Question

If $6x-x^2=1$, then the value of $(\sqrt{x}-\frac{1}{\sqrt{x}})$ is

• A. $2$
• B. $3$
• C. $1$
• D. $-1$

Answer 1

Answer: (A)

$2$

Given:

$6x-x^2=1\Rightarrow 6x=x^2+1\Rightarrow \frac{x^2}{x}+\frac{1}{x}=\frac{6x}{6}\Rightarrow x+\frac{1}{x}=6\Rightarrow (\sqrt{x}{)}^2-2.\sqrt{x}.\frac{1}{\sqrt{x}}+(\frac{1}{\sqrt{x}}{)}^2=6-2$

...... $(a-b{)}^2=a^2-2ab+b^2$

$\Rightarrow (\sqrt{x}-\frac{1}{\sqrt{x}}{)}^2=4\Rightarrow (\sqrt{x}-\frac{1}{\sqrt{x}})=\pm \sqrt{4}\therefore (\sqrt{x}-\frac{1}{\sqrt{x}})=2$