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Bunuel
Which of the following could be n if |2n – 3| < 1?

A. -4/3
B. -1
C. 4/3
D. 1
E. 0


sq. both sides to get
4n^2+9-12n<1
4n^2-12n+8<0
(2n-2)(n-4)<0

so break even points are 1 & 4 (shown below)
any number between 1& 4 willsatisfy the equation asked

thus 4/3 lies in between 1 & 4 will be answer

Ans C
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Bunuel
Which of the following could be n if |2n – 3| < 1?

A. -4/3
B. -1
C. 4/3
D. 1
E. 0

Let’s solve the given inequality for when (2n - 3) is positive and then when (2n - 3) is negative.

When (2n - 3) is positive, we have:

2n - 3 < 1

2n < 4

n < 2

Now we can solve for when (2n - 3) is negative:

-(2n - 3) < 1

-2n + 3 < 1

-2n < -2

n > 1

Thus, 1 < n < 2, so n could equal 4/3.

Answer: C
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kanusha
-1 <2n - 3< 1
3-1 < 2n < 1+3
\(\frac{2}{2} < n < \frac{4}{2}\)
1 < n < 2
C


This is by far the fastest method and should take no more than 30 seconds.
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Bunuel
Which of the following could be n if |2n – 3| < 1?

A. -4/3
B. -1
C. 4/3
D. 1
E. 0

Using POE:

A & B are out as they will add with 3 to get the value larger than 1.

D & E are out as they will yield |-1| = 1, & |-3| = 3

Hence (C)
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|2n-3| < 1

For cases when 2n-3>0 and 2n-3<0 we have two cases -
Step 1) -1 < 2n-3 < 1 (writing both inequalities together)
Step 2) simplify by adding 3 to all sides and dividing by 2.
2 < 2n < 4
1< n < 2

Hence C) fits best.
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