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TS EAMCET 9 May 2024 Shift 1 Solved Paper

Section: Mathematics

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Question : 21 of 160

Marks: +1, -0

If

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

f(14) + 2A - B =

[TG EAPCET 9 May 2024 Shift 1]

5 C
4C
6C
7C

Solution:

We have,

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

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

f(x) [(x^2+1)(x-2)]

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

\Rightarrow \frac{x^4}{(x^2+1)(x-2)}

f(x) [x^3 - 2x^2 + x - 2]

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

Here,

f(x)

will be linear function only as highest power in LHS is 4

So, let

f(x) = p x + q

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

(p x + q) [x^3 - 2x^2 + x - 2]

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

\Rightarrow \frac{x^4}{(x^2+1)(x-2)}

p x^4 + (q-2p) x^3 + (p-2q+A+C) x^2

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

On comparing both sides, we get

p=1, q-2p=0 \Rightarrow q=2

p-2q+A+C=0 \Rightarrow A+C=3

q-2p+B-2A=0 \Rightarrow B=2A

C-2B-2q=0 \Rightarrow C-2B=4

From Eqs. (i), (ii) and (iii), we get,

A = \frac{-1}{5}, C = \frac{16}{5}, B = \frac{-2}{5}, f(x) = x+2

So,

f(14) + 2A - B = 14 + 2 + 0 = 16 = 5C

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