Question

Write the following sets in the roaster form.

(i) A = {x | x is a positive integer less than 10 and 2x – 1 is an odd number}

(ii) C = {x : x2 + 7x – 8 = 0, x ∈ R}​

Answer 1

${\huge\red{A}\blue{n}\green{s}\pink{w}\purple{e}\red{r}{\huge{\mathcal{:-}}}}$

Solution:

(i) 2x – 1 is always an odd number for all positive integral values of x since 2x is an even number.

In particular, 2x – 1 is an odd number for x = 1, 2, … , 9.

Therefore, A = {1, 2, 3, 4, 5, 6, 7, 8, 9}

(ii) x2 + 7x – 8 = 0

(x + 8) (x – 1) = 0

x = – 8 or x = 1

Therefore, C = {– 8, 1}

Thank You

Answer 2

Correct Question:-

(i) A = {x | x is a positive integer less than 10 and 2x – 1 is an odd number}

(ii) C = {x : x² + 7x – 8 = 0, x ∈ R}

Answer:-

(i) x is a positive integer according to question and is less than 10.

Now, while substituting positive values less than 10 in 2x - 1, the answers we may get are : 1, 3, 5, 7, 9, 11, 13, 15, 17. All the results we get are odd.

So, the set in roaster form will be :-

A = {1, 3, 5, 7, 9, 11, 13, 15, 17}

(ii) x² + 7x - 8 = 0, Solving the Given quadratic equation.

x² + 8x - x + 8 = 0

x (x + 8) -1 (x + 8) = 0

(x - 1) (x + 8) =0

x = 1 or -8

So, both of them are Real numbers.

Hence, C = {1, -8}