Question

Match the elements of List I with elements of List II.

Answer 1

Element A:
If $ta{n}^{-1}(x^2+x+\frac{a}{4})$
$x^2+x+\frac{a}{4}>0$
$\Rightarrow {(x+\frac{1}{2})}^2+\left(\frac{a-1}{4}\right)>0,\forall xϵR$
$\Rightarrow a-1>0$
$\Rightarrow a>1$
Therefore, smallest $\left[a\right]$=$1$
Element B:
$\frac{x^2-3x+2}{x^2-7x+10}<0$
$\Rightarrow \frac{(x-1)(x-2)}{(x-2)(x-5)}<0,x\ne 2$
$\to xϵ(1,5)-2$
$\Rightarrow x=3$
Element C:
$[\overrightarrow{a}\times \overrightarrow{b}\overrightarrow{a}\overrightarrow{b}]=[\overrightarrow{a}\overrightarrow{b}\overrightarrow{a}\times \overrightarrow{b}]=(\overrightarrow{a}\times \overrightarrow{b})\cdot (\overrightarrow{a}\times \overrightarrow{b})$
$=|\overrightarrow{a}\times \overrightarrow{b}{|}^2=4$
Element D:
$f\left(x\right)=|x-0.5|+|x^2-1|+|x-2|$
For $x\in (-2,2)$ we have,
$f\left(x\right)=-x+0.5+x^2-1-x+2$ for $-2<x<-1$
$f\left(x\right)=-x+0.5-x^2+1-x+2$ for $-1<x<0.5$
$f\left(x\right)=x-0.5-x^2+1-x+2$ for $0.5<x<1$
$f\left(x\right)=x-0.5+x^2-1-x+2$ for $1<x<2$
By calculating the derivative we find that $f^{\prime }\left(x\right)$ is discontinuous at $x=-1,0.5,1$,
Hence, $3$ points.