QJEEM21Ma30cf: JEE Main 2021 Mathematics - Class 12 JEEM21Ma30cf Solution: If the function f ( x ) = { 1 x log e ( 1 + x a 1 − x b ) , x 0 …

Question

Verified Solution • 2025–26 Session moderate

If the function

f (x) = {\begin{matrix} \frac{1}{x} \log_{e} (\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}) & , & x < 0 \\ k & , & x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2} + 1} - 1} & , & x > 0 \end{matrix}

is continuous

at x = 0, then

{1 \over a} + {1 \over b} + {4 \over k}

is equal to :

Accepted Answer

If f(x) is continuous at x = 0, RHL = LHL = f(0)

\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)

\mathop {\lim }\limits_{x \to {0^ + }} - {{2{{\sin }^2}x} \over {{x^2}}}.\left( {\sqrt {{x^2} + 1} + 1} \right) = - 4

\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)

\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}

So,

{1 \over a} + {1 \over b} = - 4 = k

\Rightarrow {1 \over a} + {1 \over b} + {4 \over k} = - 4 - 1 = - 5

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