The polynomial $p(x)$ is defined by $$ p(x) = ax^3 - ax^2 - 15x + 18 $$ where $a$ is a constant. It is given that $(x + 2)$ is a factor of $p(x)$. (a) Find the value of $a$. (b) Hence factorise $p(x)$ completely. (c) Solve the equation $p(\text{cosec}^2 \theta) = 0$ for $-90^\circ < \theta < 90^\circ$.

(a) a = 4 · (b) p(x) = (x + 2)(2x - 3)^2 · (c) θ = 54.7° or θ = -54.7°

A curve has parametric equations $$ x = \frac{e^{2t} - 2}{e^{2t} + 1}, \quad y = e^{3t} + 1 $$ (a) Find an expression for $\frac{dy}{dx}$ in terms of $t$. (b) Find the exact gradient of the curve at the point where the curve crosses the $y$-axis.

(a) dy/dx = 1/2 e^t (e^{2t}+1)^2 · (b) 9√2/2

9709/21 Oct/Nov 2024 — Mathematics Solved Past Paper

Q1Medium-EasyLogarithmic and Exponential Functions

The variables $x$ and $y$ satisfy the equation $a^{2y} = e^{3x+k}$ , where $a$ and $k$ are constants.
The graph of $y$ against $x$ is a straight line.

(a)

Use logarithms to show that the gradient of the straight line is $\frac{3}{2 \ln a}$ .

1M Similar

DifficultyEasy

Worked solution

Approach

Take natural logarithms of both sides of $a^{2y}=e^{3x+k}$ , use the power law of logarithms, and rearrange into the form $y=mx+c$ to identify the gradient.

Working

Take $\ln$ of both sides:

\ln\left(a^{2y}\right)=\ln\left(e^{3x+k}\right)

Using $\ln(e^u)=u$ and the power law $\ln(a^{2y})=2y\ln a$ :

2y\ln a=3x+k

Divide by $2\ln a$ :

y=\frac{3}{2\ln a}x+\frac{k}{2\ln a}

This is a straight line with gradient $\frac{3}{2\ln a}$ .

Answer

\text{gradient}=\frac{3}{2\ln a}

Final answer

3/(2 ln a)

Detailed explanation

Walkthrough

The equation $a^{2y}=e^{3x+k}$ involves an unknown exponent on the left and an exponential on the right. Taking natural logarithms of both sides is the natural first step because logarithms turn exponents into multipliers. This gives $\ln(a^{2y})=\ln(e^{3x+k})$ . Since $\ln(e^u)=u$ , the right-hand side becomes $3x+k$ . On the left, the power law of logarithms gives $\ln(a^{2y})=2y\ln a$ . The equation is now $2y\ln a=3x+k$ . Dividing by $2\ln a$ writes it as $y=\frac{3}{2\ln a}x+\frac{k}{2\ln a}$ , which is the equation of a straight line. The coefficient of $x$ is therefore the gradient.

Key Takeaways

Logarithms convert exponential equations into linear equations. The power law $\ln(a^b)=b\ln a$ is essential when the variable appears in an exponent. Once an equation is written in the form $y=mx+c$ , the gradient can be read directly as the coefficient of $x$ .

Common Mistakes

Forgetting to apply the power law to $a^{2y}$ and writing $\ln(a^{2y})$ incorrectly.
Not showing the intermediate line $2y\ln a=3x+k$ . Part (a) is a "show that" question, so the necessary detail must be shown.
Confusing the gradient with the intercept when reading the rearranged equation.

Things to Be Careful About

$\ln a$ is a constant because $a$ is a constant, so dividing by $2\ln a$ is valid.
The gradient is the coefficient of $x$ , not the constant term $\frac{k}{2\ln a}$ .
The answer must be stated as $\frac{3}{2\ln a}$ exactly.

Techniques used

take natural logarithms of both sidesuse the logarithm power lawrearrange into straight-line formidentify the gradient

(b)

Given that the straight line passes through the points $(0.4, 0.95)$ and $(3.3, 3.80)$ , find the values of $a$ and $k$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the two given points to find the gradient of the straight line. Equate this to the gradient found in part (a), solve for $a$ , then substitute one point into the original equation to find $k$ .

Working

The gradient of the line through $(0.4,0.95)$ and $(3.3,3.80)$ is

m=\frac{3.80-0.95}{3.3-0.4}=\frac{2.85}{2.9}

From part (a), $m=\frac{3}{2\ln a}$ , so

\frac{3}{2\ln a}=\frac{2.85}{2.9}

Therefore

2\ln a=\frac{3\times 2.9}{2.85}=\frac{58}{19}

so

\ln a=\frac{29}{19},\qquad a=e^{29/19}\approx 4.6

Substitute $(0.4,0.95)$ into $a^{2y}=e^{3x+k}$ :

a^{1.9}=e^{1.2+k}

Using $a=e^{29/19}$ ,

e^{(29/19)(1.9)}=e^{1.2+k}

Since $\frac{29}{19}\times 1.9=2.9$ ,

2.9=1.2+k

Hence

k=1.7

Answer

a=e^{29/19}\approx 4.6,\qquad k=1.7

Final answer

a = e^(29/19) ≈ 4.6, k = 1.7

Detailed explanation

Walkthrough

First compute the gradient of the straight line from the two given points. The gradient is the change in $y$ divided by the change in $x$ : $\frac{3.80-0.95}{3.3-0.4}=\frac{2.85}{2.9}$ . From part (a), the gradient is also $\frac{3}{2\ln a}$ . Equating these gives an equation involving $\ln a$ . Cross-multiply to solve for $\ln a$ , then exponentiate to find $a$ . Finally, substitute one of the given points into the original equation $a^{2y}=e^{3x+k}$ . Since $a$ is now known, the only remaining unknown is $k$ . Comparing the exponents gives $k=1.7$ .

Key Takeaways

A straight-line graph of $y$ against $x$ can be used to find unknown constants in an exponential relationship. The gradient of the line gives information about the base $a$ , and substituting a point gives the remaining constant $k$ . This question connects coordinate geometry with logarithms and exponentials.

Common Mistakes

Reversing the gradient formula and writing $\frac{3.3-0.4}{3.80-0.95}$ .
Solving for $\ln a$ but forgetting to exponentiate to find $a$ .
Substituting the point into the linear form but making a sign error when finding $k$ .
Rounding intermediate values too early; the mark scheme allows greater accuracy, so exact fractions are safer.

Things to Be Careful About

The gradient is $\frac{2.85}{2.9}$ , not $\frac{2.9}{2.85}$ .
Keep exact fractions such as $\frac{29}{19}$ while working to avoid rounding errors.
When substituting $(0.4,0.95)$ , use $x=0.4$ and $y=0.95$ in the correct positions.
Check that the final values of $a$ and $k$ satisfy both given points.

Techniques used

calculate the gradient from two pointsequate the gradient to the coefficient from the linear formsolve a logarithmic equation for asubstitute a point to find k

Q24MMediumAlgebra

Solve the inequality $|x - 7| > 4x + 3$ .

DifficultyMedium

Worked solution

Approach

Use the definition of the modulus function to split the inequality into two cases, depending on the sign of $x - 7$ . Solve the resulting linear inequality in each case and keep only solutions consistent with the case condition.

Working

Case 1: $x \geq 7$

Here $x - 7 \geq 0$ , so $|x - 7| = x - 7$ . The inequality becomes

x - 7 > 4x + 3

Subtract $x$ and subtract $3$ :

-10 > 3x

x < -\frac{10}{3}

But this contradicts $x \geq 7$ , so there are no solutions from this case.

Case 2: $x < 7$

Here $x - 7 < 0$ , so $|x - 7| = 7 - x$ . The inequality becomes

7 - x > 4x + 3

Add $x$ and subtract $3$ :

4 > 5x

x < \frac{4}{5}

This is consistent with $x < 7$ , so the solutions from this case are $x < \frac{4}{5}$ .

Since Case 1 gives no solutions, the complete solution is

x < \frac{4}{5}

Answer

x < \frac{4}{5}

(Equivalently, $(-\infty, \frac{4}{5})$ .)

Final answer

x < 4/5

Detailed explanation

Walkthrough

The modulus $|x - 7|$ behaves differently depending on whether $x - 7$ is non-negative or negative. So split at $x = 7$ .

For $x \geq 7$ , $|x - 7| = x - 7$ . Substituting this gives the linear inequality $x - 7 > 4x + 3$ . Solving it gives $x < -\frac{10}{3}$ . However, this contradicts the assumption $x \geq 7$ : a number cannot be both at least 7 and less than $-\frac{10}{3}$ . Therefore this case contributes no solutions.

For $x < 7$ , $|x - 7| = 7 - x$ . This gives $7 - x > 4x + 3$ , which simplifies to $4 > 5x$ , so $x < \frac{4}{5}$ . This is compatible with $x < 7$ , so every $x$ below $\frac{4}{5}$ is a solution.

Because the first case had no solutions, the final answer is just $x < \frac{4}{5}$ .

Key Takeaways

To solve a modulus inequality, split by the sign of the expression inside the modulus.
Always check that the solution found in each case satisfies the case condition.
A case may produce an algebraic solution that must be discarded because it contradicts the assumed range of $x$ .
The final solution is the union of the valid solutions from all cases.

Common Mistakes

Forgetting to intersect the solution of each case with the case condition; this would incorrectly include values such as $x < -\frac{10}{3}$ .
Squaring both sides without checking signs. Squaring can introduce extra values, and extra checks are needed because $4x + 3$ can be negative.
Reversing the inequality sign when subtracting or adding; here no division by a negative number is needed, so the sign should stay the same.
Not explicitly stating that the first case gives no solutions, losing the 'no other value' mark.

Things to Be Careful About

The inequality is strict, so $x = \frac{4}{5}$ is not included.
The final interval is $(-\infty, \frac{4}{5})$ , not a bounded interval.
When using the case method, the condition $x < 7$ is automatically satisfied by $x < \frac{4}{5}$ , so no further restriction is needed.
If using the squaring method, you must verify solutions against the original inequality because squaring is not reversible when the right-hand side may be negative.

Techniques used

split modulus into cases by signsolve linear inequality in each caseintersect each solution with its case conditiondiscard the inconsistent case

Approach

Differentiate $f(x) = \tan^2(\frac{1}{2}x)$ using the chain rule, then substitute $x = \frac{2}{3}\pi$ and evaluate using exact trigonometric values.

Working

Let $u = \tan(\frac{1}{2}x)$ . Then $f(x) = u^2$ .

\frac{du}{dx} = \frac{1}{2}\sec^2\left(\frac{1}{2}x\right)

\frac{df}{dx} = 2u \cdot \frac{du}{dx} = 2\tan\left(\frac{1}{2}x\right) \cdot \frac{1}{2}\sec^2\left(\frac{1}{2}x\right) = \tan\left(\frac{1}{2}x\right)\sec^2\left(\frac{1}{2}x\right)

At $x = \frac{2}{3}\pi$ , we have $\frac{1}{2}x = \frac{\pi}{3}$ .

\tan\left(\frac{\pi}{3}\right) = \sqrt{3}, \quad \sec^2\left(\frac{\pi}{3}\right) = \frac{1}{\cos^2(\frac{\pi}{3})} = \frac{1}{(\frac{1}{2})^2} = 4

f'\left(\frac{2}{3}\pi\right) = \sqrt{3} \cdot 4 = 4\sqrt{3}

Answer

4\sqrt{3}

Final answer

4√3

Detailed explanation

Walkthrough

We need to differentiate $f(x) = \tan^2(\frac{1}{2}x)$ . This is a composite function: the outer function is squaring, and the inner function is $\tan(\frac{1}{2}x)$ . The chain rule says: differentiate the outer function, keep the inner function unchanged, then multiply by the derivative of the inner function.

Let $u = \tan(\frac{1}{2}x)$ . Then $f = u^2$ , so $\frac{df}{du} = 2u = 2\tan(\frac{1}{2}x)$ .

The derivative of $u = \tan(\frac{1}{2}x)$ with respect to $x$ is $\frac{du}{dx} = \frac{1}{2}\sec^2(\frac{1}{2}x)$ . The $\frac{1}{2}$ appears because of the chain rule applied to the argument $\frac{1}{2}x$ .

Multiplying these together:

\frac{df}{dx} = 2\tan\left(\frac{1}{2}x\right) \cdot \frac{1}{2}\sec^2\left(\frac{1}{2}x\right) = \tan\left(\frac{1}{2}x\right)\sec^2\left(\frac{1}{2}x\right)

Now substitute $x = \frac{2}{3}\pi$ . Then $\frac{1}{2}x = \frac{\pi}{3}$ .

From standard exact values: $\tan(\frac{\pi}{3}) = \sqrt{3}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$ , so $\sec^2(\frac{\pi}{3}) = \frac{1}{\cos^2(\frac{\pi}{3})} = \frac{1}{(\frac{1}{2})^2} = 4$ .

Therefore $f'(\frac{2}{3}\pi) = \sqrt{3} \times 4 = 4\sqrt{3}$ .

Key Takeaways

The chain rule is essential for differentiating composite functions such as $\tan^2(\frac{1}{2}x)$ .
The derivative of $\tan(ax + b)$ is $a\sec^2(ax + b)$ .
Exact trigonometric values at standard angles must be recalled accurately.

Common Mistakes

Forgetting the $\frac{1}{2}$ factor from the chain rule on $\frac{1}{2}x$ . The mark scheme's first mark accepts the form $k\tan\frac{1}{2}x\sec^2\frac{1}{2}x$ , so a wrong constant still earns the method mark but loses the accuracy marks.
Confusing $\sec^2\theta$ with $\frac{1}{\tan^2\theta}$ — the correct relationship is $\sec^2\theta = \frac{1}{\cos^2\theta}$ .
Evaluating $\sec^2(\frac{\pi}{3})$ incorrectly, e.g. forgetting to square the reciprocal of $\cos(\frac{\pi}{3})$ .

Things to Be Careful About

The domain $0 \le x < \pi$ includes $\frac{2}{3}\pi$ , so no domain restriction applies.
The mark scheme allows using identities before differentiation; an alternative is to rewrite $\tan^2(\frac{1}{2}x)$ as $\sec^2(\frac{1}{2}x) - 1$ first, then differentiate.
Give the exact surd form $4\sqrt{3}$ , not a decimal approximation.

Techniques used

apply the chain rule to a composite trigonometric functionevaluate the derivative at a given point using exact trigonometric values

(b)

Find the exact value of

\int_0^{\frac{1}{2}\pi} (f(x) + \sin x) \, dx

4M Similar

DifficultyMedium

Worked solution

Approach

Rewrite $\tan^2(\frac{1}{2}x)$ using the identity $\tan^2\theta = \sec^2\theta - 1$ , integrate term by term, then apply the limits of integration.

Working

\tan^2\left(\frac{1}{2}x\right) + \sin x = \sec^2\left(\frac{1}{2}x\right) - 1 + \sin x

Integrate term by term:

\int \left(\sec^2\left(\frac{1}{2}x\right) - 1 + \sin x\right) dx = 2\tan\left(\frac{1}{2}x\right) - x - \cos x + C

Evaluate from $0$ to $\frac{\pi}{2}$ :

At $x = \frac{\pi}{2}$ :

2\tan\left(\frac{\pi}{4}\right) - \frac{\pi}{2} - \cos\left(\frac{\pi}{2}\right) = 2(1) - \frac{\pi}{2} - 0 = 2 - \frac{\pi}{2}

At $x = 0$ :

2\tan(0) - 0 - \cos(0) = 0 - 0 - 1 = -1

Subtract:

\left(2 - \frac{\pi}{2}\right) - (-1) = 3 - \frac{\pi}{2}

Answer

3 - \frac{\pi}{2}

Final answer

3 - π/2

Detailed explanation

Walkthrough

We need to evaluate $\int_0^{\frac{1}{2}\pi} (\tan^2(\frac{1}{2}x) + \sin x) \, dx$ .

The term $\tan^2(\frac{1}{2}x)$ cannot be integrated directly. However, the Pythagorean identity $\tan^2\theta = \sec^2\theta - 1$ rewrites it as $\sec^2(\frac{1}{2}x) - 1$ , which is integrable.

So the integrand becomes $\sec^2(\frac{1}{2}x) - 1 + \sin x$ .

Now integrate term by term:

$\int \sec^2(\frac{1}{2}x) \, dx = 2\tan(\frac{1}{2}x)$ , because the derivative of $\tan(\frac{1}{2}x)$ is $\frac{1}{2}\sec^2(\frac{1}{2}x)$ , so we need a factor of 2 to compensate.
$\int -1 \, dx = -x$ .
$\int \sin x \, dx = -\cos x$ .

So the antiderivative is $2\tan(\frac{1}{2}x) - x - \cos x$ .

Now apply the limits from 0 to $\frac{\pi}{2}$ .

At the upper limit $x = \frac{\pi}{2}$ :

2\tan\left(\frac{\pi}{4}\right) - \frac{\pi}{2} - \cos\left(\frac{\pi}{2}\right) = 2(1) - \frac{\pi}{2} - 0 = 2 - \frac{\pi}{2}

At the lower limit $x = 0$ :

2\tan(0) - 0 - \cos(0) = 0 - 0 - 1 = -1

Subtract the lower-limit value from the upper-limit value:

\left(2 - \frac{\pi}{2}\right) - (-1) = 2 - \frac{\pi}{2} + 1 = 3 - \frac{\pi}{2}

Key Takeaways

The identity $\tan^2\theta = \sec^2\theta - 1$ is the key to integrating $\tan^2$ of a linear function of $x$ .
The integral of $\sec^2(ax + b)$ is $\frac{1}{a}\tan(ax + b)$ .
Definite integrals are evaluated by substituting the upper limit, substituting the lower limit, and subtracting.

Common Mistakes

Trying to integrate $\tan^2$ directly without applying the identity. The mark scheme awards B1 specifically for expressing the integrand as $\sec^2\frac{1}{2}x - 1 + \sin x$ .
Getting the coefficient wrong when integrating $\sec^2(\frac{1}{2}x)$ — the correct antiderivative is $2\tan(\frac{1}{2}x)$ , not $\frac{1}{2}\tan(\frac{1}{2}x)$ .
Sign error on $\int \sin x \, dx$ : it is $-\cos x$ , so at the lower limit 0 it contributes $-1$ , not $+1$ .

Things to Be Careful About

When subtracting the lower-limit value, watch for double negatives: $(2 - \frac{\pi}{2}) - (-1) = 3 - \frac{\pi}{2}$ .
Give the exact answer $3 - \frac{\pi}{2}$ ; the mark scheme also accepts exact equivalents such as $\frac{6 - \pi}{2}$ .
The domain $0 \le x < \pi$ ensures $\tan(\frac{1}{2}x)$ is well-defined throughout the interval of integration.

Techniques used

apply the Pythagorean identity to rewrite the integrandintegrate term by termapply limits of integration

Q4MediumAlgebraTrigonometry

The polynomial $p(x)$ is defined by

p(x) = ax^3 - ax^2 - 15x + 18

where $a$ is a constant. It is given that $(x + 2)$ is a factor of $p(x)$ .

(a)

Find the value of $a$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $(x + 2)$ is a factor of $p(x)$ , the factor theorem states that $p(-2) = 0$ . Substitute $x = -2$ into the polynomial and solve the resulting linear equation for $a$ .

Working

p(-2) = a(-2)^3 - a(-2)^2 - 15(-2) + 18

p(-2) = -8a - 4a + 30 + 18

p(-2) = -12a + 48

Since $(x + 2)$ is a factor, $p(-2) = 0$ :

-12a + 48 = 0

-12a = -48

a = 4

Answer

a = 4

Final answer

a = 4

Detailed explanation

Walkthrough

The factor theorem tells us that if $(x + 2)$ is a factor of $p(x)$ , then substituting $x = -2$ into $p(x)$ must give zero. We substitute $x = -2$ into each term of the polynomial, being careful with the signs of the powers of $-2$ : $(-2)^3 = -8$ and $(-2)^2 = 4$ . This gives $-8a - 4a + 30 + 18 = -12a + 48$ . Setting this equal to zero and solving gives $a = 4$ .

Key Takeaways

The factor theorem: $(x - c)$ is a factor of $p(x)$ if and only if $p(c) = 0$ . Here the factor $(x + 2)$ corresponds to the root $x = -2$ . Substituting the root into the polynomial converts the factor condition into an equation we can solve for the unknown coefficient $a$ .

Common Mistakes

Substituting $x = 2$ instead of $x = -2$ (confusing the sign of the root).
Sign errors when computing powers of $-2$ : $(-2)^3 = -8$ , not $8$ , and $(-2)^2 = 4$ , not $-4$ .
Arithmetic errors when combining like terms: $-8a - 4a = -12a$ , and $30 + 18 = 48$ .

Things to Be Careful About

The factor $(x + 2)$ means the root is $x = -2$ , since $x + 2 = 0$ gives $x = -2$ . Always double-check the signs when substituting negative values into polynomials with odd and even powers. The final equation $-12a + 48 = 0$ must be solved correctly: $a = \frac{48}{12} = 4$ .

Techniques used

apply the factor theorem to substitute the rootsolve a linear equation for the unknown coefficient

(b)

Hence factorise $p(x)$ completely.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

With $a = 4$ , the polynomial is $p(x) = 4x^3 - 4x^2 - 15x + 18$ . Divide by the known factor $(x + 2)$ using polynomial long division, then factorise the resulting quadratic completely.

Working

p(x) = 4x^3 - 4x^2 - 15x + 18

Divide by $(x + 2)$ . First term of the quotient: $4x^3 \div x = 4x^2$ .

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

Subtract:

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

Next term: $-12x^2 \div x = -12x$ .

(x + 2)(-12x) = -12x^2 - 24x

Subtract:

-12x^2 - 15x - (-12x^2 - 24x) = 9x

Next term: $9x \div x = 9$ .

(x + 2)(9) = 9x + 18

Subtract:

9x + 18 - (9x + 18) = 0

The quotient is $4x^2 - 12x + 9$ , so:

p(x) = (x + 2)(4x^2 - 12x + 9)

Factorise the quadratic:

4x^2 - 12x + 9 = (2x - 3)^2

Therefore:

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

Answer

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

Final answer

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

Detailed explanation

Walkthrough

We substitute $a = 4$ into the polynomial to get $p(x) = 4x^3 - 4x^2 - 15x + 18$ . Since $(x + 2)$ is a known factor, we divide the cubic by $(x + 2)$ using polynomial long division. The first term of the quotient is $4x^2$ (since $4x^2 \times x = 4x^3$ ). Multiplying $(x + 2)$ by $4x^2$ gives $4x^3 + 8x^2$ , which we subtract, leaving $-12x^2$ . The next term is $-12x$ (since $-12x \times x = -12x^2$ ). Multiplying $(x + 2)$ by $-12x$ gives $-12x^2 - 24x$ , which we subtract, leaving $9x + 18$ . The final term is $9$ (since $9 \times x = 9x$ ). Multiplying $(x + 2)$ by $9$ gives $9x + 18$ , which subtracts to leave remainder $0$ . The quotient is $4x^2 - 12x + 9$ , which factors as $(2x - 3)^2$ since it is a perfect square. The complete factorisation is $p(x) = (x + 2)(2x - 3)^2$ .

Key Takeaways

Polynomial long division breaks a higher-degree polynomial into a known factor times a lower-degree polynomial. Recognising $4x^2 - 12x + 9$ as a perfect square $(2x - 3)^2$ requires spotting that $4x^2 = (2x)^2$ , $9 = 3^2$ , and the middle term $-12x = 2 \times 2x \times (-3)$ .

Common Mistakes

Arithmetic errors in the subtraction steps of long division, especially with signs.
Not recognising $4x^2 - 12x + 9$ as a perfect square and leaving the factorisation incomplete.
Forgetting to include the $+9$ term at the end of the quotient.

Things to Be Careful About

Check the division by multiplying back: $(x + 2)(4x^2 - 12x + 9) = 4x^3 - 4x^2 - 15x + 18$ . The quadratic $4x^2 - 12x + 9$ is a perfect square trinomial; the factorisation $(2x - 3)^2$ is the correct complete form. The final answer must have integer coefficients.

Techniques used

perform polynomial long divisionfactorise a quadratic expressionwrite the complete factorisation

(c)

Solve the equation $p(\text{cosec}^2 \theta) = 0$ for $-90^\circ < \theta < 90^\circ$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Substitute $x = \text{cosec}^2 \theta$ into the factorised form of $p(x)$ . Set each factor to zero and solve. Use the identity $\text{cosec}^2 \theta = \frac{1}{\sin^2 \theta}$ to convert to an equation in $\sin^2 \theta$ , then find all angles in the given range.

Working

From part (b), the factorised form is:

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

Set $x = \text{cosec}^2 \theta$ :

p(\text{cosec}^2 \theta) = (\text{cosec}^2 \theta + 2)(2\text{cosec}^2 \theta - 3)^2 = 0

Case 1: $\text{cosec}^2 \theta + 2 = 0$

\text{cosec}^2 \theta = -2

This is impossible since $\text{cosec}^2 \theta \geq 1$ for all $\theta$ .

Case 2: $2\text{cosec}^2 \theta - 3 = 0$

\text{cosec}^2 \theta = \frac{3}{2}

Using $\text{cosec}^2 \theta = \frac{1}{\sin^2 \theta}$ :

\frac{1}{\sin^2 \theta} = \frac{3}{2}

\sin^2 \theta = \frac{2}{3}

\sin \theta = \pm \sqrt{\frac{2}{3}}

\theta = \pm \arcsin\left(\sqrt{\frac{2}{3}}\right)

\arcsin\left(\sqrt{\frac{2}{3}}\right) \approx 54.7^\circ

For $-90^\circ < \theta < 90^\circ$ , both the positive and negative angles are in range:

\theta = 54.7^\circ \quad \text{or} \quad \theta = -54.7^\circ

Answer

\theta = 54.7^\circ \text{ or } \theta = -54.7^\circ

Final answer

θ = 54.7° or θ = -54.7°

Detailed explanation

Walkthrough

We substitute $x = \text{cosec}^2 \theta$ into the factorised form $p(x) = (x + 2)(2x - 3)^2$ . The product equals zero, so either factor is zero. The first factor gives $\text{cosec}^2 \theta = -2$ , which is impossible because $\text{cosec}^2 \theta = \frac{1}{\sin^2 \theta}$ and $\sin^2 \theta \leq 1$ , so $\text{cosec}^2 \theta \geq 1$ . The second factor gives $\text{cosec}^2 \theta = \frac{3}{2}$ , so $\sin^2 \theta = \frac{2}{3}$ . Taking square roots, $\sin \theta = \pm \sqrt{\frac{2}{3}}$ . In the range $-90^\circ < \theta < 90^\circ$ , the arcsine of $\sqrt{\frac{2}{3}}$ gives approximately $54.7^\circ$ , and the negative angle $-54.7^\circ$ is also in range. No other angles in this interval satisfy the equation.

Key Takeaways

The identity $\text{cosec}^2 \theta = \frac{1}{\sin^2 \theta}$ converts a cosecant equation into a sine equation. When solving $\sin^2 \theta = k$ , remember that $\sin \theta = \pm \sqrt{k}$ , giving both positive and negative angles. The restricted domain $-90^\circ < \theta < 90^\circ$ is the principal interval for arcsine, so both $\theta$ and $-\theta$ are included.

Common Mistakes

Forgetting that $\sin \theta = \pm \sqrt{\frac{2}{3}}$ has two signs, so missing the negative angle.
Discarding the negative angle because it falls outside $0^\circ$ to $90^\circ$ — but the given range includes negative angles.
Not checking that $\text{cosec}^2 \theta = -2$ has no solution.

Things to Be Careful About

The range $-90^\circ < \theta < 90^\circ$ includes angles in the fourth quadrant (negative angles). Both $54.7^\circ$ and $-54.7^\circ$ lie in this open interval. No angles beyond $\pm 90^\circ$ are allowed, so there are exactly two solutions. The value $\arcsin\left(\sqrt{\frac{2}{3}}\right) \approx 54.7356^\circ$ , so $54.7^\circ$ is an appropriate rounding.

Techniques used

substitute into the factorised polynomialsolve a product of factors equal to zeroapply the cosecant identitysolve a trigonometric equation in a restricted domain

Q5MediumIntegrationLogarithmic and Exponential FunctionsNumerical Solution of Equations

It is given that $\int_a^{a^3} \frac{10}{2x + 1} \, dx = 7$ , where $a$ is a constant greater than 1.

(a)

Show that $a = \sqrt[3]{0.5e^{1.4}(2a + 1) - 0.5}$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Integrate $\frac{10}{2x+1}$ with respect to $x$ , apply the limits $a$ and $a^3$ , equate the result to 7, then use logarithm laws to rearrange into the required form for $a$ .

Working

Integrate the function:

\int \frac{10}{2x+1} \, dx = 5\ln(2x+1) + C

Apply the limits $x = a$ and $x = a^3$ and equate to 7:

5\ln(2a^3 + 1) - 5\ln(2a + 1) = 7

Combine the logarithms using $\ln A - \ln B = \ln\left(\frac{A}{B}\right)$ :

5\ln\left(\frac{2a^3 + 1}{2a + 1}\right) = 7

Divide both sides by 5:

\ln\left(\frac{2a^3 + 1}{2a + 1}\right) = \frac{7}{5} = 1.4

Exponentiate both sides:

\frac{2a^3 + 1}{2a + 1} = e^{1.4}

Multiply both sides by $(2a + 1)$ :

2a^3 + 1 = e^{1.4}(2a + 1)

Subtract 1 from both sides:

2a^3 = e^{1.4}(2a + 1) - 1

Divide by 2:

a^3 = 0.5e^{1.4}(2a + 1) - 0.5

Take the cube root of both sides:

a = \sqrt[3]{0.5e^{1.4}(2a + 1) - 0.5}

Answer

a = \sqrt[3]{0.5e^{1.4}(2a + 1) - 0.5}

as required.

Final answer

a = cube_root(0.5e^1.4(2a + 1) - 0.5)

Detailed explanation

Walkthrough

The problem asks us to show that $a = \sqrt[3]{0.5e^{1.4}(2a + 1) - 0.5}$ given that $\int_a^{a^3} \frac{10}{2x+1} \, dx = 7$ and $a > 1$ .

Step 1: Integrate the function. We need the antiderivative of $\frac{10}{2x+1}$ . Recall the standard result $\int \frac{1}{ax+b} \, dx = \frac{1}{a}\ln|ax+b| + C$ . Here $a = 2$ , so $\int \frac{1}{2x+1} \, dx = \frac{1}{2}\ln(2x+1) + C$ . Multiplying by the coefficient 10 gives $5\ln(2x+1) + C$ . This earns the M1 and A1 marks.

Step 2: Apply the limits. The definite integral is $F(a^3) - F(a) = 5\ln(2a^3+1) - 5\ln(2a+1)$ . We set this equal to 7, which earns the first DM1 mark.

Step 3: Combine the logarithms. Using $\ln A - \ln B = \ln(A/B)$ , we write this as $5\ln\left(\frac{2a^3+1}{2a+1}\right) = 7$ . Dividing by 5: $\ln\left(\frac{2a^3+1}{2a+1}\right) = 1.4$ .

Step 4: Exponentiate. Raising both sides as powers of $e$ eliminates the logarithm: $\frac{2a^3+1}{2a+1} = e^{1.4}$ . This is the key step that earns the second DM1 mark.

Step 5: Rearrange. Multiply both sides by $2a+1$ , subtract 1, divide by 2, and take the cube root. This gives the required form and earns the final A1 mark.

Key Takeaways

The integral of $\frac{1}{ax+b}$ is $\frac{1}{a}\ln|ax+b| + C$ . This is a fundamental result for P2.
Logarithm laws such as $\ln A - \ln B = \ln(A/B)$ are essential for simplifying expressions with logarithms.
To solve an equation involving $\ln$ , exponentiate both sides to remove the logarithm.
In an "AG" (Answer Given) question, every step must be shown — you cannot skip working.

Common Mistakes

Forgetting the factor of $\frac{1}{2}$ when integrating $\frac{1}{2x+1}$ . The integral of $\frac{10}{2x+1}$ is $5\ln(2x+1)$ , not $10\ln(2x+1)$ .
Applying the limits in the wrong order — it must be $F(a^3) - F(a)$ .
Incorrectly applying logarithm laws, e.g., writing $\ln A - \ln B = \ln(A-B)$ instead of $\ln(A/B)$ .
Not showing enough detail for an AG question.

Things to Be Careful About

Since $a > 1$ , both $2a+1$ and $2a^3+1$ are positive, so the logarithms are well-defined without absolute value signs.
The value $1.4$ comes from $\frac{7}{5}$ , which is used in the exponent $e^{1.4}$ .
When exponentiating, apply $e$ to both sides of the equation simultaneously.

Techniques used

integrate a reciprocal linear functionapply definite integral limitscombine logarithms using logarithm lawssolve an equation involving exponentials

(b)

Use an iterative formula, based on the equation in part (a), to find the value of $a$ correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the iterative formula $a_{n+1} = \sqrt[3]{0.5e^{1.4}(2a_n + 1) - 0.5}$ starting from $a_0 = 2$ , computing each iteration to 5 significant figures until the value stabilises.

Working

The iterative formula from part (a) is:

a_{n+1} = \sqrt[3]{0.5e^{1.4}(2a_n + 1) - 0.5}

Starting with $a_0 = 2$ :

a_1 = \sqrt[3]{0.5e^{1.4}(2(2) + 1) - 0.5} = \sqrt[3]{9.6380} \approx 2.1281

a_2 = \sqrt[3]{0.5e^{1.4}(2(2.1281) + 1) - 0.5} = \sqrt[3]{10.1575} \approx 2.1657

a_3 = \sqrt[3]{0.5e^{1.4}(2(2.1657) + 1) - 0.5} = \sqrt[3]{10.3099} \approx 2.1764

a_4 = \sqrt[3]{0.5e^{1.4}(2(2.1764) + 1) - 0.5} = \sqrt[3]{10.3533} \approx 2.1795

a_5 = \sqrt[3]{0.5e^{1.4}(2(2.1795) + 1) - 0.5} = \sqrt[3]{10.3659} \approx 2.1804

a_6 = \sqrt[3]{0.5e^{1.4}(2(2.1804) + 1) - 0.5} = \sqrt[3]{10.3696} \approx 2.1807

a_7 = \sqrt[3]{0.5e^{1.4}(2(2.1807) + 1) - 0.5} = \sqrt[3]{10.3708} \approx 2.1807

The sequence has converged to $a \approx 2.1807$ to 5 significant figures.

Answer

a = 2.18 \text{ (to 3 significant figures)}

Final answer

a = 2.18 (to 3 significant figures)

Detailed explanation

Walkthrough

The equation from part (a) gives $a = \sqrt[3]{0.5e^{1.4}(2a + 1) - 0.5}$ , which is already in the form $a = F(a)$ — perfect for iteration.

Step 1: Set up the iteration. Write $a_{n+1} = \sqrt[3]{0.5e^{1.4}(2a_n + 1) - 0.5}$ with starting value $a_0 = 2$ as specified in the question.

Step 2: Perform iterations. Substitute each value into the formula to get the next. Continue until successive values agree to 5 significant figures. The sequence converges to approximately 2.1807.

Step 3: Round the final answer. The converged value 2.1807, rounded to 3 significant figures, is 2.18.

Key Takeaways

An equation of the form $x = F(x)$ can be solved iteratively using $x_{n+1} = F(x_n)$ .
The sequence converges when successive approximations agree to the required precision.
Each iteration should be shown to more significant figures than the final answer requires (5 sf for a 3 sf answer).

Common Mistakes

Rounding intermediate values too early, which can cause the sequence to converge to the wrong value.
Not showing enough iterations to justify the final answer.
Giving the answer to the wrong number of significant figures (e.g., 2.181 instead of 2.18).

Things to Be Careful About

The mark scheme requires either showing sufficient iterations to 5 sf OR showing a sign change in the interval $[2.175, 2.185]$ to justify the answer.
The final answer must be exactly 3 significant figures: 2.18.
Use the initial value $a_0 = 2$ as stated in the question.

Techniques used

apply an iterative formulacompute successive approximationsround to the required precision

Q6MediumDifferentiationLogarithmic and Exponential Functions

A curve has parametric equations

x = \frac{e^{2t} - 2}{e^{2t} + 1}, \quad y = e^{3t} + 1

(a)

Find an expression for $\frac{dy}{dx}$ in terms of $t$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

For parametric equations, use

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

Differentiate $x$ with respect to $t$ using the quotient rule, differentiate $y$ , then divide and simplify.

Working

Differentiate $y = e^{3t} + 1$ :

\frac{dy}{dt} = 3e^{3t}.

Differentiate $x = \frac{e^{2t}-2}{e^{2t}+1}$ using the quotient rule:

\frac{dx}{dt} = \frac{(e^{2t}+1)(2e^{2t}) - (e^{2t}-2)(2e^{2t})}{(e^{2t}+1)^2}.

Simplify the numerator:

(e^{2t}+1)(2e^{2t}) - (e^{2t}-2)(2e^{2t}) = 2e^{2t}((e^{2t}+1)-(e^{2t}-2)) = 6e^{2t}.

So

\frac{dx}{dt} = \frac{6e^{2t}}{(e^{2t}+1)^2}.

Therefore

\frac{dy}{dx} = \frac{3e^{3t}}{\frac{6e^{2t}}{(e^{2t}+1)^2}} = \frac{3e^{3t}(e^{2t}+1)^2}{6e^{2t}} = \frac12 e^t(e^{2t}+1)^2.

Answer

\frac{dy}{dx} = \frac12 e^t(e^{2t}+1)^2.

Final answer

dy/dx = 1/2 e^t (e^{2t}+1)^2

Detailed explanation

Walkthrough

We are given $x$ and $y$ in terms of a parameter $t$ . To find the gradient of the curve, use the parametric formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ .

First differentiate $y = e^{3t} + 1$ . The derivative of $e^{3t}$ is $3e^{3t}$ , and the derivative of the constant $1$ is $0$ , so $\frac{dy}{dt} = 3e^{3t}$ .

For $x$ , use the quotient rule because $x$ is a quotient of two functions of $t$ . If $u = e^{2t} - 2$ and $v = e^{2t} + 1$ , then $u' = v' = 2e^{2t}$ , and

\frac{dx}{dt} = \frac{vu' - uv'}{v^2}.

Substituting gives the numerator shown in the working. Simplify it by factoring out $2e^{2t}$ ; the bracket simplifies to $3$ , so the numerator becomes $6e^{2t}$ .

Finally divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . This is done by multiplying by the reciprocal of $\frac{dx}{dt}$ and cancelling the common factor $e^{2t}$ , leaving $\frac12 e^t(e^{2t}+1)^2$ . The mark scheme requires this simplification to be shown and the final answer to have no fraction within a fraction.

Key Takeaways

Parametric differentiation: differentiate each coordinate with respect to the parameter, then divide $dy/dt$ by $dx/dt$ .
The quotient rule is needed when the parametric expression for $x$ is a quotient.
Simplifying the numerator before dividing avoids compound fractions.

Common Mistakes

Forgetting to use the quotient rule for $x$ and differentiating the numerator and denominator separately.
Writing the quotient rule numerator in the wrong order.
Failing to simplify $(e^{2t}+1)2e^{2t} - (e^{2t}-2)2e^{2t}$ ; the mark scheme requires this simplification to be seen.
Leaving the answer as a fraction inside a fraction, which is not accepted.
Forgetting the factor $3$ when differentiating $e^{3t}$ .

Things to Be Careful About

The quotient rule numerator is $vu' - uv'$ , not $uv' - vu'$ .
Both $u'$ and $v'$ equal $2e^{2t}$ , so the difference in brackets is $3$ .
When dividing by $\frac{dx}{dt}$ , multiply by its reciprocal and cancel powers of $e^{2t}$ correctly.
The final expression must be fully simplified, with no fractions within fractions.

Techniques used

differentiate parametric equationsapply quotient ruledivide parametric derivativessimplify compound fraction

(b)

Find the exact gradient of the curve at the point where the curve crosses the $y$ -axis.

3M Similar

DifficultyMedium

Worked solution

Approach

The curve crosses the $y$ -axis when $x = 0$ . Set the parametric expression for $x$ equal to zero, solve for $t$ , then substitute this value into the gradient expression from part (a).

Working

Set $x = 0$ :

\frac{e^{2t}-2}{e^{2t}+1} = 0

so

e^{2t} - 2 = 0 \quad \Rightarrow \quad e^{2t} = 2.

Taking natural logarithms:

2t = \ln 2 \quad \Rightarrow \quad t = \frac12 \ln 2.

Substitute into

\frac{dy}{dx} = \frac12 e^t (e^{2t}+1)^2.

At $t = \frac12 \ln 2$ :

e^t = e^{\frac12 \ln 2} = \sqrt{2}, \qquad e^{2t} = 2.

Hence

\frac{dy}{dx} = \frac12 \cdot \sqrt{2} \cdot (2+1)^2 = \frac12 \cdot \sqrt{2} \cdot 9 = \frac92 \sqrt{2}.

Answer

\frac{dy}{dx} = \frac92\sqrt{2}.

Final answer

9√2/2

Detailed explanation

Walkthrough

The curve crosses the $y$ -axis where $x = 0$ . Since $x$ is given in terms of $t$ , set the parametric expression for $x$ equal to zero:

\frac{e^{2t}-2}{e^{2t}+1} = 0.

A fraction is zero when its numerator is zero, so $e^{2t} - 2 = 0$ , giving $e^{2t} = 2$ . Taking natural logarithms gives $2t = \ln 2$ , so $t = \frac12 \ln 2$ . This is the parameter value at the crossing point.

Now substitute this value into the gradient expression from part (a). Use the exponential-logarithm rules: $e^{\frac12 \ln 2} = \sqrt{2}$ and $e^{2t} = 2$ . Then

\frac{dy}{dx} = \frac12 \cdot \sqrt{2} \cdot (2+1)^2 = \frac92\sqrt{2}.

This is the exact gradient at the point where the curve crosses the $y$ -axis.

Key Takeaways

The $y$ -axis is the line $x = 0$ ; a curve crosses it when its $x$ -coordinate is zero.
To solve $e^{2t} = 2$ , take natural logarithms: $2t = \ln 2$ .
$e^{a \ln b} = b^a$ , so $e^{\frac12 \ln 2} = \sqrt{2}$ .
The parametric derivative is evaluated by substituting the parameter value, not by substituting $x$ or $y$ .

Common Mistakes

Confusing the $y$ -axis with the $x$ -axis and setting $y = 0$ instead of $x = 0$ .
Forgetting that a fraction equals zero only when its numerator is zero.
Using $t = \ln 2$ instead of $t = \frac12 \ln 2$ .
Substituting $t$ into $x$ or $y$ instead of into $\frac{dy}{dx}$ .
Giving a decimal instead of the exact surd $\frac92\sqrt{2}$ .

Things to Be Careful About

From $e^{2t} = 2$ , the correct parameter is $t = \frac12 \ln 2$ , not $\ln 2$ .
$e^{\frac12 \ln 2} = (e^{\ln 2})^{1/2} = 2^{1/2} = \sqrt{2}$ .
The mark scheme awards a method mark for substituting the non-zero value of $t$ into their gradient expression and attempting simplification.
Leave the final gradient in exact form unless a decimal is requested.

Techniques used

set x equal to zero to locate y-axis crossingsolve exponential equation using natural logarithmssubstitute parameter value into gradient expressionsimplify surd expression

Q7MediumTrigonometry

(a)

Prove that $\cos(\theta + 30^\circ) \cos(\theta + 60^\circ) \equiv \frac{1}{4}\sqrt{3} - \frac{1}{2}\sin 2\theta$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the compound-angle formulae to expand each cosine product, substitute the exact values of $\cos 30^\circ$ , $\sin 30^\circ$ , $\cos 60^\circ$ and $\sin 60^\circ$ , then simplify using $\cos^2\theta + \sin^2\theta = 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$ .

Working

Expand each factor using $\cos(A+B) = \cos A\cos B - \sin A\sin B$ :

\cos(\theta + 30^\circ)\cos(\theta + 60^\circ) = (\cos\theta\cos 30^\circ - \sin\theta\sin 30^\circ)(\cos\theta\cos 60^\circ - \sin\theta\sin 60^\circ)

Substitute the exact values:

\cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}, \quad \cos 60^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}

So the product becomes:

\left(\frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta\right)\left(\frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta\right)

Expand:

\frac{\sqrt{3}}{4}\cos^2\theta - \frac{3}{4}\sin\theta\cos\theta - \frac{1}{4}\sin\theta\cos\theta + \frac{\sqrt{3}}{4}\sin^2\theta

Combine the middle terms:

\frac{\sqrt{3}}{4}(\cos^2\theta + \sin^2\theta) - \sin\theta\cos\theta

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

\frac{\sqrt{3}}{4} - \frac{1}{2}\sin 2\theta

Answer

\cos(\theta + 30^\circ)\cos(\theta + 60^\circ) \equiv \frac{1}{4}\sqrt{3} - \frac{1}{2}\sin 2\theta

Final answer

cos(θ + 30°)cos(θ + 60°) ≡ √3/4 - 1/2 sin 2θ

Detailed explanation

Walkthrough

We need to prove the given identity. Start by writing each cosine of a sum using the compound-angle formula. This turns the product into a product of two binomials. Then replace the trigonometric constants with their exact values. Expanding the binomials gives terms in $\cos^2\theta$ , $\sin^2\theta$ and $\sin\theta\cos\theta$ . Use $\cos^2\theta + \sin^2\theta = 1$ to combine the first and last terms, and use the double-angle formula $\sin 2\theta = 2\sin\theta\cos\theta$ to replace the remaining product. The result matches the right-hand side.

Key Takeaways

This question tests the compound-angle formula for cosine, exact values of standard angles, and the double-angle formula for sine. It also shows how an identity can be proved by transforming one side into the other using known identities.

Common Mistakes

A common mistake is expanding the product incorrectly or forgetting to distribute the negative signs. Another is using the wrong exact values, especially confusing $\cos 60^\circ$ and $\sin 30^\circ$ . A proof question requires showing the intermediate simplification; simply stating the identity is not enough.

Things to Be Careful About

Remember that $\cos(A+B) = \cos A\cos B - \sin A\sin B$ , not $\cos A\cos B + \sin A\sin B$ . Also, when combining terms, $\sin\theta\cos\theta$ must be doubled to obtain $\frac{1}{2}\sin 2\theta$ . The mark scheme requires the detail of showing the expansion and simplification.

Techniques used

expand compound angle expressionsapply exact trigonometric valuesuse double angle formula for sine

(b)

Solve the equation $5 \cos(2\alpha + 30^\circ) \cos(2\alpha + 60^\circ) = 1$ for $0^\circ < \alpha < 90^\circ$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the identity from part (a) with $\theta = 2\alpha$ to replace the product of cosines, then solve the resulting equation for $\sin 4\alpha$ . Finally find all values of $4\alpha$ in the interval $0^\circ < 4\alpha < 360^\circ$ and divide by 4.

Working

From part (a):

\cos(\theta + 30^\circ)\cos(\theta + 60^\circ) = \frac{\sqrt{3}}{4} - \frac{1}{2}\sin 2\theta

Put $\theta = 2\alpha$ :

\cos(2\alpha + 30^\circ)\cos(2\alpha + 60^\circ) = \frac{\sqrt{3}}{4} - \frac{1}{2}\sin 4\alpha

Substitute into the equation:

5\left(\frac{\sqrt{3}}{4} - \frac{1}{2}\sin 4\alpha\right) = 1

\frac{5\sqrt{3}}{4} - \frac{5}{2}\sin 4\alpha = 1

\frac{5}{2}\sin 4\alpha = \frac{5\sqrt{3}}{4} - 1

\sin 4\alpha = \frac{\sqrt{3}}{2} - \frac{2}{5}

Numerically:

\sin 4\alpha \approx 0.4660

Since $0^\circ < \alpha < 90^\circ$ , we have $0^\circ < 4\alpha < 360^\circ$ . The principal solution is:

4\alpha = \sin^{-1}(0.4660) \approx 27.8^\circ

The other solution in the range is:

4\alpha = 180^\circ - 27.8^\circ = 152.2^\circ

Divide by 4:

\alpha = 6.95^\circ \quad \text{or} \quad \alpha = 38.05^\circ

Answer

\alpha = 6.9^\circ \text{ or } 38.1^\circ \quad (0^\circ < \alpha < 90^\circ)

Final answer

α = 6.9° or 38.1°

Detailed explanation

Walkthrough

First apply the identity from part (a) by replacing $\theta$ with $2\alpha$ . This converts the product of cosines into an expression involving $\sin 4\alpha$ . Substitute this into the given equation and solve algebraically for $\sin 4\alpha$ . The value is about $0.466$ . Because $\alpha$ is between $0^\circ$ and $90^\circ$ , $4\alpha$ is between $0^\circ$ and $360^\circ$ . A sine equation has two solutions in this range: the principal angle and its supplement $180^\circ - \theta$ . Divide both by 4 to obtain the two values of $\alpha$ .

Key Takeaways

This question combines an algebraic identity with trigonometric equation solving. It is important to scale the domain correctly when the variable is multiplied by a constant. The general solution for $\sin x = k$ in $0^\circ < x < 360^\circ$ is $x = \sin^{-1} k$ and $x = 180^\circ - \sin^{-1} k$ .

Common Mistakes

A common mistake is forgetting to multiply the domain by 4 when solving for $4\alpha$ . Another is only giving the principal solution and missing the supplementary solution. Also, if the question asks for exact or rounded values, make sure the final answers are given to the required accuracy.

Things to Be Careful About

The mark scheme accepts answers to greater accuracy, so $6.9^\circ$ and $38.1^\circ$ are acceptable. There are no other solutions between $0^\circ$ and $90^\circ$ because the next solutions for $4\alpha$ would be $360^\circ + 27.8^\circ$ and $360^\circ + 152.2^\circ$ , both outside $360^\circ$ . Ensure the calculator is in degree mode.

Techniques used

apply proven identitysolve trigonometric equationfind all solutions in a given interval

(c)

Show that the exact value of $\cos 20^\circ \cos 50^\circ + \cos 40^\circ \cos 70^\circ$ is $\frac{1}{2}\sqrt{3}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the identity from part (a) twice: once with $\theta = -10^\circ$ to evaluate $\cos 20^\circ\cos 50^\circ$ , and once with $\theta = 10^\circ$ to evaluate $\cos 40^\circ\cos 70^\circ$ . Then add the results, using $\sin(-20^\circ) = -\sin 20^\circ$ .

Working

From part (a), for any $\theta$ :

\cos(\theta + 30^\circ)\cos(\theta + 60^\circ) = \frac{\sqrt{3}}{4} - \frac{1}{2}\sin 2\theta

Set $\theta = -10^\circ$ :

\cos 20^\circ\cos 50^\circ = \frac{\sqrt{3}}{4} - \frac{1}{2}\sin(-20^\circ)

Since $\sin(-20^\circ) = -\sin 20^\circ$ :

\cos 20^\circ\cos 50^\circ = \frac{\sqrt{3}}{4} + \frac{1}{2}\sin 20^\circ

Set $\theta = 10^\circ$ :

\cos 40^\circ\cos 70^\circ = \frac{\sqrt{3}}{4} - \frac{1}{2}\sin 20^\circ

Add the two results:

\cos 20^\circ\cos 50^\circ + \cos 40^\circ\cos 70^\circ = \left(\frac{\sqrt{3}}{4} + \frac{1}{2}\sin 20^\circ\right) + \left(\frac{\sqrt{3}}{4} - \frac{1}{2}\sin 20^\circ\right)

The sine terms cancel:

= \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}

Answer

\cos 20^\circ\cos 50^\circ + \cos 40^\circ\cos 70^\circ = \frac{1}{2}\sqrt{3}

Final answer

cos 20° cos 50° + cos 40° cos 70° = √3/2

Detailed explanation

Walkthrough

We need to evaluate the sum of two cosine products. Notice that each product matches the form in part (a) if we choose appropriate values of $\theta$ . For the first product, $\theta + 30^\circ = 20^\circ$ and $\theta + 60^\circ = 50^\circ$ both hold when $\theta = -10^\circ$ . For the second product, $\theta + 30^\circ = 40^\circ$ and $\theta + 60^\circ = 70^\circ$ hold when $\theta = 10^\circ$ . Substitute these values into the identity. The terms involving $\sin 20^\circ$ have opposite signs because $\sin(-20^\circ) = -\sin 20^\circ$ , so they cancel when the two expressions are added. The remaining $\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4}$ gives $\frac{\sqrt{3}}{2}$ .

Key Takeaways

This question shows how a proven identity can be applied strategically by choosing substitutions that match the angles in the problem. It also reinforces the odd symmetry of sine, $\sin(-x) = -\sin x$ .

Common Mistakes

A common mistake is using $\theta = 10^\circ$ for both products, which would give $\cos 40^\circ\cos 70^\circ$ twice instead of the required first product. Another is forgetting that $\sin(-20^\circ) = -\sin 20^\circ$ , which prevents the cancellation.

Things to Be Careful About

Make sure the substitutions are consistent: $\theta = -10^\circ$ gives $20^\circ$ and $50^\circ$ , while $\theta = 10^\circ$ gives $40^\circ$ and $70^\circ$ . The mark scheme requires clear indication that $\sin(-20^\circ) = -\sin 20^\circ$ for full credit.

Techniques used

substitute into proven identityuse odd symmetry of sineadd simplified terms

Mathematics 9709/21 — October/November 2024

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