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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