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TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Section: Mathematics

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Question : 19 of 160

Marks: +1, -0

If

\frac{1}{(x-1)(x-2)(x-3)}

\frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}

\frac{x}{(x-1)(x-2)(x-3)}

\frac{P}{x-1} + \frac{Q}{x-2} + \frac{R}{x-3}

A+2B+3C

=

$P+Q+R$
$P+2Q+3R$
$3P+2Q+R$
$AP+BQ+CR$

Solution: 👈: Video Solution

\;\frac{1}{(x-1)(x-2)(x-3)} = \;\frac{A}{x-1} + \;\frac{B}{x-2} + \;\frac{C}{x-3}

1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)

ZXWhen

x=2 \Rightarrow 1 = -B

When

x=3 \Rightarrow 1 = 2C

When

x=1 \Rightarrow 1 = 2A

A = \;\frac{1}{2}, B = -1

and

C = \;\frac{1}{2}

\;\frac{x}{(x-1)(x-2)(x-3)} = \;\frac{P}{x-1} + \;\frac{Q}{x-2} + \;\frac{R}{x-3}

x = P(x-2)(x-3) + Q(x-1)(x-3) + R(x-1)(x-2)

When

x=2 \Rightarrow 2 = -Q

When

x=3 \Rightarrow 3 = 2R

When

x=1 \Rightarrow 1 = 2P

P = \;\frac{1}{2}, Q = -2

and

R = \;\frac{3}{2}

A+2 B+3 C = \;\frac{1}{2} + (-2) + \;\frac{3}{2} = 0

P+Q+R = \;\frac{1}{2} - 2 + \;\frac{3}{2} = 0

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