Question

If the function $f$ defined on $(\frac{\pi }{6},\frac{\pi }{3})$ by $f\left(x\right)=\{\begin{array}{cc}\frac{\sqrt{2}\cos x-1}{\cot x-1},& x\ne \frac{\pi }{4}\\ k,& x=\frac{\pi }{4}\end{array}$, is continuous, then $k$ is equal to:

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

Answer 1

The correct option is B $\frac{1}{2}$

If the function is continuous at $x=\frac{\pi }{4}$,
$\Rightarrow \underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)=f\left(\frac{\pi }{4}\right)\Rightarrow \underset{x\to \frac{\pi }{4}}{lim}\frac{\sqrt{2}\cos x-1}{\cot x-1}=k$
As L.H.S. is of the form $\frac{0}{0},$
so, we can apply L Hospital's Rule.
$\Rightarrow \underset{x\to \frac{\pi }{4}}{lim}\frac{-\sqrt{2}\sin x}{-cose{c}^{2}x}=k\Rightarrow \underset{x\to \frac{\pi }{4}}{lim}\sqrt{2}{\sin }^{3}x=k\Rightarrow k=\sqrt{2}(\frac{1}{\sqrt{2}}{)}^3=\frac{1}{2}$