If the function f defined on ( π 6, π 3) by f(x)= … is continuous, then k is equal to

$\because f(x)$∵f(x) is continuous at $x=\frac{\pi }{4}$x=4π​

$\therefore f(\frac{\pi }{4})=k=\underset{x\rightarrow \frac{\pi }{4}}{\mathrm{lim}}(\frac{\sqrt{2}\mathrm{cos}x-1}{\mathrm{cot}x-1})\rightarrow (\frac{0}{0})$∴f(4π​)=k=x→4π​lim​(cotx−12​cosx−1​)→(00​) form

Applying L' Hospital Rule

$$k=\underset{x\rightarrow \frac{\pi }{4}}{lim}(\frac{\sqrt{2}(-\mathrm{sin}x)}{-{\mathrm{cosec}}^{2}x})$$k=x→4π​lim​(−cosec2x2​(−sinx)​)

$$\Rightarrow k=\frac{\sqrt{2}\mathrm{sin}\frac{\pi }{4}}{{\mathrm{cosec}}^{2}\frac{\pi }{4}}=\frac{\sqrt{2}\times (\frac{1}{\sqrt{2}})}{{(\sqrt{2})}^{2}}=\frac{1}{2}.$$⇒k=cosec24π​2​sin4π​​=(2​)22​×(2​1​)​=21​.