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Question: If \(\frac{x}{(x - 1)(x^{2} + 1)^{2}} = \frac{1}{4}\left\lbrack \frac{1}{(x - 1)} - \frac{x + 1}{x^{...

If x(x−1)(x2+1)2=14[1(x−1)−x+1x2+1]+Ax+B(x2+1)2\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 + \frac{Ax + B}{(x^{2} + 1)^{2}}then x belongs to the interval.

A

4A+2=04A + 2 = 0

B

4B−4A=44B - 4A = 4

C

A=−12A = \frac{- 1}{2}

D

None of these

Answer

A=−12A = \frac{- 1}{2}

Explanation

Solution

⇒\Rightarrow …..(i)

For log to be defined

1=Ar.(−1)rr!.(n−r)!1 = A_{r}.( - 1)^{r}r!.(n - r)!

From (i), ∴\therefore

⇒\Rightarrow x+1(x−1)(x−2)(x−3)=Ax−1+Bx−2+Cx−3⇒x+1=A(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)\frac{x + 1}{(x - 1)(x - 2)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3} \Rightarrow x + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)

⇒\Rightarrow x1+x−log⁡(1+x)=1−11+x−log⁡(1+x)B=−3x=3,C=2\frac{x}{1 + x} - \log(1 + x) = 1 - \frac{1}{1 + x} - \log(1 + x)B = - 3x = 3,C = 2

⇒\Rightarrow 1x−1−3x−2+2x−3\frac{1}{x - 1} - \frac{3}{x - 2} + \frac{2}{x - 3} ex+2=−3(2ex−3)+B(ex−1)⇒e^{x} + 2 = - 3(2e^{x} - 3) + B(e^{x} - 1) \Rightarrow, 1=−6+B1 = - 6 + B.