Resolve into partial fraction x3-3x-2-x2-x-1x-12

The correct option is D 3x 1/x^2+x+1+2/(x+1)^2 3/(x+1) Let nbsp; x^ 3 3x 2 /( x^ 2 +x+1 ) ( x+1 ) ^ 2 = A /( x+1 ) nbsp; + B /( x+1 ) nbsp; ...

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Standard XII

Mathematics

Finding Remainder of Numbers of the Form a^b

Resolve into ...

Question

Resolve into partial fraction $\frac{x^{3} - 3 x - 2}{(x^{2} + x + 1) (x + 1)^{2}}$

\frac{3 x - 1}{x^{2} - x + 1} + \frac{1}{(x + 1)^{2}} - \frac{3}{(x + 1)}

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\frac{3 x - 1}{x^{2} - x + 1} + \frac{2}{(x + 1)^{2}} - \frac{3}{(x + 1)}

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\frac{3 x - 1}{x^{2} + x + 1} + \frac{2}{(x + 1)^{2}} - \frac{3}{(x + 1)}

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\frac{3 x - 1}{x^{2} + x + 1} + \frac{2}{(x + 1)^{2}} + \frac{3}{(x + 1)}

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Solution

The correct option is D $\frac{3 x - 1}{x^{2} + x + 1} + \frac{2}{(x + 1)^{2}} - \frac{3}{(x + 1)}$
Let $\frac{x^{3} - 3 x - 2}{(x^{2} + x + 1) {(x + 1)}^{2}} = \frac{A}{(x + 1)} + \frac{B}{{(x + 1)}^{2}} + \frac{C x + D}{(x^{2} + x + 1)}$
$\Rightarrow x^{3} - 3 x - 2 = A (x + 1) (x^{2} + x + 1) + B (x^{2} + x + 1) + (C x + D) {(x + 1)}^{2} \Rightarrow x^{3} - 3 x - 2 = A (x^{3} + 2 x^{2} + 2 x + 1) + B (x^{2} + x + 1) + C (x^{3} + 2 x^{2} + x) + D (x^{2} + 2 x + 1)$
On comparing coefficients
$A + C = 1, 2 A + B + 2 C + D = 0, 2 A + B + C + 2 D = - 3, A + B + D = - 2 \Rightarrow A = - 3, B = 2, C = 3, D = - 1$
Hence
$\frac{x^{3} - 3 x - 2}{(x^{2} + x + 1) {(x + 1)}^{2}} = \frac{- 3}{(x + 1)} + \frac{2}{{(x + 1)}^{2}} + \frac{3 x - 1}{(x^{2} + x + 1)}$
Hence, option 'C' is correct.

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Similar questions

Q. Resolve into partial fraction

\frac{x^{3} - 3 x - 2}{(x^{2} + x + 1) (x + 1)^{2}}

Q. Which of the following are quadratic equations?

(i) x² + 6x − 4 = 0
(ii)

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

(iii)

x^{2} + \frac{1}{x^{2}} = 5

(iv)

x - \frac{3}{x} = x^{2}

(v)

2 x^{2} - \sqrt{3 x} + 9 = 0

(vi)

x^{2} - 2 x - \sqrt{x} - 5 = 0

(vii) 3x² − 5x + 9 = x² − 7x + 3
(viii)

x + \frac{1}{x} = 1

(ix) x² − 3x = 0
(x)

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

(xi) (2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)

x + \frac{1}{x} = x^{2}, x \neq 0

(xiii) 16x² − 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)³ = x³ − 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)

Q. Resolve

\frac{x^{2} - x + 1}{(x + 1) (x - 1)^{2}}

into partial fractions

Differentiate each the following from first principles :

$(i) \frac{2}{x}$

$(i i) \frac{1}{\sqrt{x}}$

$(i i i) \frac{1}{x^{3}}$

$(i v) \frac{x^{2} + 1}{x}$

$(v) \frac{x^{2} - 1}{x}$

$(v i) \frac{x + 1}{x + 2}$

$(v i i) \frac{x + 2}{3 x + 5}$

$(v i i i) k x^{n}$

$(i x) \frac{1}{\sqrt{3 - x}}$

$(x) x^{2} + x + 3$

$(x i) (x + 2)^{3}$

$(x i i) x^{3} + 4 x^{2} + 3 x + 2$

$(x i i i) (x^{2} + 1) (x - 5)$

$(x i v) \sqrt{2 x^{2} + 1}$

$(x v) \frac{2 x + 3}{x - 2}$

Q. Let

f (x) ⎧ ⎨ ⎩ \begin{matrix} x \forall x < 1 2 - x \forall 1 \leq x \leq 2 {- x}^{2} + 3 x - 2 \forall x > 2 \end{matrix}

, then

f (x)

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Resolve into partial fraction x3−3x−2(x2+x+1)(x+1)2

Resolve into partial fraction $\frac{x^{3} - 3 x - 2}{(x^{2} + x + 1) (x + 1)^{2}}$