Question

If ${mx}^2-9mx+5m+1>0$ $\forall X\in R,$ then m lies in the interval.

• A. $(-\frac{4}{61},0)$
• B. $(0,\frac{4}{61})$
• C. $(\frac{4}{61},\frac{61}{4})$
• D. $(-\frac{61}{4},0)$

Answer 1

Answer: (B)

$(0,\frac{4}{61})$

Discriminant$=\Delta <0$

$\Rightarrow {\left(9m\right)}^2-4\times m\times (5m+1)<0$

$\Rightarrow 81m^2-4m(5m+1)<0$

$\Rightarrow 81m^2-20m^2-4m<0$

$\Rightarrow 61m^2-4m<0$

$\Rightarrow 61m(m-\frac{4}{61})<0$

$\Rightarrow m(m-\frac{4}{61})<0$

$\Rightarrow m<0,m<\frac{4}{61}$

$\therefore m\in (0,\frac{4}{61})$