Let f:(- (π)/4,(π)/4) → R be defined as f(x) = \ (1 + | sin x|)^(3a/(| sin x|)), - (π)/4 < x < 0 b, x = 0 e^(cot 4x/ cot 2x), 0 < x < (π)/4. If f is continuous at x = 0, then the value of 6a + b² is equal to:

Limit, Continuity and Differentiability: JEE Main 2021 Mathematics question

JEE Main 2021 · 27 Jul, Morning shift · Q4149

JEE Main 202127 Jul, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

Let

f:\left( { - {\pi \over 4},{\pi \over 4}} \right) \to R

be defined as

f(x) = \left\{ {\matrix{ {{{(1 + |\sin x|)}^{{{3a} \over {|\sin x|}}}}} & , & { - {\pi \over 4} < x < 0} \cr b & , & {x = 0} \cr {{e^{\cot 4x/\cot 2x}}} & , & {0 < x < {\pi \over 4}} \cr } } \right.

If f is continuous at x = 0, then the value of 6a + b² is equal to :

JEE Main 2021 · 27 Jul, Morning shift · Q4149

JEE Main 202127 Jul, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

Let

f:\left( { - {\pi \over 4},{\pi \over 4}} \right) \to R

be defined as

f(x) = \left\{ {\matrix{ {{{(1 + |\sin x|)}^{{{3a} \over {|\sin x|}}}}} & , & { - {\pi \over 4} < x < 0} \cr b & , & {x = 0} \cr {{e^{\cot 4x/\cot 2x}}} & , & {0 < x < {\pi \over 4}} \cr } } \right.

If f is continuous at x = 0, then the value of 6a + b² is equal to :