Full Solution - Test No 10 (English)
Test No. 10 - Detailed Solution
Q1: Simplify \( \sqrt[3]{27x^6y^9z^3} \).
Step 1: Convert the radical form (cube root) to exponential form with power \( \frac{1}{3} \).
\( (27x^6y^9z^3)^{\frac{1}{3}} \)
Step 2: Write 27 as a power of 3 (\( 27 = 3^3 \)).
\( (3^3 x^6 y^9 z^3)^{\frac{1}{3}} \)
Step 3: Apply the power \( \frac{1}{3} \) to every term inside the bracket.
\( 3^{3 \times \frac{1}{3}} \cdot x^{6 \times \frac{1}{3}} \cdot y^{9 \times \frac{1}{3}} \cdot z^{3 \times \frac{1}{3}} \)
Step 4: Simplify the exponents.
\( 3^1 \cdot x^2 \cdot y^3 \cdot z^1 \)
Answer: \( 3x^2y^3z \)
Q2: Write the following in exponential form.
Rule: If \( \log_a x = y \), then \( a^y = x \).
(Base remains Base, the other two swap positions).
(i) \( 5 = \log_{10} 100000 \)
Base is 10, Exponent is 5.
\( 10^5 = 100000 \)
(ii) \( \frac{1}{2} = \log_9 3 \)
Base is 9, Exponent is \( \frac{1}{2} \).
\( 9^{\frac{1}{2}} = 3 \)
(iii) \( \log_2 16 = 4 \)
Base is 2, Exponent is 4.
\( 2^4 = 16 \)
Q3: If \( U=\{1,2,3 \dots 20\} \) and \( A=\{1,3,5 \dots 19\} \), prove that \( A \cup A' = U \).
Given:
\( U = \{1, 2, 3, \dots, 20\} \) (All integers from 1 to 20)
\( A = \{1, 3, 5, \dots, 19\} \) (Odd numbers from 1 to 20)
Step 1: Find \( A' \) (A Complement)
\( A' = U - A \) (Elements in U but not in A).
Since A contains all odd numbers, A' will contain all even numbers from 1 to 20.
\( A' = \{2, 4, 6, \dots, 20\} \)
Step 2: Find \( A \cup A' \)
Union means combining elements of both sets.
\( \{1, 3, 5, \dots\} \cup \{2, 4, 6, \dots\} \)
Combining odd and even numbers gives us all integers from 1 to 20.
\( = \{1, 2, 3, 4, \dots, 20\} \)
Conclusion:
Since this result is exactly equal to set U, hence proven:
\( A \cup A' = U \)
Q4: Factorize \( x^2 - x - 2 \).
Expression: \( x^2 - x - 2 \)
Logic: We need two numbers that:
1. Multiply to give -2.
2. Add to give -1.
Finding Factors:
The numbers are -2 and +1.
\( -2 \times 1 = -2 \) (Check)
\( -2 + 1 = -1 \) (Check)
Step 1: Split the middle term.
\( = x^2 - 2x + 1x - 2 \)
Step 2: Factor by grouping.
\( = x(x - 2) + 1(x - 2) \)
Step 3: Take \( (x-2) \) as common.
\( = (x - 2)(x + 1) \)
Final Answer: \( (x - 2)(x + 1) \)