Consdier the equaiton `2 + |x^(2) + 4x + 3_= m , m in R` Set of all real values of m so that given equation have four distinct solutions, is
A
`5`
B
`6`
C
`7`
D
`8`
Text Solution
Verified by Experts
The correct Answer is:
C
`(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}`
