Test No. 2 - Solution
Q1: Represent the following numbers on the number line.
(i) \( \sqrt{2} \)
Method (Decimal): Calculate the value of \( \sqrt{2} \).
\( \sqrt{2} \approx 1.414 \)
This number lies between integers 1 and 2. We divide the space between 1 and 2 into 10 smaller parts (0.1, 0.2...).
Since the value is approx 1.4, we mark the point at the 4th small mark after 1.
(ii) \( 4\frac{1}{3} \)
Convert to decimal:
\( 1 \div 3 = 0.333... \implies 4 + 0.33 = 4.33 \)
This number lies between 4 and 5. We mark the point roughly at the 3rd small mark (4.3).
(iii) \( \frac{5}{8} \)
Divide to find decimal:
\( 5 \div 8 = 0.625 \)
This lies between 0 and 1. Mark it slightly after the 6th small mark (0.6).
Q2: Write the following in ordinary notation.
Rules for Ordinary Notation:
1. If Power is Positive (+): Move the decimal point to the RIGHT.
2. If Power is Negative (-): Move the decimal point to the LEFT.
1. If Power is Positive (+): Move the decimal point to the RIGHT.
2. If Power is Negative (-): Move the decimal point to the LEFT.
(i) \( 8.04 \times 10^2 \)
The power is +2. Move the decimal point 2 places to the right.
\( 8.04 \xrightarrow{\text{1 jump}} 80.4 \xrightarrow{\text{2 jumps}} 804. \)
Answer: \( 804 \)
(ii) \( 4 \times 10^{-5} \)
The power is -5. Move the decimal point 5 places to the left. The decimal starts after 4 (\( 4.0 \)).
Start: 4
1 jump left: .4
2 jumps left: .04
3 jumps left: .004
4 jumps left: .0004
5 jumps left: .00004
1 jump left: .4
2 jumps left: .04
3 jumps left: .004
4 jumps left: .0004
5 jumps left: .00004
Answer: \( 0.00004 \)
Q3: Write two proper subsets of the following sets.
Definition: A Proper Subset of a set A is a subset that contains some elements of A but is not equal to A.
(i) \( \{0, 1\} \)
Possible subsets are: \( \phi, \{0\}, \{1\}, \{0, 1\} \).
Excluding the set itself (\(\{0,1\}\)), we can choose:
Answer: \( \{0\}, \{1\} \) (or \( \phi \))
(ii) \( \mathbb{Q} \) (Set of Rational Numbers)
Rational numbers (\( \mathbb{Q} \)) include all Integers (\( \mathbb{Z} \)) and Natural numbers (\( \mathbb{N} \)). Since these sets are contained within \( \mathbb{Q} \) and are smaller, they are proper subsets.
Answer: \( \mathbb{Z} \) (Integers), \( \mathbb{N} \) (Natural Numbers)
(iii) \( \mathbb{R} \) (Set of Real Numbers)
Real numbers (\( \mathbb{R} \)) contain both Rational (\( \mathbb{Q} \)) and Irrational numbers.
Answer: \( \mathbb{Q} \) (Rational Numbers), \( \mathbb{Z} \) (Integers)
Q4: Factorize \( x^2 - 11x + 24 \).
Expression: \( x^2 - 11x + 24 \)
Logic: We need to find two numbers that:
1. Multiply to give +24.
2. Add to give -11.
Checking Factors of 24:
• \( -1 \times -24 = 24 \) (Sum = -25) ✖
• \( -2 \times -12 = 24 \) (Sum = -14) ✖
• \( -3 \times -8 = 24 \) (Sum = -11) ✔ (Correct)
Step 1: Split the middle term \( -11x \) into \( -3x \) and \( -8x \).
\( = x^2 - 3x - 8x + 24 \)
Step 2: Group terms and take out common factors.
Take \( x \) common from first two terms: \( x(x - 3) \)
Take \( -8 \) common from last two terms: \( -8(x - 3) \)
(Note: Sign of +24 changes to -3 because we took -8 outside)
(Note: Sign of +24 changes to -3 because we took -8 outside)
Step 3: Combine the factors.
\( = (x - 3)(x - 8) \)
Final Answer: \( (x - 3)(x - 8) \)
No comments:
Post a Comment