Question

Find the value of $k$ for which function

$f\left(x\right)=\left\{\begin{array}{l}\frac{(\sin x-\cos x)}{4\left(\mathrm{x}-\frac{\mathrm{\pi }}{4}\right)},x\ne \frac{\mathrm{\pi }}{4}\\ k,x=\frac{\mathrm{\pi }}{4}\end{array}\right.$. where $x$ is continuous at $x=\frac{\pi }{4}$

Answer 1

Find the value of $k$

$$\begin{gathered} \begin{array}{rcl}f\left(x\right)& =& \left\{\begin{array}{l}\frac{(\sin x-\cos x)}{4\left(x-\frac{\pi }{4}\right)},x\ne \frac{\pi }{4} \\ \\ k,x=\frac{\pi }{4}\end{array}\right.\end{array} \end{gathered}$$

Where $x$ is continuous at $\frac{\pi }{4}$

$$\begin{gathered} \therefore k=\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }f\left(x\right) \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\sin \left(x\right)-\cos \left(x\right)}{4\left(x-{\displaystyle \frac{\pi }{4}}\right)} \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\sqrt{2}\left[{\displaystyle \frac{1}{\sqrt{2}}}\sin \left(x\right)-{\displaystyle \frac{1}{\sqrt{2}}}\cos \left(x\right)\right]}{4\left(x-{\displaystyle \frac{\pi }{4}}\right)} \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{{\displaystyle \sqrt{2}\left[\cos \left(\frac{\mathrm{\pi }}{4}\right)\sin \left(x\right)-\sin \left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 4}}\right)\cos \left(x\right)\right]}}{4\left(x-{\displaystyle \frac{\pi }{4}}\right)}\left[\because cos\left(\frac{\pi }{4}\right)=sin\left(\frac{{\displaystyle \pi }}{{\displaystyle 4}}\right)=\frac{1}{\sqrt{2}}\right] \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{{\displaystyle \sqrt{2}\sin \left(x-\frac{\mathrm{\pi }}{4}\right)}}{4\left(x-{\displaystyle \frac{\pi }{4}}\right)}\left[\because cos\left(y\right)sin\left(x\right)-sin\left(y\right)cos\left(x\right)=sin\left(x-y\right)\right] \\ =\underset{x-\frac{\mathrm{\pi }}{4}\to 0}{\lim }\frac{{\displaystyle \sqrt{2}\sin \left(x-\frac{\mathrm{\pi }}{4}\right)}}{4\left(x-{\displaystyle \frac{\pi }{4}}\right)} \\ =\frac{\sqrt{2}}{4}\left[\underset{y\to 0}{\lim }\frac{\sin \left(y\right)}{y}=1\right] \end{gathered}$$

Hence the value of $k$is $\frac{\sqrt{2}}{4}$