Consdier the equaiton `2 + |x^(2) + 4x + 2_= m , m in R`
Set of all real values of m so that given equation have four distinct solutions, is

Options

A. `5`
B. `6`
C. `7`
D. `8`

Answer

C

Explanation

`(c )` Case I : When `Q` is a quadratic equation
`D_(P)=(m+3)^(2)` and `D_(Q)=(m-2)^(2)`
roots of `1^(st)` equation are `2`, `-(m+1)`
roots of `2^(nd)` equation are `-1`, `(1)/(1-m)`,
For exactly there elements in `P uu Q` two of the roots must be same
So we have following possibilities ltbr. `2=-(m+1)impliesm=-3`
`2=(1)/(1-m)impliesm=1//2`
`-m-1=-1impliesm=0`
`-(m+1)=(1)/(1-m)implies1-m^(2)=-1impliesm=+-sqrt(2)`
`(1)/(1-m)=-1impliesm=2`
Case II : Now if `m=1`, then `Q` becomes linear
roots of `Q` as `x=-1`
roots of `P` are `2` and `-2`
`implies3` elements in common
`:.` all permissible values of `m` are `{-3,(1)/(2),sqrt(2),-sqrt(2),2,0,1}`