Question

For all $xϵR$, if $m{x}^2-9mx+(5m+1)>0$, then $m$ lies in the interval.

• A. $(0,\frac{4}{61})$
• B. $[\frac{4}{61},\frac{4}{61},)$
• C. $(-\frac{61}{4},0]$
• D. None of these

Answer 1

The correct option is A $(0,\frac{4}{61})$

R.E.F image

$m{x}^{2}9mx(5m+1)>0$

As the expression $m{x}^2-9mx+(5m+1)>0$ xeuer becomes 0, it doesn't have a solution

So $(9m{)}^2-4m(5m+1)<0$ ($b^{2}4ac<o$ for no solution)

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

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

$\Rightarrow m(61m-4)<0$

$\Rightarrow m<0$ and $m<4/16$

$\therefore mϵ(0,\frac{4}{61})$