Question

Plz solve the questions 4,5 and 6.​

Answer 1

Q4- option d is the write answer because

In A = {-1,1}

For Given value of x = - 1

${x}^{2} = 1 \\ lhs = {x}^{2} = { - 1}^{2} = 1$

Therefore x = - 1 satisfy the equation

For Given value of x = 1

${x}^{2} = 1 \\ lhs = {x}^{2} = {1}^{2} = 1$

Therefore x = 1 satisfy the equation

Hence option d is the right answer

Q5- Set

Q6- option b is the empty set because the given equation is

${x}^{2} + 3 = 0 \\ {x}^{2} = - 3 \\ x = \sqrt{ - 3}$

Which is not possible in case of real numbers

Therefore option b is the empty set