If 1x-1x-2x-3=Ax-1+Bx-2+Cx-3and xx-1x-2x-3=Px-1+Qx-2+Rx-3 then A+2B+3C=

1(x-1) (x-2) (x-3)=Ax-1+Bx-2+Cx-31=A(x-2) (x-3)+B(x-1) (x-3)+C(x-1) (x-2)x=1 ⇒ 1=A(-1) (-2) ⇒ A=12x=2 ⇒ 1=B1(-1) ⇒ B=-1x=3 ⇒ 1=C(1)2 ⇒ C=12x(x-1) (x-2) (x-3)=Px-1+Qx-2+Rx-3x=P(x-2) (x-3)+Q(x-1) (x-3)+R(x-1) (x-2)x=1 ⇒ 1=P(-1) (-2) ⇒ P=12x=2 ⇒ 2=Q(-1) (-1) ⇒ Q=-2x=3 ⇒ 3=R(2) (1) ⇒ R=32A+2B+3C=12-2+32 =P+Q+R

If 1x-1x-2x-3=Ax-1+Bx-2+Cx-3and xx-1x-2x-3=Px-1+Qx-2+Rx-3 then A+2B+3C=

1(x-1) (x-2) (x-3)=Ax-1+Bx-2+Cx-31=A(x-2) (x-3)+B(x-1) (x-3)+C(x-1) (x-2)x=1 ⇒ 1=A(-1) (-2) ⇒ A=12x=2 ⇒ 1=B1(-1) ⇒ B=-1x=3 ⇒ 1=C(1)2 ⇒ C=12x(x-1) (x-2) (x-3)=Px-1+Qx-2+Rx-3x=P(x-2) (x-3)+Q(x-1) (x-3)+R(x-1) (x-2)x=1 ⇒ 1=P(-1) (-2) ⇒ P=12x=2 ⇒ 2=Q(-1) (-1) ⇒ Q=-2x=3 ⇒ 3=R(2) (1) ⇒ R=32A+2B+3C=12-2+32 =P+Q+R

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$If \frac{1}{(x - 1) (x - 2) (x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3} and \frac{x}{(x - 1) (x - 2) (x - 3)} = \frac{P}{x - 1} + \frac{Q}{x - 2} + \frac{R}{x - 3} then A + 2 B + 3 C =$

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$P + Q + R$

$P + 2 Q + 3 R$

$3 P + 2 Q + R$

$AP + BQ + CR$

answer is A.

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Detailed Solution

$\begin{array}{rcl} \frac{1}{(x - 1) (x - 2) (x - 3)} & = & \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3} \\ 1 & = & A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2) \\ x & = & 1 \Rightarrow 1 = A (- 1) (- 2) \Rightarrow A = \frac{1}{2} \\ x & = & 2 \Rightarrow 1 = B 1 (- 1) \Rightarrow B = - 1 \\ x & = & 3 \Rightarrow 1 = C (1) 2 \Rightarrow C = \frac{1}{2} \\ \frac{x}{(x - 1) (x - 2) (x - 3)} & = & \frac{P}{x - 1} + \frac{Q}{x - 2} + \frac{R}{x - 3} \\ x & = & P (x - 2) (x - 3) + Q (x - 1) (x - 3) + R (x - 1) (x - 2) \\ x & = & 1 \Rightarrow 1 = P (- 1) (- 2) \Rightarrow P = \frac{1}{2} \\ x & = & 2 \Rightarrow 2 = Q (- 1) (- 1) \Rightarrow Q = - 2 \\ x & = & 3 \Rightarrow 3 = R (2) (1) \Rightarrow R = \frac{3}{2} \\ A + 2 B + 3 C & = & \frac{1}{2} - 2 + \frac{3}{2} \\ = & P + Q + R \end{array}$