If the function
$$f(x) = \left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)} & , & {x < 0} \cr k & , & {x = 0} \cr {{{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}} & , & {x > 0} \cr } } \right.$$ is continuous

at x = 0, then ${1 \over a} + {1 \over b} + {4 \over k}$ is equal to :

If f(x) is continuous at x = 0, RHL = LHL = f(0)<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}.{{\sqrt {{x^2} + 1} + 1} \over {\sqrt {{x^2} + 1} + 1}}$$ (Rationalisation)<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} - {{2{{\sin }^2}x} \over {{x^2}}}.\left( {\sqrt {{x^2} + 1} + 1} \right) = - 4$$<br><br>$$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} {1 \over x}\ln \left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)$$<br><br>$$\mathop {\lim }\limits_{x \to {0^ - }} {{\ln \left( {1{x \over a}} \right)} \over {\left( {{x \over a}} \right).\,a}} + {{\ln \left( {1 - {x \over b}} \right)} \over {\left( { - {x \over b}} \right)\,.\,b}}$$$= {1 \over a} + {1 \over b}$<br><br>So, ${1 \over a} + {1 \over b} = - 4 = k$<br><br>$\Rightarrow {1 \over a} + {1 \over b} + {4 \over k} = - 4 - 1 = - 5$