Consdier the equaiton `2 + |x^(2) + 4x + 3|= m , m in R` Set of all values of m so that the given equaition have two solutions is

Options

A. `(3, oo)`
B. `(2,oo)`
C. `{2} uu (3,oo)`
D. None of these

Answer

3

Explanation

Given equations is `2 + |x^(2) + 4x + 3|= m`
` therefore |x^(2) + 4x + 3| = m -2`
Given equation is meaningful if `m ge 2" "(1)`
`therefore x^(2) + 4x + 3= pm (m-2)`
`therefore x^(2) + 4x + 5-m =0 or x^(2) + 4x + 1 + m =0`
`therefore x = (-4pmsqrt(16-4(5-m)))/(2) or x = (-4pmsqrt(16-4(1+m)))/(2)`
`therefore x = -2 pm sqrt(m-1) or x = - 2 pm sqrt(3-m)`
For four distinct solutions `m - 1 ge 0` and `3-m ge 0`
`therefore m in (2,3) " "["Using (1)"]`
For two solutions `m-1 ge 0 and 3 - m le 0 rArr m le 0 rArr m ge 3`
or `m - 1 le 0 and 3 - m le 0`, not possible
or `m =2" "["Using (1)"]`
`therefore m in {2} uu (3,oo)`
Form three solutions, exactly one equation must given equal roots.
`therefore m = 1 and 3 - m ge 0 rArr` no values of m `("as " m ge 2)`
or `m = 3 and m ge 1 rArr m = 3`
Alternate method:
Let us draw the graph of `f(x) = |x^(2) + 4x+3|`

Now equations is `|(x+1) (x+3)|=m-2`
From the graph.
(1) for three solution `m - 2 =1 rArr m = 3`
(2) for four solutions
`0 lt m -2 lt 1 rArr 2 lt m lt 3`
(3) for only tow solutions
`m-2=0 or n - gt 1`
`rArr m = 2 or m gt 3`