Solveeit Logo
Login

Question

Question: If \(\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}\), then \(\sin^{- 1}\lef...

If (x+1)2x3+x=Ax+Bx+Cx2+1\frac{(x + 1)^{2}}{x^{3} + x} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}, then sin⁡−1(AC)=\sin^{- 1}\left( \frac{A}{C} \right) =

A

π6\frac{\pi}{6}

B

π4\frac{\pi}{4}

C

π3\frac{\pi}{3}

D

π2\frac{\pi}{2}

Answer

π6\frac{\pi}{6}

Explanation

Solution

(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 ⇒ B=0B = 0

Therefore sin⁡−1(AC)=sin⁡−1(12)=30o=π6\sin^{- 1}\left( \frac{A}{C} \right) = \sin^{- 1}\left( \frac{1}{2} \right) = 30^{o} = \frac{\pi}{6}.