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Basic of Logarithms
easy
${{{x^2} + 1} \over {(2x - 1)\,({x^2} - 1)}} = $
A
${{ - 5} \over {3(2x - 1)}} + {3 \over {(x + 1)}} + {1 \over {(x - 1)}}$
B
${{ - 5} \over {3(2x - 1)}} + {1 \over {3(x + 1)}} + {1 \over {(x - 1)}}$
C
${1 \over {2x - 1}} + {5 \over {(x + 1)}} - {3 \over {(x - 1)}}$
D
એકપણ નહીં
Solution
(b) ${{{x^2} + 1} \over {(2x – 1)\,({x^2} – 1)}} = {A \over {(2x – 1)}} + {B \over {x + 1}} + {C \over {x – 1}}$
==> ${x^2} + 1 = A({x^2} – 1) + B(2x – 1)\,(x – 1) + C(x + 1)\,(2x – 1)$
For $x = 1,$ $2 = 2C \Rightarrow C = 1$
For $x = – 1$, $2 = 6B \Rightarrow B = {1 \over 3}$
For $x = {1 \over 2}$, ${5 \over 4} = – {3 \over 4}A \Rightarrow $$A = – {5 \over 3}$
$\therefore $ Given expression = $ – {5 \over 3}{1 \over {(2x – 1)}} + {1 \over 3}{1 \over {x + 1}} + {1 \over {x – 1}}$
Standard 11
Mathematics
