TS EAMCET · Maths · Basic of Mathematics
If \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{\mathrm{A}}{x+2}+\frac{\mathrm{B} x+\mathrm{C}}{\left(x^2+1\right)}\) then \(3 \mathrm{~A}+2 \mathrm{~B}-\mathrm{C}=\)
- A \(\frac{8}{5}\)
- B \(\frac{16}{5}\)
- C \(\frac{3}{5}\)
- D \(\frac{19}{5}\)
Answer & Solution
Correct Answer
(D) \(\frac{19}{5}\)
Step-by-step Solution
Detailed explanation
\(x^2-3 = A(x^2+1) + (Bx+C)(x+2)\) For \(x=-2\): \( (-2)^2-3 = A((-2)^2+1) \) \( 1 = 5A \implies A = \frac{1}{5} \) Comparing coefficients of \(x^2\): \( A+B = 1 \implies \frac{1}{5}+B=1 \implies B = \frac{4}{5} \) Comparing constant terms:…
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