Question 1
Topic – ALV: 1.5
Solve the equation
\[sec^2 \theta + 5 \tan^2 \theta = 9 + 17 \sec \theta\]
for $0^{\circ} < \theta < 360^{\circ}$.
▶️Answer/Explanation
Solution :-
1. Attempt to express equation in terms of $sec~\theta$ only
Obtain $6sec^{2}\theta-17~sec~\theta-14(=0)$
Attempt solution of 3-term quadratic equation to find one value of $\theta$, from
\[\cos~\theta=…\]
Obtain 73.4
Obtain 286.6
Question 2
Topic – ALV: 2.2
The variables $x$ and $y$ satisfy the equation $y = Ae^{(A-B)x}$, where $A$ and $B$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $(0.4, 3.6)$ and $(2.9, 14.1)$, as shown in the diagram.
Find the values of $A$ and $B$ correct to 3 significant figures.
▶️Answer/Explanation
Solution :-
2. State or imply equation $\ln~y = \ln~A + (A-B)x$
Equate $A-B$ to gradient of line
Obtain $A-B=4.2$
Substitute appropriate values to find value of $\ln~A$
Obtain $\ln~A=1.92$ and hence $A=6.82$ and $B=2.62$
Alternative Method 1
State or imply equation $\ln~y = \ln~A + (A-B)x$
Use of coordinates to obtain equation of line
\[\frac{\ln~y-3.6}{14.1-3.6}=\frac{x-0.4}{2.9-0.4}\]
Obtain gradient equal to 4.2
Substitute appropriate values to find value of $\ln~A$
Obtain $\ln~A=1.92$ and hence $A=6.82$ and $B=2.62$
Alternative Method 2
State or imply equation $\ln~y = \ln~A + (A-B)x$
\[3.6=\ln~A+0.4(A-B)\]
\[14.1=\ln~A+2.9(A-B)\]
Obtain $A=6.82$ and $B=2.62$
Question 3
Topic – ALV: 2.2
The diagram shows part of the curve $y = \frac{6}{2x+3}$. The shaded region is bounded by the curve and the
lines $x = 6$ and $y = 2$.
Find the exact area of the shaded region, giving your answer in the form $a – b\ln c$, where $a$ and $b$ are
integers.
▶️Answer/Explanation
Solution :-
3. Integrate to obtain the form $k \ln(2x+3)$
Obtain correct $3\ln(2x+3)$
Apply limits 0 and 6 correctly to obtain $k \ln 15 – k \ln 3$
Apply relevant logarithm properties correctly to obtain form $\ln b$
Obtain $12 – \ln 125$
Question 4
Topic – ALV: 2.2
The diagram shows the graph of $y = 3 – e^{-\frac{1}{2}x}.$
On the diagram, sketch the graph of $y = |5x-4|$, and show that the equation $3 – e^{-\frac{1}{2}x} = |5x-4|$
has exactly two real roots. [2]
It is given that the two roots of $3 – e^{-\frac{1}{2}x} = |5x-4|$ are denoted by
$\alpha$ and $\beta$, where $\alpha < \beta.$
(b) Show by calculation that $\alpha$ lies between 0.36 and 0.37. [2]
(c) Use the iterative formula
\[x_{n+1} = \frac{1}{5}(7 – e^{-\frac{1}{2}x_{n}})\]
to find $\beta$ correct to 4 significant figures. Give the
result of each iteration to 6 significant figures.
▶️Answer/Explanation
Solution :-
4(a) Draw (more or less) correct sketch with vertex on positive x-axis
Indicate in some way the two roots
4(b) Consider sign of $3-e^{-\frac{1}{2}x}+5x-4$ or of $3-e^{-\frac{1}{2}x}-|5x-4|$ for 0.36 and 0.37
Obtain -0.035… and 0.018…, or equivalents, and justify conclusion
4(c) Use iteration process correctly at least once
Obtain final answer 1.295
Show sufficient iterations to 6 sf to justify answer or show sign change in
interval [1.2945, 1.2955]
Question 5
Topic – ALV: 1.7
The diagram shows the curve with equation $y = e^{-\frac{1}{2}x}(x^{2} – 5x + 4)$. The curve crosses the x-axis at the
points A and B, and has a maximum at the point C.
(a) Find the exact gradient of the curve at B.
(b) Find the exact coordinates of C.
▶️Answer/Explanation
Solution :-
5(a) Attempt use of product rule to find first derivative
Obtain $-\frac{1}{2}e^{-\frac{1}{2}x}(x^{2}-5x+4)+e^{-\frac{1}{2}x}(2x-5)$
Obtain $x = 4$ for point B
Substitute $x = 4$ to find the value of the derivative
Obtain $3e^{-2}$
5(b) Equate their first derivative to zero and simplify as far as quadratic equation
Obtain at least $x^{2}-9x+14=0$
Solve to find relevant x value and substitute to find the value of y
Obtain $x = 7$ and $y = 18e^{-\frac{7}{2}}$
Question 6
Topic – ALV: 1.5
6(a) Show that $4 \sin(\theta + \frac{1}{3}\pi) \cos(\theta – \frac{1}{3}\pi) \equiv \sqrt{3} + 2 \sin 2\theta.$
(b) Find the exact value of $4 \sin \frac{17}{24}\pi \cos \frac{1}{24}\pi$.
(c) Find the exact value of $\int_{0}^{\frac{1}{8}\pi} 4 \sin(2x+\frac{1}{3}\pi) \cos(2x-\frac{1}{3}\pi) \ dx.$
▶️Answer/Explanation
Solution :-
6(a) Obtain at least either $\left(\frac{1}{2}\sin~\theta+\frac{1}{2}\sqrt{3}\cos~\theta\right)$ or $\left(\frac{1}{2}\cos~\theta+\frac{1}{2}\sqrt{3}\sin~\theta\right)$
Expand and simplify with correct use of $\sin^{2}\theta+\cos^{2}\theta=1$
Use $\sin~\theta~\cos~\theta=\frac{1}{2}\sin~2~\theta$
Confirm given result $\sqrt{3}+2~\sin~2\theta$
6(b) Identify value of $\theta$ is $\frac{3}{8}\pi$
Obtain $\sqrt{3}+2~\sin\frac{3}{4}\pi$ and conclude $\sqrt{3}+\sqrt{2}$
6(c) Identify integrand as $\sqrt{3}+2~\sin~4x$
Integrate to obtain form $k_{1}x+k_{2}\cos~4x$
Obtain correct $\sqrt{3}x-\frac{1}{2}\cos~4x$
$\frac{1}{8}\pi\sqrt{3}+\frac{1}{2}$
Question 7
Topic – ALV: 1.7
7. A curve has parametric equations
\[x = \frac{2t+3}{t+2}, \quad y = t^2 + at + 1,\]
where $a$ is a constant. It is given that, at the point P on the curve, the gradient is 1.
(a) Show that the value of t at P satisfies the equation
\[2t^3 + (a+8)t^2 + (4a+8)t + 4a – 1 = 0.\]
(b) It is given that $(t+1)$ is a factor of
\[2t^{3}+(a+8)t^{2}+(4a+8)t+4a-1.\]
Find the value of a.
(c) Hence show that P is the only point on the curve at which the gradient is 1.
▶️Answer/Explanation
Solution :-
7(a) Attempt use of quotient rule (or equivalent) to find $\frac{dx}{dt}$
Obtain $\frac{1}{(t+2)^{2}}$
Equate $\frac{dy}{dx}$ to 1
Obtain $(2t+a)(t+2)^{2}=1$ and expand to confirm given result
7(b) Substitute $t=-1$ , equate to zero and attempt solution for a
Obtain $a=3$
7(c) Divide their cubic by $t+1$ at least as far as the x term
Obtain $2t^{2}+9t+11$
Calculate discriminant, obtain $-7$ and conclude no further value of $t$