Question

Match the expression/statements on the left with the expression on the right.

Answer 1

A) Putting $x+2=t$, where $t>0$. Then
$\frac{x^2+2x+4}{x+2}=\frac{(t-2{)}^2+2(t-2)+4}{t}=\frac{t^2-2t+4}{t}=t+\frac{4}{t}-2={[\sqrt{t}-\frac{2}{\sqrt{t}}]}^2+2\ge 2$
$\therefore$ least value of $\frac{x^2+2x+4}{x+2}$ is 2 which is attained when $x=0$.
B) $(A+B)(A+B)=(A+B)(A-B)\Rightarrow AB=BA$
Now, $(AB{)}^{\prime }=B{}^{\prime }{A}^{\prime }=(-B)A=-BA=-AB$.
$\therefore (-1{)}^k=-1\Rightarrow k=1,3$
C) $1<2^{(-k+3^{-a})}<2\Rightarrow 0<-k+3^{-a}<1$
$\Rightarrow k<3^{-a}<k+1\Rightarrow k<({\log }_{3}2{)}^{-1}<k+1$
$\Rightarrow k<{\log }_{2}3<k+1\Rightarrow 2^k<3<2^{k+1}\Rightarrow k=1$
D) $k^2-3=\frac{(n-1)!}{r!(n-r-1)!}\frac{(r+1)!(n-r-1)!}{n!}=\frac{r+1}{n}$
But $0<\frac{r+1}{n}\le 1$.
Thus, $0<k^2-3\le 1\Rightarrow k=2$.