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Question 1

Topic – ALV: 2.2

Solve the equation

$3e^{2x} – 4e^{-2x} = 5.$

Give the answer correct to 3 decimal places.

▶️Answer/Explanation

Solution :-

1. $3(e^{2x})^2 – 5(e^{2x}) – 4 = 0$

$e^{2x} = \frac{5 \pm \sqrt{73}}{6}, \quad x = \frac{1}{2} \ln \left( \frac{5 + \sqrt{73}}{6} \right)$

$x = 0.407$

Question 2

Topic – ALV: 1.2

(a) Sketch the graph of $y = |2x + 3|$.

(b) Solve the inequality $3x + 8 > |2x + 3|$.

▶️Answer/Explanation

Solution :-

(a)

2(b) Find x-coordinate of intersection with $y=3x+8$

Obtain $x = -11/5$

State final answer $x > -11/5$ only

Alternative Method 1

Solve the linear inequality $3x+8 > -(2x+3)$, or corresponding linear
equation

Obtain critical value $x = -11/5$

State final answer $x > -11/5$ only

Alternative Method 2

Solve the quadratic inequality $(3x+8)^{2} > (2x+3)^{2}$, or corresponding
quadratic equation

Obtain critical value $x = -11/5$

State final answer $x > -11/5$ only

Question 3

Topic – ALV: 1.6

Find the coefficient of $x^3$ in the binomial expansion of $(3+x)\sqrt{1+4x}.$

▶️Answer/Explanation

Solution :-

3. State unsimplified term in $x^3$, or its coefficient, in the expansion of $(1+4x)^{\frac{1}{2}}$

State unsimplified term in $x^2$, or its coefficient, in the expansion of $(1+4x)^{\frac{1}{2}}$

Multiply by $(3+x)$ and combine terms in $x^3$, or their coefficients

Obtain answer 10

Question 4

Topic – ALV: 1.5

(a) Show that the equation $\sin 2\theta + \cos 2\theta = 2 \sin^2 \theta$ can be expressed in the form

\[\cos^2 \theta + 2 \sin \theta \cos \theta – 3 \sin^2 \theta = 0.\]

(b) Hence solve the equation $\sin 2\theta + \cos 2\theta = 2 \sin^2 \theta$ for $0^{\circ} < \theta < 180^{\circ}$.

▶️Answer/Explanation

Solution :-

(a) Use correct double angle formulae

Obtain \(cos^{2}\theta+2~sin~\theta~cos~\theta-3~sin^{2}\theta=0\) from full and correct working

4(b) Factorise to obtain $(\cos \theta – \sin \theta)(\cos \theta + 3 \sin \theta) = 0$

Solve a quadratic in $\sin \theta$ and $\cos \theta$ to obtain a value for $\theta$.

Obtain one correct value e.g. $45^{\circ}$

Obtain a second correct value e.g. $161.6^{\circ}$ and no others in the interval

Alternative Method 1

Obtain $3 \tan^2 \theta – 2 \tan \theta – 1 = 0$

Solve a 3 term quadratic in $\tan \theta$ to obtain a value for $\theta$.

Obtain one correct value e.g. $45^{\circ}$

Obtain a second correct value e.g. $161.6^{\circ}$ and no others in the interval

Alternative Method 2

Obtain $(\cos \theta + \sin \theta)^2 = (2 \sin \theta)^2$

Solve to obtain a value for $\theta$.

Obtain one correct value e.g. $45^{\circ}$

Obtain a second correct value e.g. $161.6^{\circ}$ and no others in the interval

Question 5

Topic – ALV: 3.4

The equation of a curve is $x^{2}y-ay^{2}=4a^{3}$, where $a$ is a non-zero constant.

(a) Show that $\frac{dy}{dx}=\frac{2xy}{2ay-x^{2}}.$

(b) Hence find the coordinates of the points where the tangent to the curve is parallel to the y-axis.

▶️Answer/Explanation

Solution :-

(a) State or imply $2xy+x^2\frac{dy}{dx}$ as derivative of $x^2y$

State or imply $2ay\frac{dy}{dx}$ as derivative of $ay^2$

Equate attempted derivative to zero and solve for $\frac{dy}{dx}$

Obtain answer $\frac{dy}{dx}=\frac{2xy}{2ay-x^2}$ from correct working

(b) State or imply $2ay-x^2=0$

Substitute into equation of curve to obtain equation in x and a or in y and a

Obtain one correct point

Obtain second correct point and no others

Question 6

Topic – ALV: 3.7

Relative to the origin O, the points A, B and C have position vectors given by

\[\overrightarrow{OA} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix}, \quad \text{and} \quad \overrightarrow{OC} = \begin{pmatrix} 3 \\ -2 \\ -4 \end{pmatrix}.\]

The quadrilateral ABCD is a parallelogram.

(a) Find the position vector of D.

(b) The angle between BA and BC is $\theta$.

Find the exact value of $\cos \theta$. 

(c) Hence find the area of ABCD, giving your answer in the form $p\sqrt{q}$, where $p$ and $q$ are integers.

▶️Answer/Explanation

Solution :-

(a) Obtain a vector for one side of the parallelogram

Correct method to obtain ±OD

Obtain $\overrightarrow{OD} = \mathbf{i} – 4\mathbf{j} – 3\mathbf{k}$

(b) Using the correct process, evaluate the scalar product $\overrightarrow{BA}.\overrightarrow{BC}$

Using the correct process for the moduli, divide the scalar product by the
product of the moduli.

Obtain answer $\frac{2}{\sqrt{62}}$

(c) State or imply $\sin \theta = \sqrt{\frac{58}{62}}$

Use correct method to find the area of ABCD

Correct unsimplified expression for the area

Obtain answer $3\sqrt{58}$

Question 7

Topic – ALV: 3.8

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

$$ \cos 2x \frac{dy}{dx} = \frac{4 \tan 2x}{\sin^2 3y}, $$

where $0 \leq x < \frac{1}{4}\pi$. It is given that $y = 0$ when $x = \frac{1}{6}\pi$.

Solve the differential equation to obtain the value of $x$ when $y = \frac{1}{6}\pi$. Give your answer correct to 3 decimal places.

▶️Answer/Explanation

Solution :-

Correct separation of variables

Integrate to obtain k sec 2x

Obtain 2 sec 2x

Use double angle formula and integrate to obtain py + q~sin~6y

Obtain $\frac{1}{2}y – \frac{1}{12}sin~6y$

Use y=0, x{pi}{6} in a solution containing terms λsec 2x and μsin 6y to find
the constant of integration

Obtain $\frac{1}{2}y – \frac{1}{12}sin~6y = 2~sec~2x – 4$

Obtain x = 0.541

Question 8

Topic – ALV: 2.5

Let $f(x) = \frac{3 – 3x^2}{(2x + 1)(x + 2)^2}$

(a) Express $f(x)$ in partial fractions.

(b) Hence find the exact value of $\int_{0}^{4} f(x) dx$, giving your answer in the form $a + b \ln c$, where $a$, $b$ and $c$ are integers.

▶️Answer/Explanation

Solution :-

(a) State or imply the form
\[ \frac{A}{2x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^{2}} \]

Use a correct method for finding a constant

Obtain one of \(A=1, B=-2, C=3\)

Obtain a second value

Obtain the third value

(b) Integrate and obtain one of
\[ \frac{1}{2}\ln(2x+1), -2\ln(x+2), \frac{-3}{x+2} \]

Obtain a second term

Obtain the third term

Substitute limits correctly in an integral with at least two terms of the form
\[ \frac{1}{2}\ln(2x+1), -2\ln(x+2) \text{ and } \frac{-3}{x+2} \]
and subtract in correct order

Obtain \(1-\ln~3\)

Question 9

Topic – ALV: 1.8

 The constant a is such that
\[ \int_{0}^{a} xe^{-2x} dx = \frac{1}{8}. \]

(a) Show that \(a = \frac{1}{2}\ln(4a + 2)\).

(b) Verify by calculation that a lies between 0.5 and 1.

(c) Use an iterative formula based on the equation in (a) to determine a correct to 2 decimal places.
Give the result of each iteration to 4 decimal places.

▶️Answer/Explanation

Solution :-

(a) Commence integration and reach \(pxe^{-2x} + q\int e^{-2x} dx\)

Obtain \(-\frac{1}{2}xe^{-2x} + \frac{1}{2}\int e^{-2x} dx\)

Complete integration and obtain \(-\frac{1}{2}xe^{-2x} – \frac{1}{4}e^{-2x}\)

Use limits correctly and equate to \(\frac{1}{8}\) having integrated twice

Obtain \(a = \frac{1}{2}\ln(4a+2)\) correctly

(b) Calculate the values of a relevant expression or pair of expressions at \(a = 0.5\)
and \(a = 1\)

Justify the given statement with correct calculated values

(c) Use the iterative process \(a_{n+1} = \frac{1}{2}\ln(4a_{n}+2)\) correctly at least once.

Obtain final answer 0.84

Show sufficient iterations to at least 4 d.p. to justify 0.84 to 2 d.p. or show
that there is a sign change in (0.835, 0.845)

Question 10

Topic – ALV: 3.9

The polynomial \(x^{3}+5x^{2}+31x+75\) is denoted by \(p(x)\).

(a) Show that \((x+3)\) is a factor of \(p(x)\).

(b) Show that \(z=-1+2\sqrt{6}i\) is a root of \(p(z)=0\).

(c) Hence find the complex numbers z which are roots of \(p(z^{2})=0.\)

▶️Answer/Explanation

Solution :-

10(a) Substitute $x = -3$ to obtain value of $p(-3)$

Obtain $p(-3) = 0$ and hence given result

Alternative method for Question 10(a)

Divide $p(x)$ by $(x + 3)$ to obtain quotient $x^2 \pm 2x + …$

Obtain quotient $x^2 + 2x + 25$, with zero remainder and hence given result

(b) Substitute $z = -1 + 2\sqrt{6}i$ and attempt expansions of $z^2$ and $z^3$

Use $i^{2} = -1$

Obtain $p(z) = 0$ and hence given result

Alternative Method 1

Use roots $z = -1 + 2\sqrt{6}i$ to form quadratic factor

Divide $p(z)$ by their quadratic factor

Obtain zero remainder and hence given result.

Alternative Method 2

Set their quadratic factor from (a) equal to zero

Solve for z

Obtain $z = -1 + 2\sqrt{6}i$ (and $z = -1 – 2\sqrt{6}i$)

Alternative Method 3

Substitute $z = -1 + 2\sqrt{6}i$ into their quadratic factor and attempt expansion of $z^2$

Use $i^{2} = -1$

Obtain 0 and hence given result

(c) State $z_{1}=\sqrt{3}i$ and $z_{2}=-\sqrt{3}i$

Expand $(x+iy)^{2}=-1+2\sqrt{6}i$ and compare real and imaginary parts

Obtain $x^{2}-y^{2}=-1$ and $xy=\sqrt{6}$

Solve to obtain x and y

Obtain $z_{3}=\sqrt{2}+\sqrt{3}i$ and $z_{4}=-\sqrt{2}-\sqrt{3}i$

Use $z^{2}=-1-2\sqrt{6i}$ to obtain $z_{5}$ and $z_{6}$

Obtain $z_{5}=\sqrt{2}-\sqrt{3}i$ and $z_{6}=-\sqrt{2}+\sqrt{3}i$

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