If the coefficient of x7 in (ax2+1bx)11 is equal to the coefficient of x&#

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If the coefficient of

x^{7}

in

(a x^{2} + \frac{1}{b x})^{11}

is equal to the coefficient of

x^{- 7}

in

(a x - \frac{1}{b x^{2}})^{11}

, then ab =

Options:

A . 1

B . 12

C . 2

D . 3

Answer: Option A
:
A
In the expansion of

(a x^{2} + \frac{1}{b x})^{11}

, the general
term is

T_{r + 1}

=

^{11} C_{r}

(a x^{2})^{11 - r}

(\frac{1}{b x})^{r}

=

^{11} C_{r}

(a)^{11 - r}

(\frac{1}{b^{r}}) x^{22 - 3 r}

For

x^{7}

, we must have 22 - 3r = 7 ⇒ r = 5, and
the coefficient of

x^{7}

=

^{11} C_{5}

.

a^{11 - 5} \frac{1}{5}

) =

^{11} C_{5}

\frac{a^{6}}{b^{5}}

Similarly, in the expansion of

(a x - \frac{1}{b x^{2}})^{11}

, the
general term is

T_{r + 1}

=

^{11} C_{r} (- 1)^{r}

\frac{a^{11 - r}}{b^{r}}

.

x^{11 - 3 r}

For

x^{- 7}

we must have, 11 - 3r = -7 ⇒ r = 6, and
the coefficient of

x^{- 7}

is

^{11} C_{6}

\frac{a^{5}}{b^{6}}

=

^{11} C_{5}

\frac{a^{5}}{b^{6}}

.
As given,

^{11} C_{5}

\frac{a^{6}}{b^{5}}

=

^{11} C_{5}

\frac{a^{5}}{b^{6}}

⇒ ab = 1.

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