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JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 9)

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?
$${\left( {2 + a} \right)^2}f''\left( 1 \right) + {\left( {2 - a} \right)^2}f''\left( { - 1} \right) = 0$$
$${\left( {2 - a} \right)^2}f''\left( 1 \right) - {\left( {2 + a} \right)^2}f''\left( { - 1} \right) = 0$$
$$f'\left( 1 \right)f'\left( { - 1} \right) = {\left( {2 - a} \right)^2}$$
$$f'\left( 1 \right)f'\left( { - 1} \right) = -{\left( {2 + a} \right)^2}$$

Explanation

$$f(x) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}}$$

for differentiation, we can write f(x) as

$$f(x) = 1 - {{2ax} \over {{x^2} + ax + 1}}$$

Now, on differentiation

$$f'(x) = {{2a({x^2} - 1)} \over {{{({x^2} + ax + 1)}^2}}}$$ ..... (i)

$$f'(1) = 0 = f'( - 1)$$

Then again differentiation (1)

$${({x^2} + ax + 1)^2}\,.\,2x - ({x^2} - 1)$$

$$f''(x) = 2a.\,{{2({x^2} + ax + 1)(2x + a)} \over {{{({x^2} + ax + 1)}^4}}}$$

$$f''( - 1) = {{ - 4a} \over {{{(2 - a)}^2}}}$$

$$f''(1) = {{ - 4a} \over {{{(2 - a)}^2}}}$$

Combining both

$$f''(1){(2 + a)^2} + f''( - 1){(2 - a)^2} = 0$$

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