SPM Practice Question 13 – 15

Posted on May 24, 2020 by Myhometuition

Question 13:
An arithmetic progression consists of 10 terms. The sum of the last 5 terms is 5 and the fourth term is 9. Find the sum of this progression.

Solution:

Question 14:
The sum of the first 6 terms of an arithmetic progression is 39 and the sum of the next 6 terms is –69. Find
(a) The first term and the common difference.
(b) The sum of all the terms from the 15th term to the 25th term.

Solution:

Question 15:
An arithmetic progression has 9 terms. The sum of the first four terms is 24 and the sum of all the odd number terms is 55. Find
(a) The first term and common difference,
(b) The seventh term.

Solution:

SPM Practice Question 4 – 6

Posted on May 24, 2020 by Myhometuition

Question 4:
It is given that -7, h, k, 20,… are the first four terms of an arithmetic progression. Find the value of h and of k.

Solution:

Question 5:
51, 58, 65,... 191 are the first n terms of an arithmetic progression. Find the value of n.

Solution:

Question 6:
Find the number of the multiples of 8 between 100 and 300.

Solution:

The nth Term of Arithmetic Progression (Example 3 & 4)

Posted on May 24, 2020 by Myhometuition

Example 3:
The volume of water in a tank is 75 litres on the first day. Subsequently, 15 litres of water is added to the tank everyday.
Calculate the volume, in litres, of water in the tank at the end of the 12th day.

Solution:

Volume of water on the first day = 75l

Volume of water on the second day = 75 + 15 = 90l

Volume of water on the third day = 90 + 15 = 105l

75, 90, 105, …..

AP, a = 75, d = 90 – 75 = 15

Volume of water on the 12^th day,

T₁₂ = a + 11d

T₁₂ = 75 + 11 (15)

T₁₂ = 240l

Example 4:

The first three terms of an arithmetic progression are 72, 65 and 58.

The nth term of this progression is negative.

Find the least value of n.

Solution:

72, 65, 58

AP, a = 72, d = 65 – 72 = –7

The nth term is negative,

T_n< 0

a + (n – 1) d < 0

72 + (n – 1) (–7) < 0

(n – 1) (–7) < –72

n – 1 > –72/ –7

n – 1 > 10.28

n > 11.28

n must be integer, n = 12, 13, 14, ….

Therefore, the least value of n = 12.

9.2.1 First Derivative for Polynomial Function (Examples)

Posted on May 24, 2020 by Myhometuition

Example:

Find

$\frac{d y}{d x}$ for each of the following functions:

(a) y = 12

(b) y = x⁴

(c) y = 3x

(d) y = 5x³

$\begin{array}{l} (e) y = \frac{1}{x} \\ (f) y = \frac{2}{x^{4}} \\ (g) y = \frac{2}{5 x^{2}} \\ (h) y = 3 \sqrt{x} \\ (i) y = 4 \sqrt{x^{3}} \end{array}$

Solution:

(a)
y = 12

$\frac{d y}{d x} = 0$

(b)
y = x⁴

$\frac{d y}{d x}$ = 4x³

(c)
y = 3x

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

(d)
y = 5x³

$\frac{d y}{d x}$ = 3(5x²) = 15x²

(e)

$\begin{array}{l} y = \frac{1}{x} = x^{- 1} \\ \frac{d y}{d x} = - x^{- 1 - 1} = - \frac{1}{x^{2}} \end{array}$

(f)

$\begin{array}{l} y = \frac{2}{x^{4}} = 2 x^{- 4} \\ \frac{d y}{d x} = - 4 (2 x^{- 4 - 1}) = - 8 x^{- 5} = - \frac{8}{x^{5}} \end{array}$

(g)

$\begin{array}{l} y = \frac{2}{5 x^{2}} = \frac{2 x^{- 2}}{5} \\ \frac{d y}{d x} = - 2 (\frac{2 x^{- 2 - 1}}{5}) = - \frac{4 x^{- 3}}{5} = - \frac{4}{5 x^{3}} \end{array}$

(h)

$\begin{array}{l} y = 3 \sqrt{x} = 3 {(x)}^{\frac{1}{2}} \\ \frac{d y}{d x} = \frac{1}{2} (3 x^{\frac{1}{2} - 1}) = \frac{3}{2} x^{- \frac{1}{2}} = \frac{3}{2 \sqrt{x}} \end{array}$

(i)

$\begin{array}{l} y = 4 \sqrt{x^{3}} = 4 {(x^{3})}^{\frac{1}{2}} = 4 x^{\frac{3}{2}} \\ \frac{d y}{d x} = \frac{3}{2} (4 x^{\frac{3}{2} - 1}) = 6 x^{\frac{1}{2}} = 6 \sqrt{x} \end{array}$

Change of Base of Logarithms – Example 5

Posted on May 23, 2020 by Myhometuition

Example 5
Given that

$\log_{2} 3 = 1.585$ and

$\log_{2} 5 = 2.322$ , evaluate the following.
(a)

$\log_{8} 15$
(b)

$\log_{5} 0.6$
(c)

$\log_{15} 30$
(d)

$\log_{16} 45$

Solution:

5.2(a) Example 1 (Logarithms)

Posted on May 22, 2020 by Myhometuition

Example 1
Find the value of each of the following :
(a)

$\log_{2} 64$
(b)

$\log_{3} 1$
(c)

$\log_{5} 5$
(d)

$\log_{\frac{1}{2}} 4$
(e)

$\log_{8} 0.25$

Example 2
Solve the following equations:
(a)

$\log_{3} 5 x = 2$
(b)

$\log_{x + 1} 81 = 2$
(c)

$\log_{x} 5 x - 6 = 2$

SPM Practice (Paper 1)

Posted on May 22, 2020 by Myhometuition

Question 14:

$Show that 6 x - 6 - 2 k x^{2} = x^{2} has no real roots if k > \frac{1}{4} .$

Solution:

Question 15:
The quadratic equation

$x^{2} + p x + q = 0$ has roots –2 and 6. Find
(a) the value of p and of q,
(b) the range of values of r for which the equation

$x^{2} + p x + q = r$ has no real roots.

Solution:

SPM Practice (Paper 1)

Posted on May 22, 2020 by Myhometuition

Question 9:
The roots of the equation

$6 x^{2} + h x + 1 = 0 are α and β, whereas 3 α and 3 β$ are the roots of the equation

$2 x^{2} - x + k = 0$ . Find the value of h and k.

Solution:

$\begin{array}{l} 6 x^{2} + h x + 1 = 0 \\ a = 6, b = h, c = 1 \\ Roots = α, β \\ sor : \\ α + β = - \frac{b}{a} \\ α + β = - \frac{h}{6} ......... (1) \\ p o r : \\ α β = \frac{c}{a} \\ α β = \frac{1}{6} ......... (2) \\ 2 x^{2} - x + k = 0 \\ a = 2, b = - 1, c = k \\ Roots = 3 α, 3 β \\ sor : \\ 3 α + 3 β = - \frac{b}{a} \\ 3 (α + β) = - \frac{(- 1)}{2} \\ α + β = \frac{1}{6} ......... (3) \\ p o r : \\ 3 α (3 β) = \frac{c}{a} \\ 9 α β = \frac{k}{2} \\ k = 18 α β ......... (4) \\ Substitute (3) into (1) \\ α + β = - \frac{h}{6} \\ \frac{1}{6} = - \frac{h}{6} \\ h = - 1 \\ Substitute (2) into (4) \\ k = 18 α β \\ k = 18 (\frac{1}{6}) \\ k = 3 \end{array}$

Question 10:
Find the range of values of p for which the equation

$2 x^{2} + 5 x + 3 - p = 0$ has two real distinct roots.

Solution:

SPM Practice (Paper 1)

Posted on May 22, 2020 by Myhometuition

Question 3:
Solve the following quadratic equations by using quadratic formula. Give your answer in four significant figures.

$\begin{array}{l} (a) (x + 1) (x - 5) = 15 \\ (b) \frac{x^{2} + 3 x - 2}{x^{2} - x - 1} = 3 \end{array}$

Solution:

Question 4:
If the roots of

$2 x^{2} + 4 x - 1 = 0 are α and β,$ find the equations whose roots are

$\begin{array}{l} (a) α^{2}, β^{2} \\ (b) α - β, β - α \end{array}$

Solution:

Question 5:
Write and simplify the equation whose roots are double the roots of

$3 x^{2} - 5 x - 1 = 0$ , without solving the given equation.

Solution:

5.5.1a Proving Trigonometric Identities Using Addition Formula And Double Angle Formulae (Part 2)

Posted on May 22, 2020 by Myhometuition

Example 3:

(a) Given that

$\sin P = \frac{3}{5} and \sin Q = \frac{5}{13},$ such that P is an acute angle and Q is an obtuse angle, without using tables or a calculator, find the value of cos (P + Q).

(b) Given that

$\sin A = - \frac{3}{5} and \sin B = \frac{12}{13},$ such that A and B are angles in the third and fourth quadrants respectively, without using tables or a calculator, find the value of sin (A – B).

Solution:

(a)

$\begin{array}{l} \sin P = \frac{3}{5}, \cos P = \frac{4}{5} \\ \sin Q = \frac{5}{13}, \cos Q = - \frac{12}{13} \\ \cos (P + Q) \\ = \cos A \cos B - \sin A \sin B \\ = (\frac{4}{5}) (- \frac{12}{13}) - (\frac{3}{5}) (\frac{5}{13}) \\ = - \frac{48}{65} - \frac{15}{65} \\ = - \frac{63}{65} \end{array}$

(b)

$\begin{array}{l} \sin A = - \frac{3}{5}, \cos A = - \frac{4}{5} \\ \sin B = - \frac{5}{13}, \cos B = \frac{12}{13} \\ s i n (A - B) \\ = s i n A \cos B - c o s A \sin B \\ = (- \frac{3}{5}) (\frac{12}{13}) - (- \frac{4}{5}) (- \frac{5}{13}) \\ = - \frac{36}{65} - \frac{20}{65} \\ = - \frac{56}{65} \end{array}$