If x21-x2x+1=Ax-1+12x-12+Bx+1then the length of the vector 4Ai+4Bj is

x2(x-1)2 (x+1)=Az-1+12(x-1)2+Bx+1 x2=2A(x2-1)+12(x+1)+B(x-1)2 x=-1 ⇒ 1=4B ⇒ B=14x2 Coeff ⇒ 1=A+B ⇒ A=34 4Ai^+4Bj^=4(Ai^+Bj^) =32+12 =10

If x21-x2x+1=Ax-1+12x-12+Bx+1then the length of the vector 4Ai+4Bj is

x2(x-1)2 (x+1)=Az-1+12(x-1)2+Bx+1 x2=2A(x2-1)+12(x+1)+B(x-1)2 x=-1 ⇒ 1=4B ⇒ B=14x2 Coeff ⇒ 1=A+B ⇒ A=34 4Ai^+4Bj^=4(Ai^+Bj^) =32+12 =10

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$If \frac{x^{2}}{{(1 - x)}^{2} (x + 1)} = \frac{A}{(x - 1)} + \frac{1}{2 {(x - 1)}^{2}} + \frac{B}{x + 1} then the length of the vector 4 A i + 4 B j is$

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$\sqrt{10}$

$\frac{\sqrt{10}}{2}$

answer is A.

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Detailed Solution

$\frac{x^{2}}{{(x - 1)}^{2} (x + 1)} = \frac{A}{z - 1} + \frac{1}{2 {(x - 1)}^{2}} + \frac{B}{x + 1} x^{2} = 2 A (x^{2} - 1) + \frac{1}{2} (x + 1) + B {(x - 1)}^{2}$

$\begin{array}{rcl} x & = & - 1 \Rightarrow 1 = 4 B \Rightarrow B = \frac{1}{4} \end{array}$

$x^{2} Coeff \Rightarrow 1 = A + B \Rightarrow A = \frac{3}{4} 4 A \hat{i} + 4 B \hat{j} = 4 (A \hat{i} + B \hat{j}) = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$