Question

The number of integral values of $m$ for which the quadratic expression,
$(1+2m)x^2-2(1+3m)x+4(1+m),x\in R,$ is always positive, is

• A. $8$
• B. $7$
• C. $6$
• D. $3$

Answer 1

The correct option is B $7$

$(1+2m)x^2-2(1+3m)x+4(1+m)>0$

$\therefore D<0$

$4(1+3m{)}^2-16(1+2m\left)\right(1+m)<0$
$\Rightarrow 1+9m^2+6m-4-12m-8m^2<0$
$\Rightarrow m^2-6m-3<0\Rightarrow (m-3{)}^2<12\Rightarrow -2\sqrt{3}<m-3<2\sqrt{3}$
$\Rightarrow 3-2\sqrt{3}<m<3+2\sqrt{3}$

Also
$2m+1>0\Rightarrow m>-\frac{1}{2}$

Possible integral values of $m$ are $0,1,2,3,4,5,6$.

Hence, number of integral values of $m$ is $7$.