- Home
- Standard 11
- Mathematics
Basic of Logarithms
easy
${{3x - 1} \over {(1 - x + {x^2})\,(2 + x)}}$ નું આંશિક અપૂર્ણાક મેળવો.
A
${x \over {({x^2} - x + 1)}}$+${1 \over {x + 2}}$
B
${1 \over {{x^2} - x + 1}} + {x \over {x + 2}}$
C
${x \over {{x^2} - x + 1}} - {1 \over {x + 2}}$
D
${{ - 1} \over {{x^2} - x + 1}} + {x \over {x + 2}}$
Solution
(c) ${{3x – 1} \over {(1 – x + {x^2})\,(2 + x)}} = {{Ax + B} \over {{x^2} – x + 1}} + {C \over {x + 2}}$
==> $(3x – 1) = (Ax + B)\,(x + 2)\, + \,C({x^2} – x + 1)$
Comparing the coefficient of like terms, we get $A + C = 0$, $2A + B – C = 3$, $2B + C = – 1$ ==> $A = 1$, $B = 0$, $C = – 1$
$\therefore {{3x – 1} \over {(1 – x + {x^2})\,(2 + x)}} = {x \over {{x^2} – x + 1}} – {1 \over {x + 2}}$.
Standard 11
Mathematics