(a) Sketch the graph of $y = 3x - 2a $, where $a$ is a positive constant. (b) Hence or otherwise solve the inequality $ 3x - 2a < x + 5a$.

(a) Sketch with vertex at (2a/3, 0) and y-intercept at (0, 2a) · (b) -3a/4 < x < 7a/2

The equation of a curve is $xy + y^2 e^{-x} = 4$. (a) Show that $$ \frac{dy}{dx} = \frac{y^2 - ye^x}{xe^x + 2y} $$ (b) Find the gradients of the tangents to the curve when $x = 0$.

(a) dy/dx = (y^2 - y e^x)/(x e^x + 2y) · (b) 1/2 and -3/2

(a) Prove the identity $\cot^2 \theta - \tan^2 \theta \equiv 4\cot 2\theta \csc 2\theta$. (b) Hence solve the equation $\cot^2 x - \tan^2 x = 5\sec 2x$ for $0^\circ < x < 90^\circ$.

(a) cot^2 θ - tan^2 θ ≡ 4cot 2θ csc 2θ · (b) x = 20.9° or x = 69.1°

9709/33 May/June 2025 — Mathematics Solved Past Paper

Q1Medium-EasyAlgebra

(a)

Sketch the graph of $y = |3x - 2a|$ , where $a$ is a positive constant.

1M Similar

DifficultyEasy

Worked solution

Approach

The graph of $y = |3x - 2a|$ is a V-shaped curve. To sketch it, we find the vertex (where the expression inside the modulus is zero) and the y-intercept.

Working

Find the vertex by setting the inside of the modulus to zero:

3x - 2a = 0

x = \frac{2a}{3}

When $x = \frac{2a}{3}$ , $y = 0$ . So the vertex is at $\left(\frac{2a}{3}, 0\right)$ .

Find the y-intercept by setting $x = 0$ :

y = |3(0) - 2a| = |-2a|

Since $a$ is a positive constant, $|-2a| = 2a$ . So the y-intercept is at $(0, 2a)$ .

The graph consists of two straight lines meeting at the vertex $\left(\frac{2a}{3}, 0\right)$ , extending upwards into the first and second quadrants.

Answer

The sketch is as shown in Fig. 1, with vertex at $\left(\frac{2a}{3}, 0\right)$ and y-intercept at $(0, 2a)$ .

Final answer

Sketch with vertex at (2a/3, 0) and y-intercept at (0, 2a)

Detailed explanation

Walkthrough

The function $y = |3x - 2a|$ is a modulus function, which always produces a V-shaped graph. The vertex of the V occurs where the expression inside the modulus equals zero. Setting $3x - 2a = 0$ gives $x = \frac{2a}{3}$ , and the corresponding y-value is 0, so the vertex is $\left(\frac{2a}{3}, 0\right)$ . Since the coefficient of $x$ inside the modulus is positive (3), the right branch has a slope of 3 and the left branch has a slope of -3. The y-intercept is found by substituting $x = 0$ , giving $y = |-2a| = 2a$ because $a > 0$ . Plotting these two points and drawing straight lines through them into the upper quadrants completes the sketch.

Key Takeaways

The graph of $y = |mx + c|$ is V-shaped with its vertex on the x-axis at $x = -c/m$ .
The y-intercept is $|c|$ .
The branches are straight lines with slopes $m$ and $-m$ .

Common Mistakes

Forgetting that $a$ is positive and writing the y-intercept as $-2a$ instead of $2a$ .
Drawing curved branches instead of straight lines.
Placing the vertex on the y-axis instead of the x-axis.

Things to Be Careful About

Ensure the graph is symmetrical about the vertical line $x = \frac{2a}{3}$ .
Both branches must extend into the first and second quadrants (i.e., $y \geq 0$ everywhere).
The question asks to ignore $y = x + 5a$ if seen; do not draw it in part (a).

Techniques used

identify vertex of modulus functioncalculate y-interceptsketch V-shaped graph

(b)

Hence or otherwise solve the inequality $|3x - 2a| < x + 5a$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The inequality $|3x - 2a| < x + 5a$ asks where the graph of $y = |3x - 2a|$ lies below the line $y = x + 5a$ . We find the intersection points by solving $|3x - 2a| = x + 5a$ , which gives two cases depending on the sign of $3x - 2a$ .

Working

Case 1: $3x - 2a \geq 0$ (i.e., $x \geq \frac{2a}{3}$ )

3x - 2a = x + 5a

2x = 7a

x = \frac{7a}{2}

This value satisfies $x \geq \frac{2a}{3}$ since $\frac{7a}{2} > \frac{2a}{3}$ for $a > 0$ .

Case 2: $3x - 2a < 0$ (i.e., $x < \frac{2a}{3}$ )

-(3x - 2a) = x + 5a

2a - 3x = x + 5a

-4x = 3a

x = -\frac{3a}{4}

This value satisfies $x < \frac{2a}{3}$ since $-\frac{3a}{4} < \frac{2a}{3}$ for $a > 0$ .

The two graphs intersect at $x = -\frac{3a}{4}$ and $x = \frac{7a}{2}$ . Since $y = |3x - 2a|$ is a V-shaped graph opening upwards and $y = x + 5a$ is a line with positive slope, the modulus graph lies below the line between the two intersection points.

-\frac{3a}{4} < x < \frac{7a}{2}

Answer

-\frac{3a}{4} < x < \frac{7a}{2}

Final answer

-3a/4 < x < 7a/2

Detailed explanation

Walkthrough

The inequality $|3x - 2a| < x + 5a$ is satisfied where the V-shaped graph from part (a) is below the straight line $y = x + 5a$ . To find where they cross, we set $|3x - 2a| = x + 5a$ . Because of the modulus, we split this into two linear equations.

In the first case, the expression inside the modulus is positive, so $|3x - 2a| = 3x - 2a$ . Setting this equal to $x + 5a$ gives $3x - 2a = x + 5a$ , which simplifies to $2x = 7a$ and $x = \frac{7a}{2}$ . We check that this is in the valid region $x \geq \frac{2a}{3}$ , which it is.

In the second case, the expression inside the modulus is negative, so $|3x - 2a| = 2a - 3x$ . Setting this equal to $x + 5a$ gives $2a - 3x = x + 5a$ , which simplifies to $-4x = 3a$ and $x = -\frac{3a}{4}$ . We check that this is in the valid region $x < \frac{2a}{3}$ , which it is.

The line $y = x + 5a$ intersects the V-shape at these two x-values. Because the V-shape opens upwards and the line has a positive slope, the line is above the V-shape between the two intersection points. Thus, the inequality holds for $-\frac{3a}{4} < x < \frac{7a}{2}$ .

Key Takeaways

Solving $|f(x)| = g(x)$ requires considering both $f(x) = g(x)$ and $-f(x) = g(x)$ .
Always check that the solutions fall within the assumed sign region for each case.
Graphical interpretation helps confirm the solution interval for inequalities.

Common Mistakes

Forgetting to check that the solutions satisfy the assumed conditions for each case.
Writing the inequality with $\leq$ instead of $<$ .
Not verifying which region satisfies the inequality (though for this standard V-shape vs line, it is always between the intersections).

Things to Be Careful About

Ensure $a > 0$ is used when simplifying $|-2a| = 2a$ and when checking validity of solutions.
The final answer must be written as a compound inequality or as two separate inequalities $x > -\frac{3a}{4}$ and $x < \frac{7a}{2}$ .
An alternative method is to square both sides: $(3x - 2a)^2 = (x + 5a)^2$ , which gives $8x^2 - 22ax - 21a^2 = 0$ . Factoring or using the quadratic formula yields the same critical values $x = -\frac{3a}{4}$ and $x = \frac{7a}{2}$ .

Techniques used

solve modulus inequality using graph intersectionsconsider two cases for absolute valueinterpret inequality from graph

Q24MMediumLogarithmic and Exponential Functions

Solve the equation $2\ln(2x + 3) - \ln(2x + 5) = \ln(3x)$ .

DifficultyMedium

Worked solution

Approach

Use the laws of logarithms to combine the left-hand side into a single logarithm, then exponentiate both sides to obtain an equation free of logarithms. Solve the resulting quadratic and reject any solution that is not in the domain of the original logarithms.

Working

Apply the power law to the first term:

2\ln(2x+3) = \ln\left((2x+3)^2\right)

Combine the two logarithms using the quotient law:

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

The equation becomes:

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

Since $\ln$ is a one-to-one function, equate the arguments:

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

Multiply both sides by $2x+5$ :

(2x+3)^2 = 3x(2x+5)

Expand both sides:

4x^2 + 12x + 9 = 6x^2 + 15x

Rearrange into a quadratic equation:

2x^2 + 3x - 9 = 0

Factorise:

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

So:

x = \frac{3}{2} \quad \text{or} \quad x = -3

The original equation contains $\ln(3x)$ , so we require $x > 0$ . Therefore $x = -3$ is not valid.

Answer

x = \frac{3}{2}

Final answer

x = 3/2

Detailed explanation

Walkthrough

We start with an equation that has logarithms on both sides. The overall strategy is to rewrite the left-hand side as a single logarithm using the laws of logarithms, then use the fact that logarithms are one-to-one to drop the logarithms and solve the resulting algebraic equation.

First, the coefficient $2$ in front of $\ln(2x+3)$ is moved inside as a power: $2\ln(2x+3) = \ln\left((2x+3)^2\right)$ . This is the power law for logarithms.

Next, we have a difference of two logarithms with the same base. The quotient law says $\ln A - \ln B = \ln\left(\frac{A}{B}\right)$ , so the left-hand side becomes $\ln\left(\frac{(2x+3)^2}{2x+5}\right)$ .

Now both sides are single logarithms with the same base. Because the natural logarithm function is one-to-one, equal logarithms must have equal arguments. This gives the logarithmic-free equation $\frac{(2x+3)^2}{2x+5} = 3x$ .

Multiplying through by $2x+5$ and expanding produces a quadratic equation. Rearranging gives $2x^2 + 3x - 9 = 0$ , which factorises as $(2x-3)(x+3)=0$ . Hence the candidate solutions are $x = \frac{3}{2}$ and $x = -3$ .

Finally, we must check the domain. The original equation contains $\ln(3x)$ , which requires $3x > 0$ , i.e. $x > 0$ . It also contains $\ln(2x+3)$ and $\ln(2x+5)$ , but these are automatically positive for $x > 0$ . Since $x = -3$ is negative, it is not in the domain and must be rejected. The only valid solution is $x = \frac{3}{2}$ .

Key Takeaways

The laws of logarithms allow sums and differences of logarithms to be combined into a single logarithm.
If two logarithms with the same base are equal, their arguments are equal.
After removing logarithms, the resulting algebraic equation may have extra roots; always check the domain of the original logarithmic expressions.
A quadratic equation can have two solutions, but not every algebraic solution is valid in a logarithmic equation.

Common Mistakes

Forgetting to apply the power law before combining the two logarithms on the left.
Using the quotient law incorrectly, e.g. writing $\ln(2x+3)^2 - \ln(2x+5) = \ln\left((2x+3)^2 - (2x+5)\right)$ instead of a quotient.
Dropping logarithms without ensuring both sides are single logarithms with the same base.
Failing to reject $x = -3$ ; the mark scheme requires the final answer to be $\frac{3}{2}$ only, with clear rejection of the negative value if it is mentioned.
Forgetting to state the domain condition $x > 0$ before rejecting the extraneous root.

Things to Be Careful About

The domain: $\ln(3x)$ requires $x > 0$ . Also $2x+3 > 0$ and $2x+5 > 0$ ; these are satisfied for $x > 0$ but should be considered if the candidate solutions differ.
When multiplying by $2x+5$ , it is positive for $x > 0$ , so no sign change is introduced; however, the equation is valid for candidate roots as long as the argument is not zero.
The mark scheme awards the final mark only for the valid solution $\frac{3}{2}$ ; if you write both roots, you must clearly reject the negative one.
Be careful with expansion: $(2x+3)^2 = 4x^2 + 12x + 9$ , not $4x^2 + 9$ .
The coefficient $2$ must be applied as a power, not simply dropped.

Techniques used

apply the power law for logarithmscombine logarithms using the quotient lawexponentiate both sides to remove logarithmsexpand and solve the resulting quadratic equationcheck the domain and reject extraneous roots

Approach

Use the double-angle identity for cosine squared to rewrite $\cos^2 5x$ as $\frac{1}{2}(1 + \cos 10x)$ , then integrate term by term and evaluate between the given limits.

Working

Start with the double-angle identity:

\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)

With $\theta = 5x$ , this gives:

\cos^2 5x = \frac{1}{2}(1 + \cos 10x)

So the integrand becomes:

3\cos^2 5x = \frac{3}{2}(1 + \cos 10x)

Integrate term by term:

\int 3\cos^2 5x\,dx = \int \left(\frac{3}{2} + \frac{3}{2}\cos 10x\right)dx = \frac{3}{2}x + \frac{3}{20}\sin 10x + C

Now evaluate from $x = \frac{\pi}{5}$ to $x = \frac{\pi}{4}$ .

At $x = \frac{\pi}{4}$ :

\frac{3}{2}\left(\frac{\pi}{4}\right) + \frac{3}{20}\sin\left(\frac{10\pi}{4}\right) = \frac{3\pi}{8} + \frac{3}{20}\sin\left(\frac{5\pi}{2}\right) = \frac{3\pi}{8} + \frac{3}{20}

since $\sin \frac{5\pi}{2} = 1$ .

At $x = \frac{\pi}{5}$ :

\frac{3}{2}\left(\frac{\pi}{5}\right) + \frac{3}{20}\sin\left(\frac{10\pi}{5}\right) = \frac{3\pi}{10} + \frac{3}{20}\sin(2\pi) = \frac{3\pi}{10}

since $\sin 2\pi = 0$ .

Therefore:

\int_{\pi/5}^{\pi/4} 3\cos^2 5x\,dx = \left(\frac{3\pi}{8} + \frac{3}{20}\right) - \frac{3\pi}{10} = \frac{3}{20} + \frac{3\pi}{40}

Answer

\frac{3}{20} + \frac{3\pi}{40}

Final answer

3/20 + 3π/40

Detailed explanation

Walkthrough

This integral contains $\cos^2 5x$ , which is not one of the basic integrals. The standard technique is to replace $\cos^2$ using the double-angle identity:

\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)

Here $\theta = 5x$ , so $2\theta = 10x$ . This rewrites the integrand as $\frac{3}{2}(1 + \cos 10x)$ , which can be integrated directly.

Then integrate term by term: the integral of the constant $\frac{3}{2}$ is $\frac{3}{2}x$ , and the integral of $\frac{3}{2}\cos 10x$ is $\frac{3}{20}\sin 10x$ because dividing by the coefficient of $x$ in the argument gives $\frac{3}{2} \cdot \frac{1}{10} = \frac{3}{20}$ .

Finally evaluate the antiderivative at the upper limit $\frac{\pi}{4}$ and lower limit $\frac{\pi}{5}$ , subtract, and simplify. The sine term at the lower limit is zero because $\sin 2\pi = 0$ , and at the upper limit it is $\frac{3}{20}$ because $\sin \frac{5\pi}{2} = 1$ .

Key Takeaways

Recognising $\cos^2$ or $\sin^2$ in an integral should trigger the double-angle identities.
After using the identity, integration is straightforward.
Always evaluate definite integrals by substituting upper and lower limits into the antiderivative and subtracting.

Common Mistakes

Forgetting to multiply the double-angle identity by the coefficient 3.
Incorrectly integrating $\cos 10x$ as $\sin 10x$ without dividing by 10.
Evaluating $\sin(10x)$ at the limits incorrectly, especially $\sin(5\pi/2)$ and $\sin(2\pi)$ .
Mixing up the order of subtraction when applying limits.

Things to Be Careful About

The identity must be applied before integrating; the mark scheme awards the first method mark for using the correct double-angle formula.
The antiderivative must be exact: $\frac{3}{2}x + \frac{3}{20}\sin 10x$ .
The final answer must be simplified exactly; $\frac{3}{20} + \frac{3\pi}{40}$ is the expected exact value.
Make sure to use radians, not degrees, when evaluating trigonometric functions.

Techniques used

apply the double-angle identity for cosine squaredintegrate a cosine of a linear argumentevaluate a definite integral at the given limits

Approach

Multiply the two complex numbers in polar form, then take the conjugate by replacing $i$ with $-i$ in the exponent. Separately form the product of the individual conjugates and show that the two results are identical.

Working

z_1 z_2 = r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1+\theta_2)}

Taking the conjugate of this product,

(z_1 z_2)^* = r_1 r_2 e^{-i(\theta_1+\theta_2)}

Now form the product of the conjugates of $z_1$ and $z_2$ :

z_1^* z_2^* = (r_1 e^{-i\theta_1})(r_2 e^{-i\theta_2}) = r_1 r_2 e^{-i\theta_1} e^{-i\theta_2} = r_1 r_2 e^{-i(\theta_1+\theta_2)}

Since both expressions equal $r_1 r_2 e^{-i(\theta_1+\theta_2)}$ , we conclude that

(z_1 z_2)^* = z_1^* z_2^*

Answer

(z_1 z_2)^* = z_1^* z_2^*

Final answer

(z1 z2)* = z1* z2*

Detailed explanation

Walkthrough

We begin with the product of the two complex numbers in polar form. Since $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$ , multiplying them gives $z_1 z_2 = r_1 r_2 e^{i(\theta_1+\theta_2)}$ .

The conjugate of a complex number in exponential form is obtained by changing the sign of the angle, because $e^{i\theta} = \cos\theta + i\sin\theta$ and its conjugate is $\cos\theta - i\sin\theta = e^{-i\theta}$ . Therefore the conjugate of the product is $r_1 r_2 e^{-i(\theta_1+\theta_2)}$ .

Next, we compute the product of the individual conjugates. The conjugate of $z_1$ is $r_1 e^{-i\theta_1}$ and the conjugate of $z_2$ is $r_2 e^{-i\theta_2}$ . Their product is $r_1 r_2 e^{-i\theta_1} e^{-i\theta_2} = r_1 r_2 e^{-i(\theta_1+\theta_2)}$ , which is exactly the same expression. This proves the required identity.

Key Takeaways

This question tests the polar form of complex numbers, multiplication of complex numbers in polar form, and the effect of conjugation on the argument. The key idea is that conjugation changes $i$ to $-i$ , so $e^{i\theta}$ becomes $e^{-i\theta}$ .

Common Mistakes

A common mistake is to stop after writing the conjugate of the product without showing that it equals the product of the conjugates. The mark scheme requires a clear conclusion. Another mistake is to forget that $r_1$ and $r_2$ are real moduli, so they are unaffected by conjugation.

Things to Be Careful About

Remember that the conjugate of $e^{i\theta}$ is $e^{-i\theta}$ , not $e^{i\theta}$ . Also, when adding the arguments, the conjugate changes the sign of the whole sum, giving $-\theta_1 - \theta_2$ . Finally, ensure the final line explicitly states that $(z_1 z_2)^* = z_1^* z_2^*$ .

Techniques used

multiply complex numbers in polar formtake conjugates by negating the argumentcompare conjugate of product with product of conjugates

(b)

$z = 3e^{\frac{1}{4}\pi i}$ is a root of the equation $z^2 + bz + c = 0$ , where $b$ and $c$ are real.

State the other root and hence find the values of $b$ and $c$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Since the quadratic has real coefficients, non-real roots must occur in conjugate pairs. Therefore the other root is the conjugate of the given root. Then use the sum and product of the roots to recover $b$ and $c$ .

Working

The given root is

z = 3e^{\frac{1}{4}\pi i}

so the other root is its conjugate:

z^* = 3e^{-\frac{1}{4}\pi i}

For the quadratic $z^2 + bz + c = 0$ , the sum of the roots is $-b$ and the product of the roots is $c$ . Compute the sum:

3e^{\frac{1}{4}\pi i} + 3e^{-\frac{1}{4}\pi i} = 3\left(2\cos\frac{\pi}{4}\right) = 6 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2}

Therefore $-b = 3\sqrt{2}$ , so

b = -3\sqrt{2}

The product is

\left(3e^{\frac{1}{4}\pi i}\right)\left(3e^{-\frac{1}{4}\pi i}\right) = 9e^{0} = 9

so

c = 9

Answer

Other root: $3e^{-\frac{1}{4}\pi i}$ ; $b = -3\sqrt{2}$ , $c = 9$ .

Final answer

Other root = 3e^(-pi i/4); b = -3 sqrt(2), c = 9

Detailed explanation

Walkthrough

The equation $z^2 + bz + c = 0$ has real coefficients $b$ and $c$ . For such an equation, if a non-real complex number is a root, then its conjugate must also be a root. Since $z = 3e^{\frac{1}{4}\pi i}$ is given as a root, the other root is $z^* = 3e^{-\frac{1}{4}\pi i}$ .

Now recall the relationship between the roots and coefficients of a quadratic. If the roots are $\alpha$ and $\beta$ , then $\alpha + \beta = -b$ and $\alpha\beta = c$ . Here the two roots are conjugates, so their sum is twice the real part. Using $e^{i\theta} + e^{-i\theta} = 2\cos\theta$ , the sum is $3(2\cos\frac{\pi}{4}) = 3\sqrt{2}$ . Since this equals $-b$ , we get $b = -3\sqrt{2}$ .

The product of a complex number and its conjugate is the square of its modulus, so $z z^* = |z|^2 = 3^2 = 9$ . Hence $c = 9$ .

Key Takeaways

This question combines the conjugate-root theorem for polynomials with real coefficients and the sum/product relationships for quadratic roots. It also reinforces that $z z^* = |z|^2$ , which is often quicker than multiplying exponential forms.

Common Mistakes

A common mistake is to forget that the other root is the conjugate, not another unrelated root. Another is to use the sum of the roots as $b$ instead of $-b$ . Also, if expanding $(z - 3e^{\frac{1}{4}\pi i})(z - 3e^{-\frac{1}{4}\pi i})$ , students sometimes make sign errors when simplifying the cross term.

Things to Be Careful About

Remember that the coefficient of $z$ in $z^2 + bz + c$ is $b$ , so the sum of the roots is $-b$ . The product of the roots is $c$ . The conjugate root may be written as $3e^{-\frac{1}{4}\pi i}$ or equivalently $3e^{\frac{7}{4}\pi i}$ . Finally, the product $z z^* = |z|^2 = 9$ is a quick way to find $c$ .

Techniques used

apply the conjugate root theoremuse sum and product of rootscompute coefficients of the quadratic

Q5MediumDifferentiation

The equation of a curve is $xy + y^2 e^{-x} = 4$ .

(a)

Show that

\frac{dy}{dx} = \frac{y^2 - ye^x}{xe^x + 2y}

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate the equation implicitly with respect to $x$ . Use the product rule for $xy$ and for $y^2e^{-x}$ , then collect the terms containing $\frac{dy}{dx}$ and solve for it.

Working

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

\frac{d}{dx}(y^2 e^{-x}) = 2y e^{-x}\frac{dy}{dx} - y^2 e^{-x}

Differentiating the whole equation gives

y + x\frac{dy}{dx} + 2y e^{-x}\frac{dy}{dx} - y^2 e^{-x} = 0

Collect the $\frac{dy}{dx}$ terms:

\left(x + 2y e^{-x}\right)\frac{dy}{dx} = y^2 e^{-x} - y

Therefore

\frac{dy}{dx} = \frac{y^2 e^{-x} - y}{x + 2y e^{-x}}

Multiply numerator and denominator by $e^x$ :

\frac{dy}{dx} = \frac{y^2 - y e^x}{x e^x + 2y}

Answer

\frac{dy}{dx} = \frac{y^2 - y e^x}{x e^x + 2y}

Final answer

dy/dx = (y^2 - y e^x)/(x e^x + 2y)

Detailed explanation

Walkthrough

The equation $xy + y^2 e^{-x} = 4$ defines $y$ implicitly as a function of $x$ , so we differentiate both sides with respect to $x$ .

For $xy$ , the product rule gives

\frac{d}{dx}(xy) = y + x\frac{dy}{dx}.

For $y^2 e^{-x}$ , the product rule gives

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

The derivative of the right-hand side, $4$ , is $0$ . Equating the sum of the two derivatives to $0$ gives the equation with $\frac{dy}{dx}$ terms. We then collect these terms on one side and factor out $\frac{dy}{dx}$ :

\left(x + 2y e^{-x}\right)\frac{dy}{dx} = y^2 e^{-x} - y.

Dividing gives the derivative in terms of $x$ and $y$ . Finally, multiplying numerator and denominator by $e^x$ removes the negative exponent and produces the required form.

Key Takeaways

This question tests implicit differentiation, the product rule, and careful algebraic rearrangement. It also shows how to handle a product in which one factor is a function of $y$ and the other is a function of $x$ .

Common Mistakes

Forgetting to multiply by $\frac{dy}{dx}$ when differentiating $y^2$ .
Using the wrong sign for the derivative of $e^{-x}$ ; it is $-e^{-x}$ , not $e^{-x}$ .
Differentiating $xy$ as just $x\frac{dy}{dx}$ and omitting the $y$ term.
Not collecting the $\frac{dy}{dx}$ terms before solving, or losing a term when rearranging.
Stopping before multiplying by $e^x$ , so the answer does not match the given form.

Things to Be Careful About

Because $y$ is a function of $x$ , every term containing $y$ must be differentiated using the chain rule. The mark scheme requires the derivative of $xy$ and the derivative of $y^2 e^{-x}$ to be stated correctly, then the equation set to zero and solved. Since this is a show that question, enough intermediate working must be shown to justify the final expression.

Techniques used

differentiate implicitlyapply product rule to xyapply product rule to y^2 e^{-x}solve for dy/dx

(b)

Find the gradients of the tangents to the curve when $x = 0$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute $x=0$ into the curve equation to find the corresponding $y$ -values. Then substitute $x=0$ and each value of $y$ into the derivative expression to obtain the tangent gradients.

Working

When $x=0$ :

0 \cdot y + y^2 e^{0} = 4 \quad \Rightarrow \quad y^2 = 4

So

y = 2 \quad \text{or} \quad y = -2

Substitute $x=0$ into the derivative:

\frac{dy}{dx} = \frac{y^2 - y e^0}{0 \cdot e^0 + 2y} = \frac{y^2 - y}{2y}

For $y=2$ :

\frac{dy}{dx} = \frac{4 - 2}{4} = \frac{1}{2}

For $y=-2$ :

\frac{dy}{dx} = \frac{4 - (-2)}{-4} = -\frac{3}{2}

Answer

The gradients of the tangents when $x=0$ are $\frac{1}{2}$ and $-\frac{3}{2}$ .

Final answer

1/2 and -3/2

Detailed explanation

Walkthrough

At $x=0$ , the curve equation becomes

0 \cdot y + y^2 e^0 = 4 \quad \Rightarrow \quad y^2 = 4,

so $y = 2$ or $y = -2$ . These are the two points on the curve with $x=0$ : $(0,2)$ and $(0,-2)$ .

Now substitute $x=0$ into the derivative formula from part (a):

\frac{dy}{dx} = \frac{y^2 - y e^0}{0 \cdot e^0 + 2y} = \frac{y^2 - y}{2y}.

For $y=2$ , this gives

\frac{dy}{dx} = \frac{4-2}{4} = \frac{1}{2}.

For $y=-2$ , it gives

\frac{dy}{dx} = \frac{4-(-2)}{-4} = -\frac{3}{2}.

These are the gradients of the two tangents at $x=0$ .

Key Takeaways

This part uses the result of implicit differentiation and applies it at specific points. It also reinforces that a vertical line $x=0$ may meet an implicit curve in more than one point, so each point must be considered separately.

Common Mistakes

Substituting $x=0$ into the derivative before finding the corresponding $y$ -values.
Forgetting that $e^0=1$ .
Making a sign error for $y=-2$ : $y^2 - y = 4 - (-2) = 6$ , not $2$ .
Dividing by $2y$ without noting that $y$ is not zero; here $y=\pm 2$ , so the division is valid.

Things to Be Careful About

The mark scheme awards one mark for each gradient. Both points must be found, and the derivative must be evaluated at both. Since the denominator is $2y$ , the simplification $\frac{y^2-y}{2y}=\frac{y-1}{2}$ is valid only because $y \neq 0$ .

Techniques used

substitute x = 0 into curve equationsolve for yevaluate derivative at each pointsimplify gradients

Q66MMedium-HardComplex Numbers

Find the complex numbers $z$ for which $\frac{z + 4}{z + 4i}$ is real and $|z| = \sqrt{10}$ . Give your answers in the form $z = x + iy$ , where $x$ and $y$ are real.

DifficultyMedium-Hard

Worked solution

Approach

Let $z = x + iy$ and write the given fraction in terms of $x$ and $y$ . The condition that the fraction is real means its imaginary part is zero; the second condition gives $x^2 + y^2 = 10$ .

Working

\frac{z+4}{z+4i} = \frac{(x+4)+iy}{x+i(y+4)}

Multiply numerator and denominator by the conjugate of the denominator, $x - i(y+4)$ :

\frac{(x+4)+iy}{x+i(y+4)} \times \frac{x-i(y+4)}{x-i(y+4)}

The numerator becomes

(x+4+iy)(x-i(y+4)) = x(x+4) + y(y+4) + i\left[xy - (y+4)(x+4)\right]

For the fraction to be real, the imaginary part of the numerator must be zero:

xy - (y+4)(x+4) = 0

Expanding,

xy - (xy + 4x + 4y + 16) = 0 \Rightarrow x + y = -4

Now use the modulus condition:

|z|^2 = x^2 + y^2 = 10

Since $y = -x - 4$ , substitute into the modulus equation:

x^2 + (-x-4)^2 = 10

x^2 + x^2 + 8x + 16 = 10 \Rightarrow x^2 + 4x + 3 = 0

(x+3)(x+1) = 0

So $x = -3$ or $x = -1$ . With $y = -x - 4$ :

if $x = -3$ , then $y = -1$ , giving $z = -3 - i$ ;
if $x = -1$ , then $y = -3$ , giving $z = -1 - 3i$ .

Answer

z = -3 - i \quad \text{or} \quad z = -1 - 3i

Final answer

z = -3 - i or z = -1 - 3i

Detailed explanation

Walkthrough

We need to find all complex numbers satisfying two conditions. First, write $z = x + iy$ , so $z+4 = (x+4)+iy$ and $z+4i = x+i(y+4)$ .

The fraction $\frac{z+4}{z+4i}$ is not in a simple form because the denominator is complex. To decide whether it is real, multiply numerator and denominator by the conjugate of the denominator, $x - i(y+4)$ . This keeps the fraction unchanged but makes the denominator real, so the real/imaginary status of the whole fraction is determined by the numerator. We expand the numerator and collect real and imaginary parts. For the fraction to be real, the imaginary part must be zero. This gives $xy-(y+4)(x+4)=0$ , which simplifies to $x+y=-4$ . This is the line on which the fraction is real.

Next, use the modulus condition $|z|=\sqrt{10}$ . Since $|z|^2=x^2+y^2$ , we have $x^2+y^2=10$ . Substitute $y=-x-4$ from the realness condition. This gives a quadratic in $x$ : $2x^2+8x+6=0$ , or $x^2+4x+3=0$ . Factorising gives $x=-3$ or $x=-1$ , and the corresponding $y$ -values are $-1$ and $-3$ . Hence the two complex numbers are $z=-3-i$ and $z=-1-3i$ .

Key Takeaways

To test whether a quotient of complex numbers is real, multiply by the conjugate of the denominator and set the imaginary part of the numerator equal to zero.
The modulus condition $|z|=\sqrt{10}$ translates to $x^2+y^2=10$ , not $x^2+y^2=\sqrt{10}$ .
A real condition plus a modulus condition often produces a system that reduces to a quadratic, leading to two solutions.

Common Mistakes

Forgetting to multiply by the conjugate of the denominator and instead trying to set the whole denominator real.
Expanding the numerator incorrectly, especially the sign in $xy-(y+4)(x+4)$ .
Writing $|z|=\sqrt{10}$ as $x^2+y^2=\sqrt{10}$ instead of $x^2+y^2=10$ .
Giving only one solution. The quadratic gives two values of $x$ , and both must be converted to the form $z=x+iy$ .
The mark scheme notes that both pairs of $x$ and $y$ correct but not written in the form $z=x+iy$ only earns partial credit (SC A1 A0).

Things to Be Careful About

The denominator $z+4i$ must not be zero; here that would require $z=-4i$ , which does not satisfy $|z|=\sqrt{10}$ , so no solution is lost.
When substituting $y=-x-4$ , it is easy to make a sign error: $(-x-4)^2=(x+4)^2$ .
The mark scheme requires a clear method: an unsupported answer is not awarded full credit.
Both solutions are required for the final A1 marks; one correct solution earns only one of the two final A1 marks.

Techniques used

multiply numerator and denominator by the conjugate of the denominatorequate the imaginary part to zeroapply the modulus conditionsolve a quadratic in x or y

Approach

Write $f(x)$ in the standard partial-fraction form for two distinct linear factors, then determine the constants by clearing denominators and comparing coefficients.

Working

Let

\frac{3a - 5x}{(3a+2x)(2a-x)} = \frac{A}{3a+2x} + \frac{B}{2a-x}.

Multiplying by $(3a+2x)(2a-x)$ :

3a - 5x = A(2a-x) + B(3a+2x).

Comparing coefficients of $x$ and the constant terms:

-A + 2B = -5, \qquad 2aA + 3aB = 3a.

Since $a > 0$ , divide the second equation by $a$ :

2A + 3B = 3.

From $-A + 2B = -5$ , $A = 2B + 5$ . Substitute:

2(2B+5)+3B = 3 \implies 7B = -7 \implies B = -1.

Then $A = 2(-1)+5 = 3$ .

Answer

f(x) = \frac{3}{3a+2x} - \frac{1}{2a-x}.

Final answer

f(x) = 3/(3a+2x) - 1/(2a-x)

Detailed explanation

Walkthrough

The denominator has two distinct linear factors, so $f(x)$ can be written as $\frac{A}{3a+2x} + \frac{B}{2a-x}$ . To find $A$ and $B$ , multiply through by the common denominator. This removes the fractions and gives an identity that holds for all $x$ . Comparing coefficients of $x$ and the constant terms gives two linear equations. Since $a$ is positive, the constant equation can be divided by $a$ safely. Solving the two equations gives $B = -1$ and then $A = 3$ .

Key Takeaways

This part tests the standard method of partial fractions for two distinct linear factors: set up the correct form, clear denominators, and compare coefficients. The constants can also be found by substituting suitable values of $x$ , but coefficient comparison is reliable here.

Common Mistakes

A common error is to forget the factor $a$ in the constant term when comparing coefficients. Another is a sign error when collecting the coefficient of $x$ , especially from the term $-x$ in $A(2a-x)$ .

Things to Be Careful About

The identity must be true for all $x$ , so comparing coefficients is valid. Because $a$ is positive, dividing by $a$ does not change the inequality or sign. Keep the partial fractions in the exact form requested, with constants $A$ and $B$ clearly identified.

Techniques used

express a rational function in partial fractionsclear denominatorscompare coefficientssolve simultaneous linear equations

(b)

Hence obtain the expansion of $f(x)$ in ascending powers of $x$ , up to and including the term in $x^2$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the partial fractions from part (a). Rewrite each fraction as a constant multiple of $(1+u)^{-1}$ or $(1-u)^{-1}$ , expand each using the binomial series up to the term in $x^2$ , then add the two expansions.

Working

From part (a),

f(x) = \frac{3}{3a+2x} - \frac{1}{2a-x}.

For the first term:

\frac{3}{3a+2x} = \frac{3}{3a}\left(1+\frac{2x}{3a}\right)^{-1} = \frac{1}{a}\left(1 - \frac{2x}{3a} + \frac{4x^2}{9a^2} - \cdots\right)

= \frac{1}{a} - \frac{2x}{3a^2} + \frac{4x^2}{9a^3} - \cdots

For the second term:

-\frac{1}{2a-x} = -\frac{1}{2a}\left(1-\frac{x}{2a}\right)^{-1} = -\frac{1}{2a}\left(1 + \frac{x}{2a} + \frac{x^2}{4a^2} + \cdots\right)

= -\frac{1}{2a} - \frac{x}{4a^2} - \frac{x^2}{8a^3} - \cdots

Adding the two expansions:

f(x) = \left(\frac{1}{a}-\frac{1}{2a}\right) + \left(-\frac{2}{3a^2}-\frac{1}{4a^2}\right)x + \left(\frac{4}{9a^3}-\frac{1}{8a^3}\right)x^2 + \cdots

= \frac{1}{2a} - \frac{11}{12a^2}x + \frac{23}{72a^3}x^2 + \cdots

Answer

\frac{1}{2a} - \frac{11}{12a^2}x + \frac{23}{72a^3}x^2

Final answer

1/(2a) - 11x/(12a^2) + 23x^2/(72a^3)

Detailed explanation

Walkthrough

After part (a), $f(x)$ is a sum of two fractions. Each fraction can be written in the form $k(1+u)^{-1}$ or $k(1-u)^{-1}$ , so the binomial expansion applies. For the first fraction, factor $3a$ out of $3a+2x$ ; for the second, factor $2a$ out of $2a-x$ . Expand each bracket up to $x^2$ using $(1+u)^{-1} = 1 - u + u^2 - \cdots$ and $(1-u)^{-1} = 1 + u + u^2 + \cdots$ . Then multiply by the outside constants and add the two series, collecting constant, $x$ and $x^2$ terms separately.

Key Takeaways

The key skill is converting a linear denominator into the form $(1+u)$ before expanding. This is a direct application of the binomial theorem for negative indices. The final answer must be simplified by combining coefficients over common denominators.

Common Mistakes

A frequent error is expanding only one of the two fractions, or stopping after the $x$ term. Another is forgetting to multiply the binomial expansion by the constant factor pulled out of the denominator, such as $\frac{3}{3a}$ or $-\frac{1}{2a}$ . Sign errors in the second expansion are also common because the denominator is $2a-x$ .

Things to Be Careful About

The mark scheme allows follow-through on the values of $A$ and $B$ from part (a), but the final simplified coefficients must be correct. Keep terms only up to $x^2$ ; higher powers can be ignored. Combine fractions carefully: $\frac{1}{a}-\frac{1}{2a}=\frac{1}{2a}$ , $-\frac{2}{3a^2}-\frac{1}{4a^2}=-\frac{11}{12a^2}$ , and $\frac{4}{9a^3}-\frac{1}{8a^3}=\frac{23}{72a^3}$ .

Techniques used

rewrite each partial fraction in binomial formexpand each binomial series up to x^2collect like powers of x

(c)

State the set of values of $x$ for which the expansion in part (b) is valid.

1M Similar

DifficultyEasy

Worked solution

Approach

The binomial expansion of $(1+u)^n$ is valid only when $|u| < 1$ for the negative indices used here. Apply this condition to each factor in part (b) and take the intersection of the two intervals.

Working

For the first factor, $u = \frac{2x}{3a}$ , so

\left|\frac{2x}{3a}\right| < 1 \implies |x| < \frac{3a}{2}.

For the second factor, $u = -\frac{x}{2a}$ , so

\left|-\frac{x}{2a}\right| < 1 \implies |x| < 2a.

The stricter condition is $|x| < \frac{3a}{2}$ .

Answer

|x| < \frac{3a}{2}

Final answer

|x| < 3a/2

Detailed explanation

Walkthrough

The binomial series for $(1+u)^n$ converges only when $|u|<1$ when $n$ is negative or fractional. In part (b), the two expansions used $u = \frac{2x}{3a}$ and $u = -\frac{x}{2a}$ . Both must satisfy $|u|<1$ for the combined expansion to be valid. This gives $|x|<\frac{3a}{2}$ and $|x|<2a$ . Since $\frac{3a}{2} < 2a$ for $a>0$ , the more restrictive condition is the validity set.

Key Takeaways

Validity of a binomial expansion is not optional: for negative or fractional $n$ , the series is valid only for $|u|<1$ . When a function is expanded as a sum of two binomial series, both series must converge, so the validity interval is the intersection of the individual intervals.

Common Mistakes

A common mistake is to use only one of the two conditions, usually the first, without checking that the second is less restrictive. Another is to write $|x|<2a$ instead of $|x|<\frac{3a}{2}$ because the second factor was forgotten or the comparison was not made.

Things to Be Careful About

Remember that $a$ is positive, so $\frac{3a}{2} < 2a$ . The condition can also be written as $-\frac{3a}{2} < x < \frac{3a}{2}$ . The answer must be a set of values of $x$ , not just a single bound.

Techniques used

apply the binomial convergence conditionintersect validity intervals

Q8MediumTrigonometry

(a)

Prove the identity $\cot^2 \theta - \tan^2 \theta \equiv 4\cot 2\theta \csc 2\theta$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Express the left-hand side in terms of $\sin\theta$ and $\cos\theta$ , combine into a single fraction, factorise the numerator using the difference of two squares, then apply the double angle formulae to obtain the right-hand side.

Working

Starting with the left-hand side:

\cot^2\theta - \tan^2\theta = \frac{\cos^2\theta}{\sin^2\theta} - \frac{\sin^2\theta}{\cos^2\theta}

Combine into a single fraction over the common denominator $\sin^2\theta\cos^2\theta$ :

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

Factorise the numerator as a difference of two squares:

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

Use $\cos^2\theta + \sin^2\theta = 1$ and the double angle formula $\cos^2\theta - \sin^2\theta = \cos 2\theta$ :

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

Since $\sin 2\theta = 2\sin\theta\cos\theta$ , squaring gives $\sin^2 2\theta = 4\sin^2\theta\cos^2\theta$ , so $\sin^2\theta\cos^2\theta = \frac{\sin^2 2\theta}{4}$ :

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

Finally, write this as a product of $\cot 2\theta$ and $\csc 2\theta$ :

\frac{4\cos 2\theta}{\sin^2 2\theta} = 4 \cdot \frac{\cos 2\theta}{\sin 2\theta} \cdot \frac{1}{\sin 2\theta} = 4\cot 2\theta \csc 2\theta

Therefore:

\cot^2\theta - \tan^2\theta \equiv 4\cot 2\theta \csc 2\theta

Answer

\cot^2\theta - \tan^2\theta \equiv 4\cot 2\theta \csc 2\theta

Final answer

cot^2 θ - tan^2 θ ≡ 4cot 2θ csc 2θ

Detailed explanation

Walkthrough

We need to prove that $\cot^2\theta - \tan^2\theta$ equals $4\cot 2\theta \csc 2\theta$ . The strategy is to express everything in terms of $\sin$ and $\cos$ , since these are the fundamental functions from which all others are defined.

Step 1 — Express in terms of sine and cosine. Write $\cot\theta = \frac{\cos\theta}{\sin\theta}$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$ . This gives the difference of two fractions.

Step 2 — Combine into one fraction. Over the common denominator $\sin^2\theta\cos^2\theta$ , the difference becomes $\frac{\cos^4\theta - \sin^4\theta}{\sin^2\theta\cos^2\theta}$ .

Step 3 — Factorise the numerator. The numerator is a difference of two squares: $\cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)$ . This factorisation is the key step that exposes the double angle structure.

Step 4 — Apply identities. Use $\cos^2\theta + \sin^2\theta = 1$ (Pythagorean identity) and $\cos^2\theta - \sin^2\theta = \cos 2\theta$ (double angle formula for cosine).

Step 5 — Relate the denominator to $\sin 2\theta$ . Since $\sin 2\theta = 2\sin\theta\cos\theta$ , squaring gives $\sin^2 2\theta = 4\sin^2\theta\cos^2\theta$ . Therefore $\sin^2\theta\cos^2\theta = \frac{\sin^2 2\theta}{4}$ .

Step 6 — Simplify. Substitute to get $\frac{4\cos 2\theta}{\sin^2 2\theta}$ , then split as $4 \cdot \frac{\cos 2\theta}{\sin 2\theta} \cdot \frac{1}{\sin 2\theta} = 4\cot 2\theta \csc 2\theta$ .

This proves the identity.

Key Takeaways

The fundamental strategy for proving trigonometric identities: express everything in terms of $\sin$ and $\cos$ .
The difference of two squares factorisation $a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$ is a powerful tool in trigonometry proofs.
The double angle formulae convert expressions in $\theta$ to expressions in $2\theta$ .
The relationship $\sin 2\theta = 2\sin\theta\cos\theta$ and its square $\sin^2 2\theta = 4\sin^2\theta\cos^2\theta$ are essential.
Recognising $\frac{\cos 2\theta}{\sin 2\theta} = \cot 2\theta$ and $\frac{1}{\sin 2\theta} = \csc 2\theta$ completes the proof.

Common Mistakes

Forgetting to square the double angle relation: $\sin^2 2\theta = 4\sin^2\theta\cos^2\theta$ , not $2\sin^2\theta\cos^2\theta$ .
Making sign errors when combining the fractions.
Not recognising the difference of two squares pattern in $\cos^4\theta - \sin^4\theta$ .
Confusing $\csc$ (reciprocal of $\sin$ ) with $\sec$ (reciprocal of $\cos$ ).
The mark scheme requires showing all intermediate steps — a correct final answer without working earns no marks.

Things to Be Careful About

The identity $\sin^2 2\theta = 4\sin^2\theta\cos^2\theta$ must be applied correctly — the factor of 4 comes from squaring the factor of 2.
The final step requires recognising both $\frac{\cos 2\theta}{\sin 2\theta} = \cot 2\theta$ and $\frac{1}{\sin 2\theta} = \csc 2\theta$ .
Each step in the proof must be justified; the mark scheme awards method marks for each stage (expressing in $\sin$ / $\cos$ , combining, factorising, applying double angle formulae).

Techniques used

express cotangent and tangent in terms of sine and cosinecombine fractions over a common denominatorfactorise the numerator using difference of two squaresapply double angle formulaerecognise cotangent and cosecant in the final expression

(b)

Hence solve the equation $\cot^2 x - \tan^2 x = 5\sec 2x$ for $0^\circ < x < 90^\circ$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Substitute the identity from part (a) into the equation, express everything in terms of $\sin 2x$ and $\cos 2x$ , solve for $\tan^2 2x$ , then find all values of $x$ in the given range.

Working

Using the identity from part (a) with $\theta = x$ :

4\cot 2x \csc 2x = 5\sec 2x

Express in terms of $\sin$ and $\cos$ :

4 \cdot \frac{\cos 2x}{\sin 2x} \cdot \frac{1}{\sin 2x} = \frac{5}{\cos 2x}

\frac{4\cos 2x}{\sin^2 2x} = \frac{5}{\cos 2x}

Cross-multiply:

4\cos^2 2x = 5\sin^2 2x

Divide by $\cos^2 2x$ (note $\cos 2x \neq 0$ in the given range, since $\cos 2x = 0$ would make $\sec 2x$ undefined):

4 = 5\tan^2 2x

\tan^2 2x = \frac{4}{5}

So:

\tan 2x = \pm \frac{2}{\sqrt{5}}

Since $0^\circ < x < 90^\circ$ , we have $0^\circ < 2x < 180^\circ$ .

For $\tan 2x = +\frac{2}{\sqrt{5}}$ :

2x = \arctan\left(\frac{2}{\sqrt{5}}\right) = 41.81^\circ

x = 20.9^\circ

For $\tan 2x = -\frac{2}{\sqrt{5}}$ :

In the range $0^\circ < 2x < 180^\circ$ , the tangent is negative in the second quadrant:

2x = 180^\circ - 41.81^\circ = 138.19^\circ

x = 69.1^\circ

Both values lie in the range $0^\circ < x < 90^\circ$ .

Answer

x = 20.9^\circ \quad \text{or} \quad x = 69.1^\circ

Final answer

x = 20.9° or x = 69.1°

Detailed explanation

Walkthrough

We use the identity proved in part (a) to simplify the equation.

Step 1 — Substitute the identity. Replace $\cot^2 x - \tan^2 x$ with $4\cot 2x \csc 2x$ , giving:

4\cot 2x \csc 2x = 5\sec 2x

Step 2 — Express in terms of sine and cosine. Write $\cot 2x = \frac{\cos 2x}{\sin 2x}$ , $\csc 2x = \frac{1}{\sin 2x}$ , and $\sec 2x = \frac{1}{\cos 2x}$ .

Step 3 — Cross-multiply. $\frac{4\cos 2x}{\sin^2 2x} = \frac{5}{\cos 2x}$ becomes $4\cos^2 2x = 5\sin^2 2x$ .

Step 4 — Solve for $\tan^2 2x$ . Divide by $\cos^2 2x$ (valid since $\cos 2x = 0$ would make $\sec 2x$ undefined in the original equation): $4 = 5\tan^2 2x$ , so $\tan^2 2x = \frac{4}{5}$ .

Step 5 — Take square roots. $\tan 2x = \pm \frac{2}{\sqrt{5}}$ . The $\pm$ is crucial — we must consider both signs.

Step 6 — Find solutions in range. Since $0^\circ < x < 90^\circ$ , the angle $2x$ ranges over $0^\circ < 2x < 180^\circ$ .

For $\tan 2x = +\frac{2}{\sqrt{5}}$ : the principal value is $2x = 41.81^\circ$ , giving $x = 20.9^\circ$ .
For $\tan 2x = -\frac{2}{\sqrt{5}}$ : tangent is negative in the second quadrant, so $2x = 180^\circ - 41.81^\circ = 138.19^\circ$ , giving $x = 69.1^\circ$ .

Both solutions lie in the required range.

Key Takeaways

A proved identity can be directly substituted to simplify an equation.
Solving $\tan^2 \theta = k$ requires considering both $\tan \theta = +\sqrt{k}$ and $\tan \theta = -\sqrt{k}$ .
When the argument is $2x$ rather than $x$ , the range of the angle doubles: $0^\circ < x < 90^\circ$ becomes $0^\circ < 2x < 180^\circ$ .
The general solution of $\tan \theta = k$ in a given range must account for the periodicity and sign of the tangent function.

Common Mistakes

Forgetting the $\pm$ sign when taking square roots of $\tan^2 2x = \frac{4}{5}$ .
Only finding the first-quadrant solution and missing the second-quadrant one.
Forgetting that the range of $2x$ is double the range of $x$ — this could lead to including extra solutions or missing valid ones.
Dividing by $\cos 2x$ without checking it is not zero. Here it is safe because $\cos 2x = 0$ (i.e. $x = 45^\circ$ ) would make $\sec 2x$ undefined in the original equation.
The mark scheme requires showing the substitution of the identity — an unsupported answer earns no marks.

Things to Be Careful About

The domain $0^\circ < x < 90^\circ$ translates to $0^\circ < 2x < 180^\circ$ for the angle in the equation.
In the second quadrant, $\tan$ is negative, giving the second solution $2x = 138.19^\circ$ .
Both solutions should be verified to satisfy the original equation.
The mark scheme awards marks for: using the identity (M1), obtaining $\tan^2 2x = \frac{4}{5}$ (A1), finding one solution (DM1), and finding the second solution with no extras in range (A1).

Techniques used

substitute the proved identityexpress in terms of sine and cosinesolve for the square of tangent of the double anglefind all solutions within the given range

Q9MediumVectors

With respect to the origin $O$ , the points $A$ , $B$ and $C$ have position vectors given by

\vec{OA} = \mathbf{i} + 2\mathbf{j}, \quad \vec{OB} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} \quad \text{and} \quad \vec{OC} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}

The line $l$ passes through $B$ and $C$ .

(a)

Find a vector equation for $l$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The line $l$ passes through $B$ and $C$ , so a direction vector for $l$ is $\vec{BC} = \vec{OC} - \vec{OB}$ . A vector equation of a line is $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ , where $\mathbf{a}$ is a point on the line and $\mathbf{d}$ is a direction vector.

Working

The direction vector is:

\vec{BC} = \vec{OC} - \vec{OB} = (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) - (\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) = \mathbf{i} - 4\mathbf{j} + 5\mathbf{k}

Using $B$ as the known point:

\mathbf{r} = \vec{OB} + \lambda \vec{BC} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} + \lambda(\mathbf{i} - 4\mathbf{j} + 5\mathbf{k})

Answer

\mathbf{r} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} + \lambda(\mathbf{i} - 4\mathbf{j} + 5\mathbf{k})

Final answer

r = i + 3j - 2k + lambda(i - 4j + 5k)

Detailed explanation

Walkthrough

We need the equation of the line through $B$ and $C$ . A vector equation of a line has the form $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ , where $\mathbf{a}$ is the position vector of a point on the line and $\mathbf{d}$ is a direction vector. Choose $\mathbf{a} = \vec{OB} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k}$ . The direction vector is $\vec{BC} = \vec{OC} - \vec{OB}$ . Subtract the components: the $\mathbf{i}$ -component is $2 - 1 = 1$ , the $\mathbf{j}$ -component is $-1 - 3 = -4$ , and the $\mathbf{k}$ -component is $3 - (-2) = 5$ . Therefore $\vec{BC} = \mathbf{i} - 4\mathbf{j} + 5\mathbf{k}$ . Substituting gives $\mathbf{r} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} + \lambda(\mathbf{i} - 4\mathbf{j} + 5\mathbf{k})$ . As $\lambda$ varies, this traces every point on the line.

Key Takeaways

The vector equation of a line requires one point on the line and one direction vector. The direction vector can be any scalar multiple of $\vec{BC}$ , including its negative, so there are many equivalent correct answers.

Common Mistakes

Writing $l = ...$ instead of $\mathbf{r} = ...$ ; the mark scheme requires a vector equation with $\mathbf{r}$ or component/column vector form.
Using the wrong direction vector, such as $\vec{CB}$ instead of $\vec{BC}$ , without adjusting the parameter; this is still acceptable if the parameter sign is handled consistently.
Arithmetic errors when subtracting components, especially with negative $\mathbf{k}$ -components.

Things to Be Careful About

The direction vector may be scaled by any non-zero constant; both $\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}$ and $-\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}$ are valid.
Use either $B$ or $C$ as the known point; the resulting equations are equivalent.

Techniques used

find direction vector between two pointswrite parametric vector equation of a line

(b)

The point $P$ is the foot of the perpendicular from $A$ to $l$ .

Find the position vector of $P$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Let $P$ be a general point on $l$ with parameter $\lambda$ . Then $\vec{AP} = \vec{OP} - \vec{OA}$ . Since $AP$ is perpendicular to $l$ , $\vec{AP} \cdot \mathbf{d} = 0$ . Solve for $\lambda$ and substitute back.

Working

A general point on $l$ is:

\vec{OP} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} + \lambda(\mathbf{i} - 4\mathbf{j} + 5\mathbf{k})

Hence:

\vec{AP} = \vec{OP} - \vec{OA} = \lambda\mathbf{i} + (1 - 4\lambda)\mathbf{j} + (-2 + 5\lambda)\mathbf{k}

Since $AP$ is perpendicular to $l$ , its scalar product with the direction vector $\mathbf{d} = \mathbf{i} - 4\mathbf{j} + 5\mathbf{k}$ is zero:

\vec{AP} \cdot \mathbf{d} = \lambda - 4(1 - 4\lambda) + 5(-2 + 5\lambda) = 0

Simplify:

\lambda - 4 + 16\lambda - 10 + 25\lambda = 42\lambda - 14 = 0

So:

\lambda = \frac{14}{42} = \frac{1}{3}

Substitute $\lambda = \frac{1}{3}$ :

\vec{OP} = \left(1 + \frac{1}{3}\right)\mathbf{i} + \left(3 - \frac{4}{3}\right)\mathbf{j} + \left(-2 + \frac{5}{3}\right)\mathbf{k}

\vec{OP} = \frac{4}{3}\mathbf{i} + \frac{5}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}

Answer

\vec{OP} = \frac{4}{3}\mathbf{i} + \frac{5}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}

Final answer

OP = (4/3)i + (5/3)j - (1/3)k

Detailed explanation

Walkthrough

Start with a general point $P$ on $l$ : $\vec{OP} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} + \lambda(\mathbf{i} - 4\mathbf{j} + 5\mathbf{k})$ . The vector from $A$ to $P$ is $\vec{AP} = \vec{OP} - \vec{OA}$ . Subtract $\vec{OA} = \mathbf{i} + 2\mathbf{j}$ component by component: $\vec{AP} = \lambda\mathbf{i} + (1 - 4\lambda)\mathbf{j} + (-2 + 5\lambda)\mathbf{k}$ . Because $P$ is the foot of the perpendicular from $A$ , $AP$ is perpendicular to $l$ , so $\vec{AP} \cdot \mathbf{d} = 0$ where $\mathbf{d} = \mathbf{i} - 4\mathbf{j} + 5\mathbf{k}$ . Expand the scalar product: $\lambda - 4(1 - 4\lambda) + 5(-2 + 5\lambda) = 0$ . This simplifies to $42\lambda - 14 = 0$ , so $\lambda = \frac{1}{3}$ . Substitute this value back into $\vec{OP}$ to get $\frac{4}{3}\mathbf{i} + \frac{5}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}$ .

Key Takeaways

The perpendicularity condition between a line and a segment is expressed by setting the scalar product of the segment vector and the line's direction vector equal to zero. Parameterising the line first lets us write the unknown foot as a single variable vector.

Common Mistakes

Using $\vec{OP}$ instead of $\vec{AP}$ in the scalar product; the mark scheme specifically requires $\vec{AP}$ .
Forgetting to set the scalar product equal to zero.
Sign errors when expanding $(-2 + 5\lambda) \cdot 5$ .
Not substituting the found $\lambda$ back into $\vec{OP}$ .

Things to Be Careful About

The direction vector may be a scalar multiple of $\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}$ ; if a different direction vector is used, the value of $\lambda$ changes but the final point $P$ is the same.
If the line is written using $C$ with parameter $\mu$ , the corresponding value is $\mu = -\frac{2}{3}$ .
Keep fractions exact; do not round.

Techniques used

parameterise a general point on a lineform vector from point to general pointapply perpendicularity condition using scalar productsolve for parameter and substitute back

(c)

The point $D$ is the reflection of $A$ in $l$ .

Find the position vector of $D$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $P$ is the foot of the perpendicular from $A$ to $l$ , $P$ is the midpoint of $AD$ when $D$ is the reflection of $A$ in $l$ . Therefore $\vec{OD} = \vec{OA} + 2\vec{AP}$ .

Working

From part (b):

\vec{AP} = \vec{OP} - \vec{OA} = \frac{1}{3}\mathbf{i} - \frac{1}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}

Then:

\vec{OD} = \vec{OA} + 2\vec{AP} = (\mathbf{i} + 2\mathbf{j}) + 2\left(\frac{1}{3}\mathbf{i} - \frac{1}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}\right)

\vec{OD} = \left(1 + \frac{2}{3}\right)\mathbf{i} + \left(2 - \frac{2}{3}\right)\mathbf{j} + \left(0 - \frac{2}{3}\right)\mathbf{k}

\vec{OD} = \frac{5}{3}\mathbf{i} + \frac{4}{3}\mathbf{j} - \frac{2}{3}\mathbf{k}

Answer

\vec{OD} = \frac{5}{3}\mathbf{i} + \frac{4}{3}\mathbf{j} - \frac{2}{3}\mathbf{k}

Final answer

OD = (5/3)i + (4/3)j - (2/3)k

Detailed explanation

Walkthrough

When $D$ is the reflection of $A$ in the line $l$ , the foot $P$ is the midpoint of $AD$ . Therefore $\vec{OD} = \vec{OA} + 2\vec{AP}$ , or equivalently $\vec{OD} = 2\vec{OP} - \vec{OA}$ . From part (b), $\vec{AP} = \frac{1}{3}\mathbf{i} - \frac{1}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}$ . Then $2\vec{AP} = \frac{2}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} - \frac{2}{3}\mathbf{k}$ . Adding $\vec{OA} = \mathbf{i} + 2\mathbf{j}$ gives $\vec{OD} = \frac{5}{3}\mathbf{i} + \frac{4}{3}\mathbf{j} - \frac{2}{3}\mathbf{k}$ .

Key Takeaways

Reflection in a line uses the perpendicular foot as the midpoint of the original point and its image. This gives a simple vector addition formula: image = original + 2 times the perpendicular vector.

Common Mistakes

Using $\vec{OD} = \vec{OA} + \vec{AP}$ instead of $\vec{OA} + 2\vec{AP}$ .
Using $\vec{AP} = \vec{OA} - \vec{OP}$ or another sign-reversed vector.
A single component slip may still earn the method mark, but the final answer must be fully correct for the accuracy mark.

Things to Be Careful About

$\vec{AP}$ means $P - A$ , not $A - P$ .
Check component arithmetic: the $\mathbf{k}$ -component of $\vec{OA}$ is $0$ , so the reflected $\mathbf{k}$ -component is $-\frac{2}{3}$ .
The formula $\vec{OD} = 2\vec{OP} - \vec{OA}$ is equivalent and often quicker.

Techniques used

use midpoint property of reflectionscale and add vectors to find reflected point

Q10Medium-HardDifferential EquationsIntegrationTrigonometry

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

\sin 4y \frac{dy}{dx} = x \sin 2y \sin 3x

It is given that $y = \frac{1}{12}\pi$ when $x = \frac{1}{2}\pi$ .

(a)

Solve the differential equation, obtaining a relation between $x$ and $y$ .

8M Similar

DifficultyMedium-Hard

Worked solution

Approach

Separate the variables so all $y$ -terms sit with $dy$ and all $x$ -terms with $dx$ . Simplify the $y$ -integral using the double angle formula $\sin 4y = 2\sin 2y\cos 2y$ , then evaluate the $x$ -integral by parts. Finally substitute the given initial condition to determine the constant of integration.

Working

Separate the variables:

\frac{\sin 4y}{\sin 2y}\,dy = x\sin 3x\,dx

Apply $\sin 4y = 2\sin 2y\cos 2y$ :

\frac{\sin 4y}{\sin 2y} = \frac{2\sin 2y\cos 2y}{\sin 2y} = 2\cos 2y

Hence:

\int 2\cos 2y\,dy = \int x\sin 3x\,dx

Left-hand side:

\int 2\cos 2y\,dy = \sin 2y

Right-hand side — integrate by parts with $u = x$ , $dv = \sin 3x\,dx$ :

\int x\sin 3x\,dx = -\frac{1}{3}x\cos 3x + \frac{1}{3}\int \cos 3x\,dx

= -\frac{1}{3}x\cos 3x + \frac{1}{9}\sin 3x

Therefore:

\sin 2y = -\frac{1}{3}x\cos 3x + \frac{1}{9}\sin 3x + C

Use $y = \frac{1}{12}\pi$ when $x = \frac{1}{2}\pi$ :

\sin 2y = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \qquad \cos 3x = \cos\left(\frac{3\pi}{2}\right) = 0, \qquad \sin 3x = \sin\left(\frac{3\pi}{2}\right) = -1

\frac{1}{2} = -\frac{1}{3}(0) + \frac{1}{9}(-1) + C

\frac{1}{2} = -\frac{1}{9} + C

C = \frac{1}{2} + \frac{1}{9} = \frac{11}{18}

Answer

\sin 2y = -\frac{1}{3}x\cos 3x + \frac{1}{9}\sin 3x + \frac{11}{18}

Final answer

sin 2y = -(1/3)x cos 3x + (1/9) sin 3x + 11/18

Detailed explanation

Walkthrough

This is a first-order separable differential equation. The first step is to rearrange it so that every $y$ -term is paired with $dy$ and every $x$ -term with $dx$ . Dividing both sides by $\sin 2y$ and multiplying by $dx$ achieves this, giving $\frac{\sin 4y}{\sin 2y}\,dy = x\sin 3x\,dx$ .

Before integrating the left-hand side, notice that $\frac{\sin 4y}{\sin 2y}$ is not immediately integrable. The double angle formula $\sin 4y = 2\sin 2y\cos 2y$ rewrites the numerator so that $\sin 2y$ cancels, leaving $2\cos 2y$ , which integrates directly to $\sin 2y$ .

The right-hand side, $\int x\sin 3x\,dx$ , is a product of a polynomial and a trigonometric function, so integration by parts is the appropriate tool. Choosing $u = x$ and $dv = \sin 3x\,dx$ gives $du = dx$ and $v = -\frac{1}{3}\cos 3x$ . Applying $\int u\,dv = uv - \int v\,du$ yields $-\frac{1}{3}x\cos 3x + \frac{1}{3}\int\cos 3x\,dx$ , and $\int\cos 3x\,dx = \frac{1}{3}\sin 3x$ completes the integration.

Equating the two sides gives $\sin 2y = -\frac{1}{3}x\cos 3x + \frac{1}{9}\sin 3x + C$ . The final step is to use the initial condition $y = \frac{\pi}{12}$ at $x = \frac{\pi}{2}$ to find $C$ . Substituting these values, $\sin 2y = \sin(\pi/6) = \frac{1}{2}$ , $\cos 3x = \cos(3\pi/2) = 0$ , and $\sin 3x = \sin(3\pi/2) = -1$ . This gives $\frac{1}{2} = -\frac{1}{9} + C$ , so $C = \frac{11}{18}$ .

Key Takeaways

Separable differential equations are solved by collecting all $y$ -terms with $dy$ and all $x$ -terms with $dx$ before integrating.
Double angle formulas often simplify ratios of trigonometric functions into directly integrable forms.
Integration by parts is the standard method for products of a polynomial and a trigonometric function.
Initial conditions determine the constant of integration, producing the particular solution.

Common Mistakes

Forgetting to simplify $\frac{\sin 4y}{\sin 2y}$ with the double angle formula before integrating.
Sign errors in integration by parts, particularly the negative sign from $v = -\frac{1}{3}\cos 3x$ .
Omitting the constant of integration.
Substituting the initial condition into the wrong side of the equation or miscomputing $\sin(\pi/6)$ , $\cos(3\pi/2)$ , or $\sin(3\pi/2)$ .

Things to Be Careful About

The double angle formula $\sin 4y = 2\sin 2y\cos 2y$ — the factor of 2 must not be lost.
When integrating by parts, the factor $\frac{1}{3}$ from $\int\cos 3x\,dx$ multiplies the existing $\frac{1}{3}$ , giving $\frac{1}{9}\sin 3x$ .
$\cos(3\pi/2) = 0$ and $\sin(3\pi/2) = -1$ — these exact values are needed for the constant calculation.
The final answer must be a relation between $x$ and $y$ with the constant evaluated.

Techniques used

separate the variables in a differential equationapply the double angle formula to simplify the integrandintegrate by partssubstitute initial conditions to find the constant of integration

(b)

Given that $0 < y < \frac{1}{2}\pi$ , find the values of $y$ when $x = 0$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Substitute $x = 0$ into the solution from part (a) to obtain a trigonometric equation in $y$ . Solve $\sin 2y = \frac{11}{18}$ over the range $0 < 2y < \pi$ , which follows from $0 < y < \frac{1}{2}\pi$ .

Working

With $x = 0$ :

\sin 2y = -\frac{1}{3}(0)\cos 0 + \frac{1}{9}\sin 0 + \frac{11}{18} = \frac{11}{18}

Since $0 < y < \frac{1}{2}\pi$ , we have $0 < 2y < \pi$ . The two solutions in this range are:

2y = \arcsin\left(\frac{11}{18}\right) \approx 0.658

y \approx 0.329

and

2y = \pi - \arcsin\left(\frac{11}{18}\right) \approx \pi - 0.658 \approx 2.484

y \approx 1.242 \approx 1.24

Answer

y \approx 0.329 \quad \text{or} \quad y \approx 1.24

Final answer

y ≈ 0.329 or y ≈ 1.24

Detailed explanation

Walkthrough

Part (b) uses the solution from part (a). Setting $x = 0$ makes the terms $-\frac{1}{3}x\cos 3x$ and $\frac{1}{9}\sin 3x$ both vanish (the first because of the factor $x$ , the second because $\sin 0 = 0$ ), leaving $\sin 2y = \frac{11}{18}$ .

The range $0 < y < \frac{1}{2}\pi$ translates to $0 < 2y < \pi$ . Within this range, the equation $\sin \theta = k$ (with $0 < k < 1$ ) has two solutions: $\theta = \arcsin k$ and $\theta = \pi - \arcsin k$ . Here $k = \frac{11}{18} \approx 0.6111$ , so $\arcsin(11/18) \approx 0.658$ . The two values of $2y$ are therefore approximately $0.658$ and $2.484$ . Dividing by 2 gives $y \approx 0.329$ and $y \approx 1.24$ , both within $0 < y < \frac{\pi}{2}$ .

Key Takeaways

Substituting into a particular solution lets you solve for a specific variable value.
The equation $\sin\theta = k$ has two solutions in $(0, \pi)$ : $\theta = \arcsin k$ and $\theta = \pi - \arcsin k$ .
Always divide by the coefficient of $y$ (here 2) after solving for $2y$ .

Common Mistakes

Finding only one of the two solutions.
Giving answers in degrees — the mark scheme explicitly disallows this.
Forgetting to halve the solutions of $2y$ to obtain $y$ .

Things to Be Careful About

The range $0 < y < \frac{\pi}{2}$ means $0 < 2y < \pi$ , so both solutions of $\sin 2y = \frac{11}{18}$ must lie in this interval.
Use radians throughout.
The values $0.329$ and $1.24$ are approximate (AWRT — allow within reason).

Techniques used

substitute a value into the particular solutionsolve a trigonometric equation over a restricted range

Q11MediumDifferentiationNumerical Solution of Equations

Fig. 1

The diagram shows the curve $y = \sqrt{x} \sin 2x$ for $0 \leq x \leq \frac{1}{2}\pi$ . The curve has a maximum point at $M$ , where $x = a$ .

(a)

Show that $\tan 2a = -4a$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the product rule to differentiate $y = \sqrt{x} \sin 2x$ . Since $M$ is a maximum point, set $\frac{dy}{dx} = 0$ at $x = a$ and rearrange the resulting equation to obtain $\tan 2a = -4a$ .

Working

Given:

y = \sqrt{x} \sin 2x = x^{1/2} \sin 2x

Differentiate using the product rule:

\frac{dy}{dx} = \frac{1}{2}x^{-1/2} \sin 2x + \sqrt{x} \cdot 2 \cos 2x

\frac{dy}{dx} = \frac{\sin 2x}{2\sqrt{x}} + 2\sqrt{x} \cos 2x

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

\frac{\sin 2a}{2\sqrt{a}} + 2\sqrt{a} \cos 2a = 0

Multiply the entire equation by $2\sqrt{a}$ to eliminate the surd in the denominator:

\sin 2a + 4a \cos 2a = 0

Rearrange to isolate $\sin 2a$ :

\sin 2a = -4a \cos 2a

Divide both sides by $\cos 2a$ :

\frac{\sin 2a}{\cos 2a} = -4a

\tan 2a = -4a

Answer

Shown that $\tan 2a = -4a$ .

Final answer

Shown

Detailed explanation

Walkthrough

First, we differentiate the given function $y = \sqrt{x} \sin 2x$ using the product rule. The derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$ and the derivative of $\sin 2x$ is $2\cos 2x$ . Setting the derivative equal to zero at the maximum point $x = a$ gives an equation involving $\sin 2a$ and $\cos 2a$ . By multiplying through by $2\sqrt{a}$ , we clear the fraction and surd, leaving a simple trigonometric equation. Dividing by $\cos 2a$ converts the sine and cosine into tangent, yielding the required result $\tan 2a = -4a$ .

Key Takeaways

The product rule is essential for differentiating products of functions.
At a maximum or minimum point, the first derivative is zero.
Basic trigonometric identities like $\tan \theta = \frac{\sin \theta}{\cos \theta}$ are useful for rearranging equations.

Common Mistakes

Forgetting the chain rule when differentiating $\sin 2x$ (must multiply by 2).
Failing to multiply the entire equation by $2\sqrt{a}$ , leaving a fraction that is harder to rearrange.
Algebraic errors when isolating $\tan 2a$ .

Things to Be Careful About

Ensure all trigonometric functions are in radians when using a calculator in later parts.
The mark scheme awards a method mark for obtaining the form $\frac{p}{\sqrt{x}}\sin 2x + q\sqrt{x}\cos 2x$ , so show the unsimplified derivative first.

Techniques used

apply product ruledifferentiate trigonometric and power functionsset derivative to zero for maximumrearrange to find tan 2a

(b)

Show by calculation that $0.9 < a < 0.95$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Rearrange $\tan 2a = -4a$ into the form $f(x) = 0$ . Evaluate $f(x)$ at $x = 0.9$ and $x = 0.95$ . If there is a sign change, then by the intermediate value theorem, a root exists in the interval $(0.9, 0.95)$ .

Working

Let $f(x) = \tan 2x + 4x$ . We need to show $f(0.9) < 0$ and $f(0.95) > 0$ .

Calculate $f(0.9)$ (in radians):

f(0.9) = \tan(1.8) + 4(0.9)

f(0.9) \approx -4.28635 + 3.6 = -0.68635

Since $-0.68635 < 0$ , $f(0.9) < 0$ .

Calculate $f(0.95)$ (in radians):

f(0.95) = \tan(1.9) + 4(0.95)

f(0.95) \approx -2.92604 + 3.8 = 0.87396

Since $0.87396 > 0$ , $f(0.95) > 0$ .

Since $f(0.9) < 0$ and $f(0.95) > 0$ , there is a sign change in the interval $(0.9, 0.95)$ . Therefore, $0.9 < a < 0.95$ .

Answer

Shown that $0.9 < a < 0.95$ .

Final answer

Shown

Detailed explanation

Walkthrough

We start with the equation $\tan 2a = -4a$ and define a function $f(x) = \tan 2x + 4x$ . To show that the root $a$ lies between 0.9 and 0.95, we evaluate $f(x)$ at these two endpoints. It is crucial to use radians on the calculator. At $x = 0.9$ , $f(0.9) \approx -0.686$ , which is negative. At $x = 0.95$ , $f(0.95) \approx 0.874$ , which is positive. Because the function changes sign between these two values and is continuous in this interval, there must be a root between them.

Key Takeaways

Locating roots by sign change is a fundamental numerical method.
Always ensure your calculator is in the correct mode (radians for calculus and trigonometry involving $x$ ).
The function must be continuous in the interval for the sign change to guarantee a root.

Common Mistakes

Using degrees instead of radians, which will give completely wrong values.
Failing to state the values of $f(0.9)$ and $f(0.95)$ clearly.
Not explicitly mentioning the sign change.

Things to Be Careful About

The mark scheme allows a smaller interval provided it contains the root, but using the given bounds is safest.
Show at least 3 decimal places in your calculated values to justify the sign.

Techniques used

define function from equationevaluate at interval endpointscheck for sign change

(c)

Show that if a sequence of values given by the iterative formula

x_{n+1} = \frac{1}{2} \left( \pi - \tan^{-1}(4x_n) \right)

converges, then it converges to $a$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Assume the sequence converges to a limit $L$ . Then as $n \to \infty$ , both $x_n$ and $x_{n+1}$ approach $L$ . Substitute $x = L$ into the iterative formula and rearrange to show it satisfies the equation $\tan 2x = -4x$ , which is satisfied by $a$ .

Working

Assume the sequence converges to a limit $x$ . Then:

x = \frac{1}{2} \left( \pi - \tan^{-1}(4x) \right)

Multiply both sides by 2:

2x = \pi - \tan^{-1}(4x)

Rearrange to isolate the inverse tangent:

\pi - 2x = \tan^{-1}(4x)

Take the tangent of both sides:

\tan(\pi - 2x) = 4x

Use the compound angle identity $\tan(\pi - \theta) = -\tan \theta$ :

-\tan 2x = 4x

\tan 2x = -4x

This is exactly the equation satisfied by $a$ (from part (a)). Therefore, if the sequence converges, it converges to $a$ .

Answer

Shown that the sequence converges to $a$ .

Final answer

Shown

Detailed explanation

Walkthrough

When an iterative formula $x_{n+1} = F(x_n)$ converges, the limit $L$ satisfies $L = F(L)$ . We substitute $x$ for the limit in the given formula and perform algebraic manipulations to recover the original equation $\tan 2x = -4x$ . The key step is taking the tangent of both sides of $\pi - 2x = \tan^{-1}(4x)$ and using the identity $\tan(\pi - \theta) = -\tan \theta$ to simplify $\tan(\pi - 2x)$ to $-\tan 2x$ .

Key Takeaways

If an iterative sequence converges, the limit is a fixed point of the iteration function.
Inverse trigonometric functions can be removed by applying the corresponding trigonometric function to both sides.
Compound angle identities are useful for simplifying expressions involving $\pi - \theta$ .

Common Mistakes

Forgetting to use the correct identity for $\tan(\pi - \theta)$ .
Algebraic errors when rearranging the iterative formula.
Not explicitly stating that the limit satisfies the original equation from part (a).

Things to Be Careful About

The question says "if a sequence ... converges", so you do not need to prove convergence, only that the limit must be $a$ .
Work in radians throughout.

Techniques used

assume convergence to a limitsubstitute limit into iterative formularearrange to original equation

(d)

Use the iterative formula in part (c) to calculate $a$ correct to 4 decimal places. Give the result of each iteration to 6 decimal places.

3M Similar

DifficultyMedium

Worked solution

Approach

Choose an initial value $x_1$ in the interval $(0.9, 0.95)$ , for example $x_1 = 0.9$ . Apply the iterative formula $x_{n+1} = \frac{1}{2}(\pi - \tan^{-1}(4x_n))$ repeatedly, calculating each result to 6 decimal places. Continue until the value is constant to 4 decimal places, and verify by checking for a sign change in the interval $(0.91825, 0.91835)$ .

Working

Let $x_1 = 0.9$ .

Iteration 1:

x_2 = \frac{1}{2} \left( \pi - \tan^{-1}(4 \times 0.9) \right) = \frac{1}{2} \left( \pi - \tan^{-1}(3.6) \right)

x_2 \approx 0.920872

Iteration 2:

x_3 = \frac{1}{2} \left( \pi - \tan^{-1}(4 \times 0.920872) \right) = \frac{1}{2} \left( \pi - \tan^{-1}(3.683488) \right)

x_3 \approx 0.917944

Iteration 3:

x_4 = \frac{1}{2} \left( \pi - \tan^{-1}(4 \times 0.917944) \right) = \frac{1}{2} \left( \pi - \tan^{-1}(3.671776) \right)

x_4 \approx 0.918347

Iteration 4:

x_5 = \frac{1}{2} \left( \pi - \tan^{-1}(4 \times 0.918347) \right) = \frac{1}{2} \left( \pi - \tan^{-1}(3.673388) \right)

x_5 \approx 0.918292

Iteration 5:

x_6 = \frac{1}{2} \left( \pi - \tan^{-1}(4 \times 0.918292) \right) = \frac{1}{2} \left( \pi - \tan^{-1}(3.673168) \right)

x_6 \approx 0.918304

Iteration 6:

x_7 = \frac{1}{2} \left( \pi - \tan^{-1}(4 \times 0.918304) \right) \approx 0.918300

The values are stabilising around $0.9183$ . To justify $a = 0.9183$ to 4 decimal places, we check for a sign change in $f(x) = \tan 2x + 4x$ between $x = 0.91825$ and $x = 0.91835$ :

f(0.91825) = \tan(1.8365) + 3.673 \approx -0.0007 < 0

f(0.91835) = \tan(1.8367) + 3.6734 \approx 0.0012 > 0

Since there is a sign change in $(0.91825, 0.91835)$ , $a = 0.9183$ correct to 4 decimal places.

Answer

a = 0.9183

Final answer

0.9183

Detailed explanation

Walkthrough

We start with an initial guess $x_1 = 0.9$ and repeatedly apply the iterative formula. Each new value is calculated using the previous value, and we keep 6 decimal places to avoid rounding errors accumulating. The sequence oscillates and converges towards the root. To confirm that $a = 0.9183$ to 4 decimal places, we don't just rely on the iterations stabilising; we check for a sign change in the original equation $f(x) = \tan 2x + 4x$ across the interval $(0.91825, 0.91835)$ . Since $f(0.91825) < 0$ and $f(0.91835) > 0$ , the root must lie in this interval, confirming the answer to 4 d.p.

Key Takeaways

Iterative formulas can be used to find roots of equations to any desired accuracy.
Always carry extra decimal places during intermediate calculations to prevent rounding errors.
Justification of the final answer to a given number of decimal places requires checking for a sign change in the appropriate interval.

Common Mistakes

Working in degrees instead of radians, which will give completely wrong iterations.
Rounding intermediate results to too few decimal places, causing the final answer to be incorrect.
Failing to justify the final answer to 4 d.p. by checking the interval or showing sufficient iterations.

Things to Be Careful About

The mark scheme requires showing sufficient iterations to at least 6 d.p. to justify the answer. Show at least 4-5 iterations.
The final answer must be $0.9183$ . If you get a different value, check your calculator mode and rounding.

Techniques used

choose initial value from intervalapply iterative formula repeatedlycheck convergence to required accuracy

Mathematics 9709/33 — May/June 2025

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