JEE Main 202131 Aug, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002If the function f(x)={1xloge(1+xa1−xb),x<0k,x=0cos2x−sin2x−1x2+1−1,x>0f(x) = \left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)} & , & {x < 0} \cr k & , & {x = 0} \cr {{{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}} & , & {x > 0} \cr } } \right.f(x)=⎩⎨⎧x1loge(1−bx1+ax)kx2+1−1cos2x−sin2x−1,,,x<0x=0x>0 is continuous at x = 0, then 1a+1b+4k{1 \over a} + {1 \over b} + {4 \over k}a1+b1+k4 is equal to :AOption A. −-−5BOption B. 5COption C. −-−4DOption D. 4CheckSkip · show solution
