The polynomial $p(x)$ is defined by $p(x) = 9x^3 + 18x^2 + 5x + 4$. (a) Find the quotient when $p(x)$ is divided by $(3x + 2)$, and show that the remainder is 6. (b) Find the value of $\int_0^2 \frac{p(x)}{3x + 2} \, dx$, giving your answer in the form $a + \ln b$ where $a$ and $b$ are integers.

(a) Quotient = 3x^2 + 4x - 1, Remainder = 6 · (b) 14 + ln 16

The diagram shows the curve with equation $y = \frac{\ln(2x + 1)}{x + 3}$. The curve has a maximum point $M$. (a) Find an expression for $\frac{dy}{dx}$. (b) Show that the $x$-coordinate of $M$ satisfies the equation $x = \frac{x + 3}{\ln(2x + 1)} - 0.5$. (c) Show by calculation that the $x$-coordinate of $M$ lies between 2.5 and 3.0. (d) Use an iterative formula based on the equation in part (b) to find the $x$-coordinate of $M$ correct to 4 significant figures. Give the…

(a) dy/dx = ((x+3)·(2/(2x+1)) - ln(2x+1)) / (x+3)^2 · (b) x = (x+3)/ln(2x+1) - 0.5 · (c) 2.5 < x < 3.0 · (d) x = 2.569

9709/22 May/June 2024 — Mathematics Solved Past Paper

Q14MMedium-EasyAlgebra

Solve the inequality $|5x + 7| > |2x - 3|$ .

DifficultyMedium-Easy

Worked solution

Approach

For an inequality involving two moduli, first find the boundary values where the two expressions are equal in magnitude. These occur when the expressions inside the moduli are equal or are negatives of each other. Then test one point in each interval to decide which intervals satisfy the inequality.

Working

Solve the two linear equations that give equality of the moduli.

Case 1: $5x+7 = 2x-3$ .

5x + 7 = 2x - 3

3x = -10

x = -\frac{10}{3}

Case 2: $5x+7 = -(2x-3)$ .

5x + 7 = -2x + 3

7x = -4

x = -\frac{4}{7}

These two values divide the real line into intervals. Test $x=0$ :

|5(0)+7| = 7 > |2(0)-3| = 3

so $x=0$ satisfies the inequality. Test $x=-1$ :

|5(-1)+7| = 2 \quad \text{and} \quad |2(-1)-3| = 5

so $2 > 5$ is false, and the interval between the critical values does not satisfy the inequality. Therefore the solution consists of the two outer intervals.

Answer

x < -\frac{10}{3} \quad \text{or} \quad x > -\frac{4}{7}

Final answer

x < -10/3 or x > -4/7

Detailed explanation

Walkthrough

We need all values of $x$ for which the distance of $5x+7$ from zero is greater than the distance of $2x-3$ from zero. The boundary between true and false occurs when $|5x+7| = |2x-3|$ . This equality holds when the expressions inside the moduli are equal or are negatives of each other.

Solving $5x+7 = 2x-3$ gives $x=-\frac{10}{3}$ ; solving $5x+7 = -(2x-3)$ gives $x=-\frac{4}{7}$ . These two critical values split the real line into three intervals: $(-\infty, -\frac{10}{3})$ , $(-\frac{10}{3}, -\frac{4}{7})$ and $(-\frac{4}{7}, \infty)$ .

We test one point in each interval. At $x=0$ , $|5(0)+7|=7$ and $|2(0)-3|=3$ , so $7>3$ is true. At $x=-1$ , $|5(-1)+7|=2$ and $|2(-1)-3|=5$ , so $2>5$ is false. Thus the middle interval is not part of the solution, while the two outer intervals are.

An alternative is to square both sides: since both sides are non-negative, $|5x+7| > |2x-3|$ is equivalent to $(5x+7)^2 > (2x-3)^2$ . Expanding gives a quadratic inequality whose boundary roots are the same two critical values; testing intervals then gives the same answer.

Key Takeaways

The main idea is that a modulus inequality can be solved by first locating the points where the two sides are equal, then testing the intervals between these points. This avoids having to split into many case branches. The final answer must be written as a union of intervals using "or", not "and".

Common Mistakes

Writing the answer as $x < -\frac{10}{3}$ and $x > -\frac{4}{7}$ instead of using "or". The marking scheme specifically disallows "and".
Forgetting to solve both equality cases $a=b$ and $a=-b$ ; solving only one case loses the second critical value.
Sign errors when expanding $-(2x-3)$ : it is $-2x+3$ , not $-2x-3$ .
Including the boundary points in the solution. Because the inequality is strict, $x=-\frac{10}{3}$ and $x=-\frac{4}{7}$ are not included.

Things to Be Careful About

The critical values themselves do not satisfy the strict inequality, so use strict inequality signs in the final answer. When testing intervals, choose simple values such as $x=0$ and $x=-1$ to avoid arithmetic errors. If using the squaring method, remember that squaring is valid here because both $|5x+7|$ and $|2x-3|$ are non-negative; the resulting quadratic inequality must still be solved with the correct sign.

Techniques used

solve equality cases of the two modulitest intervals around the critical valuesstate the solution as a union of intervals

Approach

Take natural logarithms of both sides so the unknown exponents become linear factors. Apply the logarithm law $\ln(ab)=\ln a+\ln b$ to the right-hand side, then solve the resulting linear equation for $x$ .

Working

Take logarithms of both sides:

\ln(6^{2x-1})=\ln(5e^{3x+2})

Use $\ln(a^b)=b\ln a$ on the left and $\ln(ab)=\ln a+\ln b$ on the right:

(2x-1)\ln 6=\ln 5+3x+2

Expand the left-hand side:

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

Collect the $x$ -terms on one side:

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

Factor out $x$ :

x(2\ln 6-3)=\ln 5+2+\ln 6

Solve for $x$ :

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

Evaluating with a calculator:

x\approx \frac{5.401197}{0.583519}\approx 9.256

Answer

x=9.256 \quad (4\text{ s.f.})

Final answer

x = 9.256

Detailed explanation

Walkthrough

The equation has the variable $x$ in the exponents, so we cannot isolate it by ordinary algebraic operations. Taking natural logarithms of both sides is the key step because logarithms convert powers into products: $\ln(a^b)=b\ln a$ . Applying this to $6^{2x-1}$ gives $(2x-1)\ln 6$ .

On the right-hand side, $\ln(5e^{3x+2})$ is a logarithm of a product, so use $\ln(ab)=\ln a+\ln b$ :

\ln(5e^{3x+2})=\ln 5+\ln(e^{3x+2})=\ln 5+3x+2

because $\ln(e^u)=u$ . The equation is now linear in $x$ :

(2x-1)\ln 6=\ln 5+3x+2

Expand the left side, move every term containing $x$ to one side and constant terms to the other, then factor out $x$ . Dividing by the coefficient of $x$ gives the exact expression for $x$ . Finally, use a calculator to evaluate the expression and round to 4 significant figures.

Key Takeaways

Logarithms are the standard tool for solving equations with the unknown in an exponent.
The power law $\ln(a^b)=b\ln a$ brings the exponent down.
The product law $\ln(ab)=\ln a+\ln b$ separates a product inside a logarithm.
Because the equation contains $e$ , natural logarithms are especially convenient since $\ln(e^k)=k$ .
After obtaining an exact expression, evaluate it accurately before rounding.

Common Mistakes

Writing $\ln(5e^{3x+2})$ as $\ln 5 \cdot (3x+2)$ instead of $\ln 5 + (3x+2)$ . The logarithm of a product is a sum, not a product.
Forgetting to multiply the whole bracket by $\ln 6$ : the left side is $(2x-1)\ln 6$ , not $2x-1\ln 6$ .
Making a sign error when moving $\ln 6$ to the other side. From $2x\ln 6 - \ln 6 = \ln 5 + 3x + 2$ , adding $\ln 6$ to both sides gives $+\ln 6$ on the right.
Rounding intermediate values too early, which can change the fourth significant figure.

Things to Be Careful About

Use natural logarithms because of the $e$ ; this makes $\ln(e^{3x+2})=3x+2$ immediate.
The denominator $2\ln 6-3$ is positive here, so the division is valid.
The final answer must be given to 4 significant figures: $9.256$ , not $9.3$ or $9.2557$ .
Show the linear equation solving steps clearly to earn the method marks; an unsupported calculator answer may not receive full credit.

Techniques used

take natural logarithms of both sidesapply logarithm laws to expand powers and productscollect x terms and solve the resulting linear equationevaluate the expression and round to significant figures

Approach

Point $B$ is where the curve crosses the $x$ -axis, so set $y = 0$ and solve for $x$ using logarithms. Once the upper limit is confirmed as $\ln 2$ , integrate the curve equation from $x = 0$ to $x = \ln 2$ to find the area of the shaded region.

Working

Finding the $x$ -coordinate of $B$ :

8e^{-x} - e^{2x} = 0

8e^{-x} = e^{2x}

Multiply both sides by $e^x$ :

8 = e^{3x}

Take the natural logarithm of both sides:

\ln 8 = 3x

x = \frac{\ln 8}{3} = \frac{\ln(2^3)}{3} = \frac{3\ln 2}{3} = \ln 2

Finding the area of the shaded region:

\text{Area} = \int_0^{\ln 2} \left(8e^{-x} - e^{2x}\right) \, dx

= \left[ -8e^{-x} - \frac{1}{2}e^{2x} \right]_0^{\ln 2}

Evaluate at the upper limit $x = \ln 2$ :

-8e^{-\ln 2} - \frac{1}{2}e^{2\ln 2} = -8\left(\frac{1}{2}\right) - \frac{1}{2}(4) = -4 - 2 = -6

Evaluate at the lower limit $x = 0$ :

-8e^{0} - \frac{1}{2}e^{0} = -8 - \frac{1}{2} = -\frac{17}{2}

Subtract the lower limit value from the upper limit value:

\text{Area} = -6 - \left(-\frac{17}{2}\right) = -6 + \frac{17}{2} = \frac{5}{2}

Answer

\frac{5}{2}

Final answer

5/2

Detailed explanation

Walkthrough

First, we find the $x$ -coordinate of point $B$ by setting $y = 0$ . The equation $8e^{-x} - e^{2x} = 0$ rearranges to $8e^{-x} = e^{2x}$ . Multiplying by $e^x$ gives $8 = e^{3x}$ . Taking the natural logarithm yields $\ln 8 = 3x$ , and since $\ln 8 = \ln(2^3) = 3\ln 2$ , we get $x = \ln 2$ . This confirms the upper limit of integration.

Next, we calculate the area under the curve from $x = 0$ to $x = \ln 2$ . The integral of $8e^{-x}$ is $-8e^{-x}$ and the integral of $-e^{2x}$ is $-\frac{1}{2}e^{2x}$ . Evaluating this antiderivative at $x = \ln 2$ requires knowing that $e^{-\ln 2} = \frac{1}{2}$ and $e^{2\ln 2} = e^{\ln 4} = 4$ . This gives a value of $-6$ . At $x = 0$ , the value is $-\frac{17}{2}$ . Subtracting the lower limit evaluation from the upper limit evaluation gives $-6 + \frac{17}{2} = \frac{5}{2}$ .

Key Takeaways

Solving exponential equations often involves isolating the exponential term and using logarithms.
Properties of logarithms and exponentials, such as $e^{\ln a} = a$ and $e^{-\ln a} = \frac{1}{a}$ , are essential for simplifying evaluation steps.
Always subtract the lower limit evaluation from the upper limit evaluation when computing definite integrals.

Common Mistakes

Forgetting to multiply by $e^x$ on both sides when solving $8e^{-x} = e^{2x}$ , leading to an incorrect equation.
Incorrectly evaluating $e^{2\ln 2}$ as $2e^{\ln 2}$ instead of $(e^{\ln 2})^2 = 4$ .
Forgetting to subtract the lower limit value, resulting in $-6 - \frac{17}{2} = -\frac{29}{2}$ instead of $\frac{5}{2}$ .
Using decimal approximations for $\ln 2$ during the working, which the mark scheme explicitly penalises (A0 if decimals used).

Things to Be Careful About

Ensure all working for showing $x = \ln 2$ is present; the mark scheme requires necessary detail and awards A0 if decimals are used.
Be careful with the signs when integrating $-e^{2x}$ , which gives $-\frac{1}{2}e^{2x}$ , not $+\frac{1}{2}e^{2x}$ .
The area is always positive; if a negative area is obtained, check the subtraction order (upper limit minus lower limit).

Techniques used

solve exponential equation using logarithmsintegrate exponential functionevaluate definite integral

Q47MMediumDifferentiation

A curve is defined by the parametric equations

x = 4\cos^2 t, \quad y = \sqrt{3}\sin 2t,

for values of $t$ such that $0 < t < \frac{1}{2}\pi$ .

Find the equation of the normal to the curve at the point for which $t = \frac{1}{6}\pi$ . Give your answer in the form $ax + by + c = 0$ where $a$ , $b$ and $c$ are integers.

DifficultyMedium

Worked solution

Approach

Differentiate $x$ and $y$ with respect to $t$ , then use

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

Evaluate this at $t = \frac{\pi}{6}$ to find the gradient of the tangent, then take its negative reciprocal to get the gradient of the normal. Finally, find the coordinates of the point and write the equation of the normal.

Working

Differentiate parametrically:

\begin{aligned} \frac{dx}{dt} &= 4 \cdot 2\cos t(-\sin t) = -8\cos t\sin t = -4\sin 2t, \\ \frac{dy}{dt} &= \sqrt{3} \cdot 2\cos 2t = 2\sqrt{3}\cos 2t. \end{aligned}

Therefore

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\sqrt{3}\cos 2t}{-8\cos t\sin t}.

At $t = \frac{\pi}{6}$ :

\frac{dy}{dx} = \frac{2\sqrt{3}\cos\left(\frac{\pi}{3}\right)}{-8\cos\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{6}\right)} = \frac{2\sqrt{3} \cdot \frac{1}{2}}{-8 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}} = -\frac{1}{2}.

So the gradient of the normal is $2$ .

The coordinates of the point are

x = 4\cos^2\left(\frac{\pi}{6}\right) = 4\left(\frac{\sqrt{3}}{2}\right)^2 = 3, \qquad y = \sqrt{3}\sin\left(\frac{\pi}{3}\right) = \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2}.

The equation of the normal is

y - \frac{3}{2} = 2(x - 3).

Multiply by $2$ :

2y - 3 = 4x - 12,

so

4x - 2y - 9 = 0.

Answer

4x - 2y - 9 = 0

Final answer

4x - 2y - 9 = 0

Detailed explanation

Walkthrough

We are given a curve in parametric form, so the gradient of the tangent is found using

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

First differentiate $x = 4\cos^2 t$ . Using the chain rule, the derivative of $\cos^2 t$ is $-2\cos t\sin t$ , so

\frac{dx}{dt} = -8\cos t\sin t.

Then differentiate $y = \sqrt{3}\sin 2t$ . The derivative of $\sin 2t$ is $2\cos 2t$ , so

\frac{dy}{dt} = 2\sqrt{3}\cos 2t.

Substitute $t = \frac{\pi}{6}$ into both derivatives. Since

\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2},

we get

\frac{dy}{dx} = \frac{2\sqrt{3} \cdot \frac{1}{2}}{-8 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}} = -\frac{1}{2}.

This is the gradient of the tangent. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal:

-\frac{1}{-\frac{1}{2}} = 2.

Next find the coordinates of the point on the curve when $t = \frac{\pi}{6}$ :

x = 4\cos^2\left(\frac{\pi}{6}\right) = 3, \qquad y = \sqrt{3}\sin\left(\frac{\pi}{3}\right) = \frac{3}{2}.

Finally, use the point-slope form of a line:

y - \frac{3}{2} = 2(x - 3).

Multiplying by $2$ and rearranging gives

4x - 2y - 9 = 0.

Key Takeaways

For parametric curves, the gradient is obtained as the ratio of the two derivatives with respect to the parameter.
Trigonometric derivatives such as those of $\cos^2 t$ and $\sin 2t$ must be applied carefully, including the chain rule.
The gradient of the normal is the negative reciprocal of the gradient of the tangent.
The final line equation must be simplified to the required integer form.

Common Mistakes

Forgetting the negative sign when differentiating $\cos^2 t$ .
Missing the factor $2$ when differentiating $\sin 2t$ .
Using the tangent gradient instead of taking its negative reciprocal for the normal.
Failing to find the actual coordinates of the point before writing the line equation.
Leaving the equation with fractions instead of multiplying through to integer coefficients.

Things to Be Careful About

The parameter range $0 < t < \frac{\pi}{2}$ ensures the point is on the given part of the curve, but the calculation itself does not depend on this restriction.
At $t = \frac{\pi}{6}$ , $\frac{dx}{dt} \neq 0$ , so the formula for $\frac{dy}{dx}$ is valid.
The mark scheme requires an attempt at substitution before the value of $\frac{dy}{dx}$ is accepted.
The final answer must be written as $ax + by + c = 0$ with integer coefficients; here that is $4x - 2y - 9 = 0$ .

Techniques used

differentiate parametric equations with respect to the parametercompute dy/dx as (dy/dt)/(dx/dt)evaluate the derivative at the given parameter valuetake the negative reciprocal to find the normal gradientfind the point coordinates and write the equation of the normal

Approach

Use polynomial long division to divide $p(x) = 9x^3 + 18x^2 + 5x + 4$ by $3x + 2$ . The quotient and remainder are read from the final division line.

Working

Divide $9x^3$ by $3x$ to get the first quotient term:

3x^2

Multiply and subtract:

(3x+2)(3x^2) = 9x^3 + 6x^2

(9x^3 + 18x^2) - (9x^3 + 6x^2) = 12x^2

Bring down $5x$ . Divide $12x^2$ by $3x$ :

4x

Multiply and subtract:

(3x+2)(4x) = 12x^2 + 8x

(12x^2 + 5x) - (12x^2 + 8x) = -3x

Bring down $4$ . Divide $-3x$ by $3x$ :

-1

Multiply and subtract:

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

(-3x + 4) - (-3x - 2) = 6

Thus the quotient is $3x^2 + 4x - 1$ and the remainder is $6$ .

Answer

\text{Quotient} = 3x^2 + 4x - 1, \quad \text{Remainder} = 6

Final answer

Quotient = 3x^2 + 4x - 1, Remainder = 6

Detailed explanation

Walkthrough

We are dividing a cubic polynomial by a linear polynomial. The goal is to write

p(x) = (3x+2)Q(x) + R

where $Q(x)$ is the quotient and $R$ is a constant remainder. Since the divisor $3x+2$ has degree 1, the remainder must be a constant.

Start by comparing the leading term $9x^3$ with $3x$ . The first quotient term is $3x^2$ . Multiply $3x+2$ by $3x^2$ to get $9x^3+6x^2$ , subtract from the first two terms, and continue with the remaining terms. Repeat this process: the next quotient term is $4x$ , then $-1$ . After the final subtraction, the leftover $6$ is the remainder.

This long-division process directly gives both the quotient and the remainder, so no separate check is needed.

Key Takeaways

Polynomial long division expresses a polynomial as divisor times quotient plus remainder.
The degree of the remainder is less than the degree of the divisor.
To show a remainder when the answer is given, you must show the division or an equivalent check.

Common Mistakes

Sign errors when subtracting the product of the divisor and the current quotient term, especially with negative terms.
Forgetting to bring down the next term before continuing.
Stopping too early; the remainder is the final leftover after all terms are used.
Using only the factor theorem to show the remainder is $6$ will not find the quotient, which is required.

Things to Be Careful About

The divisor is $3x+2$ , not $x+2$ , so the quotient coefficients differ from division by $x+2$ .
Because the question says the remainder is $6$ , the mark scheme requires enough working to confirm it.
Keep the subtraction aligned by place value to avoid arithmetic slips.

Techniques used

perform polynomial long divisiondetermine the quotient and remainderverify the remainder

(b)

Find the value of $\int_0^2 \frac{p(x)}{3x + 2} \, dx$ , giving your answer in the form $a + \ln b$ where $a$ and $b$ are integers.

5M Similar

DifficultyMedium

Worked solution

Approach

Use the quotient from part (a) to rewrite the integrand as a polynomial plus a term of the form $\frac{6}{3x+2}$ . Then integrate term by term and apply the limits.

Working

From part (a):

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

Integrate:

\int_0^2 \frac{p(x)}{3x+2}\,dx = \int_0^2 \left(3x^2 + 4x - 1 + \frac{6}{3x+2}\right)dx

= \left[ x^3 + 2x^2 - x + 2\ln(3x+2) \right]_0^2

Evaluate at $x = 2$ :

2^3 + 2(2^2) - 2 + 2\ln(8) = 8 + 8 - 2 + 2\ln 8 = 14 + 2\ln 8

Evaluate at $x = 0$ :

0^3 + 2(0^2) - 0 + 2\ln(2) = 2\ln 2

Subtract:

(14 + 2\ln 8) - (2\ln 2) = 14 + 2\ln 8 - 2\ln 2

Use logarithm laws:

2\ln 8 - 2\ln 2 = \ln(8^2) - \ln(2^2) = \ln 64 - \ln 4 = \ln\left(\frac{64}{4}\right) = \ln 16

Therefore:

\int_0^2 \frac{p(x)}{3x+2}\,dx = 14 + \ln 16

Answer

14 + \ln 16

Final answer

14 + ln 16

Detailed explanation

Walkthrough

From part (a), the division identity is

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

Dividing both sides by $3x+2$ gives

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

This is the key step: it turns a rational integrand into a simple polynomial plus a reciprocal term.

Now integrate term by term. The polynomial part integrates normally: $3x^2 \to x^3$ , $4x \to 2x^2$ , $-1 \to -x$ . For the reciprocal term, recall

\int \frac{1}{3x+2}\,dx = \frac{1}{3}\ln(3x+2)

so $6/(3x+2)$ integrates to $2\ln(3x+2)$ .

Apply the limits $0$ and $2$ . At $x=2$ the antiderivative is $14+2\ln 8$ ; at $x=0$ it is $2\ln 2$ . Subtract the lower value from the upper value and combine the logarithms using

2\ln 8 - 2\ln 2 = \ln 64 - \ln 4 = \ln 16

Thus the definite integral is $14+\ln 16$ .

Key Takeaways

A division identity can simplify an integral of a rational function.
The integral of $\frac{k}{ax+b}$ is $\frac{k}{a}\ln|ax+b|$ .
Logarithmic terms from definite integrals must be combined using logarithm laws.

Common Mistakes

Integrating $\frac{6}{3x+2}$ as $6\ln(3x+2)$ instead of $2\ln(3x+2)$ ; the factor $\frac{1}{3}$ is essential.
Forgetting to subtract the value of the antiderivative at the lower limit.
Combining $2\ln 8 - 2\ln 2$ incorrectly. It equals $\ln 16$ , not $\ln 6$ or $\ln 4$ .
Not using the quotient from part (a); the mark scheme follows through from the candidate's quotient.

Things to Be Careful About

On the interval $[0,2]$ , $3x+2>0$ , so the absolute value in $\ln|3x+2|$ is unnecessary.
The final answer must be in the form $a+\ln b$ with integers $a$ and $b$ ; here $a=14$ and $b=16$ .
Keep the limits attached to the whole antiderivative, not to individual terms separately.

Techniques used

rewrite the integrand using the division identityintegrate a polynomial and a reciprocal termapply the limits of integrationcombine logarithmic terms using logarithm laws

Q6MediumDifferentiationNumerical Solution of Equations

Fig. 2

The diagram shows the curve with equation $y = \frac{\ln(2x + 1)}{x + 3}$ . The curve has a maximum point $M$ .

(a)

Find an expression for $\frac{dy}{dx}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The function $y = \frac{\ln(2x+1)}{x+3}$ is a quotient, so the quotient rule is the natural tool. The numerator $\ln(2x+1)$ is a composition of $\ln$ with a linear function, so use the chain rule for its derivative.

Working

Let $u = \ln(2x+1)$ and $v = x+3$ .

By the chain rule:

\frac{du}{dx} = \frac{1}{2x+1} \cdot 2 = \frac{2}{2x+1}.

And trivially $\frac{dv}{dx} = 1$ .

Applying the quotient rule $\frac{dy}{dx} = \frac{v\,u' - u\,v'}{v^2}$ :

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

Answer

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

Final answer

dy/dx = ((x+3)·(2/(2x+1)) - ln(2x+1)) / (x+3)^2

Detailed explanation

Walkthrough

The function $y = \frac{\ln(2x+1)}{x+3}$ is a quotient of two expressions, so the quotient rule is the standard technique:

\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \cdot u' - u \cdot v'}{v^2}.

For the top function $u = \ln(2x+1)$ , this is a composition of the natural log with the linear function $2x+1$ . The chain rule says: differentiate the outer function ( $\ln$ ) evaluated at the inner, then multiply by the derivative of the inner:

\frac{du}{dx} = \frac{1}{2x+1} \cdot 2 = \frac{2}{2x+1}.

The factor of $2$ comes from the inner derivative; forgetting it is a very common slip.

For the bottom function $v = x+3$ , we have $\frac{dv}{dx} = 1$ .

Substituting into the quotient rule:

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

This is the required expression. An important observation for what follows: the denominator $(x+3)^2$ is always positive wherever the function is defined, so the sign of the derivative is determined entirely by the numerator.

Key Takeaways

The quotient rule differentiates $\frac{f(x)}{g(x)}$ by computing $g f' - f g'$ and dividing by $g^2$ .
Composite functions like $\ln(2x+1)$ require the chain rule: differentiate the outer function, then multiply by the inner derivative.
For this curve, the denominator of $\frac{dy}{dx}$ is always positive, so stationary points occur exactly when the numerator is zero.

Common Mistakes

Forgetting the chain rule when differentiating $\ln(2x+1)$ , writing $\frac{1}{2x+1}$ instead of $\frac{2}{2x+1}$ .
Misapplying the quotient rule (e.g., $\frac{u'v + uv'}{v^2}$ instead of $\frac{vu' - uv'}{v^2}$ ); the order of subtraction is the minus sign in the middle, not a plus.

Things to Be Careful About

The chain rule contributes a factor of $2$ because the inner function is $2x+1$ , not $x+1$ .
The expression can be rewritten by distributing $(x+3)$ over $\frac{2}{2x+1}$ to give $\frac{2(x+3)/(2x+1) - \ln(2x+1)}{(x+3)^2}$ , which is algebraically equivalent; either is acceptable.

Techniques used

apply quotient ruledifferentiate composite logarithmic function using chain rule

(b)

Show that the $x$ -coordinate of $M$ satisfies the equation $x = \frac{x + 3}{\ln(2x + 1)} - 0.5$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

At the maximum $M$ , the gradient is zero: $\frac{dy}{dx} = 0$ . Since the denominator of $\frac{dy}{dx}$ (from part (a)) is $(x+3)^2$ , which is positive, the numerator alone must vanish. Rearrange this equation to obtain the form $x = \frac{x+3}{\ln(2x+1)} - 0.5$ .

Working

Set the numerator of $\frac{dy}{dx}$ to zero:

\frac{2(x+3)}{2x+1} - \ln(2x+1) = 0.

Rearrange:

\frac{2(x+3)}{2x+1} = \ln(2x+1).

Cross-multiply:

2(x+3) = (2x+1)\ln(2x+1).

Divide both sides by $2\ln(2x+1)$ :

\frac{x+3}{\ln(2x+1)} = \frac{2x+1}{2} = x + 0.5.

Subtract $0.5$ from both sides:

\frac{x+3}{\ln(2x+1)} - 0.5 = x.

Answer

x = \frac{x+3}{\ln(2x+1)} - 0.5

Final answer

x = (x+3)/ln(2x+1) - 0.5

Detailed explanation

Walkthrough

A maximum point is where the gradient is zero, so $\frac{dy}{dx} = 0$ . Because the denominator $(x+3)^2$ is always positive, this means the numerator from part (a) must vanish:

\frac{2(x+3)}{2x+1} - \ln(2x+1) = 0.

The target equation $x = \frac{x+3}{\ln(2x+1)} - 0.5$ is in 'fixed-point' form (with $x$ on the right). To reach it, isolate $\frac{x+3}{\ln(2x+1)}$ step by step:

Move the log term to the right: $\frac{2(x+3)}{2x+1} = \ln(2x+1)$ .
Cross-multiply: $2(x+3) = (2x+1)\ln(2x+1)$ .
Divide by $2\ln(2x+1)$ : $\frac{x+3}{\ln(2x+1)} = \frac{2x+1}{2}$ .

Now $\frac{2x+1}{2} = x + 0.5$ , so the right-hand side is $x + 0.5$ . Subtracting $0.5$ isolates $x$ :

x = \frac{x+3}{\ln(2x+1)} - 0.5.

This rearrangement is not just algebraic curiosity — it produces the fixed-point form that drives the iteration in part (d).

Key Takeaways

Stationary points occur when $\frac{dy}{dx} = 0$ , which for this curve means the numerator is zero.
Rearranging the stationary-point equation into a fixed-point form $x = g(x)$ is the standard way to set up an iteration.
The chain of equivalences (dividing by a positive quantity, cross-multiplying) is valid because $\ln(2x+1) > 0$ on the relevant domain ( $x > -0.5$ ).

Common Mistakes

Forgetting that the denominator $(x+3)^2$ is never zero — a student who sets it to zero is looking for an undefined point, not a stationary point.
Dividing by $\ln(2x+1)$ without first checking it is non-zero (it is positive for $x > -0.5$ , which is satisfied by the relevant part of the curve).
Sign error in the step that converts $\frac{2x+1}{2}$ to $x + 0.5$ .

Things to Be Careful About

The rearranged form has $x$ on both sides. The 'show that' wording means we are not solving for $x$ , only verifying the equivalence.
A 'show that' question demands all working, since the final answer is already given.

Techniques used

set derivative to zero at stationary pointrearrange equation into fixed-point form

(c)

Show by calculation that the $x$ -coordinate of $M$ lies between 2.5 and 3.0.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the sign change method. Define $h(x) = x - \frac{x+3}{\ln(2x+1)} + 0.5$ , so that $h(x) = 0$ is exactly the equation from part (b). Evaluate $h$ at $x = 2.5$ and $x = 3$ . If the signs differ, the root of $h$ — and hence the $x$ -coordinate of $M$ — lies in the interval.

Working

At $x = 2.5$ :

h(2.5) = 2.5 - \frac{2.5+3}{\ln(2(2.5)+1)} + 0.5 = 3 - \frac{5.5}{\ln 6}.

Using $\ln 6 = 1.79176$ :

h(2.5) = 3 - \frac{5.5}{1.79176} = 3 - 3.0696 = -0.0696 \approx -0.07.

At $x = 3$ :

h(3) = 3 - \frac{3+3}{\ln(2(3)+1)} + 0.5 = 3.5 - \frac{6}{\ln 7}.

Using $\ln 7 = 1.94591$ :

h(3) = 3.5 - \frac{6}{1.94591} = 3.5 - 3.0834 = 0.4166 \approx 0.4.

Since $h(2.5) \approx -0.07 < 0$ and $h(3) \approx 0.4 > 0$ , the values have opposite signs. The function $h$ is continuous on $[2.5, 3]$ (since $\ln(2x+1) > 0$ throughout), so by the intermediate value theorem, $h(x) = 0$ has a root in the interval $(2.5, 3.0)$ . Therefore the $x$ -coordinate of $M$ lies between $2.5$ and $3.0$ .

Answer

The $x$ -coordinate of $M$ lies between $2.5$ and $3.0$ (sign change of $h$ at the endpoints).

Final answer

2.5 < x < 3.0

Detailed explanation

Walkthrough

To prove that a root lies in a given interval using algebra, the standard approach is the sign change (or 'change of sign') method. We define a function whose root is the quantity we want, evaluate it at the two endpoints, and check whether the signs differ.

The equation from part (b) is $x = \frac{x+3}{\ln(2x+1)} - 0.5$ . Rearranging gives $x - \frac{x+3}{\ln(2x+1)} + 0.5 = 0$ , so we set

h(x) = x - \frac{x+3}{\ln(2x+1)} + 0.5.

A root of $h$ is exactly the $x$ -coordinate of $M$ .

Computing $h(2.5)$ :

Numerator inside: $2.5 + 3 = 5.5$ .
Logarithm: $\ln(2(2.5)+1) = \ln 6 \approx 1.79176$ .
Fraction: $5.5 / 1.79176 \approx 3.0696$ .
$h(2.5) = 2.5 - 3.0696 + 0.5 = -0.0696$ .

Computing $h(3)$ :

Numerator inside: $3 + 3 = 6$ .
Logarithm: $\ln(2(3)+1) = \ln 7 \approx 1.94591$ .
Fraction: $6 / 1.94591 \approx 3.0834$ .
$h(3) = 3 - 3.0834 + 0.5 = 0.4166$ .

The values have opposite signs: $h(2.5) < 0 < h(3)$ . Since $h$ is continuous on $[2.5, 3]$ (the only potential issue would be $\ln(2x+1) = 0$ , which occurs at $x = -0.5$ , well outside the interval), the intermediate value theorem guarantees a root in $(2.5, 3.0)$ .

Key Takeaways

The 'change of sign' (or interval halving) technique locates a root by showing $h(a) < 0 < h(b)$ for continuous $h$ .
This is the algebraic counterpart of looking for where a graph crosses the $x$ -axis.
The justification is the intermediate value theorem, which requires the function to be continuous on the interval.

Common Mistakes

Using the wrong equation (e.g., trying to find where $\frac{dy}{dx} = 0$ directly via the original derivative rather than via the rearranged equation $x = g(x)$ ).
Making a sign error when defining $h$ .
Quoting only the sign of the value at one endpoint without stating the other.
Not justifying the conclusion: simply observing opposite signs is not enough; you must invoke the intermediate value theorem (or 'continuous function has a root between sign changes').

Things to Be Careful About

The function $h$ is continuous on $[2.5, 3]$ because $\ln(2x+1) > 0$ throughout the interval.
The two approximations $-0.07$ and $0.4$ are correct to 1 decimal place; the more precise values $-0.0696$ and $0.4166$ are also acceptable.
This is a 'show that' question: the answer is given, so you must show all working clearly.

Techniques used

define auxiliary function h(x)evaluate h at endpoints of intervalapply intermediate value theorem

(d)

Use an iterative formula based on the equation in part (b) to find the $x$ -coordinate of $M$ correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

3M Similar

DifficultyMedium

Worked solution

Approach

The equation from part (b) is already in fixed-point form $x = g(x)$ , so the iterative formula is

x_{n+1} = \frac{x_n+3}{\ln(2x_n+1)} - 0.5.

Start with $x_0 = 3$ (in the interval from part (c)) and iterate until two consecutive values agree to 6 significant figures.

Working

Start with $x_0 = 3$ .

Iteration 1:

x_1 = \frac{3+3}{\ln(2(3)+1)} - 0.5 = \frac{6}{\ln 7} - 0.5 = \frac{6}{1.94591} - 0.5 = 3.08339 - 0.5 = 2.58339.

Iteration 2:

x_2 = \frac{2.58339+3}{\ln(2(2.58339)+1)} - 0.5 = \frac{5.58339}{\ln 6.16678} - 0.5.

Using $\ln 6.16678 = 1.81918$ :

x_2 = \frac{5.58339}{1.81918} - 0.5 = 3.06918 - 0.5 = 2.56918.

Iteration 3:

x_3 = \frac{2.56918+3}{\ln(2(2.56918)+1)} - 0.5 = \frac{5.56918}{\ln 6.13836} - 0.5.

Using $\ln 6.13836 = 1.81456$ :

x_3 = \frac{5.56918}{1.81456} - 0.5 = 3.06917 - 0.5 = 2.56917.

Iteration 4:

x_4 = \frac{2.56917+3}{\ln(2(2.56917)+1)} - 0.5 = \frac{5.56917}{\ln 6.13834} - 0.5 = \frac{5.56917}{1.81455} - 0.5 = 2.56917.

Since $x_3 = x_4 = 2.56917$ to 6 significant figures, the iteration has converged.

Answer

x = 2.569 \quad (\text{to 4 s.f.})

Final answer

x = 2.569

Detailed explanation

Walkthrough

The equation from part (b) has the form $x = g(x)$ with $g(x) = \frac{x+3}{\ln(2x+1)} - 0.5$ . This is exactly the shape required for fixed-point iteration: $x_{n+1} = g(x_n)$ .

A good starting value is one that lies in the interval we just located — $x_0 = 3$ is the upper bound and works well.

Iteration 1: Substitute $x_0 = 3$ into $g$ :

$2x_0 + 1 = 7$ , $\ln 7 = 1.94591$ .
$\frac{x_0+3}{\ln(2x_0+1)} = \frac{6}{1.94591} = 3.08339$ .
$x_1 = 3.08339 - 0.5 = 2.58339$ .

Iteration 2: Substitute $x_1 = 2.58339$ into $g$ :

$2x_1 + 1 = 6.16678$ , $\ln 6.16678 = 1.81918$ .
$\frac{x_1+3}{\ln(2x_1+1)} = \frac{5.58339}{1.81918} = 3.06918$ .
$x_2 = 3.06918 - 0.5 = 2.56918$ .

Iteration 3: Substitute $x_2 = 2.56918$ into $g$ :

$2x_2 + 1 = 6.13836$ , $\ln 6.13836 = 1.81456$ .
$\frac{x_2+3}{\ln(2x_2+1)} = \frac{5.56918}{1.81456} = 3.06917$ .
$x_3 = 3.06917 - 0.5 = 2.56917$ .

Iteration 4: Substitute $x_3 = 2.56917$ into $g$ :

$2x_3 + 1 = 6.13834$ , $\ln 6.13834 = 1.81455$ .
$\frac{x_3+3}{\ln(2x_3+1)} = \frac{5.56917}{1.81455} = 3.06917$ .
$x_4 = 3.06917 - 0.5 = 2.56917$ .

Because $x_3$ and $x_4$ agree to 6 significant figures, the iteration has converged: subsequent iterations will not change the 6th digit. To 4 significant figures, the $x$ -coordinate of $M$ is $2.569$ .

The convergence is also visible in the successive differences: $x_2 - x_1 \approx -0.0142$ , $x_3 - x_2 \approx -0.00001$ , $x_4 - x_3 = 0$ . The map is a strong contraction near the fixed point (its derivative is close to zero), so successive iterates get closer very quickly.

Key Takeaways

An equation of the form $x = g(x)$ gives the fixed-point iteration $x_{n+1} = g(x_n)$ .
The starting value should lie in an interval known to contain the root.
Convergence is judged by repeated iterates agreeing to the required number of significant figures (here, 6 sf for intermediate, 4 sf for the final answer).
The logarithm $\ln(2x+1)$ is well-defined and positive for $x > -0.5$ , so the iteration is safe once $x_0 \geq 2.5$ .

Common Mistakes

Using the wrong iterative formula (e.g., rearranging differently and getting a formula that does not match the equation in part (b)).
Stopping after only one or two iterations, which is not enough to justify 4 sf accuracy.
Not giving intermediate values to 6 significant figures, even though the question explicitly asks for it.
Rounding errors: if you truncate each iterate, you can introduce bias; ideally carry full accuracy in the calculator and only write the rounded value.

Things to Be Careful About

The sign of the log is positive throughout, so no sign issues arise.
The iteration must continue at least until two consecutive values match to 6 sf — three or four iterations is the norm here.
The starting value $x_0 = 3$ is the upper bound from part (c); $x_0 = 2.5$ would also work and converges similarly. The chosen start does not affect the final answer (to 4 sf), only how quickly the iterates settle.

Techniques used

apply fixed-point iterationiterate until convergencerecord values to 6 significant figures

Q7Medium-HardTrigonometryIntegration

(a)

Prove that $2\sin\theta \csc 2\theta \equiv \sec\theta$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Rewrite $\csc 2\theta$ as $\frac{1}{\sin 2\theta}$ , use the double-angle formula $\sin 2\theta = 2\sin\theta\cos\theta$ , cancel, and simplify to $\sec\theta$ .

Working

2\sin\theta\csc 2\theta = 2\sin\theta \cdot \frac{1}{\sin 2\theta}

Using $\sin 2\theta = 2\sin\theta\cos\theta$ :

2\sin\theta\csc 2\theta = \frac{2\sin\theta}{2\sin\theta\cos\theta} = \frac{1}{\cos\theta}

Since $\frac{1}{\cos\theta} = \sec\theta$ , the identity is proved:

2\sin\theta\csc 2\theta \equiv \sec\theta

Answer

2\sin\theta\csc 2\theta \equiv \sec\theta

Final answer

sec θ

Detailed explanation

Walkthrough

We start with the left-hand side $2\sin\theta\csc 2\theta$ . The cosecant function is defined as the reciprocal of sine, so $\csc 2\theta = \frac{1}{\sin 2\theta}$ . This lets us write the whole expression as a single fraction. Next, we replace $\sin 2\theta$ using the double-angle formula $\sin 2\theta = 2\sin\theta\cos\theta$ . The factor $2\sin\theta$ then cancels with the numerator, leaving $\frac{1}{\cos\theta}$ , which is exactly $\sec\theta$ . Because the question says the identity is given, we must show enough working to justify the cancellation and the final equivalence.

Key Takeaways

This question tests the definitions of reciprocal trigonometric functions and the double-angle formula for sine. It also reinforces the idea that proving an identity means transforming one side into the other using known formulae, with every step justified.

Common Mistakes

Forgetting that $\csc 2\theta = \frac{1}{\sin 2\theta}$ , not $\frac{1}{\cos 2\theta}$ .
Incorrectly writing $\sin 2\theta = \sin\theta\cos\theta$ instead of $2\sin\theta\cos\theta$ , which prevents the cancellation.
Since the identity is given, an unsupported statement such as " $2\sin\theta\csc 2\theta = \sec\theta$ " without intermediate steps may not receive full credit.

Things to Be Careful About

The identity is defined only where $\sin\theta$ and $\sin 2\theta$ are non-zero, so $\theta$ cannot be a multiple of $\pi$ or a half-multiple where $\sin 2\theta = 0$ . In a proof, we assume these values are excluded. Also, be consistent with notation: $\csc x = \frac{1}{\sin x}$ , while $\sec x = \frac{1}{\cos x}$ .

Techniques used

rewrite csc 2θ in terms of sin 2θapply the double-angle formula for sin 2θsimplify the product to sec θ

(b)

Solve the equation $\tan^2\theta + 7\sin\theta \csc 2\theta = 8$ for $-\pi < \theta < \pi$ .

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Use the identity from part (a) to replace $\sin\theta\csc 2\theta$ with $\frac{1}{2}\sec\theta$ . Then use $\tan^2\theta = \sec^2\theta - 1$ to obtain a quadratic in $\sec\theta$ . Solve the quadratic, convert to $\cos\theta$ , and find all angles in $-\pi < \theta < \pi$ .

Working

From part (a):

\sin\theta\csc 2\theta = \frac{1}{2\cos\theta} = \frac{1}{2}\sec\theta

Substitute into the equation and use $\tan^2\theta = \sec^2\theta - 1$ :

\sec^2\theta - 1 + \frac{7}{2}\sec\theta = 8

Multiply by 2:

2\sec^2\theta + 7\sec\theta - 18 = 0

Let $s = \sec\theta$ :

2s^2 + 7s - 18 = 0

Using the quadratic formula:

s = \frac{-7 \pm \sqrt{7^2 - 4(2)(-18)}}{4} = \frac{-7 \pm \sqrt{193}}{4}

Since $\sqrt{193} \approx 13.892$ :

s \approx 1.723 \quad \text{or} \quad s \approx -5.223

Now $\cos\theta = \frac{1}{s}$ .

For $s \approx 1.723$ :

\cos\theta \approx 0.5804 \quad \Rightarrow \quad \theta \approx \pm 0.952

For $s \approx -5.223$ :

\cos\theta \approx -0.1915 \quad \Rightarrow \quad \theta \approx \pm 1.76

All four values lie in $-\pi < \theta < \pi$ , so:

\theta \approx \pm 0.952, \ \pm 1.76

Final answer

θ ≈ ±0.952, ±1.76 radians

Detailed explanation

Walkthrough

We first use the identity proved in part (a): $\sin\theta\csc 2\theta = \frac{1}{2}\sec\theta$ . This converts the term $7\sin\theta\csc 2\theta$ into $\frac{7}{2}\sec\theta$ . The equation still contains $\tan^2\theta$ , so we use the Pythagorean identity $\tan^2\theta = \sec^2\theta - 1$ . This gives a quadratic equation in $\sec\theta$ only. Multiplying by 2 gives $2\sec^2\theta + 7\sec\theta - 18 = 0$ . We solve this with the quadratic formula, obtaining two values of $\sec\theta$ . Since $\cos\theta = \frac{1}{\sec\theta}$ , each value of $\sec\theta$ gives a value of $\cos\theta$ . Because $\cos$ is even, each positive or negative value of $\cos\theta$ gives two angles in $(-\pi, \pi)$ : one positive and one negative. Hence there are four solutions. The mark scheme requires all four and no extra values in the interval.

Key Takeaways

This problem combines trigonometric identities with solving a quadratic equation. It also tests the ability to find all solutions of $\cos\theta = k$ in a given interval, remembering that $\cos(-\theta) = \cos\theta$ .

Common Mistakes

Forgetting to use $\tan^2\theta = \sec^2\theta - 1$ and trying to solve an equation with two different trig ratios.
Solving the quadratic in $\sec\theta$ but forgetting to convert to $\cos\theta$ before finding angles.
Only giving the positive angles and missing the negative counterparts, since $\cos\theta$ is even.
Not checking that all found angles lie in $-\pi < \theta < \pi$ .

Things to Be Careful About

The quadratic formula gives two values of $\sec\theta$ ; each must be handled separately. When $\sec\theta$ is negative, $\cos\theta$ is negative, and the corresponding angles are in the second and third quadrants; in the interval $(-\pi, \pi)$ these are the negative and positive angles with the same reference angle. Use sufficient accuracy, e.g. $\theta \approx \pm 0.952$ and $\theta \approx \pm 1.76$ , and do not include any angles outside the interval.

Techniques used

substitute the identity from part (a)use the Pythagorean identity to eliminate tan^2form and solve a quadratic in sec θfind all angles in the given interval

(c)

Find $\int 8\sin^2 \frac{1}{2}x \csc^2 x \, dx$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Simplify the integrand using $\csc^2 x = \frac{1}{\sin^2 x}$ and the double-angle formula $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ . Then integrate the resulting $2\sec^2\frac{x}{2}$ .

Working

\int 8\sin^2\frac{x}{2}\csc^2 x\,dx = \int \frac{8\sin^2\frac{x}{2}}{\sin^2 x}\,dx

Using $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ :

\sin^2 x = 4\sin^2\frac{x}{2}\cos^2\frac{x}{2}

So the integrand becomes:

\frac{8\sin^2\frac{x}{2}}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}} = \frac{2}{\cos^2\frac{x}{2}} = 2\sec^2\frac{x}{2}

Therefore:

\int 8\sin^2\frac{x}{2}\csc^2 x\,dx = \int 2\sec^2\frac{x}{2}\,dx

Since $\frac{d}{dx}\left(\tan\frac{x}{2}\right) = \frac{1}{2}\sec^2\frac{x}{2}$ :

\int 2\sec^2\frac{x}{2}\,dx = 4\tan\frac{x}{2} + C

Answer

\int 8\sin^2\frac{x}{2}\csc^2 x\,dx = 4\tan\frac{x}{2} + C

Final answer

4 tan(x/2) + C

Detailed explanation

Walkthrough

We need to integrate $8\sin^2\frac{x}{2}\csc^2 x$ . Since $\csc^2 x = \frac{1}{\sin^2 x}$ , the integrand is $\frac{8\sin^2(x/2)}{\sin^2 x}$ . The denominator contains $\sin x$ , so we use the double-angle formula $\sin x = 2\sin(x/2)\cos(x/2)$ . Squaring this gives $\sin^2 x = 4\sin^2(x/2)\cos^2(x/2)$ . The $\sin^2(x/2)$ factors cancel, leaving $\frac{2}{\cos^2(x/2)} = 2\sec^2(x/2)$ . This is the key simplification. Now we integrate: since the derivative of $\tan(x/2)$ is $\frac{1}{2}\sec^2(x/2)$ , multiplying by 2 gives $2\sec^2(x/2)$ , so the integral is $4\tan(x/2)$ . We add the constant of integration.

Key Takeaways

This question shows how a trigonometric identity can simplify an apparently complicated integrand into a standard form. It also tests the standard integral $\int \sec^2(ax+b)\,dx = \frac{1}{a}\tan(ax+b) + C$ .

Common Mistakes

Forgetting that $\csc^2 x = \frac{1}{\sin^2 x}$ and trying to integrate directly.
Incorrectly expanding $\sin x = 2\sin(x/2)\cos(x/2)$ ; the factor 2 must be squared when squaring.
Forgetting the factor 2 when integrating $\sec^2(x/2)$ : $\int \sec^2(x/2)\,dx = 2\tan(x/2) + C$ .
Omitting the constant of integration, though the mark scheme condones this.

Things to Be Careful About

Remember that $\int \sec^2(ax+b)\,dx = \frac{1}{a}\tan(ax+b) + C$ . Here $a = \frac{1}{2}$ , so the factor is 2. Also, the simplification requires $\sin(x/2) \neq 0$ ; at such points the original integrand is undefined. The final answer is $4\tan\frac{x}{2} + C$ .

Techniques used

rewrite csc^2 x in terms of sin xapply the double-angle formula to sin xsimplify the integrand to a sec^2 formintegrate sec^2(ax+b)

Mathematics 9709/22 — May/June 2024

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