If |x2-2x-8|+|x2+x-2|=3|x+2| then the set of all real values of x is

Explanation:

Given that, 

|x2-2x-8|+|x2+x-2|=3|x+2|

|x2-4x+2x-8|+|x2+2x-x-2|=3|x+2|

|x(x-4)+2(x-4)|+|x(x+2)-1(x+2)|=3|x+2|

|(x+2)(x-4)|+|(x+2)(x-1)|-3|x+2|=0

|x+2|.|x-4|+|x+2|.|x-1|-3|x+2|=0

|x+2|{|x-4|+|x-1|-3}=0

Either, |x+2|=0x+2=0x=-2

Or, 

|x-4|+|x-1|-3=0

Above is possible when |x-4|=4-x and |x-1|=x-1 then 

|x-4|+|x-1|-3=4-x+x-1-3=0

|x-4|=4-xx4

|x-1|=x-1x1

Therefore 1x4x[1,4]

So the set of all values of x is given by, x[1,4]{-2}

Final Answer:

Hence the correct option is (A).i.e x[1,4]{-2}