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JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 8)

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
$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$
Let $$g\left( x \right) = \int\limits_0^{{e^x}} {{{f'\left( t \right)} \over {1 + {t^2}}}} \,dt.$$

Which of the following is true?
$$g'(x)$$ is positive on $$\left( { - \infty ,0} \right)$$ and negative on $$\left( {0,\infty } \right)$$
$$g'(x)$$ is negative on $$\left( { - \infty ,0} \right)$$ and positive on $$\left( {0,\infty } \right)$$
$$g'(x)$$ changes sign on both $$\left( { - \infty ,0} \right)$$ and $$\left( {0,\infty } \right)$$
$$g'(x)$$ does not change sign on $$\left( { - \infty ,0} \right)$$

Explanation

Given $$g'(x) = \int\limits_0^{{e^x}} {{{f'(t)dt} \over {1 + {t^2}}}} $$

$$g''(x) = {{dg(x)} \over {d({e^x})}}.\,{{d({e^x})} \over {dx}}$$

$$ = {{f'({e^x})} \over {1 + {e^{2x}}}}\,.\,{e^x}$$

$$ = {{{e^x}} \over {1 + {e^{2x}}}}\,.\,{{2a({e^{2x}} - 1)} \over {{{({e^{2x}} + a{e^x} + 1)}^2}}} > 0$$

when $$x \in (0,\infty )$$

$$g'(x)$$ < 0 when $$x \in ( - \infty ,0)$$

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