8.2b Standard Normal Distribution Tables (Example 1)

Posted on May 21, 2020 by Myhometuition

8.2b Standard Normal Distribution Tables (Example 1)

Example 1:

Find the value of each of the following probabilities by reading the standard normal distribution tables.

(a) P (Z > 0.600)

(b) P (Z < –1.24)

(d) P (Z < 0.76)

Solution:

Standard Normal Distribution Table

*When reading the standard normal distribution tables, it involves subtraction of values.

(a)

From the standard normal distribution table, P (Z > 0.600) = 0.2743

(b)

P (Z < –1.24)

= P (Z > 1.24)

= Q (1.24)

= 0.1075 ← (reading from the standard normal distribution table)

(*In the standard normal distribution table, all the values of z are positive. As the curve is symmetrical about the vertical axis, the area of the shaded region in both of the graphs are the same.)

(c)

P (Z > –1.1)

= 1 – Area P

= 1 – Q (–1.1)

= 1 – 0.1357 ← (reading from the standard normal distribution table)

= 0.8643

(d)

P (Z < 0.76)

= 1 – Area P

= 1 – Q (0.76)

= 1 – 0.2236 ← (reading from the standard normal distribution table)

= 0.7764

4.3.1a Addition of Vectors (Examples)

Posted on May 21, 2020 by Myhometuition

Example 1:

Find

(a) The resultant vector of the addition of the two parallel vectors above

(b) The magnitude of the resultant vector.

Solution:

(a)

Resultant vector

= addition of the two vectors

$= \vec{P Q} + \vec{R S}$

(b)

Magnitude of the resultant vector

$\begin{array}{l} = | \vec{P Q} | + | \vec{R S} | \\ = | \sqrt{6^{2} + 8^{2}} | + | \sqrt{6^{2} + 8^{2}} | \\ = 10 + 10 \\ = 20 units \end{array}$

Example 2:

The diagram above shows a parallelogram OABC. M is the midpoint of BC. Vector

$\vec{O A} = \underset{˜}{a} and \vec{O C} = \underset{˜}{c} .$ Find the following vectors in terms of

$\underset{˜}{a} and \underset{˜}{c} .$

$\begin{array}{l} (a) \vec{O B} \\ (b) \vec{M B} \\ (c) \vec{O M} \end{array}$

Solution:

(a)

$\begin{array}{l} \vec{O B} = \vec{O A} + \vec{A B} \leftarrow Triangle Law of addition \\ = \vec{O A} + \vec{O C} \leftarrow \vec{A B} = \vec{O C} \\ = \underset{˜}{a} + \underset{˜}{c} \end{array}$

(b)

$\begin{array}{l} \vec{M B} = \frac{1}{2} \vec{C B} \leftarrow M is the midpoint of C B \\ = \frac{1}{2} \vec{O A} \\ = \frac{1}{2} \underset{˜}{a} \end{array}$

(c)

$\begin{array}{l} \vec{O M} = \vec{O C} + \vec{C M} \leftarrow Triangle Law of addition \\ = \vec{O C} + \vec{M B} \leftarrow \vec{C M} = \vec{M B} \\ = \underset{˜}{c} + \frac{1}{2} \underset{˜}{a} \end{array}$

3.6.1 Integration as the Summation of Volumes – Examples

Posted on May 21, 2020 by Myhometuition

Example 1:

Find the volume generated for the following diagram when the shaded region is revolved through 360° about the x-axis.

Solution:
Volume generated, V_x

$\begin{array}{l} V_{x} = π \int_{a}^{b} y^{2} d x \\ V_{x} = π \int_{2}^{4} {(3 x - \frac{8}{x})}^{2} d x \\ V_{x} = π \int_{2}^{4} (3 x - \frac{8}{x}) (3 x - \frac{8}{x}) d x \\ V_{x} = π \int_{2}^{4} (9 x^{2} - 48 + \frac{64}{x^{2}}) d x \\ V_{x} = π {[\frac{9 x^{3}}{3} - 48 x + \frac{64 x^{- 1}}{- 1}]}_{2}^{4} \\ V_{x} = π {[3 x^{3} - 48 x - \frac{64}{x}]}_{2}^{4} \\ V_{x} = π [(3 {(4)}^{3} - 48 (4) - \frac{64}{4}) - (3 {(2)}^{3} - 48 (2) - \frac{64}{2})] \\ V_{x} = π (- 16 + 104) \\ V_{x} = 88 π u n i t^{3} \end{array}$

Example 2:

Find the volume generated for the following diagram when the shaded region is revolved through 360° about the y-axis.

Solution:
Volume generated, V_y

$\begin{array}{l} V_{y} = π \int_{a}^{b} x^{2} d y \\ V_{y} = π \int_{1}^{2} {(\frac{2}{y})}^{2} d y \\ V_{y} = π \int_{1}^{2} (\frac{4}{y^{2}}) d y \\ V_{y} = π \int_{1}^{2} 4 y^{- 2} d y \\ V_{y} = π {[\frac{4 y^{- 1}}{- 1}]}_{1}^{2} = π {[- \frac{4}{y}]}_{1}^{2} \\ V_{y} = π [(- \frac{4}{2}) - (- \frac{4}{1})] \\ V_{y} = 2 π u n i t^{3} \end{array}$

3.5.1 Integration as the Summation of Areas – Examples

Posted on May 21, 2020 by Myhometuition

Example 1
Find the area of the shaded region.

Solution:

$\begin{array}{l} Area of the shaded region \\ = \int_{a}^{b} y d x \\ = \int_{0}^{4} (6 x - x^{2}) d x \\ = {[\frac{6 x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{4} \\ = [3 {(4)}^{2} - \frac{{(4)}^{3}}{3}] - 0 \\ = 26 \frac{2}{3} {unit}^{2} \end{array}$

Example 2
Find the area of the shaded region.

Solution:

y = x -----(1)

x = 8y – y²-----(2)

Substitute (1) into (2),

y = 8y – y²

y²– 7y = 0

y (y – 7) = 0

y = 0 or 7

From (1), x = 0 or 7

Therefore the intersection points of the curve and the straight line is (0, 0) and (7, 7).

Intersection point of the curve and y-axis is,

x = 8y – y²

At y-axis, x = 0

0 = 8y – y²

y (y – 8) = 0

y = 0, 8

Area of shaded region = (A₁) Area of triangle + (A₂) Area under the curve from y = 7 to y = 8.

$\begin{array}{l} = \frac{1}{2} \times base \times height + \int_{7}^{8} x d y \\ = \frac{1}{2} \times (7) (7) + \int_{7}^{8} (8 y - y^{2}) d y \\ = \frac{49}{2} + {[\frac{8 y^{2}}{2} - \frac{y^{3}}{3}]}_{7}^{8} \\ = 24 \frac{1}{2} + [4 {(8)}^{2} - \frac{{(8)}^{3}}{3}] - [4 {(7)}^{2} - \frac{{(7)}^{3}}{3}] \\ = 24 \frac{1}{2} + 85 \frac{1}{3} - 81 \frac{2}{3} \\ = 28 \frac{1}{6} {unit}^{2} \end{array}$

3.4b Laws of Definite Integrals

Posted on May 21, 2020 by Myhometuition

3.4b Laws of Definite Integrals

Example:
Given that

$\int_{3}^{7} f (x) d x = 5$ , find the values for each of the following:

$\begin{array}{l} (a) \int_{3}^{7} 6 f (x) d x \\ (b) \int_{3}^{7} [3 - f (x)] d x \\ (c) \int_{7}^{3} 2 f (x) d x \\ (d) \int_{3}^{4} f (x) d x + \int_{4}^{5} f (x) d x + \int_{3}^{7} f (x) d x \\ (e) \int_{3}^{7} \frac{f (x) + 7}{2} d x \end{array}$

Solution:

$\begin{array}{l} (a) \int_{3}^{7} 6 f (x) d x = 6 \int_{3}^{7} f (x) d x \\ = 6 (5) = 30 \\ (b) \int_{3}^{7} [3 - f (x)] d x = \int_{3}^{7} 3 d x - \int_{3}^{7} f (x) d x \\ = {[3 x]}_{3}^{7} - 5 \\ = [3 (7) - 3 (3)] - 5 \\ = 7 \\ (c) \int_{7}^{3} 2 f (x) d x = - \int_{3}^{7} 2 f (x) d x \\ = - 2 \int_{3}^{7} f (x) d x \\ = - 2 (5) \\ = - 10 \\ (d) \int_{3}^{4} f (x) d x + \int_{4}^{5} f (x) d x + \int_{3}^{7} f (x) d x \\ = \int_{3}^{7} f (x) d x \\ = 5 \\ (e) \int_{3}^{7} \frac{f (x) + 7}{2} d x = \int_{3}^{7} [\frac{1}{2} f (x) + \frac{7}{2}] d x \\ = \int_{3}^{7} \frac{1}{2} f (x) d x + \int_{3}^{7} \frac{7}{2} d x \\ = \frac{1}{2} \int_{3}^{7} f (x) d x + {[\frac{7 x}{2}]}_{3}^{7} \\ = \frac{1}{2} (5) + [\frac{7 (7)}{2} - \frac{7 (3)}{2}] \\ = \frac{5}{2} + 14 \\ = 16 \frac{1}{2} \end{array}$

Short Questions (Question 10)

Posted on May 18, 2020 by Myhometuition

Question 10 (4 marks):
Diagram 10 shows the position of three campsites A, B and C at a part of a riverbank drawn on a Cartesian plane, such that A and B lie on the same straight riverbank.

Diagram 10

Sam wants to cross the river from campsite C to the opposite riverbank where the campsites A and B are located.
Find the shortest distance, in m, that he can take to cross the river. Give your answer correct to four decimal places.

Solution:

$\begin{array}{l} Let the shortest distance from campsite C \\ to opposite riverbank is C D . \\ Area of △ A B C \\ = \frac{1}{2} | \begin{array}{l} - 2 7 4 \\ 35 - 3 \end{array} \begin{array}{l} - 2 \\ 3 \end{array} | \\ = \frac{1}{2} | (- 2) (5) + (7) (- 3) + (4) (3) \\ - (3) (7) - (5) (4) - (- 3) (- 2) | \\ = \frac{1}{2} | - 66 | \\ = 33 {units}^{2} \\ Distance of A B \\ = \sqrt{{(- 2 - 7)}^{2} + {(3 - 5)}^{2}} \\ = \sqrt{85} m \\ Area of △ A B C = 33 \\ \frac{1}{2} (A B) (C D) = 33 \\ \frac{1}{2} (\sqrt{85}) (C D) = 33 \\ C D = \frac{33 (2)}{\sqrt{85}} \\ C D = 7.1587 (4 d .p .) \\ Thus, the shortest distance is 7 .1587 m . \end{array}$

Short Question 9 & 10

Posted on May 18, 2020 by Myhometuition

Question 9 (2 marks):

$\begin{array}{l} (a) Given {}^{6}C_{n} > 1, list out all the \\ possible values of n . \\ (b) Given {}^{y}C_{m} = {}^{y}C_{n}, express y in \\ terms of m and n . \end{array}$

Solution:
(a)
n = 1, 2, 3, 4, 5

(b)
y = m + n

Question 10 (4 marks):
Danya has a home decorations shop. One day, Danya received 14 sets of cups from a supplier. Each set contained 6 pieces of cups of different colours.

(a) Danya chooses 3 sets of cups at random to be checked.
Find the number of different ways that Danya uses to choose those sets of cups.

(b) Danya takes a set of cups to display by arranging it in a row.
Find the number of different ways the cups can be arranged such that the blue cup is not displaced next to the red cup.

Solution:
(a)
Number of different ways 3 sets of cups at random to be checked
= ¹⁴C₃=364

(b)

Number of ways (Blue cup and red cup are next to each other)
= 5! × 2!
= 240

Number of different ways the cups can be arranged such that the blue cup is not displaced next to the red cup
= 6! – 240
= 720 – 240
= 480

Short Question 19

Posted on May 18, 2020 by Myhometuition

Question 19 (4 marks):
It is given that cos α = t where t is a constant and 0^o≤ α ≤ 90^o.
Express in terms of t
(a) sin (180^o + α),
(b) sec 2α.

Solution:
(a)

$\begin{array}{l} sin (180^{o} + α) \\ = \sin 180 \cos α + \cos 180 \sin α \\ = 0 - \sin α \\ = - \sin α \\ = - \sqrt{1 - t^{2}} \end{array}$

(b)

$\begin{array}{l} \sec 2 α = \frac{1}{\cos 2 α} \\ = \frac{1}{2 \cos^{2} α - 1} \\ = \frac{1}{2 t^{2} - 1} \end{array}$

Short Questions (Question 5 & 6)

Posted on May 18, 2020 by Myhometuition

Question 5 (4 marks):
Diagram shows a circle with centre O.

Diagram

PR and QR are tangents to the circle at points P and Q respectively. It is given that the length of minor arc PQ is 4 cm and

$O R = \frac{5}{α} cm .$
Express in terms of α,
(a) the radius, r, of the circle,
(b) the area, A, of the shaded region.

Solution:
(a)

$\begin{array}{l} Given s_{P Q} = 4 \\ r α = 4 \\ r = \frac{4}{α} cm \end{array}$

(b)

$\begin{array}{l} P R = \sqrt{{(\frac{5}{α})}^{2} - {(\frac{4}{α})}^{2}} \\ P R = \sqrt{\frac{9}{α^{2}}} \\ P R = \frac{3}{α} \\ A = Area of shaded region \\ A = Area of quadrilateral O P R Q - \\ Area of sector O P Q \\ = 2 (Area of △ O P R) - \frac{1}{2} r^{2} θ \\ = 2 [\frac{1}{2} \times \frac{3}{α} \times \frac{4}{α}] - [\frac{1}{2} \times {(\frac{4}{α})}^{2} \times α] \\ = \frac{12}{α^{2}} - \frac{8}{α} \\ = \frac{12 - 8 α}{α^{2}} {cm}^{2} \end{array}$

Question 6 (3 marks):
Diagram shows two sectors AOD and BOC of two concentric circles with centre O.

Diagram

The angle subtended at the centre O by the major arc AD is 7α radians and the perimeter of the whole diagram is 50 cm.
Given OB = r cm, OA = 2OB and ∠BOC = 2α, express r in terms of α.

Solution:

$\begin{array}{l} Length of major arc A O D \\ = 2 r \times 7 α \\ = 14 r α \\ Length of minor arc B O C \\ = r \times 2 α \\ = 2 r α \\ Perimeter of the whole diagram \\ = 50 cm \\ 14 r α + 2 r α + r + r = 50 \\ 16 r α + 2 r = 50 \\ 8 r α + r = 25 \\ r (8 α + 1) = 25 \\ r = \frac{25}{8 α + 1} \end{array}$

SPM Practice Question 15 & 16

Posted on May 18, 2020 by Myhometuition

Question 15 (2 marks):
It is given that the n^th term of a geometric progression is

$T_{n} = \frac{3 r^{n - 1}}{2}, r \neq k .$
State
(a) the value of k,
(b) the first term of progression.

Solution:
(a)
k = 0, k = 1 or k = -1 (Any one of these answer).

(b)

$\begin{array}{l} T_{n} = \frac{3}{2} r^{n - 1} \\ T_{1} = \frac{3}{2} r^{1 - 1} \\ = \frac{3}{2} r^{0} \\ = \frac{3}{2} (1) \\ = \frac{3}{2} \end{array}$

Question 16 (4 marks):
It is given that p, 2 and q are the first three terms of a geometric progression.
Express in terms of q
(a) the first term and the common ratio of the progression.
(b) the sum to infinity of the progression.

Solution:
(a)

$\begin{array}{l} T_{1} = p, T_{2} = 2, T_{3} = q \\ \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{2}} \\ \frac{2}{p} = \frac{q}{2} \\ p = \frac{4}{q} \\ First term, T_{1} = p = \frac{4}{q} \\ Common ratio = \frac{q}{2} \end{array}$

(b)

$\begin{array}{l} a = \frac{4}{q}, r = \frac{q}{2} \\ S_{\infty} = \frac{a}{1 - r} \\ = \frac{\frac{4}{q}}{1 - \frac{q}{2}} \\ = \frac{4}{q} \div [1 - \frac{q}{2}] \\ = \frac{4}{q} \div [\frac{2 - q}{2}] \\ = \frac{4}{q} \times \frac{2}{2 - q} \\ = \frac{8}{2 q - q^{2}} \end{array}$