Let `f(x)=(1+m)x^2-2(1+3m)x+4(1+2m).` Numbe of interval values of `m` for which given qudratic expression is always positive is (A) `8` (B) `7` (C) `8` (D) `9`

Options

A. 6
B. 8
C. 7
D. 3

Answer

C

Explanation

The quadratic expression
`ax^(2)+bx+c,x in R` is always positive,
if `a gt 0 and Dlt0.`
So, the quadratic expression
`(1+2m)x^(2)-2(1+3m)^(2)-4(2m+1)(1+m),x inR` will be
always positive, if `1+2n gt 0 " "…(i)`
and `D=4(1+3m)^(2)-4(2m+m)lt0" "(ii)`
Form inequality Eq. (1), we get
`m gt-1/2" "...(iii)`
From inequatity Eq. (ii), we get
`1+9m^(2)+6m-(2m^(2)+3m+1)lt0`
`impliesm^(2)-6m-3lt0`
`implies[m-(3+sqrt12)][m-(3sqrt12)]lt0`
`"["becausem^(2)-6m-3=0impliesm=(6+-sqrt(36+12))/(2)=3+-sqrt12"]"`
`implies3-sqrt12klt3+sqrt12" "...(iv)`
From inequalities Eqs. (iii) and (iv), the integral values of m are `0,1,2,3,4,5,6`
Hence, the number of integral values of m is 7.