If the function f defined on \left(\right. \frac{\pi}{6} , \frac{\pi}{3} \left.\right) by f \left(\right. x \left.\right) = \left{\right. \frac{\sqrt{2} cosx - 1}{cotx - 1} & , x \neq \frac{\pi}{4} \\ k & , x = \frac{\pi}{4} is continuous, then k is equal to :

Engineering

Mathematics

Basic Definition of Continuity

Question

If the function f defined on $(\frac{π}{6}, \frac{π}{3})$ by $f (x) = {\begin{matrix} \frac{\sqrt{2} cosx - 1}{cotx - 1} & , x \neq \frac{π}{4} \\ k & , x = \frac{π}{4} \end{matrix}$ is continuous, then k is equal to :

$\frac{1}{2}$

$\frac{1}{\sqrt{2}}$

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∴ function should be continuous at $x = \frac{π}{4}$

$∴ \lim_{x \to \frac{π}{4}} f (x) = f (\frac{π}{4})$

$\Rightarrow \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} cosx - 1}{cotx - 1} = k$

$\Rightarrow \lim_{x \to \frac{π}{4}} \frac{- \sqrt{2} sinx}{- {cosec}^{2} x} = k (Using L Hospital Rule)$

$\lim_{x \to \frac{π}{4}} \sqrt{2} \sin^{3} x = k$

$\Rightarrow k = \sqrt{2} {(\frac{1}{\sqrt{2}})}^{3} = \frac{1}{2}$