- #1
mmaismma
- 18
- 1
Summary:: The set of values of ##b## for which the origin and the point ##(1, 1)## lie on the same side of the straight line ##a^2x+aby+1=0## ##\forall~a\in\mathbb{R},~b>0##.(a) ##a\geq1## or ##a\leq-3##
(b) ##a\in~(-3,~0)\cup(\frac13,~1)##
(c) ##a\in~(0,~1)##
(d) ##a\in~(-\infty,~0)##
I tried solving it but I didn't get an answer:
##f(x)=a^2+aby+1=0\\
f(0, 0)=0+0+1>0\\
So,~f(1, 1)>0\\
a^2+ab+1>0\\
{}^A/_Q~a\in\mathbb{R}\\
So,~D>0\\
b^2-4.1>0\\
b^2>4\\
b\in(2,~\infty)
##
(b) ##a\in~(-3,~0)\cup(\frac13,~1)##
(c) ##a\in~(0,~1)##
(d) ##a\in~(-\infty,~0)##
I tried solving it but I didn't get an answer:
##f(x)=a^2+aby+1=0\\
f(0, 0)=0+0+1>0\\
So,~f(1, 1)>0\\
a^2+ab+1>0\\
{}^A/_Q~a\in\mathbb{R}\\
So,~D>0\\
b^2-4.1>0\\
b^2>4\\
b\in(2,~\infty)
##