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Question 1

Topic – ALV: 1.2

The diagram shows the graph of $y = f(x)$, which consists of the two straight lines AB and BC. The lines A’B’ and B’C’ form the graph of $y = g(x)$, which is the result of applying a sequence of two transformations, in either order, to $y = f(x)$.

State fully the two transformations .

▶️Answer/Explanation

Solution :-

Translation will be $\begin{pmatrix}
{0}\\
{-2}
\end{pmatrix}$

 

Question 2

Topic – ALV: 2.1

The function f is defined for $x \in R$ by $f(x) = x^2 – 6x + c$, where c is a constant. It is given that $f(x) > 2$
for all values of x.

Find the set of possible values of c.

▶️Answer/Explanation

Solution :-

\begin{aligned}
& x^2 – 6x + c > 2 \quad \text{leading to} \quad (x-3)^2 – 9 + c > 2 \\
& c > 11 – (x-3)^2 \quad \text{and} \quad (x-3)^2 \geq 0 \\
& c > 11 \\
& \\
& \text{Alternative Method 1} \\
& \frac{dy}{dx} = 2x – 6 = 0 \\
& x = 3 \\
& \text{When } x = 3, \quad y = 9 – 18 + c \\
& [-9 + c > 2] \quad c > 11 \\
& \\
& \text{Alternative Method 2} \\
& x^2 – 6x + c > 2 \quad \text{leading to} \quad x^2 – 6x + c – 2 > 0 \quad \text{then use of } b^2 – 4ac’ \\
& 36 – 4(1)(c-2) < 0 \\
& c > 11
\end{aligned}

Question 3

Topic – ALV: 2.1

(a) Give the complete expansion of $\left(x + \frac{2}{x}\right)^5$.

(b) In the expansion of $\left(a + bx^2 + \frac{2}{x}\right)^5$, the coefficient of x is zero and the coefficient of $\frac{1}{x}$ is 80.

Find the values of the constants a and b.

▶️Answer/Explanation

Solution :-

(a) $x^5 + 10x^3 + 40x + \frac{80}{x} + \frac{80}{x^3} + \frac{32}{x^5}$

$OR$

$x^5 + 10x^3 + 40x + 80x^{-1} + 80x^{-3} + 32x^{-5}$

(b) $40 \times a + (\text{their coefficient of } x^{-1}) \times b = 0$

$(\text{their coefficient of } x^{-1}) \times a + (\text{their coefficient of } x^{-3}) \times b = 80$

$a = 2 \quad b = -1$

Question 4

Topic – ALV: 3.2

(a) Show that the equation

$3 \tan^2 x – 3 \sin^2 x – 4 = 0$

may be expressed in the form $a \cos^4 x + b \cos^2 x + c = 0$, where a, b and c are constants to be
found.

(b) Hence solve the equation $3 \tan^2 x – 3 \sin^2 x – 4 = 0$ for $0^{\circ} \leq x \leq 180^{\circ}$.

▶️Answer/Explanation

Solution :-

(a) $3\sin^2 x – 3\sin^2 x \cos^2 x – 4\cos^2 x = 0$

$3(1-\cos^2 x) – 3(1-\cos^2 x)\cos^2 x – 4\cos^2 x = 0$

$3\cos^4 x – 10\cos^2 x + 3 = 0 \quad \text{or} \quad -3\cos^4 x + 10\cos^2 x – 3 = 0$

(b) $(3\cos^2 x – 1)(\cos^2 x – 3) = 0$

$\cos x = \pm \frac{1}{\sqrt{3}}$

$125.3^{\circ}$

Question 5

Topic – ALV: 3.1

A circle has equation $(x-1)^2 + (y+4)^2 = 40$. A line with equation $y = x – 9$ intersects the circle at
points A and B.

(a) Find the coordinates of the two points of intersection.

(b) Find an equation of the circle with diameter AB.

▶️Answer/Explanation

Solution :-

(a)

$(x-1)^2 + (x-9+4)^2 = 40$

$x^2 – 6x – 7 = 0$ leading to $(x+1)(x-7) = 0$

$(-1, -10), (7, -2)$

$OR$

$ x = -1 \text{ and } 7, y = -10 \text{ and } -2$

(b)

$\text{C is mid-point}$

$\left( \frac{{x_1} +{x_2}}{2}, \frac{{y_1} +{y_2}}{2}\right)$

$\text{Radius} = \sqrt{{(x -3)^2} + {(y -(-6))^2}}$

$\text{OR}$

$=\sqrt{{\left( \frac{7 – (-1)}{2} \right)^2 + \left( \frac{-2 – (-10)}{2} \right)^2}}$

$(x-3)^2 + (y+6)^2 = 32$

Question 6

Topic – ALV: 3.3

The diagram shows a sector OAB of a circle with centre O and radius r cm. Angle AOB = \theta radians.
It is given that the length of the arc AB is 9.6 cm and that the area of the sector OAB is 76.8 cm².

(a) Find the area of the shaded region.

(b) Find the perimeter of the shaded region.

▶️Answer/Explanation

Solution :-

(a) $\frac{\frac{1}{2}r^2 \theta}{r\theta} = \frac{76.8}{9.6}$ 

$OR$

$\frac{1}{2} \left( \frac{9.6}{\theta} \right)^2 \theta = 76.8$

$r = 16$

$\theta = 0.6$

$\Delta OAB = \frac{1}{2} \times \text{their } 16^2 \times \sin \text{ their } 0.6$

$\text{Area} = 76.8 – 72.27= 4.53$

(b) $AB = 2 \times 16 \times \sin 0.3$ \quad OR \quad $AB^{2} = 16^{2} + 16^{2} – 2 \times 16^{2} \cos 0.6$

$\text{Perimeter} = 9.6 + 9.46 = 19.1$

Question 7

Topic – ALV: 2.3

The function f is defined by $f(x) = 2 – \frac{5}{x+2}$ for $x > -2$.

(a) State the range of f.

(b) Obtain an expression for $f^{-1}(x)$ and state the domain of $f^{-1}$.

The function g is defined by $g(x) = x + 3$ for $x > 0$.

(c) Obtain an expression for $fg(x)$ giving your answer in the form $\frac{ax + b}{cx + d}$, where a, b, c and d are
integers.

▶️Answer/Explanation

Solution :-

(a) $y < 2 \text{OR} f(x) < 2$

7(b) $y = 2 – \frac{5}{x+2} \text{leading to} y(x+2) = 2(x+2) – 5\text{leading to } xy + 2y = 2x – 1$

$2y + 1 = 2x – xy \text{leading to} 2y + 1 = x(2 – y)$

$x = \frac{2y + 1}{2 – y}\rightarrow {f^{-1}(x)} = \frac{2x + 1}{2 – x}$

$\text{Domain is } x < 2$

(c) $fg(x) = 2 – \frac{5}{x+3+2}$

= $\frac{2(x+5)-5}{x+5} \text{OR}\frac{2(x+5)}{x+5} – \frac{5}{x+5}$

= $\frac{2x+5}{x+5}$

Question 8

Topic – ALV: 2.4

A progression has first term \(a\) and second term \(\frac{a^2}{a+2}\), where \(a\) is a positive constant.

(a) For the case where the progression is geometric and the sum to infinity is 264, find the value.

(b) For the case where the progression is arithmetic and \(a = 6\), determine the least value of \(n\) required for the sum of the first \(n\) terms to be less than \(-480a\).

▶️Answer/Explanation

Solution :-

(a) \(r = \frac{a^2}{a+2}\)

\(\frac{a}{1-\frac{a}{a+2}}=264\)

\(\frac{a(a+2)}{a+2-a}=264 \quad \text{leading to} \quad \frac{a(a+2)}{2}=264 \quad \text{leading to} \quad a^{2}+2a-528 = 0\)

\((a-22)(a+24)=0\)

\(a=22\) (only)

(b) \(d = \frac{6^{2}}{6+2}-6=-\frac{3}{2}\)

\(\frac{n}{2} \left[12+(n-1) \left( -\frac{3}{2} \right) \right] < -480\)

\(3 \left[ n^2 – 9n – 640 \right] > 0\)

\(n = \frac{9 \pm \sqrt{81 + 2560}}{2}\)

31 only

Question 9

Topic – ALV: 3.8

A curve which passes through (0, 3) has equation $y = f(x)$. It is given that $f'(x) = 1 – \frac{2}{(x-1)^3}$.

(a) Find the equation of the curve.

The tangent to the curve at (0, 3) intersects the curve again at one other point, P.

(b) Show that the x-coordinate of P satisfies the equation $(2x+1)(x-1)^2-1=0$.

(c) Verify that $x = \frac{3}{2}$ satisfies this equation and hence find the y-coordinate of P.

▶️Answer/Explanation

Solution:-

(a) $y = x + (x-1)^{-2} + c$

Sub $x = 0, y = 3$ leading to $3 = 0 + 1 + c$

$y = x + (x-1)^{-2} + 2$ or $f(x) = x + (x-1)^{-2} + 2$

9(b) Gradient of tangent = $f'(0) = 3$

Equation of tangent is $y – 3 = $ their gradient at $x = 0 (x – 0)$

Intersection given by $3x + 3 = x + (x-1)^{-2} + 2$

$2x + 1 = \frac{1}{(x-1)^{2}} \rightarrow (2x+1)(x-1)^{2} – 1 = 0$ or solve equation before given

form reached and show solution $(x = 3/2)$ satisfies given result

(c) Substitute $x = \frac{3}{2}$ leading to $(2x+1)(x-1)^{2} – 1$ leading to $4 \times \frac{1}{4} – 1 = 0$.

Hence $x = \frac{3}{2}$

If shown in (b) must be referenced here (in part (c))

When $x = \frac{3}{2}$ $y = 7\frac{1}{2}$

 

Question 10

Topic – ALV: 3.8

The diagram shows the points A (1\frac{1}{2}, 5\frac{1}{2}) and B (7\frac{3}{4}, 3\frac{1}{2}) lying on the curve with equation

$y = 9x – (2x + 1)^{\frac{3}{2}}$.

(a) Find the coordinates of the maximum point of the curve.

(b) Verify that the line AB is the normal to the curve at A.

(c) Find the area of the shaded region.

▶️Answer/Explanation

Solution:-

(a) $ \frac{dy}{dx} = \left[ 9 + \frac{3}{2}(2x+1)^{\frac{1}{2}} \right] \times 2$

$9 – 3(2x+1)^{\frac{1}{2}} = 0$ leading to $2x+1 = 9$

Max point = (4, 9)

(b) When $x = 1\frac{1}{2}$, shows substitution or $\frac{dy}{dx} = 3$

Gradient of AB is $\frac{5\frac{1}{2}-3\frac{1}{2}}{1\frac{1}{2}-7\frac{1}{2}} = \frac{-1}{3}$

$-\frac{1}{3} \times 3 = -1$. [Hence AB is the normal]

Alternative method for Question 10(b)

When $x = 1\frac{1}{2}$ $\frac{dy}{dx} = 3$ [perpendicular gradient is -1/3]

Perpendicular through A has equation $y = \frac{-x}{3} + 6$ which contains B(7.5, 3.5)

leading to AB is a normal to the curve at A

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