A) (2, 12)
B) \[(-2,12)\]
C) \[(2,\,\,-12)\]
D) None of these
Correct Answer: A
Solution :
\[{{(a+bx)}^{-2}}=\frac{1}{{{a}^{2}}}{{\left( 1+\frac{b}{a}x \right)}^{-2}}=\frac{1}{{{a}^{2}}}\left[ a+\frac{(-2)}{1!}\left( \frac{b}{a} \right)x+.... \right]\] Equating it to \[\frac{1}{4}-3x+....,\]we get\[a=2,b=12\].
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