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TS EAMCET 18-July-2022 Shift 1 Solved Paper

Section: Mathematics

Question : 20 of 160

Marks: +1, -0

\;\frac{x^{2}+3}{(x^{2}+1)(x^{2}+2)} = \;\frac{A x + B}{x^{2}+1} + \;\frac{C x + D}{x^{2}+2}

A + B + C + D =

Solution: 👈: Video Solution

Given,

\;\frac{x^{2}+3}{(x^{2}+1)(x^{2}+2)} = \;\frac{AX+B}{x^{2}+1} + \;\frac{CX+D}{x^{2}+2}

\Rightarrow x^{2}+3 = (x^{2}+2)(A x - B) + (x^{2}+1)(C x + D)

\Rightarrow x^{2}+3 = A x^{3} + B x^{2} + 2 A x + 2 B + C x^{3} + x^{2} D + C x + D

\Rightarrow \;\; x^{2}+3 = (A+C)x^{3} + (B+D)x^{2} + (2A+C)x + 2B + D

On comparing,

\;A+C=0\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)

\;B+D=1\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(ii)

\;2A+C=0\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(iii)

\;2B+D=3\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(iv)

From Eqs. (i) and (iii),

A = C = 0

From Eqs. (ii) and (iv),

B = 2

and

D = -1

\therefore \;\; A + B + C + D = 1

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