Answer:
Well, we have to see for three cases- i. x<1 , ii. 1<x<2 and iii. x>2 .
We need not consider 1 and 2 since we can see that the equation is not satisfied for x=1 or x=2 .
Case i.
x<1⟹x−1<0⟹|x−1|=−(x−1)=1−x and |x−2|=−(x−2)=2−x ( ∵x<1,∴x<2 and x−2<0 )
So here we get the equation as 1−x+2–x=4 , which gives 2x=−1.
∴x=−12=−0.5 .
What is the solution of |x-1|+|x-2|≥4?
Well, we have to see for three cases- i. x<1 , ii. 1<x<2 and iii. x>2 .
We need not consider 1 and 2 since we can see that the equation is not satisfied for x=1 or x=2 .
Case i.
x<1⟹x−1<0⟹|x−1|=−(x−1)=1−x and |x−2|=−(x−2)=2−x ( ∵x<1,∴x<2 and x−2<0 )
So here we get the equation as 1−x+2–x=4 , which gives 2x=−1.
∴x=−12=−0.5 .
Case ii.
1<x<2⟹x−1>0 but x−2<0 . ∴|x−1|=x−1 and |x−2|=2−x
So here we get the equation as x−1+2−x=4⟹1=4 which is not possible.
Hence we do not have any solution for the equation in the interval (1,2) .
Case iii.
x>2⟹x>1 . Hence x−1>0 and x−2>0 .
∴|x−1|=x−1 and |x−2|=x−2
So the equation becomes x−1+x−2=4 , which gives 2x=7 .
∴x=72=3.5 .
Hence we get two solutions for the given equation:
x=−12 and x=72
Why did we divide the domain (i.e. the set of real numbers) into three such parts?
If we want to define f(x)=|x−1|+|x−2| without using modulus symbols we would have to do it in the following way.
In other words, f(x) is not differentiable at x=1 and x=2 .
Step-by-step explanation:
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