Question 1
Topic – ALV: 3.5
Find the exact value of $\int_{0}^{\frac{1}{2\pi}}2\tan^{2}(\frac{1}{2}x)dx$.
▶️Answer/Explanation
Solution: –
Express integrand as $2\sec^{2}\frac{1}{2}x-2$
Integrate to obtain form $a \tan \frac{1}{2}x + bx$
Obtain correct $4 \tan \frac{1}{2}x – 2x$
Obtain $4-\pi$
Question 2
Topic – ALV: 3.2
Solve the equation $\tan(\theta – 60^{\circ}) = 3 \cot \theta$ for $-90^{\circ} < \theta < 90^{\circ}$.
▶️Answer/Explanation
Solution: –
State $\frac{\tan \theta – \tan 60}{1 + \tan \theta \tan 60} = \frac{3}{\tan \theta}$
Attempt arrangement of equation to quadratic form
Obtain $\tan^2 \theta – 4\sqrt{3} \tan \theta – 3 = 0$
Solve 3-term quadratic equation to obtain at least one value of $\theta$
Obtain -22.2 and 82.2
Question 3
Topic – ALV: 2.1
The polynomial p(x) is defined by
\[p(x)=ax^{3}-ax^{2}+ax+b\]
where a and b are constants. It is given that $(x+2)$ is a factor of p(x), and that the remainder is 35 when p(x) is divided by $(x-3)$.
(a) Find the values of a and b.
(b) Hence factorise p(x) and show that the equation p(x) = 0 has exactly one real root.
▶️Answer/Explanation
Solution: –
(a) Substitute $x=-2$ and equate to zero
Substitute $x=3$ and equate to 35
Obtain $-8a-4a-2a+b=0$ and $27a-9a+3a+b=35$
Solve a pair of relevant simultaneous linear equations to find a or $b$
Obtain $a=1$ and $b=14$
(b) Divide by $x+2$ at least as far as the x term
Obtain $[(x+2)](x^{2}-3x+7)$
Conclude with reference to -2, and discriminant is 9-28 and hence no root
Question 4
Topic – ALV: 2.1
(a) Sketch, on the same diagram, the graphs of \(y=|2x-11|\) and \(y=3x-3\).
(b) Solve the inequality |2x – 11| < 3x – 3.
(c) Find the smallest integer N satisfying the inequality $|2 \ln N – 11| < 3 \ln N – 3$.
▶️Answer/Explanation
Solution: –
(a) Draw V-shaped graph with vertex on positive x-axis
Draw approximately correct graph of $y=3x-3$ with greater gradient
(b) Attempt solution of linear equation where signs of 2x and 3x are different
Solve $-2x+11=3x-3$ to obtain $x=\frac{14}{5}$
Conclude $x>\frac{14}{5}$
Alternative method for Question 4(b)
Attempt solution of 3-term equation $(2x-11)^{2}=(3x-3)^{2}$ to obtain at least one value of x
Obtain at least $x=\frac{14}{5}$
Conclude $x>\frac{14}{5}$
(c) Attempt value of N (maybe non-integer at this stage) using logarithms and their answer to part (b).
Conclude with single integer 17
Question 5
Topic – ALV: 3.5
It is given that
$ \int_{1}^{a} \left( \frac{4}{1+2x} + \frac{3}{x} \right) dx = \ln 10, $
where a is a constant greater than 1.
(a) Show that
$ a = \sqrt[3]{90(1+2a)^{-2}}. $
(b) Use an iterative formula, based on the equation in (a), to find the value of a correct to 3 significant figures. Use an initial value of 1.7 and give the result of each iteration to 5 significant figures.
▶️Answer/Explanation
Solution: –
(a) Integrate to obtain form $k_{1}ln(1+2x)+k_{2}ln~x$
Obtain correct $2~ln(1+2x)+3~ln~x$
Use limits correctly and equate to ln10
Apply relevant logarithm properties correctly and arrange as far as
\[a^{3}=…\]
Confirm given result $a=\sqrt[3]{90(1+2a)^{-2}}$ with sufficient detail
(b) Use iteration process correctly at least once
Obtain final answer 1.68
Show sufficient iterations to 5 sf to justify answer or show a sign
change in the interval [1.675, 1.685]
Question 6
Topic – ALV: 3.4
The diagram shows the curve with equation
$y = \frac{4e^{2x} + 9}{e^x + 2}$.
The curve has a minimum point M and crosses the y-axis at the point P.
(a) Find the exact value of the gradient of the curve at P.
(b) Find the exact coordinates of M.
▶️Answer/Explanation
Solution: –
(a) Differentiate using quotient rule
Obtain $\frac{(e^{x}+2)8e^{2x}-(4e^{2x}+9)e^{x}}{(e^{x}+2)^{2}}$
Substitute $x=0$ in first derivative and attempt evaluation
Obtain $\frac{11}{9}$
(b) Equate first derivative to zero and attempt factorisation or
equivalent
Solve a three-term quadratic equation in $e^{x}$ to obtain $e^{x}=…$
Obtain x-coordinate $\ln\frac{1}{2}$ or -ln2
Obtain y-coordinate 4
Question 7
Topic – ALV: 3.4
The diagram shows the curve with parametric equations
$ x = k \tan t, \quad y = 3 \sin 2t – 4 \sin t, $
for 0 < t < π/2. It is given that k is a positive constant. The curve crosses the x-axis at the point P.
(a) Find the value of cos t at P, giving your answer as an exact fraction.
(b) Express $\frac{dy}{dx}$ in terms of k and cos t.
(c) Given that the normal to the curve at P has gradient $-\frac{9}{10}$, find the value of k, giving your answer as an exact fraction.
▶️Answer/Explanation
Solution: –
(a) Use identity $sin~2t=2~sin~t~cos~t$
Attempt solution of $y=0$ for $cos~t$
Obtain $cos~t=\frac{2}{3}$
(b) Differentiate to obtain at least one of $\frac{dx}{dt}$ and $\frac{dy}{dt}$ correct
Obtain $\frac{dy}{dx}=\frac{6~cos~2t-4~cos~t}{k~sec^{2}t}$
Attempt to express $\frac{dy}{dx}$ in terms of $cos~t$
Obtain $\frac{dy}{dx}=\frac{6(2~cos^{2}t-1)~cos^{2}t-4~cos^{3}t}{k}$
(c) Substitute value from part (a) in expression for $\frac{dy}{dx}$ involving k
and $cos~t$
Equate to $-\frac{10}{9}$ and solve for k
Obtain $k=\frac{4}{3}$