Let f : R $\to$ R be defined as
$$f(x) = \left\{ {\matrix{
{{{\lambda \left| {{x^2} - 5x + 6} \right|} \over {\mu (5x - {x^2} - 6)}},} & {x < 2} \cr
{{e^{{{\tan (x - 2)} \over {x - [x]}}}},} & {x > 2} \cr
{\mu ,} & {x = 2} \cr
} } \right.$$
where [x] is the greatest integer is than or equal to x. If f is continuous at x = 2, then $\lambda$ + $\mu$ is equal to :
Solution
$$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} {e^{{{\tan (x - 2)} \over {x - 2}}}} = {e^1}$$<br><br>$$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} {{ - \lambda (x - 2)(x - 3)} \over {\mu (x - 2)(x - 3)}} = - {\lambda \over \mu }$$<br><br>For continuity $\mu = e = - {\lambda \over \mu } \Rightarrow \mu = e,\lambda = - {e^2}$<br><br>$\lambda + \mu = e( - e + 1)$
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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