Question

If $1\le x\le 2$ then $\sqrt{(x+2\sqrt{x-1})}+\sqrt{(x-2\sqrt{x-1})}=$

• A. $2\sqrt{x-1}$
• B. $2$
• C. $0$
• D. none of these

Answer 1

Answer: (B)

$2$

$(x+2\sqrt{x-1}{)}^{\frac{1}{2}}+(x-2\sqrt{x-1}{)}^{\frac{1}{2}}$

$=\left(\right(\sqrt{x-1}+1{)}^2{)}^{\frac{1}{2}}+\left(\right(\sqrt{x-1}-1{)}^2{)}^{\frac{1}{2}}$
$=\left(\right(\sqrt{x-1}+1{)}^{\frac{1}{2}}{)}^2+\left(\right(\sqrt{x-1}-1{)}^{\frac{1}{2}}{)}^2$
$=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|$

$=\sqrt{x-1}+1-\sqrt{x-1}+1$ ........... $(\because 1\le x\le 2)$
$=2$