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Basic of Logarithms
easy
${{{x^2} - 5} \over {{x^2} - 3x + 2}}$ નું આંશિક અપૂર્ણાક મેળવો.
A
$1 + {1 \over {(x - 1)}} - {1 \over {{{(x - 2)}^2}}}$
B
${1 \over {(x - 1)}} + {1 \over {{{(x - 2)}^2}}}$
C
${1 \over {(x - 1)}} - {1 \over {{{(x - 2)}^2}}}$
D
$1 + {4 \over {(x - 1)}} - {1 \over {(x - 2)}}$
Solution
(d) ${{{x^2} – 5} \over {{x^2} – 3x + 2}} = {{{x^2} – 5} \over {(x – 1)\,(x – 2)}} = {A \over {x – 1}} + {B \over {x – 2}} + C$
$ \Rightarrow $ ${x^2} – 5 = A(x – 2) + B(x – 1) + C(x – 1)(x – 2)$
$ \Rightarrow $ $C = 1,\,A + B – 3C = 0,\, – 2A – B + 2C = – 5$
$\therefore A = 4,\,B = – 1,\,C = 1$
$\therefore $ Given expression = $1 + {4 \over {x – 1}} – {1 \over {x – 2}}$
Standard 11
Mathematics
