{{3{x^2} + 5} over {{{({x^2} + 1)}^2}}} = {a over {{x^2} + 1}} + {b over {{{({x^2} + 1)}^2}}} , then

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Basic of Logarithms

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${{3{x^2} + 5} \over {{{({x^2} + 1)}^2}}} = {a \over {{x^2} + 1}} + {b \over {{{({x^2} + 1)}^2}}}$, then $(a,b) = $

A

$(2, 3)$

B

$(3, 2)$

C

$(-2,3)$

D

$(-3, 2)$

Solution

(b) $3{x^2} + 5 = a\,({x^2} + 1) + b$

$ \Rightarrow $ $a = 3,\,a + b = 5 \Rightarrow b = 2$;

$\therefore (a,\,b) = (3,\,2)$.

Standard 11

Mathematics

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