If the function \(f\) defined on \(\left( {\frac{\pi }{6},\frac{\pi }{3}} \right)\) by

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\sqrt 2 {\rm{cos}}x - 1}}{{{\rm{cot}}x - 1}},}&{x \ne \frac{\pi }{4}}\\ {{\rm{k}},}&{x = \frac{\pi }{4}} \end{array}} \right.\) is continuous, then k is equal to:This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 09 Apr 2019 Shift 1)

Option 2 : \(\frac{1}{2}\)

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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Since,f(x) is continuous, then

\(\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{4}} f\left( x \right) = f\left( {\frac{\pi }{4}} \right)\)

\(\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{4}} \frac{{\sqrt 2 {\rm{cos}}x - 1}}{{{\rm{cot}}x - 1}} = k\)

Now by L-hospital's rule

\(\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{4}} \frac{{-\sqrt 2 {\rm{sin}}x}}{{{-\rm{cose}}{{\rm{c}}^2}x}} = k\)

\(\Rightarrow \frac{{\sqrt 2 {\rm{sin}}\left( {\frac{\pi }{4}} \right)}}{{{\rm{cose}}{{\rm{c}}^2}\left( {\frac{\pi }{4}} \right)}} = k\)

\(\Rightarrow \frac{{\sqrt 2 \left( {\frac{1}{{\sqrt 2 }}} \right)}}{{{{(\sqrt 2 )}^2}}} = k\;\;\;\;\;\;\;\;\;\;\;\;\left[\because {{\rm{sin\;}}45^\circ = \frac{1}{{\sqrt 2 }},{\rm{\;cosec\;}}45^\circ = \sqrt 2 } \right]\)

\(\Rightarrow k = \frac{1}{2}\)