26) If log2 x2 = 2, then x=
A) 2
B) -2
C) 3
D) -3
Explanation:To solve the equation 
, let’s follow these steps:
Step 1: Rewrite the Logarithmic Equation in Exponential Form
Given:
This means:
Step 2: Solve for x
Take the square root of both sides:
Step 3: Verify the Solutions
Both x = 2 and x = -2 satisfy the original equation because:
Final Answer
Both A. 2 and B. -2 are correct solutions. However, if the question expects a single answer, either 
or 
is acceptable.
If the options are to be interpreted as allowing multiple correct answers, both A and B are correct.
But typically, such questions expect one correct choice, so you can select either A. 2 or B. -2 based on the context.
However, looking at the options, both A and B are correct. If only one option is to be selected, the primary answer is: A. 2
(and B. -2 is also correct)
If the question implies selecting all correct options, then both A and B are correct.
But in most standard cases, the simplest answer is:
A. 2
*(Note: Since logarithms of squared numbers are defined for both positive and negative values, both solutions are valid.)
Conclusion
The correct options are A and B, but if only one answer is expected, A. 2 is the standard choice.
Final Answer: A. 2 (and B. -2 is also correct)
27) Set of even prime numbers is
A) {3,4}
B) {4,6,8}
C) {8,10}
D) {2}
28) If A∩B = B, then the correct statement is
A) A⊆B
B) B⊂A
C) A≠Φ
D) B≠Φ
29) Which of the following sets are finite
A) set of all natural numbers
B) set of all prime numbers
C) set of months in a year
D) none of these
30) The number of zeroes, a biquadratic polynomial can have at most is
A) 1
B) 2
C) 3
D) 4
Explanation:A biquadratic polynomial is a polynomial of degree 4, typically of the form:
Even though it’s written in terms of 
, its highest power is 4.
Shortcut:
A polynomial of degree n can have at most n real or complex zeroes.
So, a biquadratic polynomial (degree 4) can have:
At most 4 zeroes
Final Answer: 4