exact value of the area of the region bounded by the curve, the $x$ -axis and the lines $x = -\frac{1}{4}\pi$ and $x = \frac{1}{4}\pi$ .

6M Similar

DifficultyMedium-Easy

Worked solution

Approach

The area under the curve is given by the definite integral $\int_{-\pi/4}^{\pi/4} \sec^2 x \sqrt{3 + 2\tan x} \, dx$ . Use the substitution $u = 3 + 2\tan x$ to simplify the integral.

Working

The area is:

A = \int_{-\pi/4}^{\pi/4} \sec^2 x \sqrt{3 + 2\tan x} \, dx

Let $u = 3 + 2\tan x$ . Then:

\frac{du}{dx} = 2\sec^2 x \quad \Rightarrow \quad dx = \frac{du}{2\sec^2 x}

Transform the limits:

When $x = -\frac{\pi}{4}$ : $u = 3 + 2\tan\left(-\frac{\pi}{4}\right) = 3 - 2 = 1$
When $x = \frac{\pi}{4}$ : $u = 3 + 2\tan\left(\frac{\pi}{4}\right) = 3 + 2 = 5$

Substitute into the integral:

A = \int_1^5 \sec^2 x \cdot \sqrt{u} \cdot \frac{du}{2\sec^2 x}

A = \int_1^5 \frac{1}{2} u^{1/2} \, du

Integrate:

A = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} \Big|_1^5 = \frac{1}{3} u^{3/2} \Big|_1^5

Evaluate at the limits:

A = \frac{1}{3} \left(5^{3/2} - 1^{3/2}\right) = \frac{1}{3} \left(5\sqrt{5} - 1\right)

A = \frac{5\sqrt{5}}{3} - \frac{1}{3}

Answer

A = \frac{5\sqrt{5} - 1}{3}

Final answer

(5√5 - 1)/3

Detailed explanation

Walkthrough

We need to find the area bounded by the curve $y = \sec^2 x \sqrt{3 + 2\tan x}$ , the x-axis, and the vertical lines $x = -\pi/4$ and $x = \pi/4$ . This is the definite integral $\int_{-\pi/4}^{\pi/4} \sec^2 x \sqrt{3 + 2\tan x} \, dx$ .

The substitution $u = 3 + 2\tan x$ is suggested. Differentiating gives $du/dx = 2\sec^2 x$ , so $dx = du/(2\sec^2 x)$ . Notice that the $\sec^2 x$ factor in the integrand cancels with the $\sec^2 x$ in the denominator from $dx$ , leaving a simple power of $u$ to integrate.

We must transform the limits: at $x = -\pi/4$ , $\tan(-\pi/4) = -1$ , so $u = 3 - 2 = 1$ . At $x = \pi/4$ , $\tan(\pi/4) = 1$ , so $u = 3 + 2 = 5$ .

The integral becomes $\int_1^5 \frac{1}{2} u^{1/2} \, du$ . Integrating gives $\frac{1}{3} u^{3/2}$ , and evaluating from 1 to 5 gives $\frac{1}{3}(5^{3/2} - 1) = \frac{5\sqrt{5} - 1}{3}$ .

Key Takeaways

Integration by substitution works well when the derivative of the substitution appears (up to a constant) as a factor in the integrand.
Always transform the limits of integration when using substitution, rather than converting back to the original variable.
The $\sec^2 x$ factor is the derivative of $\tan x$ (up to a constant), making it ideal for substitution with $u = 3 + 2\tan x$ .

Common Mistakes

Forgetting to transform the limits of integration and instead trying to convert back to $x$ at the end.
Making arithmetic errors when evaluating $5^{3/2}$ (it equals $5\sqrt{5}$ , not $5\sqrt{2}$ or $25$ ).
Forgetting the factor of $1/2$ from $du = 2\sec^2 x \, dx$ .

Things to Be Careful About

The mark scheme requires an exact answer; decimal approximations are not accepted.
$5^{3/2} = (\sqrt{5})^3 = 5\sqrt{5}$ , not $25$ or $\sqrt{15}$ .
The answer can be written as $\frac{5\sqrt{5} - 1}{3}$ or equivalently $\frac{\sqrt{125} - 1}{3}$ .

Techniques used

set up definite integral for areaapply substitution u = 3 + 2tan xtransform limits of integrationintegrate power functionevaluate at bounds

Mathematics 9709/33 — October/November 2025

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