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TS EAMCET · Maths · Quadratic Equation

If \(\frac{3 x^2+x+1}{(x-1)^4}=\frac{a}{(x-1)}+\frac{b}{(x-1)^2}\) \(+\frac{c}{(x-1)^3}+\frac{d}{(x-1)^4}\) then \(\left[\begin{array}{ll}a & b \ c & d\end{array}\right]\) is equal to

  1. A \(\left[\begin{array}{ll}3 & 7 \ 5 & 0\end{array}\right]\)
  2. B \(\left[\begin{array}{ll}0 & 3 \ 7 & 5\end{array}\right]\)
  3. C \(\left[\begin{array}{ll}0 & 7 \ 3 & 5\end{array}\right]\)
  4. D \(\left[\begin{array}{ll}3 & 5 \ 7 & 0\end{array}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left[\begin{array}{ll}0 & 3 \ 7 & 5\end{array}\right]\)

Step-by-step Solution

Detailed explanation

\(\frac{3 x^2+x+1}{(x-1)^4}=\frac{a}{(x-1)}+\frac{b}{(x-1)^2}\) \(+\frac{c}{(x-1)^3}+\frac{d}{(x-1)^4}\) \(3 x^2+x+1=a(x-1)^3+b(x-1)^2\) \(+c(x-1)+d\) \(3 x^2+x+1=a\left(x^3-1+3 x-3 x^2\right)\) \(+b\left(x^2+1-2 x\right)+c(x-1)+d\) \(3 x^2+x+1=a x^3+(-3 a+b) x^2\)…
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