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If the coefficient of $x^{7}$ in $\left [ ax^{2}+\left ( \frac{1}{bx} \right ) \right ]^{11}$ equals the coefficient of $x^{-7}$ in $\left [ ax-\left ( \frac{1}{bx^{2}} \right ) \right ]^{11}$ , then $a\; and\; b$ satisfy the relation

Option 1)
$a+b=1\;$

Option 2)
$\; a-b=1\;$

Option 3)
$\; ab=1\;$

Option 4)
$\; \frac{a}{b}=1$

Answers (1)

best_answer

As we learnt in

General Term in the expansion of (x+a)^n -

$T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}$

- wherein

Where $r\geqslant 0 \, and \, r\leqslant n$

$r= 0,1,2,----n$

in $\left[ax^{2}+\frac{1}{bx} \right ]^{11}$

General term $=\ ^{11}C_{r}(ax^{2})^{11-r}\left(\frac{1}{bx} \right )^{r}=\ ^{11}C_{r}\frac{a^{11-r}}{b^{r}}.x^{22-3r}$

For coefficient of x⁷, Put 22 - 3r = 7, r = 5

Also, in $\left[ax^{2}-\frac{1}{bx} \right ]^{11};$ General term $=\ ^{11}C_{r}(ax)^{11-r}\left(\frac{-1}{bx^{2}} \right )^{r}=\ ^{11}C_{r}\frac{a^{11-r}}{b^{r}}\times x^{11-3r}\times (-1)^{r}$

For coefficient of x^-7, Put 11 - 3r = - 7 $\Rightarrow\ \;r = 6$

We get $^{11}C_{5}\ \frac{a^{6}}{b^{5}}=\ ^{11}C_{6}\ \frac{a^{5}}{b^{6}}(-1)^{6}$

ab = 1

Correct option is 3.

Option 1)

$a+b=1\;$

This is an incorrect option.

Option 2)

$\; a-b=1\;$

This is an incorrect option.

Option 3)

$\; ab=1\;$

This is the correct option.

Option 4)

$\; \frac{a}{b}=1$

This is an incorrect option.

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