Question

If $\alpha$ is positive root of the equation, $\begin{array}{rcl}p\left(x\right)& =& x^2-x-2=0\end{array}$, $\underset{x\to {\alpha }^{+}}{\lim }\frac{\sqrt{1-\cos p\left(x\right)}}{x+\alpha -4}$ is equal to:

• A. $\frac{1}{2}$
• B. $\frac{3}{\sqrt{2}}$
• C. $\frac{3}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B

$\frac{3}{\sqrt{2}}$

Find the value of $\underset{x\to {\alpha }^{+}}{\lim }\frac{\sqrt{1-\cos p\left(x\right)}}{x+\alpha -4}$:

Given, $\begin{array}{rcl}{x}^2-x-2& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{x}^2-2x+x-2& =& 0\end{array}$

$\Rightarrow$$x\left(x-2\right)+1\left(x-2\right)=0$

$\Rightarrow$ $\left(x+1\right)\left(x-2\right)=0$

$\Rightarrow$ $x=2,-1$

$\begin{array}{rcl}\therefore \underset{x\to {\alpha }^{+}}{\lim }\frac{\sqrt{1-\cos p\left(x\right)}}{x+\alpha -4}& =& \underset{x\to 2^{+}}{\lim }\frac{\sqrt{1-\cos \left(x^2-x-2\right)}}{x+2-4}\\ & =& \underset{x\to 2^{+}}{\lim }\frac{\sqrt{2{\sin }^2{\displaystyle \frac{\left(x^2-x-2\right)}{2}}}}{x-2}\\ & =& \underset{x\to 2^{+}}{\lim }\frac{\sqrt{2}\sin {\displaystyle \frac{\left(x-2\left)\right(x+1\right)}{2}}}{x-2}\\ & =& \frac{\mathbf{3}}{\sqrt{\mathbf{2}}}\end{array}$

Hence, Option ‘B’ is Correct.