13 Most SAQ’s of Partial Fractions Chapter in Inter 2nd Year Maths-2A (TS/AP)

4 Marks

SAQ-1 : Resolve (3x+7)/(x²-3x+2) into partial fractions

$$G.E = 3x + \frac{7}{x^2} – 3x + 2 = 3x + \frac{7}{(x – 1)(x – 2)}$$

Let $$\frac{3x + 7}{(x – 1)(x – 2)} = \frac{A}{x- 1} + \frac{B}{x – 2} = \frac{A(x – 2) + B(x – 1)}{(x – 1)(x – 2)}$$

$$A(x-2) + B(x-1) = 3x + 7…(1)$$

Putting x = 2 in (1) we get $$A(2-2) + B(2-1) = 3(2) + 7 \Rightarrow B = 13$$

Putting x = 1 in (1) we get $$A(1-2) + B(1-1) = 3(1) + 7 \Rightarrow -A = 10 \Rightarrow A = -10$$

$$3x + \frac{7}{x^2} – 3x + 2 = \frac{A}{(x-1)} + \frac{B}{(x-2)} = \frac{-10}{(x-1)} + \frac{13}{(x-2)} = \frac{10}{(1-x)} + \frac{13}{(x-2)}$$

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SAQ-2 : Resolve (x+4)/(x²-4)(x+1) into partial fractions

$$G.E = x + \frac{4}{(x^2 – 4)(x + 1)} = x + \frac{4}{(x+ 2)(x – 2)(x + 1)}$$

Let $$\frac{x + 4}{(x + 2)(x – 2)(x + 1)} = \frac{A}{x + 2} + \frac{B}{x – 2} + \frac{C}{x + 1}$$

$$= \frac{A(x-2)(x+1) + B(x+2)(x+1) + C(x^2-4)}{(x+2)(x-2)(x+1)}$$

$$A(x-2)(x+1) + B(x+2)(x+1) + C(x^2-4) = x+4…(1)$$

Putting x = −2 in (1) we get $$A(-2-2)(-2+1) + B(0) + C(0) = -2 + 4 \Rightarrow 4A = 2 \Rightarrow A = 1/2$$

Putting x = 2 in (1) we get $$A(0) + B(2+2)(2+1) + C(0) = 2 + 4 \Rightarrow 12B = 6 \Rightarrow B = 1/2$$

Putting x = −1 in (1) we get $$A(0) + B(0) + C(1-4) = -1 + 4 \Rightarrow -3C = 3 \Rightarrow C = -1$$

$$x + \frac{4}{(x^2 – 4)(x + 1)} = \frac{A}{(x+2)} + \frac{B}{(x-2)} + \frac{C}{(x+1)} = \frac{1}{2}(x+2) + \frac{1}{2}(x-2) – \frac{1}{(x+1)}$$

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SAQ-3 : Resolve (x²-x+1)/(x+1)(x-1)² into partial fractions

Let $$x^2 – x + \frac{1}{(x + 1)(x – 1)^2} = \frac{A}{x + 1} + \frac{B}{x – 1} + \frac{C}{(x – 1)^2}$$

$$x^2 – x + \frac{1}{(x+1)(x-1)^2} = \frac{A(x-1)^2 + B(x+1)(x-1) + C(x+1)}{(x+1)(x-1)^2}$$

$$A(x-1)^2 + B(x+1)(x-1) + C(x+1) = x^2-x+1…(1)$$

Putting x = 1 in (1) we get $$1 = A(1-1)^2 + B(2)(0) + C(1+1) \Rightarrow 2C = 1 \Rightarrow C = 1/2$$

Putting x = −1 in (1) we get $$= A(1-(-1))^2 + B(-1+1)(-1-1) + C(-1+1) = 3 \Rightarrow 4A = 3 \Rightarrow A = 3/4$$

Equating the coefficients of x² we get $$A+B = 1 \Rightarrow B = 1 – A = 1 – 3/4 = 1/4$$

$$x^2 – x + \frac{1}{(x + 1)(x – 1)^2} = \frac{A}{x + 1} + \frac{B}{x – 1} + \frac{C}{(x – 1)^2} = \frac{3}{4}(x + 1) + \frac{1}{4}(x – 1) + \frac{1}{2}(x – 1)^2$$

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SAQ-4 : Resolve 2x²+2x+1/x³+x² into partial fractions

Let $$2x^2 + 2x + \frac{1}{x^3} + x^2 = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1}$$

$$= \frac{Ax(x+1) + B(x+1) + Cx^2}{x^2(x+1)}$$

$$Ax(x+1) + B(x+1) + Cx^2 = 2x^2 + 2x + 1…(1)$$

Putting x = 0 in (1) we get: $$A(0) + B(1) + C(0) = 1 \Rightarrow B = 1$$

Putting x = −1 in (1) we get: $$A(0) + B(0) + C(-1)^2 = 2(-1)^2 + 2(-1) + 1 \Rightarrow C = 1$$

Equating the coefficients of x² we get: $$2 = A + C \Rightarrow A = 2 – C = 2 – 1 = 1$$

$$2x^2 + 2x + \frac{1}{x^3} + x^2 = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x + 1}$$

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SAQ-5 : Resolve (3x-18)/(x³(x+3) into partial fractions

Let $$3x – \frac{18}{x^3(x + 3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x + 3}$$

$$\Rightarrow 3x – \frac{18}{x^3(x + 3)} = \frac{Ax^2(x + 3) + Bx(x + 3) + C(x + 3) + Dx^3}{x^3(x + 3)}$$

$$Ax^2(x + 3) + Bx(x + 3) + C(x + 3) + Dx^3 = 3x – 18….(1)$$

Putting x = −3 in (1) we get: $$A(0) + B(0) + C(0) + D(-3)^3 = 3(-3) – 18 \Rightarrow -27D = -27 \Rightarrow D = 1$$

Putting x = 0 in (1) we get: $$A(0) + B(0) + C(0+3) + D(0) = 3(0) – 18 \Rightarrow 3C = -18 \Rightarrow C = -6$$

Equating the coefficient of x³ we get: $$A + D = 0 \Rightarrow A = -D \Rightarrow A = -1$$

Equating the coefficient of x² we get: $$3A + B = 0 \Rightarrow B = -3A = -3(-1) \Rightarrow B = 3$$

$$3x – \frac{18}{x^3(x + 3)} = \frac{-1}{x} + \frac{3}{x^2} – \frac{6}{x^3} + \frac{1}{x + 3}$$

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SAQ-6 : Resolve (x²+5x+7)/(x-3)³ into partial fractions

Put $$x – 3 = y$$ then $$x = y + 3$$

$$\frac{x^2 + 5x + 7}{(x – 3)^3} = \frac{(y + 3)^2 + 5(y + 3) + 7}{y^3} = \frac{y^2 + 6y + 9 + 5y + 15 + 7}{y^3} = \frac{y^2 + 11y + 31}{y^3}$$

$$= \frac{1}{y} + \frac{11}{y^2} + \frac{31}{y^3} = \frac{1}{x – 3} + \frac{11}{(x – 3)^2} + \frac{31}{(x – 3)^3}$$

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SAQ-7 : Resolve 2x²+1/x³-1into partial fractions

Let $$2x^2 + \frac{1}{x^3 – 1} = 2x^2 + \frac{1}{(x – 1)(x^2 + x + 1)}$$

$$2x^2 + \frac{1}{(x – 1)(x^2 + x + 1)} = \frac{A}{x – 1} + \frac{Bx + C}{x^2 + x + 1} = \frac{A(x^2 + x + 1) + (Bx + C)(x – 1)}{(x – 1)(x^2 + x + 1)}$$

$$A(x^2 + x + 1) + (Bx + C)(x – 1) = 2x^2 + 1…(1)$$

Putting x = 1 in (1) we get $$A(1 + 1 + 1) + (B + C)(0) = 2(1)^2 + 1 \Rightarrow 3A = 3 \Rightarrow A = 1$$

Comparing the coefficients of x² in (1) we get $$A + B = 2 \Rightarrow B = 2 – A = 2 – 1 \Rightarrow B = 1$$

Comparing the constant terms in (1) we get $$A + C = 1 \Rightarrow C = 1 – A = 1 – 1 \Rightarrow C = 0$$

$$2x^2 + \frac{1}{x^3 – 1} = \frac{A}{x – 1} + \frac{Bx + C}{x^2 + x + 1} = \frac{1}{x – 1} + \frac{x}{x^2 + x + 1}$$

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SAQ-8 : Resolve 2x²+3x+4/(x-1)(x²+2) into partial fractions

Let $$2x^2 + 3x + 4/(x – 1)(x^2 + 2) = \frac{A}{x – 1} + \frac{Bx + C}{x^2 + 2} = \frac{A(x^2 + 2) + (Bx + C)(x – 1)}{(x – 1)(x^2 + 2)}$$

$$A(x^2 + 2) + (Bx + C)(x – 1) = 2x^2 + 3x + 4…(1)$$

Putting x = 1 in (1) we get $$A(1^2 + 2) + (B + C)(0) = 2(1)^2 + 3(1) + 4$$

$$\Rightarrow 3A = 9 \Rightarrow A = 3$$

Putting x = 0 in (1) we get $$A(0^2 + 2) + (0 + C)(0 – 1) = 2(0)^2 + 3(0) + 4 \Rightarrow 2A – C = 4$$

$$\Rightarrow C = 2A – 4 = 2(3) – 4 = 2$$

Comparing the coefficient of x² in (1) we get $$A + B = 2 \Rightarrow 3 + B = 2 \Rightarrow B = -1$$

$$2x^2 + 3x + 4/(x – 1)(x^2 + 2) = \frac{3}{x – 1} + \frac{-1x + 2}{x^2 + 2} = \frac{3}{x – 1} + \frac{x}{x^2 + 2} + \frac{2}{x^2 + 2}$$

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SAQ-9 : Resolve x²-3/(x+2)(x²+1) into partial fractions

Let $$x^2 – \frac{3}{(x + 2)(x^2 + 1)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1} = \frac{A(x^2+1) + (Bx + C)(x + 2)}{(x+2)(x^2+1)}$$

$$\Rightarrow A(x^2+1) + (Bx+C)(x+2) = x^2-3…(1)$$

Putting x = −2 in (1) we get $$A(4+1) + (B(-2)+C)(0) = 4 – 3 \Rightarrow 5A = 1 \Rightarrow A = 1/5$$

Putting x = 0 in (1) we get $$A(0^2 + 1) + (0 + C)(2) = 0^2 – 3 \Rightarrow 1A + 2C = -3 \Rightarrow C = -\frac{8}{5}$$

Comparing the coefficients of x² we get $$A + B = 1 \Rightarrow B = 1 – A = 1 – \frac{1}{5} = \frac{4}{5}$$

$$x^2 – \frac{3}{(x+2)(x^2+1)} = \frac{1}{5}(x+2) + \left(\frac{4}{5}x – \frac{8}{5}\right)/(x^2+1)$$

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SAQ-10 : Resolve x³/(x-1)(x+2) into partial fractions

Here the degree of numerator 3 ≥ degree of denominator 2 So it is an improper function

Also $$(x-1)(x+2) = x^2 + x – 2$$

Now on dividing x³ by x² + x − 2 we have:

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

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

$$= \frac{A}{x-1} + \frac{B}{x+2} = \frac{A(x+2)+B(x-1)}{(x-1)(x+2)}$$

$$\Rightarrow A(x+2) + B(x-1) = 3x – 2…(1)$$

Putting x = 1 in (1) we get $$A(3) + B(0) = 3 – 2 \Rightarrow 3A = 1 \Rightarrow A = 1/3$$

Putting x = −2 in (1) we get $$A(0) + B(-3) = 3(-2)-2 \Rightarrow -3B = -8 \Rightarrow B = 8/3$$

$$\Rightarrow \frac{3x-2}{x^2+x-2} = \frac{A}{x-1} + \frac{B}{x+2} = \frac{1}{3}(x-1) + \frac{8}{3}(x+2)$$

$$x^3/(x-1)(x+2) = x-1 + \frac{1}{3}(x-1) + \frac{8}{3}(x+2)$$

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SAQ-11 : Resolve x⁴/(x-1)(x-2) into partial fractions

$$(x-1)(x-2) = x^2-3x+2$$ Now on dividing x⁴ by $$x^2-3x+2$$ we have:

$$x^4/(x-1)(x-2)$$

$$= (x^2 + 3x + 7) + \frac{15x-14}{x^2-3x+2}$$

Now, $$\frac{15x-14}{x^2-3x+2} = \frac{15x-14}{(x-1)(x-2)}$$

$$= \frac{A}{(x-1)} + \frac{B}{(x-2)} = \frac{A(x-2) + B(x-1)}{(x-1)(x-2)}$$

$$\Rightarrow A(x-2) + B(x-1) = 15x-14…(1)$$

Putting x = 1 in (1) we get $$A(1-2) + B(0) = 15(1) – 14 = 1 \Rightarrow A = -1$$

Putting x = 2 in (1) we get $$A(0) + B(2-1) = 15(2) – 14 = 16 \Rightarrow B = 16$$

$$x^4/(x-1)(x-2) = x^2 + 3x + 7 + \frac{-1}{x-1} + \frac{16}{x-2}$$

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SAQ-12 : Resolve x³/(2x-1)(x-1)² into partial fractions

Let $$x^3/(2x-1)(x-1)^2 = \frac{1}{2} + \frac{A}{2x-1} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$$

$$= \frac{(2x-1)(x-1)^2+2A(x-1)^2+2B(2x-1)(x-1)+2C(2x-1)}{2(2x-1)(x-1)^2}$$

$$\Rightarrow (2x-1)(x-1)^2+2A(x-1)^2+2B(2x-1)(x-1)+2C(2x-1)=2x^3…(1)$$

Putting $$x=\frac{1}{2}$$ in (1) we get $$2A\left(\frac{1}{2}-1\right)^2=2\left(\frac{1}{2}\right)^3 \Rightarrow A = \frac{1}{2}$$

Putting x = 1 in (1) we get $$2C(2\cdot1-1) = 2(1)^3 \Rightarrow C = 1$$

Putting x = 0 in (1) gives:

$$(-1)(-1)^2+2A(-1)^2+2B(2\cdot0-1)(-1)+2C(2\cdot0-1)=0$$

$$2A+2B-2C=0$$

Given $$A = \frac{1}{2}$$ and $$C = 1$$ we solve for B:

$$2B=2-2A$$

$$2B=2-1$$

$$B=1/2$$

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

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SAQ-13 : Resolve x³/(x-a)(x-b)(x-c) into partial fractions

Let $$x^3/(x-a)(x-b)(x-c) = 1 + \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}$$

$$= \frac{(x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)}{(x-a)(x-b)(x-c)}$$

$$\Rightarrow (x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b) = x^3…(1)$$

Putting x = a in (1) we get $$0+A(a-b)(a-c)+0+0 = a^3 \Rightarrow A = \frac{a^3}{(a-b)(a-c)}$$

Similarly, by putting $$x=b$$ and $$x=c$$ we get $$B = \frac{b^3}{(b-a)(b-c)}$$ $$C = \frac{c^3}{(c-a)(c-b)}$$

$$x^3/(x-a)(x-b)(x-c) = 1 + \frac{a^3}{(a-b)(a-c)}(x-a) + \frac{b^3}{(b-c)(b-a)}(x-b) + \frac{c^3}{(c-a)(c-b)}(x-c)$$