Answer :

Given:

< 2, x ϵ R.

Intervals of |x - 3|

|x – 3| = -(x – 3) or (x – 3)

When |x - 3| = x – 3

x – 3 ≥ 0

Therefore, x ≥ 3

When |x - 3| = -(x – 3)

(x – 3) < 0

Therefore, x < 3

Intervals: x ≥ 3 or x < 3

Domain of < 2:

is not defined for x = 0

Therefore, x > 0 or x < 0

Now, combining intervals and domain:

x < 0 or 0 < x < 3 or x ≥ 3

For x = 0

→

Now, subtracting 2 from both the sides

Signs of 3 – 4x:

3 – 4x = 0 →

(Subtracting 3 from both the sides and then dividing both sides by -1)

3 – 4x > 0 →

(Subtracting 3 from both the sides and then multiplying both sides by -1)

3 – 4x < 0 →

(Subtracting 3 from both the sides and then multiplying both sides by -1)

Signs of x:

x = 0

x < 0

x > 0

Intervals satisfying the required condition: < 0

x < 0 or

Combining the intervals:

x < 0 or and x < 0

Merging the overlapping intervals:

x < 0

Similarly, for 0 < x < 3

x < 0 or and 0 < x < 3

Merging the overlapping intervals:

< x < 3

For, x ≥ 3

→

Now, subtracting 2 from both the sides

Signs of -3 – 2x:

-3 – 2x = 0 →

(Adding 3 to both the sides and then dividing both sides by -2)

-3 – 2x > 0 →

(Adding 3 to both the sides and then multiplying both sides by -1)

-3 – 2x < 0 →

(Adding 3 to both the sides and then multiplying both sides by -1)

Signs of x:

x = 0

x < 0

x > 0

Intervals satisfying the required condition: < 0

or x > 0

Combining the intervals:

or x > 0 and x ≥ 3

Merging the overlapping intervals:

x ≥ 3

Combining all the intervals:

x < 0 or or x ≥ 3

Merging overlapping intervals:

x < 0 and

Therefore,

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