Basic of Logarithms
medium

જો ${{{{(x + 1)}^2}} \over {{x^3} + x}} = {A \over x} + {{Bx + C} \over {{x^2} + 1}}$, તો ${\sin ^{ - 1}}\left( {{A \over C}} \right) = $

A

${\pi \over 6}$

B

${\pi \over 4}$

C

${\pi \over 3}$

D

${\pi \over 2}$

Solution

(a) ${{{{(x + 1)}^2}} \over {{x^3} + x}} = {A \over x} + {{Bx + C} \over {{x^2} + 1}}$

==> ${(x + 1)^2} = A({x^2} + 1) + (Bx + C)\,x$

==> $A + B = 1$, $C = 2$,$A = 1$ ==> $B=0$

${\sin ^{ – 1}}\left( {\frac{A}{C}} \right) = {\sin ^{ – 1}}\left( {\frac{1}{2}} \right) = 30^\circ  = \frac{\pi }{6}$

Standard 11
Mathematics

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