The values of x satisfying |x2-1|/|x2+x+1|<1 are
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Step-by-step explanation:
Answered by
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Step-by-step explanation:
Let mod (x-2) = A & mod ( x-3) = B, both A& B are positive.
And since A + B = 1
So A & B can hold values from 0 to 1
ie 0 < ,= A <,=1
Also, 0 < , = B < , = 1
=> 0 < mod (x - 2) < 1 ……………..(1)
And 0 < mod ( x-3) < 1 ……………(2)
=> x-2 = 0 through 1 ( by (1)
=> x = 2 through 3 ●●●●●●●(3)
Now, x-3 = 0 through 1
=> x = 3 through 4 ●●●●●●●●(4)
Now we select the values of x from ●●●(3) & ●●●(4) , which satisfy equations …(1) & ….(2) both
And those are
x = 2 through 3
ie, 2 <,= x <,= 3
So, values of x are all real numbers lying between 2 & 3 , including 2 & 3.
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