Question

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Basic of Logarithms

easy

The partial fractions of ${{3{x^3} - 8{x^2} + 10} \over {{{(x - 1)}^4}}}$ is

${3 \over {(x - 1)}} + {1 \over {{{(x - 1)}^2}}} + {7 \over {{{(x - 1)}^3}}} + {5 \over {{{(x - 1)}^4}}}$

${3 \over {(x - 1)}} + {1 \over {{{(x - 1)}^2}}} - {7 \over {{{(x - 1)}^3}}} + {5 \over {{{(x - 1)}^4}}}$

None of these

Solution

(c) ${{3{x^3} – 8{x^2} + 10} \over {{{(x – 1)}^4}}} = {A \over {x – 1}} + {B \over {{{(x – 1)}^2}}} + {C \over {{{(x – 1)}^3}}} + {D \over {{{(x – 1)}^4}}}$

==>$3{x^3} – 8{x^2} + 10 = A{(x – 1)^3} + B{(x – 1)^2} + C(x – 1) + D$

Equating coefficients of different powers of $x$, $3 = A$

$ – 8 = – 3A + B \Rightarrow B = 1$

$0 = 3A – 2B + C \Rightarrow C = – 7$

$10 = – A + B – C + D \Rightarrow D = 5$

$\therefore$ Given expression

= ${3 \over {x – 1}} + {1 \over {{{(x – 1)}^2}}} – {7 \over {{{(x – 1)}^3}}} + {5 \over {{{(x – 1)}^4}}}$.

Std 11

Mathematics

If ${{{{\sin }^2}x + 1} \over {2{{\sin }^2}x – 5\sin x + 3}}$=${A \over {(2\sin x – 3)}} + {B \over {(\sin x – 1)}} + C$, then

easy

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

easy

If ${{{{(x – 1)}^2}} \over {{x^3} + x}} = {A \over x} + {{Bx + C} \over {{x^2} + 1}}$, then

easy

If ${1 \over {x(x + 1)\,(x + 2)….(x + n)}} = {{{A_0}} \over x} + {{{A_1}} \over {x + 1}} + {{{A_2}} \over {x + 2}} + …. + {{{A_n}} \over {x + n}}$ then ${A_r} = $

easy

The coefficient of ${x^n}$ in the expression ${{5x + 6} \over {(2 + x)\,(1 – x)}}$ when expanded in ascending order is

medium

The partial fractions of ${{3{x^3} - 8{x^2} + 10} \over {{{(x - 1)}^4}}}$ is

Solution

Similar Questions

If ${{{{\sin }^2}x + 1} \over {2{{\sin }^2}x – 5\sin x + 3}}$=${A \over {(2\sin x – 3)}} + {B \over {(\sin x – 1)}} + C$, then

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

If ${{{{(x – 1)}^2}} \over {{x^3} + x}} = {A \over x} + {{Bx + C} \over {{x^2} + 1}}$, then

If ${1 \over {x(x + 1)\,(x + 2)….(x + n)}} = {{{A_0}} \over x} + {{{A_1}} \over {x + 1}} + {{{A_2}} \over {x + 2}} + …. + {{{A_n}} \over {x + n}}$ then ${A_r} = $

The coefficient of ${x^n}$ in the expression ${{5x + 6} \over {(2 + x)\,(1 – x)}}$ when expanded in ascending order is