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Question 1

Topic – ALV: 1.3

A line has equation $y = 3x – 2k$ and a curve has equation $y = x^2 – kx + 2$, where $k$ is a constant.

Show that the line and the curve meet for all values of $k$.

▶️Answer/Explanation

Solution: –

$x^2-kx+2=3x-2k$ leading to $x^2-x(k+3)+(2+2k)[=0]$

$b^2-4ac=(k+3)^2-8(1+k)$ (ignore ‘=0’ at this stage)

$=(k-1)^2$ accept $(k-1)(k-1)$

$\ge0$ Hence will meet for all values of k

Question 2

Topic – ALV: 1.2

A function f is defined by $f(x) = x^2 – 2x + 5$ for $x \in \mathbb{R}$. A sequence of transformations is applied in the following order to the graph of $y = f(x)$ to give the graph of $y = g(x)$:

* Stretch parallel to the x-axis with scale factor $\frac{1}{2}$
* Reflection in the y-axis
* Stretch parallel to the y-axis with scale factor 3

Find $g(x)$, giving your answer in the form $ax^2 + bx + c$, where $a$, $b$, and $c$ are constants.

▶️Answer/Explanation

Solution: –

Stretch: $(2x)^2-2(2x)+5~or~(x-1)^2+4$ leading to $(2x-1)^2+4$

Reflection: $(-2x)^2-2(-2x)+5~or~(-2x-1)^2+4$

Stretch: $3\{(-2x)^2-2(-2x)+5\}~or~3\{(-2x-1)^2+4\}$

$12x^2+12x+15$

Question 3

Topic – ALV: 1.7

A curve has equation $y = \frac{1}{60}(3x + 1)^2$ and a point is moving along the curve.

Find the x-coordinate of the point on the curve at which the x- and y-coordinates are increasing at the same rate.

▶️Answer/Explanation

Solution: –

$\frac{dy}{dx}=\left\{\frac{1}{60}(3x+1)\times2\right\}\times\{3\}$

$\frac{1}{10}(3x+1)=1$

$x=3$

Question 4

Topic – ALV: 1.6

The circumference round the trunk of a large tree is measured and found to be 5.00 m. After one year, the circumference is measured again and found to be 5.02 m.

(a) Given that the circumferences at yearly intervals form an arithmetic progression, find the circumference 20 years after the first measurement. 

(b) Given instead that the circumferences at yearly intervals form a geometric progression, find the circumference 20 years after the first measurement.

▶️Answer/Explanation

Solution: –

4(a)

$5.00+20\times0.02$ or $5.02+19\times0.02$

$5.40$

4(b)

$r=\frac{5.02}{5}=1.004$ or $\frac{251}{250}$

$5.00\times(their~1.004)^{20}$ or $5.02\times(their~1.004)^{19}$

$5.42$

Question 5

Topic – ALV: 1.3

Points A(7, 12) and B lie on a circle with center (-2, 5). The line AB has equation y = -2x + 26.

Find the coordinates of B.

▶️Answer/Explanation

Solution: –

$r^{2}=(7+2)^{2}+(12-5)^{2}$

Equation of circle is $(x+2)^{2}+(y-5)^{2}=130$

$(x+2)^{2}+(-2x+21)^{2}=130$

$5x^{2}-80x+315[=0]$ leading to $[5](x-9)(x-7)$

$x=9$

$y=8$ OR $(9,8)$

Question 6

Topic – ALV: 1.6

In the expansion of $\left(\frac{x}{a} + \frac{a}{x^2}\right)^7$, it is given that

$\frac{\text{the coefficient of } x^4}{\text{the coefficient of } x} = 3$.

Find the possible values of the constant $a$.

▶️Answer/Explanation

Solution: –

${7}C_{1}\left(\frac{x}{a}\right)^{6}\left(\frac{a}{x^{2}}\right)^{1}$ or ${7}C_{6}\left(\frac{x}{a}\right)^{1}\left(\frac{a}{x^{2}}\right)^{6}$ ${7}C_{2}\left(\frac{x}{a}\right)^{5}\left(\frac{a}{x^{2}}\right)^{2}$ or ${7}C_{5}\left(\frac{x}{a}\right)^{2}\left(\frac{a}{x^{2}}\right)^{5}$

$\frac{\left(\frac{7}{a^{1}}\right)}{\left(\frac{21}{a^{3}}\right)}=3$

$a^{2}=\frac{1}{9}$

$a=\pm\frac{1}{3}$

Question 7

Topic – ALV: 1.5

(a) By first obtaining a quadratic equation in $\cos \theta$, solve the equation

$\tan \theta \sin \theta = 1$

for $0^{\circ} < \theta < 360^{\circ}$.

(b) Show that $\frac{\tan \theta}{\sin \theta} – \frac{\sin \theta}{\tan \theta} = \tan \theta \sin \theta$.

▶️Answer/Explanation

Solution: –

7(a)

$tan~\theta~sin~\theta=1$ leading to $sin^{2}\theta=cos~\theta$

$1-cos^{2}\theta=cos~\theta$ or $cos^{2}\theta+cos~\theta-1[=0]$

$[cos~\theta=]\frac{-1\pm\sqrt{5}}{2}$

$51.8°, 308.2°$

7(b)

$\frac{tan~\theta}{sin~\theta}-\frac{sin~\theta}{tan~\theta}=\frac{sin~\theta}{sin~\theta~cos~\theta}-\frac{sin~\theta~cos~\theta}{sin~\theta}=\frac{1}{cos~\theta}-cos~\theta$

$=\frac{1-cos^{2}\theta}{cos~\theta}=\frac{sin^{2}\theta}{cos~\theta}$

$=tan~\theta~sin~\theta$

Question 8

Topic – ALV: 1.4

The diagram shows triangle ABC in which angle B is a right angle. The length of AB is 8 cm and the length of BC is 4 cm. The point D on AB is such that $AD=5~cm$. The sector DAC is part of a circle with centre D.

(a) Find the perimeter of the shaded region.

(b) Find the area of the shaded region.

▶️Answer/Explanation

Solution: –

8(a)

$tan~BDC=\frac{4}{3}$ or $sin~BDC=\frac{4}{5}$ or $cos~BDC=\frac{3}{5}$ used to find ADC

$BDC=0.927[3]~\rightarrow~ADC=\pi-0.927[3]~[=2.214to2.215]$

$Arc~AC=5\times their~2.214$

$AC=\sqrt{8^{2}+4^{2}}$ or $2\times5\times sin~1.107$

$[Perimeter=11.07+8.94]=20.0$

8(b)

$Sector~ACD=\frac{1}{2}\times5^{2}\times their~2.214$

$Subtracting~the~area~of~\Delta ADC=\frac{1}{2}\times5\times4$ or $\frac{1}{2}5^{2}~sin~their~2.214$ or $\frac{1}{2}\times8\times4-\frac{1}{2}\times3\times4$

$Shaded~area=27.7-10=17.7$

Question 9

Topic – ALV: 1.2

The function f is defined by $f(x) = -3x^2 + 2$ for $x \leq -1$.

(a) State the range of f.

(b) Find an expression for $f^{-1}(x)$.

The function g is defined by $g(x) = -x^2 – 1$ for $x \leq -1$.

(c) Solve the equation $fg(x) – gf(x) + 8 = 0$.

▶️Answer/Explanation

Solution: –

9(a)

$[y]\leq-1$

9(b)

$y=-3x^{2}+2$ rearranged to $3x^{2}=2-y$, leading to $x^{2}=\frac{2-y}{3}$ or $y^{2}=\frac{2-x}{3}$

$x=[\pm]\sqrt{\frac{2-y}{3}}~~\rightarrow~~[f^{-1}(x)]=\{\pm\}\left\{\sqrt{\frac{2-x}{3}}\right\}$

9(c)

$fg(x)=-3(-x^{2}-1)^{2}+2$

$gf(x)=-(-3x^{2}+2)^{2}-1$

$fg(x)-gf(x)+8=0$ leading to $6x^{4}-18x^{2}+12[=0]$

$[6](x^{2}-1)(x^{2}-2)[=0]$ or formula or completion of the square

$x=-1,~\sqrt{2}$ only these two solutions

Question 10

Topic – ALV: 1.5

At the point (4, -1) on a curve, the gradient of the curve is $\frac{3}{2}$. It is given that $\frac{dy}{dx} = x^{-\frac{1}{2}} + k$, where $k$ is a constant.

(a) Show that $k = -2$.

(b) Find the equation of the curve.

(c) Find the coordinates of the stationary point.

(d) Determine the nature of the stationary point.

▶️Answer/Explanation

Solution: –

10(a)

$-\frac{3}{2}=\frac{1}{2}+k$ leading to $k=-2$

10(b)

$[y]=2x^{\frac{1}{2}}-2x$ $[+c]$

$-1=4-8+c$

$y=2x^{\frac{1}{2}}-2x+3$ or $y=2\sqrt{x}-2x+3$

10(c)

$x^{-\frac{1}{2}}-2=0$

$x=\frac{1}{4}$

$(\frac{1}{4},3\frac{1}{2})$

10(d)

$\frac{d^{2}y}{dx^{2}}=-\frac{1}{2}x^{-\frac{3}{2}}$

$<0$ (or -4) hence Maximum

Question 11

Topic – ALV: 1.8

The diagram shows the curve with equation $x = y^2 + 1$. The points A(5, 2) and B(2, -1) lie on the curve.

(a) Find an equation of the line AB. 

(b) Find the volume of revolution when the region between the curve and the line AB is rotated through 360° about the y-axis. 

▶️Answer/Explanation

Solution: –

11(a)

Gradient of AB = $\frac{2-(-1)}{5-2}$

Equation of AB is $y-2=1(x-5)$ or $y+1=l(x-2)$

11(b)

$[\pi]\int x^{2}dy=[\pi]\int[(y+1)^{2}+1]dy=[\pi]\int[(y^{2}+2y+1)+1]dy$

$[\pi]\int\left[\frac{y^{3}}{3}+\frac{2y^{2}}{2}+y\right]$

$[\pi]\int(y+3)^{2}dy=[\pi]\int(y^{2}+6y+9)dy$

$[\pi]\left[\frac{y^{3}}{3}+3y^{2}+9y\right]$ or $[\pi]\left[\frac{(y+3)^{3}}{3}\right]$

$[\pi]\left\{\frac{8}{3}+12+18-\left(-\frac{1}{3}+3-9\right)\right\}$ or $[\pi]\left\{\frac{32}{5}+\frac{16}{3}+2-\left(-\frac{1}{5}-\frac{2}{3}-1\right)\right\}$

Volume = $\pi \left[ \left( 39 – 15 \frac{3}{5} \right) \right]$

= $23 \frac{2}{5} \pi$ or $\frac{117}{5} \pi$ or awrt $73.5$ [1327]

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