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TS EAMCET 7 May 2018 Shift 1 Solved Paper

Section: Mathematics

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Question : 16 of 160

Marks: +1, -0

The coefficient of

x^{3}

in the expansion of

rac{1-2x}{(2x+1)(2-x)}

is

- $rac{509}{80}$
$rac{509}{80}$
$-\frac{103}{16}$
$rac{103}{16}$

Solution:

Given:

rac{1-2 x}{(2 x+1)(2-x)}

The expansion is,

= \frac{(1-2 x)(1+2 x)^{-1}}{2} \left(1- \frac{x}{2} \right)^{-1}

= \frac{(1-2 x)}{2}\left[ 1-2 x+ \frac{(-1)(-1-1)}{2}(2 x)^{2}+ \cdots \right]

\frac{(-1)(-1-1)(-1-2)}{6}(2x)^6 \ldots

1+(-1) \left( -\frac{x}{2} \right)+ \frac{(-1)(-1-1)}{2} \left( -\frac{x}{2} \right)^{2}

\left[ + \frac{(-1)(-1-1)(-1-2)}{6} \left( -\frac{x}{2} \right) \ldots \right]

= \frac{1}{2}(1-2 x) (1-2 x+4 x^{2}-8 x^{3} )

\left(1+ \frac{x}{2}+ \frac{x^{2}}{4}+ \frac{x^{3}}{8} \right)

= \frac{1}{2} \left[1-4 x+8 x^{2}-16 x^{3} \right]

\left[1+ \frac{x}{2}+ \frac{x^{2}}{4}+ \frac{x^{3}}{8} \right]

Coefficient of

x^{3}

is,

= \frac{1}{2} \left[ \frac{1}{8}-1+4-16 \right]

= - \frac{103}{16}

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