Let `f(x)=x^2-2x-1 AA x in R` Let `f:(-oo, a]->[b, oo)`, where a is the largest real number for which f(x) is bijective. If `f : R->R` , `g(x) = f(x) + 3x-1` , then the least value of function `y = g(|x|)` is

Options

A. `-9//4`
B. `-5//4`
C. `-2`
D. `-1`

Answer

C

Explanation

`f(x)=(x-1)^(2)-2`
`rArr" "a=1, b=-2`
Now, `g(x)=x^(2)+x-2=(x+(1)/(2))^(2)-(9)/(4)`
`rArr" "g(|x|)=(|x|+(1)/(2))^(2)-(9)/(4)`
`therefore" "g_("min")=g(0)=-2`
Now, `f:[1,oo)rarr[-2,oo),`
`rArr" "f^(-1):[-2,oo)rarr [1,oo),`
`f(x)=y`
`rArr" "x^(2)-2x-(1+y)=0`
`rArr" "x=(2pi sqrt(4+4(1+y)))/(2)`
`rArr" "x=1 pm sqrt(2+y)`
`"So, "f^(-1)(y)=1+ sqrt(2+y),`
`" "f^(-1)(x)=1+sqrt(2+x)`
Graph of `y=f(|x|)`

From graph `f(|x|)=k` has four roots if `k in (-2,-1).`