For - x - - 1- the constant term in the expansion of 1 x - 12 x - 2 is

The correct option is D 12 B=( x 1) ^ 2 (x 2) ^ 1 Take, a = 1 #160; amp; #160; b = 2 ∴ B= (x+a) ^ b (x+b) ^ a = (a+x ...

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Standard XII

Mathematics

Algebra of Derivatives

For | x | <...

Question

For $| x | < 1$ , the constant term in the expansion of $\frac{1}{(x - 1)^{2} (x - 2)}$ is

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- \frac{1}{2}

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Solution

The correct option is D $- \frac{1}{2}$
$B = ({x - 1)}^{- 2} {(x - 2)}^{- 1}$
Take, $a = - 1$ & $b = - 2$
$∴ B = {(x + a)}^{b} {(x + b)}^{a} = {(a + x)}^{b} {(b + x)}^{a}$
$∴ T_{r + 1} = [{^{b} C}_{r} . a^{b - r} x^{r}] [{^{a} C}_{r} {. b}^{(a - r)} x^{r}]$
For constant term : $2 r = 0$
$\Rightarrow$ $r = 0$
$∴ T_{r + 1} = {^{a} C}_{0} . {^{b} C}_{0} . a^{b} . b^{a}$
$= - 1.1. ({- 1)}^{- 2} . {(- 2)}^{- 1}$
$= \frac{1}{2}$ .
Hence the answer is $\frac{1}{2} .$

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For |x|<1, the constant term in the expansion of 1(x−1)2(x−2) is

The correct option is D −12B=(x−1)−2(x−2)−1Take, a=−1 & b=−2∴B=(x+a)b(x+b)a=(a+x)b(b+x)a∴Tr+1=[bCr.ab−rxr][aCr.b(a−r)xr]For constant term : 2r=0⇒ r=0∴Tr+1=aC0.bC0.ab.ba =−1.1.(−1)−2.(−2)−1 =12.Hence the answer is 12.

For $| x | < 1$ , the constant term in the expansion of $\frac{1}{(x - 1)^{2} (x - 2)}$ is