Question

Let `f(x)=x+2|x+1|+x-1|dotIf f(x)=k` has exactly one real solution, then the value of `k` is3 (b) 0 (c) 1(d) 2
A. 3
B. 0
C. 1
D. 2

Answer 1

Correct Answer - A
Let `f(x)=x+2|x+1|+2|x-1|`
`={(x-2(x+1)-2(x-1)",", x lt -1),(x+2(x+1)-2(x-1)",", -1 le x le 1),(x+2(x+1)+2(x-1)",", x gt 1):}`
`={(-3x",", x lt -1),(x+4",", -1 le x le 1),(5x",", x gt 1):}`