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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

Answer:-

Attempt to express equation in terms of \(sec~\theta\) only

Obtain $6~sec^{2}\theta-17~sec~\theta-14(=0)$

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

Answer:-

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 – ln b, where a and b are integers.

▶️Answer/Explanation

Answer:- 

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

(a)

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.

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.

(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

Answer:- 

(a) Draw (more or less) correct sketch with vertex on positive x-axis

Indicate in some way the two roots

(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

(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

Answer:- 

(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 $3e^{-2}$

(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

(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

Answer:- 

(a) Obtain at least either $(\frac{1}{2}\sin~\theta+\frac{1}{2}\sqrt{3}\cos~\theta)$ or $(\frac{1}{2}\cos~\theta+\frac{1}{2}\sqrt{3}\sin~\theta)$

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 $\sqrt{3}+2~\sin~2\theta$

(b) Identify value of $\theta$ is $\frac{3}{8}\pi$

Obtain $\sqrt{3}+2~\sin\frac{3}{4}\pi$ and conclude $\sqrt{3}+\sqrt{2}$

(c) Identify integrand as $\sqrt{3}+2~\sin~4x$

Integrate to obtain form $k_{1}x+k_{2}\cos~4x$

Obtain correct $\sqrt{3}x-\frac{1}{2}\cos~4x$

Obtain $\frac{1}{8}\pi\sqrt{3}+\frac{1}{2}$

Question 7

Topic – ALV: 1.7

A curve has parametric equations:

\[x=\frac{2t+3}{t+2}\] \[y=t^{2}+at+1\]

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

Answer:- 

(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

(b) Substitute $t=-1$, equate to zero and attempt solution for a

Obtain $a=3$

(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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