A) a = 2, b = 3
B) a = 2, b = -6
C) a = 3, b = 2
D) a = -3, b = 2
Correct Answer: B
Solution :
| [b] Given expansion is \[{{(a+bx)}^{-3}}\] which can be written as |
| \[{{\left[ a\left( 1+\frac{b}{a}x \right) \right]}^{-3}}={{a}^{-3}}{{\left( 1+\frac{b}{a}x \right)}^{-3}}\] |
| \[={{a}^{-3}}\left( 1-\frac{3b}{a}x+6{{\left( \frac{b}{a}x \right)}^{2}}-...... \right)\] |
| \[(By\,\,using{{(1+x)}^{-3}}=1-3x+6{{x}^{2}}-.........)\] |
| But given that: \[{{(a+bx)}^{-3}}=\frac{1}{8}+\frac{9}{8}x+.......\] |
| \[\therefore \,\,\,\,{{a}^{-3}}\left[ 1-\frac{3b}{a}x+6\frac{{{b}^{2}}}{{{a}^{2}}}.{{x}^{2}}-.... \right]=\frac{1}{8}+\frac{9}{8}x+....\] |
| \[\Rightarrow \,{{a}^{-3}}=\frac{1}{8}={{2}^{-3}}\Rightarrow a=2\] |
| and \[-3b{{a}^{-4}}={{9.2}^{-3}}\Rightarrow b=-6\] |
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