Question

Given below are Matching type questions,with two columns(each having some items) each.Each item of Column$I$ has to be matched with the items of Column$II$, by encircling the correct match(es).
Note:An item of column$I$ can be matched with more than one item of column$II$.All the items of column$II$ have to be matched.
Observe the following columns:

Answer 1

$\left(A\right)$ We have ${\lambda }^2+{(\lambda +2)}^2<4$
$\Rightarrow 2{\lambda }^2+4\lambda <0$
$\Rightarrow \lambda (\lambda +2)<0$
$\Rightarrow -2<\lambda <0$
$\therefore -\frac{1}{2}\in (-2,0)$
and $-1\in (-2,0)$
Options $(1,2)$ are correct.
$\left(B\right)$ We have
${\lambda }^2+{(\lambda +2)}^2-2\lambda +4(\lambda +2)>0$
$\Rightarrow 2{\lambda }^2+6\lambda +12>0$
$\Rightarrow {\lambda }^2+3\lambda +6>0$
$\Rightarrow {\lambda }^2+3\lambda +6=0$
$\Rightarrow {(\lambda +\frac{3}{2})}^2+\frac{15}{4}>0$
$\therefore$ All options are correct.
$\left(c\right)$ For both the circles $r>0$
$\Rightarrow \sqrt{{\lambda }^2-4}>0$ and $\sqrt{(2{\lambda }^2-8)}>0$
Squaring both sides, we get
${\lambda }^2-4>0$ and ${\lambda }^2-4>0$
$\therefore {\lambda }^2>4$
$\Rightarrow \lambda <-2$ and $\lambda >2$
$\therefore \lambda \in (-\infty ,-2)\cup (2,\infty )$
Options $(4,5)$ are correct.