JEE MAIN - Mathematics (2021 - 25th July Morning Shift - No. 4)
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 :
$$f(x) = \left\{ {\matrix{ {{{\lambda \left| {{x^2} - 5x + 6} \right|} \over {\mu (5x - {x^2} - 6)}},} & {x < 2} \cr {{e^{{{\tan (x - 2)} \over {x - [x]}}}},} & {x > 2} \cr {\mu ,} & {x = 2} \cr } } \right.$$
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 :
e($$-$$e + 1)
e(e $$-$$ 2)
1
2e $$-$$ 1
Explanation
$$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} {e^{{{\tan (x - 2)} \over {x - 2}}}} = {e^1}$$
$$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} {{ - \lambda (x - 2)(x - 3)} \over {\mu (x - 2)(x - 3)}} = - {\lambda \over \mu }$$
For continuity $$\mu = e = - {\lambda \over \mu } \Rightarrow \mu = e,\lambda = - {e^2}$$
$$\lambda + \mu = e( - e + 1)$$
$$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} {{ - \lambda (x - 2)(x - 3)} \over {\mu (x - 2)(x - 3)}} = - {\lambda \over \mu }$$
For continuity $$\mu = e = - {\lambda \over \mu } \Rightarrow \mu = e,\lambda = - {e^2}$$
$$\lambda + \mu = e( - e + 1)$$
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