Question

1/mod x - 3 ≤ 1/2, x € R.

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Answer 1

Answer:

Consider the inequality, ∣x−1∣+∣x+2∣≥3

⟹x−1+x+2≥3 and x−1+x+2≤−3

⟹2x+1≥3 and 2x+1≤−3

⟹2x≥3−1 and 2x≤−3−1

⟹2x≥2 and 2x≤−4

⟹x≥1 and x≤−2

The solution set is,

From x≥1 we get [1,∞)

From x≤−2 we get (−∞,−2]

Solution : A and B.

sorry

Answer 2

Step-by-step explanation:

1/mod x - 3 ≤ 1/2, x € R.

Method-1

As you can see that the difficulty arises due to the terms |x-2| and |x-3|. So, we’ll solve it by simplifying our problem and breaking it down for various cases of x. Though this particular problem has an easy solution but I’m showing you the method that you can apply even when there are inequalities instead of equal signs or even if there are more than two modulus terms

R=(−∞,2)∪(2,3)∪(3,∞)∪{2,3}

Hence we have-

⋆Case-1–––––––x∈(−∞,2)

Hence in this interval |x-2|<0 and |x-3|<0

Thus the equation becomes-

⟹−(x−2)−(x−3)=1

⟹−x+2−x+3=1

⟹−2x=−4

⟹x=2which is not possible as x<2

Hence no possible solution from this interval

⋆Case-2–––––––x∈(2,3)

Hence in this interval |x-2|>0 and |x-3|<0

Thus the equation becomes-

⟹(x−2)−(x−3)=1

⟹x−2−x+3=1

⟹1=1which is true

Hence every x belonging to this interval is a solution

⋆Case-3–––––––x∈(3,∞)

Hence in this interval |x-2|>0 and |x-3|>0

Thus the equation becomes-

⟹(x−2)+(x−3)=1

⟹x−2+x−3=1

⟹2x=6

⟹x=3which is not possible as x>3

Hence no possible solution from this interval

⋆Case-4–––––––x∈{2,3}

Let first x=2, then |x-2|=0 and|x-3|=|2-3|=1

Thus the equation becomes-⟹0+1=0which is true

Hence one possible solution is x=2

Now let x=3, then |x-2|=|3-2|=1 and|x-3|=0

Thus the equation becomes-

⟹1+0=0which is true

Hence one possible solution is x=3

So, combining the 3 cases we see that possible solutions are x∈[2,3]

Method-2

Another way of solving this particular problem is the graphical method:

|x−2|+|x−3|=1⟹|x−2|=1−|x−3|⟹|x−2|=(−|x−3|)+1