JEE Main 202122 Jul, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002Let f : R →\to→ R be defined as f(x)={x3(1−cos2x)2loge(1+2xe−2x(1−xe−x)2),x≠0α,x=0f(x) = \left\{ {\matrix{ {{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr {\alpha ,} & {x = 0} \cr } } \right.f(x)={(1−cos2x)2x3loge((1−xe−x)21+2xe−2x),α,x=0x=0 If f is continuous at x = 0, then α\alphaα is equal to :AOption A. 1BOption B. 3COption C. 0DOption D. 2CheckSkip · show solution
