Let the given quadratic expression (1+2m)x2−2(1+3m)x+4(1+m), is positive for all x∈R then 1+2m>0 . . .(i) D<0 ⇒4(1+3m)2−4(1+2m)4(1+m)<0 ⇒1+9m2+6m−4[1+2m2+3m]<0 ⇒m2−6m−3<0 ⇒m∈(3−2√3,3+2√3) From (i) ∴m>−
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m∈(3−2√3,3+2√3) Then, integral values of m={0,1,2,3,4,5,6} Hence, number of integral values of m=7