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Question: The set of real values of x satisfying \(\therefore\) is....

The set of real values of x satisfying ∴\therefore is.

A

(x+1)2x3+x=Ax+Bx+Cx2+1\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}

B

(x+1)2=A(x2+1)+(Bx+C)x(x + 1)^{2} = A(x^{2} + 1) + (Bx + C)x

C

A+B=1A + B = 1

D

None of these

Answer

(x+1)2=A(x2+1)+(Bx+C)x(x + 1)^{2} = A(x^{2} + 1) + (Bx + C)x

Explanation

Solution

A(2)2A(2)^{2} …..(i)

For log to be defined, A=74A = \frac{7}{4}

3x−1(1−x+x2)(2+x)=Ax+Bx2−x+1+Cx+2(3x−1)=(Ax+B)(x+2)+C(x2−x+1)\frac{3x - 1}{(1 - x + x^{2})(2 + x)} = \frac{Ax + B}{x^{2} - x + 1} + \frac{C}{x + 2}(3x - 1) = (Ax + B)(x + 2) + C(x^{2} - x + 1), which is true A+C=0A + C = 0.

From (i), C=−1C = - 1

⇒\Rightarrow 3x−1(1−x+x2)(2+x)=xx2−x+1−1x+2\frac{3x - 1}{(1 - x + x^{2})(2 + x)} = \frac{x}{x^{2} - x + 1} - \frac{1}{x + 2} (x+1)2x3+x=Ax+Bx+Cx2+1\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1} (x+1)2=A(x2+1)+(Bx+C)x(x + 1)^{2} = A(x^{2} + 1) + (Bx + C)x

A+B=1A + B = 1 C=2C = 2 A=1A = 1 x2+3x−7x^{2} + 3x - 7; ∴\therefore x(x−1)(x2+1)2=14[1(x−1)−x+1x2+1]+y\frac{x}{(x - 1)(x^{2} + 1)^{2}} = \frac{1}{4}\left\lbrack \frac{1}{(x - 1)} - \frac{x + 1}{x^{2} + 1} \right\rbrack + y.