The partial fractions of x^2 (x - 1) ^3 (x - 2) are - Vidyadip

MCQ

The partial fractions of ${{{x^2}} \over {{{(x - 1)}^3}(x - 2)}}$ are

A
${{ - 1} \over {{{(x - 1)}^3}}} + {3 \over {{{(x - 1)}^2}}} - {4 \over {(x - 1)}} + {4 \over {(x - 2)}}$
B
${{ - 1} \over {{{(x - 1)}^3}}} - {3 \over {{{(x - 1)}^2}}} + {4 \over {(x - 1)}} + {4 \over {(x - 2)}}$
✓
${{ - 1} \over {{{(x - 1)}^3}}} + {{ - 3} \over {{{(x - 1)}^2}}} + {{ - \,4} \over {(x - 1)}} + {4 \over {(x - 2)}}$
D
None of these

✓

Answer

Correct option: C.

${{ - 1} \over {{{(x - 1)}^3}}} + {{ - 3} \over {{{(x - 1)}^2}}} + {{ - \,4} \over {(x - 1)}} + {4 \over {(x - 2)}}$

c
(c) Put the repeated factor

$(x - 1) = y \Rightarrow x = y + 1$

$\therefore {{{x^2}} \over {{{(x - 1)}^3}(x - 2)}} = {{{{(1 + y)}^2}} \over {{y^3}(y - 1)}} = {{1 + 2y + {y^2}} \over {{y^3}( - 1 + y)}}$

Dividing the numerator,

$(1 + 2y + {y^2})$ by $( - 1 + y)$ till ${y^3}$ appears as factor,

we get ${{1 + 2y + {y^2}} \over { - 1 + y}} = ( - 1 - 3y - 4{y^2}) + {{4{y^3}} \over { - 1 + y}}$

Given expression = ${{ - 1} \over {{y^3}}} - {3 \over {{y^2}}} - {4 \over y} + {4 \over { - 1 + y}}$

= ${{ - 1} \over {{{(x - 1)}^3}}} + {{ - 3} \over {{{(x - 1)}^2}}} + {{ - 4} \over {(x - 1)}} + {4 \over {(x - 2)}}$.

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