We'll use the underlying symmetry of an inequality with absolute value.
This is a Logical approach.

All our answers are of the form |x + a| <b.
This can be written as -b < x+a < b.
So, we'd like to rewrite our original expression, -3 < x < 4, so that the number on the left and that on the right are the same.
We can SEE that if we subtract 1/2 from all sides we get -3.5 < x - 1/2 < 3.5 which is what we want.
Then |x - 1/2| <3.5 is our answer.
(A) is our answer.
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Easiest way i think is by inputting values
input 0

A. \(|x - \frac{1}{2}| < \frac{7}{2}\)
Pass
B. \(|x - \frac{1}{2}| > \frac{7}{2}\)
Fails
C. \(|x - \frac{1}{2}| \leq \frac{7}{2}\)
Pass
D. \(|x - \frac{1}{2}| \geq \frac{7}{2}\)
Fails
Fails as -3 is not in the range and this option gives RHS = LHS at -3.
E. \(|2x - \frac{1}{2}| < \frac{7}{2}\)
Pass

Input -3 in A, C and E

A. \(|x - \frac{1}{2}| < \frac{7}{2}\)
Pass

C. \(|x - \frac{1}{2}| \leq \frac{7}{2}\)
Fails

E. \(|2x - \frac{1}{2}| < \frac{7}{2}\)
Fails
\(|2*-3 - \frac{1}{2}| = |-6 - \frac{1}{2} | = |\frac{-13}{2}| = \frac{13}{2}which is greater than \frac{7}{2}\) proving the option wrong.

Please give Kudos in case post was helpful.
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Mid point = \(\frac{1}{2}\)
Distance from either side = \(\frac{7}{2}\)
End points exclusive

\(|x-\frac{1}{2}| < \frac{7}{2}\)

IMO A
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