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Let f(x)=x+2|x+1|+x-1|dotIf f(x)=k has e...

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

Text Solution

Verified by Experts

The correct Answer is:
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):}`
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