Aspire Faculty ID #15883 · Topic: JEE Main 2019 (12 April Evening Shift) · Just now
JEE Main 2019 (12 April Evening Shift)

Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x $ \in $ R. If f(x) attains maximum value at $\alpha $ and g(x) attains minimum value at $\beta $, then $\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}$ is equal to :

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