TS EAMCET · Maths · Basic of Mathematics
If \(\frac{x+3}{(x+1)\left(x^2+2\right)}=\frac{a}{x+1}+\frac{b x+c}{x^2+2}\) then \(a-b+c=\)
- A \(0\)
- B \(1\)
- C \(3\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
\( x+3 = a(x^2+2) + (bx+c)(x+1) \) \( x+3 = (a+b)x^2 + (b+c)x + (2a+c) \) For \( x=-1 \): \( -1+3 = a((-1)^2+2) \Rightarrow 2 = 3a \Rightarrow a = \frac{2}{3} \) Compare coeff. of \( x^2 \): \( 0 = a+b \Rightarrow b = -a = -\frac{2}{3} \) Compare constant terms:…
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