Question Id : 15778 |
Context : JEE Main 2019 (9 April Morning Shift)
If the function $f$ defined on $\left(\dfrac{\pi}{6}, \dfrac{\pi}{3}\right)$ by
$f(x) =
\begin{cases}
\dfrac{\sqrt{2}\cos x - 1}{\cot x - 1}, & x \ne \dfrac{\pi}{4} \
k, & x = \dfrac{\pi}{4}
\end{cases}$
is continuous, then $k$ is equal to
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