Short Questions (Question 8 – 10)

Posted on May 16, 2020 by Myhometuition

Question 8:

Find

\frac{d s}{d t}

for each of the following functions.

\begin{array}{l} (a) s = {(t - \frac{3}{t})}^{2} \\ (b) s = \frac{(t + 1) (3 - 5 t)}{t^{2}} \end{array}

Solution:
(a)

\begin{array}{l} s = {(t - \frac{3}{t})}^{2} \\ s = (t - \frac{3}{t}) (t - \frac{3}{t}) \\ s = t^{2} - 6 + \frac{9}{t^{2}} \\ s = t^{2} - 6 + 9 t^{- 2} \\ \frac{d s}{d t} = 2 t - 18 t^{- 3} = 2 t - \frac{18}{t^{3}} \end{array}

(b)

\begin{array}{l} s = \frac{(t + 1) (3 - 5 t)}{t^{2}} \\ s = \frac{3 t - 5 t^{2} + 3 - 5 t}{t^{2}} = \frac{- 5 t^{2} - 2 t + 3}{t^{2}} \\ s = - 5 - \frac{2}{t} + \frac{3}{t^{2}} = - 5 - 2 t^{- 1} + 3 t^{- 2} \\ \frac{d s}{d t} = 2 t^{- 2} - 6 t^{- 3} = \frac{2}{t^{2}} - \frac{6}{t^{3}} \end{array}

Question 9:

Given that y = \frac{1 - 5 x^{4}}{x - 3}, find \frac{d y}{d x} .

Solution:

\begin{array}{l} \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} = \frac{(x - 3) . - 20 x^{3} - (1 - 5 x^{4}) .1}{{(x - 3)}^{2}} \\ \frac{d y}{d x} = \frac{- 20 x^{4} + 60 x^{3} - 1 + 5 x^{4}}{{(x - 3)}^{2}} \\ \frac{d y}{d x} = \frac{- 15 x^{4} + 60 x^{3} - 1}{{(x - 3)}^{2}} \end{array}

Question 10:

Given that f (x) = \frac{{(x^{2} - 3)}^{5}}{1 - 3 x}, find f' (0) .

Solution:

\begin{array}{l} f (x) = \frac{{(x^{2} - 3)}^{5}}{1 - 3 x} \\ f' (x) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} = \frac{(1 - 3 x) .5 {(x^{2} - 3)}^{4} .2 x - {(x^{2} - 3)}^{5} . - 3}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{10 x (1 - 3 x) {(x^{2} - 3)}^{4} + 3 {(x^{2} - 3)}^{5}}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{{(x^{2} - 3)}^{4} [10 x - 30 x^{2} + 3 (x^{2} - 3)]}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{{(x^{2} - 3)}^{4} [- 27 x^{2} + 10 x - 9]}{{(1 - 3 x)}^{2}} \\ ∴ f' (0) = \frac{{(0^{2} - 3)}^{4} [- 27 {(0)}^{2} + 10 (0) - 9]}{{(1 - 3 (0))}^{2}} \\ f' (0) = \frac{81 \times (- 9)}{1} = - 729 \end{array}

Long Questions (Question 8)

Posted on May 16, 2020 by Myhometuition

Question 8:

In the diagram above, AXB is an arc of a circle centre O and radius 10 cm with ∠AOB = 0.82 radian. AYB is an arc of a circle centre P and radius 5 cm with ∠APB = θ.
Calculate:

(a) the length of the chord AB,

(b) the value of θ in radians,

(c) the difference in length between the arcs AYB and AXB.

Solution:
(a)

\begin{array}{l} \frac{1}{2} A B = \sin 0.41 \times 10 \to (Change the calculator to Rad mode) \\ \frac{1}{2} A B = 3.99 \\ ∴ The length of chord A B = 3.99 \times 2 = 7.98 c m . \end{array}

(b)

\begin{array}{l} Let \frac{1}{2} θ = α, θ = 2 α \\ \sin α = \frac{3.99}{5} \\ α = 0.924 rad \\ ∴ θ = 0.924 \times 2 = 1.848 radian . \end{array}

(c)

Using s = rθ

Arcs AXB = 10 × 0.82 = 8.2 cm

Arcs AYB = 5 × 1.848 = 9.24 cm

Difference in length between the arcs AYB and AXB

= 9.24 – 8.2

= 1.04 cm

Long Questions (Question 7)

Posted on May 16, 2020 by Myhometuition

Question 7:

Diagram below shows a circle PQRT, centre O and radius 5 cm. AQB is a tangent to the circle at Q. The straight lines, AO and BO, intersect the circle at P and R respectively.

OPQR is a rhombus. ACB is an arc of a circle at centre O.

Calculate

(a) the angle x , in terms of π ,

(b) the length , in cm , of the arc ACB ,

(c) the area, in cm²,of the shaded region.

Solution:

(a)

Rhombus has 4 equal sides, therefore OP = PQ = QR = OR = 5 cm

OR is radius to the circle, therefore OR = OQ = 5 cm

Triangles OQR and OQP are equilateral triangle,

Therefore, ∠ QOR= ∠QOP = 60^o

∠ POR = 120^o

x = 120^o × π/180^o

x = 2π/ 3 rad

(b)

cos ∠ AOQ= OQ / OA

cos 60^o = 5 / OA

OA = 10 cm

Length of arc, ACB,

s = r θ

Arc ACB = (10) (2π / 3)

Arc ACB = 20.94 cm

(c)

\begin{array}{l} Area of shaded region \\ = \frac{1}{2} r^{2} (θ - \sin θ) (change calculator to Rad mode) \\ = \frac{1}{2} {(10)}^{2} (\frac{2 π}{3} - \sin \frac{2 π}{3}) \\ = 50 (2.094 - 0.866) \\ = 61.40 {cm}^{2} \end{array}

Long Questions (Question 6)

Posted on May 16, 2020 by Myhometuition

Question 6:
Diagram below shows a circle PQT with centre O and radius 7 cm.

QS is a tangent to the circle at point Q and QSR is a quadrant of a circle with centre Q. Q is the midpoint of OR and QP is a chord. OQR and SOP are straight lines.
[Use π = 3.142]
Calculate
(a) angle θ, in radians,
(b) the perimeter, in cm ,of the shaded region,
(c) the area, in cm² ,of the shaded region.

Solution:
(a)

\begin{array}{l} O Q = Q R = Q S = 7 cm \\ \tan θ = 1 \\ θ = 45^{o} \\ = 45^{o} \times \frac{π}{180^{o}} \\ = 0.7855 rad \end{array}

(b)

\begin{array}{l} Length of arc R S \\ = 7 \times (0.7855 \times 2) \to \begin{array}{l} π rad = 180^{o} \\ 45^{o} = 0.7855 rad \\ 90^{o} = 0.7855 \times 2 rad \end{array} \\ = 7 \times 1.571 \\ = 10.997 cm \\ Length of arc Q P \\ = 7 \times (0.7855 \times 3) \\ = 7 \times 2.3565 \\ = 16.496 cm \\ Length of chord Q P \\ = \sqrt{7^{2} + 7^{2} - 2 (7) (7) \cos 135^{o}} \leftarrow \begin{array}{l} refer form 4 chapter 10 \\ (solution of triangle) for cosine rule \end{array} \\ = \sqrt{167.30} \\ = 12.934 cm \\ Perimeter of the shaded region \\ = 7 + 7 + 10.997 + 16.496 + 12.934 \\ = 54.427 cm \end{array}

(c)

\begin{array}{l} Area of shaded region \\ = (\frac{1}{2} \times 7^{2} \times 1.571) + (\frac{1}{2} \times 7^{2} \times 2.3565) \\ - (\frac{1}{2} \times 7 \times 7 \times sin 135^{o}) \leftarrow \begin{array}{l} refer form 4 chapter 10 \\ (solution of triangle) for area rule \end{array} \\ = 38.4895 + 57.7343 - 17.3241 \\ = 78.8997 {cm}^{2} \end{array}

Long Questions (Question 5)

Posted on May 16, 2020 by Myhometuition

Question 5:
Diagram below shows a semicircle PTS, centre O and radius 8 cm. PTR is a sector of a circle with centre P and Q is the midpoint of OS.

[Use π = 3.142]
Calculate
(a) ∠TOQ, in radians,
(b) the length , in cm , of the arc TR ,
(c) the area, in cm² ,of the shaded region.

Solution:
(a)

\begin{array}{l} \cos ∠ T O Q = \frac{4}{8} = \frac{1}{2} \\ ∠ T O Q = 60^{o} \\ = 60 \times \frac{π}{180} \\ = 1.047 radians \end{array}

(b)

\begin{array}{l} ∠ T P O = 30^{o} \\ = 30 \times \frac{π}{180} \\ = 0.5237 \\ P T^{2} = 8^{2} + 8^{2} - 2 (8) (8) \cos 120 \\ P T^{2} = 192 \\ P T = \sqrt{192} \\ P T = 13.86 cm \\ Length of arc T R = 13.86 \times 0.5237 \\ = 7.258 cm \end{array}

(c)

\begin{array}{l} Area of sector P T R \\ = \frac{1}{2} \times {13.86}^{2} \times 0.5237 \\ = 50.30 {cm}^{2} \\ Length T Q \\ = \sqrt{P T^{2} - P Q^{2}} \\ = \sqrt{{13.86}^{2} - 12^{2}} \\ = 6.935 cm \\ Area of △ P T Q \\ = \frac{1}{2} \times 12 \times 6.935 \\ = 41.61 {cm}^{2} \\ Area of shaded region \\ = 50.30 - 41.61 \\ = 8.69 {cm}^{2} \end{array}

Long Questions (Question 4)

Posted on May 16, 2020 by Myhometuition

Question 4:

Diagram below shows two circles. The larger circle has centre A and radius 20 cm. The smaller circle has centre B and radius 12 cm. The circles touch at point R. The straight line PQ is a common tangent to the circles at point P and point Q.

[Use π = 3.142]

Given that angle PAR = θ radians,

(a) show that θ = 1.32 ( to two decimal places),

(b) calculate the length, in cm, of the minor arc QR,

(c) calculate the area, in cm², of the shaded region.

Solution:
(a)

\begin{array}{l} In △ B S A \\ \cos θ = \frac{8}{32} = \frac{1}{4} \\ θ = 1.32 rad (2 d .p .) \end{array}

(b)

Angle QBR = 3.142 – 1.32 = 1.822 rad

Length of minor arc QR

= 12 × 1.822

= 21.86 cm

(c)

P Q = \sqrt{32^{2} - 8^{2}} = 30.98 cm

Area of the shaded region

= Area of trapezium PQBA– Area of sector QBR – Area of sector PAR

= ½ (12 + 20) (30.98) – ½ (12)² (1.822) – ½ (20)²(1.32)

= 495.68 – 131.18 – 264

= 100.5 cm²

Short Questions (Question 5 & 6)

Posted on May 16, 2020 by Myhometuition

Question 5:

Diagram below shows a sector QOR of a circle with centre O.

It is given that PS = 8 cm and QP = PO= OS = SR = 5 cm.

Find

(a) the length, in cm, of the arc QR,

(b) the area, in cm², of the shaded region.

Solution:

(a) Length of arc QR = r θ = 10 (1.75) = 17.5 cm

(b)

Area of the shaded region

= Area of sector QOR – Area of triangle POS

= ½ (10)² (1.75) – ½ (5) (5) sin 1.75 (change calculator to Rad mode)

= 87.5 – 12.30

= 75.2 cm²

Question 6:

Diagram below shows a circle with centre O and radius 12 cm.

Given that A, B and C are points such that OA = AB and ∠OAC = 90°, find

(a) ∠BOC, in radians,

(b) the area, in cm², of the shaded region.

Solution:

(a) For triangle OAC,

cos ∠AOC = 6/12

ÐAOC = 1.047 rad (change calculator to Rad mode)

ÐBOC = 1.047 rad

(b)

Area of the shaded region

= Area of BOC – Area of triangle AOC

= ½ (12)² (1.047) – ½ (6) (12) sin 1.047 (change calculator to Rad mode)

= 75.38 – 31.17

= 44.21 cm²

Long Questions (Question 7 & 8)

Posted on May 16, 2020 by Myhometuition

Question 7:

The diagram shows a straight line PQ which meets a straight line RS at the point Q. The point P lies on the y-axis.

(a) Write down the equation of RS in the intercept form.

(b) Given that 2RQ = QS, find the coordinates of Q.

(c) Given that PQ is perpendicular to RS, find the y-intercept of PQ.

Solution:

(a)

Equation of RS

\begin{array}{l} \frac{x}{12} + \frac{y}{- 6} = 1 \\ \frac{x}{12} - \frac{y}{6} = 1 \end{array}

(b)

\begin{array}{l} Given 2 R Q = Q S \\ \frac{R Q}{Q S} = \frac{1}{2} \\ Lets coordinates of Q = (x, y) \\ (\frac{(0) (2) + (12) (1)}{1 + 2}, \frac{(- 6) (2) + (0) (1)}{1 + 2}) = (x, y) \\ x = \frac{12}{3} = 4 \\ y = - \frac{12}{3} = - 4 \\ Q = (4, - 4) \end{array}

(c)

\begin{array}{l} Gradient of R S, m_{R S} \\ = - (\frac{- 6}{12}) = \frac{1}{2} \\ m_{P Q} = - \frac{1}{m_{R S}} = - \frac{1}{\frac{1}{2}} = - 2 \end{array}

Point Q = (4, –4), m = –2

Using y = mx+ c

–4 = –2 (4) + c

c = 4

y–intercept of PQ = 4

Question 8:

In the diagram, the equation of FMG is y = – 4. A point P moves such that its distance from E is always half of the distance of E from the straight line FG. Find

(a) The equation of the locus of P,

(b) The x-coordinate of the point of intersection of the locus and the x-axis.

Solution:

(a)

Gradient of the straight line FMG = 0

EM is perpendicular to FMG, so gradient of EM also = 0, equation of EM is x = 2

Thus, coordinates of point M = (2, –4).

Let coordinates of point P= (x, y).

Given PE = ½ EM

2PE = EM

2 [(x – 2)²+ (y – 4)²]^½ = [(2 – 2)² + (4 – (–4))²]^½

4 (x² – 4x + 4 + y² – 8y +16) = (0 + 64) → (square for both sides)

4x² – 16x + 16 + 4y² – 32y + 64 = 64

4x² + 4y² – 16x – 32y + 16 = 0

x² + y² – 4x – 8y + 4 = 0

(b)

x² + y² – 4x – 8y + 4 = 0

At x axis, y = 0.

x² + 0 – 4x – 8(0) + 4 = 0

x² – 4x+ 4 = 0

(x – 2) (x – 2) = 0

x = 2

The x-coordinate of the point of intersection of the locus and the x-axis is 2.

Long Questions (Question 6)

Posted on May 16, 2020 by Myhometuition

Question 7:

Solutions by scale drawing will not be accepted.

Diagram below shows a triangle OPQ. Point S lies on the line PQ.

(a) A point Y moves such that its distance from point S is always 5 uints.

Find the equation of the locus of Y.

(b) It is given that point P and point Q lie on the locus of Y .

Calculate

(i) the value of k,

(ii) the coordinates of Q.

(c) Hence, find the area, in uint², of triangle OPQ.

Solution:
(a)

\begin{array}{l} The equation of the locus Y (x, y) is given by Y S = 5 units \\ \sqrt{{(x - 5)}^{2} + {(y - 3)}^{2}} = 5 \\ x^{2} - 10 x + 25 + y^{2} - 6 y + 9 = 25 \\ x^{2} + y^{2} - 10 x - 6 y + 9 = 0 \end{array}

(b)(i)

Given P (2, k) lies on the locus of Y.

(2)² + (k)²– 10(2) – 6(k) + 9 = 0

4 + k²– 20 – 6k + 9 = 0

k² – 6k – 7 = 0

(k – 7) (k + 1) = 0

k = 7 or k = – 1
Based on the diagram, k = 7.

(b)(ii)

As P and Q lie on the locus of Y, S is the midpoint of PQ. P = (2, 7), S = (5, 3).

Let the coordinates of Q = (x, y),

\begin{array}{l} (\frac{2 + x}{2}, \frac{7 + y}{2}) = (5, 3) \\ \frac{2 + x}{2} = 5 and \frac{7 + y}{2} = 3 \\ 2 + x = 10 and       7 + y = 6 \\ x = 8 and y = - 1 \end{array}

Coordinates of point Q = (8, –1).

(c)

\begin{array}{l} Area of △ O P Q \\ = \frac{1}{2} | \begin{array}{l} 0 8     2 \\ 0 - 17 \end{array} \begin{array}{l} 0 \\ 0 \end{array} | \\ = \frac{1}{2} | 0 + (8) (7) + 0 - 0 - (- 1) (2) - 0 | \\ = \frac{1}{2} | 58 | \\ = 29 {units}^{2} \end{array}

Long Questions (Question 5)

Posted on May 16, 2020 by Myhometuition

Question 5:
Diagram below shows a quadrilateral ABCD. Point C lies on the y-axis.

The equation of a straight line AD is 2y = 5x – 21
(a) Find
(i) the equation of the straight line AB,
(ii) the coordinates of A,
(b) A point P moves such that its distance from point D is always 5 units.
Find the equation of the locus of P.

Solution:
(a)(i)

\begin{array}{l} 2 y = 5 x - 21 \\ y = \frac{5}{2} x - \frac{21}{2} \\ m_{A D} = \frac{5}{2} \\ m_{A B} \times m_{A D} = - 1 \\ m_{A B} \times \frac{5}{2} = - 1 \\ m_{A B} = - \frac{2}{5} \\ Equation of A B \\ y - y_{1} = m_{A B} (x - x_{1}) \\ y + 1 = - \frac{2}{5} (x + 2) \\ 5 y + 5 = - 2 x - 4 \\ 5 y = - 2 x - 9 \end{array}

(a)(ii)

\begin{array}{l} 2 y = 5 x - 21 .......... (1) \\ 5 y = - 2 x - 9 .......... (2) \\ (1) \times 5 : 10 y = 25 x - 105 .......... (3) \\ (2) \times 2 : 10 y = - 4 x - 18 .......... (4) \\ (2) - (4) : 0 = 29 x - 87 \\ x = 3 \\ From (1), \\ 2 y = 15 - 21 \\ 2 y = - 6 \\ y = - 3 \\ A = (3, - 3) \end{array}

(b)

\begin{array}{l} y = 2, \\ 4 = 5 x - 21 \\ 5 x = 25 \\ x = 5 \\ Point D = (5, 2) \\ P D = 5 \\ \sqrt{{(x - 5)}^{2} + {(y - 2)}^{2}} = 5 \\ {(x - 5)}^{2} + {(y - 2)}^{2} = 25 \\ x^{2} - 10 x + 25 + (y^{2} - 4 y + 4) = 25 \\ x^{2} + y^{2} - 10 x - 4 y + 4 = 0 \end{array}