If the function ƒ defined on, ((π)/6,(π)/3) by f(x) = \ 2 cosx - 1 cot x - 1, x ≠ (π)/4 k, x = (π)/4. is continuous, then k is equal to

Limit, Continuity and Differentiability: JEE Main 2019 Mathematics question

JEE Main 2019 · 9 Apr, Morning shift · Q4135

JEE Main 20199 Apr, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

If the function ƒ defined on ,

\left( {{\pi \over 6},{\pi \over 3}} \right)

f(x) = \left\{ {\matrix{ {{{\sqrt 2 {\mathop{\rm cosx}\nolimits} - 1} \over {\cot x - 1}},} & {x \ne {\pi \over 4}} \cr {k,} & {x = {\pi \over 4}} \cr } } \right.

is continuous, then k is equal to

JEE Main 2019 · 9 Apr, Morning shift · Q4135

JEE Main 20199 Apr, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

If the function ƒ defined on ,

\left( {{\pi \over 6},{\pi \over 3}} \right)

f(x) = \left\{ {\matrix{ {{{\sqrt 2 {\mathop{\rm cosx}\nolimits} - 1} \over {\cot x - 1}},} & {x \ne {\pi \over 4}} \cr {k,} & {x = {\pi \over 4}} \cr } } \right.

is continuous, then k is equal to