If the function f(x) = \ (log e (1 - x + x²) + log e (1 + x + x²))/(sec x - cos x), x ((- π)/2,(π)/2) - 0 k, x = 0. is continuous at x = 0, then k is equal to:

Limit, Continuity and Differentiability: JEE Main 2022 Mathematics question

JEE Main 2022 · 26 Jul, Morning shift · Q4262

JEE Main 202226 Jul, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

If the function

f(x) = \left\{ {\matrix{ {{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \cr k & , & {x = 0} \cr } } \right.

is continuous at x = 0, then k is equal to:

JEE Main 2022 · 26 Jul, Morning shift · Q4262

JEE Main 202226 Jul, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

If the function

f(x) = \left\{ {\matrix{ {{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \cr k & , & {x = 0} \cr } } \right.

is continuous at x = 0, then k is equal to: