Given $f\left( x \right) = \frac{{ax + b}}{{x + 1}},\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 1$ and $\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 2$, then $f\left( { - 2} \right)$ is

JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Application +4 | -1

Given

f\left( x \right) = \frac{{ax + b}}{{x + 1}},\lim\limits_{x \to \infty } f\left( x \right) = 1

and

\lim\limits_{x \to 0} f\left( x \right) = 2

, then

f\left( { - 2} \right)

View answer and explanation

Correct option: A

Option A (0) is correct because by using the given limits, we find a=1, b=2, and substituting these values into f(-2) results in 0.

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