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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 $t$

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