If (a + bx)^{-2} = frac{1}{4} - 3x + dots,the | vedclass

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(A) Given $(a + bx)^{-2} = \frac{1}{4} - 3x + \dots$We can rewrite the expression as $\frac{1}{a^2} (1 + \frac{b}{a}x)^{-2} = \frac{1}{4} - 3x + \dots

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$(2, 12)$

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(A) Given $(a + bx)^{-2} = \frac{1}{4} - 3x + \dots$We can rewrite the expression as $\frac{1}{a^2} (1 + \frac{b}{a}x)^{-2} = \frac{1}{4} - 3x + \dots$Using the binomial expansion $(1 + z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots$,we get:$\frac{1}{a^2} [1 + (-2)(\frac{b}{a}x) + \dots] = \frac{1}{4} - 3x + \dots$$\frac{1}{a^2} - \frac{2b}{a^3}x + \dots = \frac{1}{4} - 3x + \dots$Comparing the constant terms: $\frac{1}{a^2} = \frac{1}{4} \implies a^2 = 4 \implies a = 2$ (assuming $a > 0$ for standard expansion).Comparing the coefficients of $x$: $-\frac{2b}{a^3} = -3$.Substituting $a = 2$: $-\frac{2b}{8} = -3 \implies -\frac{b}{4} = -3 \implies b = 12$.Thus,$(a, b) = (2, 12)$.

If $(a + bx)^{-2} = \frac{1}{4} - 3x + \dots$,then $(a, b) = $

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