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Manipal Entrance Test 2018 Math Paper

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Question : 7 of 70

Marks: +1, -0

The coefficient of

x^{-9}

in the expansion of

\left(\frac{x^2}{2} - \frac{2}{x}\right)^{9}

is

512
-512
521
251

Solution:

Let the coefficient of

x^{-9}

is in the

(r+1)

th term in the expansion of

\left(\frac{x^2}{2} - \frac{2}{x}\right)^{9},

then

T_{r+1}={}^{9}C_{r}\left(\frac{x^2}{2}\right)^{9-r}\left(-\frac{2}{x}\right)^{r}

={}^{9}C_{r}\frac{x^{18-2r}}{2^{9-r}}\cdot\frac{(-1)^r 2^r}{x^r}

={}^{9}C_{r}\frac{x^{18-3r}}{2^{9-2r}}(-1)^r

For coefficient of

x^{-9},

put

x^{18-3r}=x^{-9}

\Rightarrow 18-3r=-9 \Rightarrow 27=3r

\Rightarrow r=9

∴ Coefficient of

x^{-9}={}^{9}C_{9}\cdot\frac{1}{2^{-9}}(-1)^9

=-2^9=-512

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