Full Solution - Test No 9 (English)
Test No. 9 - Detailed Solution
Q1: Simplify \( \sqrt[4]{16x^4y^8} \).
Step 1: Convert the radical form to exponential form.
\( (16x^4y^8)^{\frac{1}{4}} \)
Step 2: Factorize the coefficient (16) into prime factors.
\( 16 = 2 \times 2 \times 2 \times 2 = 2^4 \)
Substitute back:
\( (2^4 x^4 y^8)^{\frac{1}{4}} \)
Step 3: Apply the power to each term inside the bracket.
\( 2^{4 \times \frac{1}{4}} \cdot x^{4 \times \frac{1}{4}} \cdot y^{8 \times \frac{1}{4}} \)
Step 4: Simplify the exponents.
\( 2^1 \cdot x^1 \cdot y^2 \)
Answer: \( 2xy^2 \)
Q2: Write the following in logarithmic form.
Rule: Converting Exponential to Logarithmic form:
If \( a^x = y \), then \( \log_a y = x \).
(i) \( 16^{-\frac{1}{4}} = \frac{1}{2} \)
Here, Base \( a = 16 \), Exponent \( x = -\frac{1}{4} \), Value \( y = \frac{1}{2} \).
\( \log_{16} (\frac{1}{2}) = -\frac{1}{4} \)
(ii) \( p = q^r \)
Here, Base is \( q \), Exponent is \( r \), and Value is \( p \).
\( \log_q p = r \)
(iii) \( 11^2 = 121 \)
Here, Base is \( 11 \), Exponent is \( 2 \), and Value is \( 121 \).
\( \log_{11} 121 = 2 \)
Q3: Write the Power Set of the following sets.
Definition: The Power Set \( P(A) \) is the set of all possible subsets of A, including the empty set and the set itself.
(i) \( \{ \phi \} \)
This set contains one element (which happens to be the null set symbol). Let's call this set A.
Number of subsets = \( 2^n = 2^1 = 2 \).
1. The empty set: \( \phi \)
2. The set itself: \( \{ \phi \} \)
Answer: \( \{ \phi, \{ \phi \} \} \)
(ii) \( \{ \{a,b\}, \{b,c\}, \{d,e\} \} \)
Let the elements be \( x=\{a,b\} \), \( y=\{b,c\} \), and \( z=\{d,e\} \).
So the set is \( \{ x, y, z \} \). Number of subsets = \( 2^3 = 8 \).
The Subsets are:
1. Empty set: \( \phi \)
2. Singleton sets: \( \{x\}, \{y\}, \{z\} \)
3. Sets with two elements: \( \{x,y\}, \{x,z\}, \{y,z\} \)
4. The set itself: \( \{x,y,z\} \)
Replacing values back:
\( P(A) = \Big\{ \phi, \)
\( \{ \{a,b\} \}, \{ \{b,c\} \}, \{ \{d,e\} \}, \)
\( \{ \{a,b\}, \{b,c\} \}, \{ \{a,b\}, \{d,e\} \}, \{ \{b,c\}, \{d,e\} \}, \)
\( \{ \{a,b\}, \{b,c\}, \{d,e\} \} \Big\} \)
Q4: Factorize \( y^2 + 4y - 12 \).
Expression: \( y^2 + 4y - 12 \)
Logic (Mid-Term Break): We need two numbers that:
1. Multiply to give -12.
2. Add to give +4.
Checking Factors of -12:
• \( 12 \times -1 \) (Sum = 11) ✖
• \( 6 \times -2 \) (Sum = 4) ✔ (Correct)
Step 1: Split the middle term \( +4y \) into \( +6y \) and \( -2y \).
\( = y^2 + 6y - 2y - 12 \)
Step 2: Factor by grouping.
Take \( y \) common from first pair, and \( -2 \) from second pair.
\( = y(y + 6) - 2(y + 6) \)
(Note: Sign of -12 changes to +6 because we took -2 outside)
Step 3: Take \( (y+6) \) as the common factor.
\( = (y + 6)(y - 2) \)
Final Answer: \( (y + 6)(y - 2) \)