Question 1
It is given that $x = \ln(2y-3) – \ln(y+4)$.
Express $y$ in terms of $x$.
▶️Answer/Explanation
Solution: –
Use law of the logarithm of a quotient or express x as In e
Remove logarithms and obtain a correct equation e.g. \(e^{x}=\frac{2y-3}{y+4}\)
Obtain answer \(y=\frac{3+4e^{x}}{2-e^{x}}\)
Question 2
(a) On an Argand diagram, shade the region whose points represent complex numbers $z$ satisfying the inequalities $-\frac{1}{3}\pi \leq arg(z-1-2i) \leq \frac{1}{3}\pi$ and $Re~z \leq 3$.
(b) Calculate the least value of $arg~z$ for points in the region from (a). Give your answer in radians correct to 3 decimal places.
▶️Answer/Explanation
Solution: –
$2(a)$ Show correct half-lines from \(l+2i,\) symmetrical about \(y=2i\)
Drawn between $\frac{\pi}{4}$ and $\frac{5\pi}{12}$.
Show the line \(x=3\) extending in both quadrants.
Shade the correct region.
Allow dashes on axes as scale.
FT If only error is one of following: FULL lines or \(x\ne3\) or one sign error
in \(\underline{1+2i}\) or angle outside tolerance or scale missing on one axis.
$2(b)$
Carry out a complete method for finding the least value of arg =
Obtain answer -0.454 3dp
Question 3
The polynomial $2x^{4}+ax^{3}+bx-1$, where $a$ and $b$ are constants, is denoted by $p(x)$. When $p(x)$ is divided by $x^{2}-x+1$ the remainder is $3x+2$.
Find the values of $a$ and $b$.
▶️Answer/Explanation
Solution: –
Commence division and reach partial quotient $2x^2 + (a \pm 2)x$
Obtain correct quotient $2x^2 + (a + 2)x + a$
Set their linear remainder equal to part of “$-3x + 2$” and solve for a or for b
Obtain answer a = -3
Obtain answer b = 5
Alternative method for Question 3
State \(2x^{4}+ax^{3}+0x^{2}+bx-1=(x^{2}-x+1)(2x^{2}+Ax+B)+3x+2\) and
form and solve equation(s) to obtain A or B
Obtain A = -1, B = -3
Form and solve equations for a or for b
Obtain answer a = -3
Obtain answer b = 5
Alternative method for Question 3
Use remainder theorem with \(x=\frac{1\pm\sqrt{3}}{2}\) or \(x=\frac{1\pm i\sqrt{3}}{2}\)
Obtain \(-a+\frac{b}{2}\pm\frac{b\sqrt{3}}{2}\mp\sqrt{3}-2=\frac{7}{2}\pm\frac{3\sqrt{3}}{2}\) or
\[-a+\frac{b}{2}\pm\frac{bi\sqrt{3}}{2}\mp i\sqrt{3}-2=\frac{7}{2}\pm\frac{3i\sqrt{3}}{2}\]
Solve simultaneous equations, or single equation, for a or for b
Obtain answer a = -3 from exact working
Obtain answer b = 5 from exact working
Question 4
Solve the equation
$\frac{5z}{1+2i} – zz^* + 30 + 10i = 0,$
giving your answers in the form $x + iy$, where $x$ and $y$ are real.
▶️Answer/Explanation
Solution: –
Substitute $z=x+iy$ and $z^{*}=x-iy$ to obtain a correct equation, horizontal
or with $(1-2i)$ seen, in x and y
Use $i^{2}=-1$ at least once and equate real and imaginary parts to zero
Obtain two correct equations
e.g. $x+2y-x^{2}-y^{2}+30=0$ and $-2x+y+10=0$
Solve quadratic equation for x or for y
Obtain answers $3-4i$ and $6+2i$
Question 5
The parametric equations of a curve are
\[x=te^{2t},\]
\[y=t^{2}+t+3.\]
(a) Show that $\frac{dy}{dx}=e^{-2t}.$
(b) Hence show that the normal to the curve, where $t = -1$, passes through the point $\left(0, 3 – \frac{1}{e^2}\right)$.
▶️Answer/Explanation
Solution: –
5(a)
Obtain $\frac{dx}{dt}=e^{2t}+2te^{2t}$
Use $\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$
Obtain the given answer $\frac{dy}{dx}=e^{-2x}$
5(b)
Obtain $x=-e^{-2}$ or $-\frac{1}{e^{2}}$ and $y=3$ at $t=-1$
Obtain gradient of normal $=-e^{-2}$ or $-\frac{1}{e^{2}}$
$x=0$ substituted into equation of normal or use of gradients to give
$y=3-\frac{1}{e^{4}}$ with no errors
Question 6
(a) Express $5 \sin \theta + 12 \cos \theta$ in the form $R \cos(\theta – \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$.
(b) Hence solve the equation $5 \sin 2x + 12 \cos 2x = 6$ for $0 \leq x \leq \pi$.
▶️Answer/Explanation
Solution: –
6(a)
State $R=13$
Use correct trig formulae to find
$\alpha=tan^{-1}(\pm5/12)=cos^{-1}(\pm12/13)=sin^{-1}(\pm5/13)$
Obtain $\alpha=0.395$
6(b)
$cos^{-1}(\frac{6}{R})$
Use correct method to find a value of 2x in the interval
Obtain answer, e.g. $x=0.743$ or 0.742
Obtain second answer, e.g. $x=2.79$ and no others in the interval
Question 7
The diagram shows a circle with centre O and radius r. The angle of the minor sector AOB of the circle is x radians. The area of the major sector of the circle is 3 times the area of the shaded region.
(a) Show that $x = \frac{3}{4}\pi – \frac{1}{2}\sin x$.
(b) Show by calculation that the root of the equation in (a) lies between 2 and 2.5.
(c) Use an iterative formula based on the equation in (a) to calculate this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
▶️Answer/Explanation
Solution: –
7(a)
State or imply area of major sector = $\frac{1}{2}r^{2}(2\pi-x)$
State or imply area of shaded segment = $\frac{1}{2}r^{2}x-\frac{1}{2}r^{2}sin~x$
State $\frac{1}{2}r^{2}(2\pi-x)=\frac{3}{2}\left[\frac{1}{2}r^{2}x-\frac{1}{2}r^{2}sin~x\right]$
Obtain the given answer $x=\frac{3}{4}sin~x+\frac{\pi}{2}$ after full and correct working
7(b)
Calculate the values of a relevant expression or pair of expressions at $x=2$ and $x=2.5$
Complete the argument correctly with correct calculated values
7(c)
Use the iterative formula correctly at least twice
Obtain final answer 2.18
Show sufficient iterations to 4 d.p. to justify 2.18 to 2 d.p. or show there is
a sign change in the interval (2.175, 2.185)
Question 8
The diagram shows the curve $y = x^3 \ln x$, for $x > 0$, and its minimum point M.
(a) Find the exact coordinates of M.
(b) Find the exact area of the shaded region bounded by the curve, the x-axis and the line $x=\frac{1}{2}$.
▶️Answer/Explanation
Solution: –
8(a)
Use the product rule correctly
Obtain the correct derivative in any form
Equate derivative to zero and solve exactly for x
Obtain answer $\left(\frac{1}{\sqrt{e}},-\frac{1}{3e}\right)$ or exact equivalent
8(b)
Integrate by parts and reach $ax^{4}ln~x+b\int(x^{4}/x)dx$
Obtain $\frac{x^{4}}{4}ln~x-\frac{1}{4}\int(x^{4}/x)dx$
Complete integration and obtain $\frac{x^{4}}{4}ln~x-\frac{x^{4}}{16}$
Use limits of $x=\frac{1}{2}$ and $x=1$ in the correct order, having integrated twice
Obtain answer $\frac{15}{256}-\frac{1}{64}ln~2$ or exact equivalent final answer
Question 9
The variables x and y satisfy the differential equation
\[\frac{dy}{dx}=e^{3y}\sin^{2}2x.\]
It is given that $y=0$ when $x=0$.
Solve the differential equation and find the value of y when $x=\frac{1}{2}.$
▶️Answer/Explanation
Solution: –
Separate variables correctly and obtain $e^{-3y}$ and $sin^{2}2x$ on the opposite sides
Obtain term $-\frac{1}{3}e^{-3y}$
Use correct double angle formula for $sin^{2}2x=\frac{1}{2}[1-cos~4x]$
Obtain terms $\left[\frac{x}{2}-\frac{1}{4}sin~4x\right]$ or e
Use $x=0, y=0$ to evaluate a constant or as limits in a solution containing
terms of the form $ax$ and $b~sin~4x$ and $ce^{\pm3y}$
Obtain correct answer in any form
e.g. $-\frac{1}{3}e^{-3y}=\left[\frac{x}{2}-\frac{1}{4}sin~4x\right]-\frac{1}{3}$
Substitute $x=\frac{1}{2}$ and obtain $y=0.175$ or $-\frac{1}{3}ln(\frac{1}{4}+\frac{3}{8}sin~2)$
Question 10
With respect to the origin O, the points A, B, C and D have position vectors given by
\[\overrightarrow{OA} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}, \quad \overrightarrow{OC} = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix}, \quad \text{and} \quad \overrightarrow{OD} = \begin{pmatrix} 5 \\ -6 \\ 11 \end{pmatrix}.\]
(a) Find the obtuse angle between the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$.
The line l passes through the points A and B.
(b) Find a vector equation for the line l.
(c) Find the position vector of the point of intersection of the line l and the line passing through C and D.
▶️Answer/Explanation
Solution: –
10(a)
Carry out correct process for evaluating the scalar product of $\overrightarrow{OA}$ and $\overrightarrow{OB}$
Using the correct process for the moduli, divide the scalar product by the
product of the moduli and obtain
$cos^{-1}\{\pm(3-2-6)[\sqrt{(3^{2}+(-1)^{2}+2^{2})}\sqrt{(1^{2}+2^{2}+(-3)^{2})}]\}$
Obtain answer $110.9^{\circ}$ or 1.94°
10(b)
Use a correct method to form an equation for line through AB
Obtain $r=3i-j+2k+\mu_{1}(2i-3j+5k)$
10(c)
Obtain a correct equation for line through CD
e.g. $[r]=i-2j+5k+\lambda_{1}(-4i+4j-6k)$
Equate two pairs of components of general points on their I and their CD
and solve for λ or for μ
Obtain e.g. $\lambda_{1}=-2$ or $\mu_{1}=3$ or $\lambda_{2}=-1$ or
$\mu_{2}=-4$
Obtain position vector $9i-10j+17k$
Question 11
Let $f(x) = \frac{5x^2 + x + 11}{(4 + x^2)(1 + x)}$.
(a) Express $f(x)$ in partial fractions.
(b) Hence show that $\int_{0}^{2} f(x) dx = \ln 54 – \frac{1}{8}\pi$.
▶️Answer/Explanation
Solution: –
11(a)
State or imply the form
\[\frac{Ax+B}{4+x^{2}}+\frac{C}{1+x}\]
Use a correct method for finding a coefficient
Obtain one of \(A=2, B=-1\) and \(C=3\)
Obtain a second value
Obtain the third value
11(b)
Integrate and obtain terms
\[\left(\frac{A}{2}\right)ln(4+x^{2})+\frac{B}{2}tan^{-1}(\frac{x}{2})+C~ln(1+x)\]
Substitute limits 0 and 2 correctly in an integral of the form
\(a~ln(4+x^{2})+b~tan^{-1}(\frac{x}{2})+c~ln(1+x)\),
where \(abc\neq0\)
Obtain answer $ln~54-\frac{\pi}{8}$ after full and correct working