SPM Practice Question 3

Posted on May 16, 2020 by Myhometuition

Question 3:
An arithmetic progression has 16 terms. The sum of the 16 terms is 188, and the sum of the even terms is 96. Find
(a) the first term and the common difference,
(b) the last term.

Solution:
(a) Let the first term = a
Common difference = d

\begin{array}{l} Given S_{16} = 188 \\ Thus, \frac{16}{2} [2 a + 15 d] = 188 \\ 8 [2 a + 15 d] = 188 \\ 2 a + 15 d = \frac{188}{8} \\ 2 a + 15 d = 23.5 - - - (1) \end{array}

Given the sum of the even terms = 96

\begin{array}{l} T_{2} + T_{4} + T_{6} + ..... + T_{16} = 96 \\ (a + d) + (a + 3 d) + (a + 5 d) + ..... + (a + 15 d) = 96 \\ \frac{8}{2} [(a + d) + (a + 15 d)] = 96 \\ 4 [2 a + 16 d] = 96 \\ 2 a + 16 d = 24 - - - (2) \end{array}

(2) – (1):
16d – 15d = 24 – 23.5
d = 0.5
Substitute d = 0.5 into (2):
2a + 16 (0.5) = 24
2a + 8 = 24
2a = 16
a = 8
Therefore, first term = 8 and common difference = 0.5.

(b) Last term
= T₂
= 8 + 15 (0.5)
= 8 +7.5
= 15.5

(Long Questions) – Question 8

Posted on May 16, 2020 by Myhometuition

Question 8:
Diagram below shows a cyclic quadrilateral PQRS.

(a) Calculate
(i) the length, in cm, of PR,
(ii) ∠PRQ.
(b) Find
(i) the area, in cm², of ∆ PRS,
(ii) the short distance, in cm, from point S to PR.

Solution:
(a)(i)

\begin{array}{l} P R^{2} = 7^{2} + 8^{2} - 2 (7) (8) \cos 80^{o} \\ P R^{2} = 113 - 19.4486 \\ P R = \sqrt{93.5514} \\ P R = 9.6722 cm \end{array}

(a)(ii)

\begin{array}{l} In cyclic quadrilateral \\ ∠ P Q R + ∠ P S R = 180 \\ ∠ P Q R + 80 = 180 \\ ∠ P Q R = 100^{o} \\ \frac{\sin ∠ Q P R}{3} = \frac{\sin 100}{9.6722} \\ \sin ∠ Q P R = 0.3055 \\ ∠ Q P R = 17^{o} 47' \\ ∠ P R Q = 180^{o} - 100^{o} - 17^{o} 47' \\ = 62^{o} 13' \end{array}

(b)(i)

\begin{array}{l} Area of △ P R S \\ = \frac{1}{2} \times 7 \times 8 \sin 80^{o} \\ = 27.5746 {cm}^{2} \end{array}

(b)(ii)

\begin{array}{l} Area of △ P R S = 27.5746 \\ \frac{1}{2} \times 9.6722 \times h = 27.5746 \\ h = \frac{27.5746 \times 2}{9.6722} \\ = 5.7018 cm \\ Shortest distance = 5.7018 cm \end{array}

(Long Questions) – Question 7

Posted on May 16, 2020 by Myhometuition

Question 7:
Diagram below shows a quadrilateral ABCD where the sides AB and CD are parallel. ∠BAC is an obtuse angle.

Given that AB = 14 cm, BC = 27 cm, ∠ACB = 30^o and AB : DC = 7 : 3.
Calculate
(a) ∠BAC.
(b) the length, in cm, of diagonal BD.
(c) the area, in cm², of quadrilateral ABCD.

Solution:
(a)

\begin{array}{l} \frac{\sin ∠ B A C}{27} = \frac{\sin 30^{o}}{14} \\ \sin ∠ B A C = \frac{\sin 30^{o}}{14} \times 27 \\ \sin ∠ B A C = 0.9643 \\ ∠ B A C = {74.64}^{o} \\ ∠ B A C (obtuse) = 180^{o} - {74.64}^{o} \\ = {105.36}^{o} \end{array}

(b)

\begin{array}{l} A B parallel with D C \\ ∠ B A C = ∠ A C D \\ ∠ B C D = {105.36}^{o} + 30^{o} \\ = {135.36}^{o} \\ \frac{D C}{A B} = \frac{3}{7} \\ D C = \frac{3}{7} \times 14 cm \\ = 6 cm \\ B D^{2} = 27^{2} + 6^{2} - 2 (27) (6) \cos ∠ B C D \\ B D^{2} = 765 - 324 \cos {135.36}^{o} \\ B D^{2} = 995.54 \\ B D = 31.55 cm \end{array}

(c)

\begin{array}{l} ∠ A B C = 180^{o} - 30^{o} - {105.36}^{o} \\ = {44.64}^{o} \\ A C^{2} = 27^{2} + 14^{2} - 2 (27) (14) \cos ∠ A B C \\ A C^{2} = 925 - 756 \cos {44.64}^{o} \\ A C^{2} = 387.08 \\ A C = 19.67 cm \\ Area △ A B C \\ = \frac{1}{2} (14) (27) \sin {44.64}^{o} \\ = 132.80 {cm}^{2} \\ Area △ A C D \\ = \frac{1}{2} (19.67) (6) \sin {105.36}^{o} \\ = 56.90 {cm}^{2} \\ Area of quadrilateral A B C D \\ = 132.80 + 56.90 \\ = 189.7 {cm}^{2} \end{array}

Short Questions (Question 17 – 19)

Posted on May 16, 2020 by Myhometuition

Question 17:

In the diagram above, the straight line PR is normal to the curve

y = \frac{x^{2}}{2} + 1

at Q. Find the value of k.

Solution:

\begin{array}{l} y = \frac{x^{2}}{2} + 1 \\ \frac{d y}{d x} = x \\ At point Q, x -coordinate = 2, \\ Gradient of the curve, \frac{d y}{d x} = 2 \\ Hence, gradient of normal to the curve, P R = - \frac{1}{2} \\ \frac{3 - 0}{2 - k} = - \frac{1}{2} \\ 6 = - 2 + k \\ k = 8 \end{array}

Question 18:
The normal to the curve y = x² + 3x at the point P is parallel to the straight line
y = –x + 12. Find the equation of the normal to the curve at the point P.

Solution:

Given normal to the curve at point P is parallel to the straight line y = –x + 12.

Hence, gradient of normal to the curve = –1.

As a result, gradient of tangent to the curve = 1

y = x² + 3x

$\frac{d y}{d x}$ = 2x + 3

2x + 3 = 1

2x = –2

x = –1

y = (–1)²+ 3(–1)

y = –2

Point P = (–1, –2).

Equation of the normal to the curve at point P is,

y – (–2) = –1 (x – (–1))

y + 2 = – x – 1

y = – x– 3

Question 19:

Given that

y = \frac{3}{4} x^{2}

, find the approximate change in x which will cause y to decrease from 48 to 47.7.

Solution:

\begin{array}{l} y = \frac{3}{4} x^{2} \\ \frac{d y}{d x} = (2) \frac{3}{4} x = \frac{3}{2} x \\ δ y = 47.7 - 48 = - 0.3 \\ Approximate change in x to y \\ \frac{δ x}{δ y} \approx \frac{d x}{d y} \\ δ x = \frac{d x}{d y} \times δ y \\ δ x = \frac{2}{3 x} \times (- 0.3) \\ δ x = \frac{2}{3 (8)} \times (- 0.3) \leftarrow \begin{array}{l} y = 48 \\ \frac{3}{4} x^{2} = 48 \\ x^{2} = 64 \\ x = 8 \end{array} \\ δ x = - 0.025 \end{array}

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