Question

If $\frac {3x+1}{(x-1)(x+3)} =\frac {A}{x-1} + \frac {B} {x+3} \sin^{-1} \frac {A}{B}$ = is equal to

• A. $\frac {\pi}{2}$
• B. $\frac {\pi}{3}$
• C. $\frac {\pi}{6}$
• D. $\frac {\pi}{4}$

Answer 1

Answer: (C)

$\frac {\pi}{6}$

Answer 2

Given, $\frac{3x+1}{\left(x-1\right)\left(x+3\right)}=\frac{A}{x-1}+\frac{B}{x+3}$
$\Rightarrow 3x + 1 = A \left(x + 3\right) + B \left(x - 1\right)$
$\Rightarrow 3x + 1 = \left(A + B\right) x + \left(3A - B\right)$
On equating the coefficient of x from both sides, we get
$A+B=3 \,\,\,\,\dots(i)$
and $3A-B =1\,\,\,\,\dots(ii)$
On adding Eqs. (i) and (ii), we get
$4A =4 \Rightarrow A=1$
$\therefore$ From E (i), we get
$B=3-A=3-1=2$
$\therefore sin^{-1} \frac{A}{B} = sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$