Q. 1: Write the following sets in the roaster form.
(i) A =
{x | x is a positive integer less than 10 and 2^(x) – 1 is an odd number}
(ii) C =
{x : x^(2) + 7x – 8 = 0, x ∈ R}
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Q. 1: Write the following sets in the roaster form.
(i) A =
{x | x is a positive integer less than 10 and 2^(x) – 1 is an odd number}
(ii) C =
{x : x^(2) + 7x – 8 = 0, x ∈ R}
PLS ANSWER FAST
THANKS FOR YOUR EFFORTS
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Answer:
.(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}
Step-by-step explanation:
given the Question:-
(i) A =
(i) A ={x | x is a positive integer less than 10 and 2^(x) – 1 is an odd number}
SOLVE the equation
We given a Set-Builder form of a Set A.
A = { x | x is positive integer less than 10 and 2^x-1 is an odd }
We need to write given set in Roaster Form.
Positive integers less than 10 = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9
Now, applying second condition we get,
we know
that 2x is always a even number.
So when 1 is subtracted from 2x
gives an odd number.
⇒ Roaster Form of
Set A = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 }
Therefore,
Set A = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 }
Given the second Question:-
Solve the Question:-
[tex] {x}^{2} +7x-8=0 \\
=> {x}^{2} +8x-1x-8=0 \\
=> x(x+8)-1(x+8)=0 \\
=>(x+8)(x-1)=0 \\
=> x+8=0 or x-1=0 \\
=> x=-8 or x=1 \\
[/tex]