Question

The set of values of $b$ for which the origin and the point $(1,1)$ lie on the same side of the straight line, $a^{2}x+aby+1=0\forall a\in R,b>0$ are:

• A. $b\in (2,4)$
• B. $b\in (0,2)$
• C. $b\in [0,2]$
• D. None of these

Answer 1

The correct option is B $b\in (0,2)$

$L=a^{2}x+aby+1$

for origin $(0,0)$

$\therefore L=0+0+1=1\Rightarrow +ve$

for $(1,1)L=a^2+ab+1$

for both points to be on same side nature of L should be same

$\therefore a^2+ab+1>0$

which is a quadratic expression in $a$

then $D<0$

$\Rightarrow b^2-\mathrm{4.1.1}<0$

$\Rightarrow b^2<4$

$\Rightarrow b^2-4<0$

$\Rightarrow -2<b<2$

$\therefore b\in (-2,2)$

but in question $b>0\left(given\right)$

$henceb\in (0,2)$