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Mathematics · 2019

JEE Main · 9 April 2019, Shift 1 · Q70

If the function f defined on ((π)/6, (π)/3) by f(x)= {(√ 2 cos x-1)/(cot x-1), x ≠ (π)/4; k, x=(π)/4} is continuous, then k is equal to:

If the function $\displaystyle f$ defined on $\displaystyle \left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by $$f(x)=\left\{\begin{array}{cc} \frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ \mathrm{k}, & x=\frac{\pi}{4} \end{array}\right. $$ is continuous, then k is equal to :
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.