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JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 1)

Consider the function $$f:\left( { - \infty ,\infty } \right) \to \left( { - \infty ,\infty } \right)$$ defined by

$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$
Consider the function $$f:\left( { - \infty ,\infty } \right) \to \left( { - \infty ,\infty } \right)$$ defined by

$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$
Consider the function $$f:\left( { - \infty ,\infty } \right) \to \left( { - \infty ,\infty } \right)$$ defined by

$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$
Which of the following is true?
$$f(x)$$ is decreasing on $$(-1,1)$$ and has a local minimum at $$x=1$$
$$f(x)$$ is increasing on $$(-1,1)$$ and has a local minimum at $$x=1$$
$$f(x)$$ is increasing on $$(-1,1)$$ but has neither a local maximum nor a local minimum at $$x=1$$
$$f(x)$$ is decreasing on $$(-1,1)$$ but has neither a local maximum nor a local minimum at $$x=1$$

Explanation

As when $$x \in ( - 1,1),f'(x) < 0$$

So, f(x) is decreasing on ($$-1,1$$) at x = 1

$$f''(1) = {{4a} \over {{{(a + 2)}^2}}} > 0$$

So, local minima at x = 1

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