Let frac{1}{(x^2-3)^2} = frac{A_1}{x-sqrt{3}} | vedclass

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(C) Given $\frac{1}{(x^2-3)^2} = \frac{A_1}{x-\sqrt{3}} + \frac{A_2}{(x-\sqrt{3})^2} + \frac{A_3}{x+\sqrt{3}} + \frac{A_4}{(x+\sqrt{3})^2}$.Multiplyin

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Only statement (iv) is false

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(C) Given $\frac{1}{(x^2-3)^2} = \frac{A_1}{x-\sqrt{3}} + \frac{A_2}{(x-\sqrt{3})^2} + \frac{A_3}{x+\sqrt{3}} + \frac{A_4}{(x+\sqrt{3})^2}$.Multiplying by $(x^2-3)^2$,we get $1 = A_1(x-\sqrt{3})(x+\sqrt{3})^2 + A_2(x+\sqrt{3})^2 + A_3(x+\sqrt{3})(x-\sqrt{3})^2 + A_4(x-\sqrt{3})^2$.Setting $x = \sqrt{3}$,we get $1 = A_2(2\sqrt{3})^2 = 12A_2 \implies A_2 = \frac{1}{12}$.Setting $x = -\sqrt{3}$,we get $1 = A_4(-2\sqrt{3})^2 = 12A_4 \implies A_4 = \frac{1}{12}$.Comparing coefficients of $x^3$: $A_1 + A_3 = 0 \implies A_3 = -A_1$.Comparing coefficients of $x^2$: $A_1(\sqrt{3}) + A_2 + A_3(-\sqrt{3}) + A_4 = 0 \implies \sqrt{3}(A_1 - A_3) + A_2 + A_4 = 0$.Since $A_3 = -A_1$,we have $2\sqrt{3}A_1 + \frac{1}{12} + \frac{1}{12} = 0 \implies 2\sqrt{3}A_1 = -\frac{1}{6} \implies A_1 = -\frac{1}{12\sqrt{3}}$.Thus,$A_3 = \frac{1}{12\sqrt{3}}$.Values are $A_1 = -\frac{1}{12\sqrt{3}}, A_2 = \frac{1}{12}, A_3 = \frac{1}{12\sqrt{3}}, A_4 = \frac{1}{12}$.$(i)$ $A_i$'s are not distinct (True,$A_2=A_4$).(ii) $A_p^2 = A_q^2$ for $p=2, q=4$ (True,$(\frac{1}{12})^2 = (\frac{1}{12})^2$).(iii) $\sum A_i = A_1+A_2+A_3+A_4 = 0 + \frac{1}{12} + \frac{1}{12} = \frac{1}{6}$ (True).(iv) $\sum A_i = 1$ (False).Thus,only statement (iv) is false.

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