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TS EAMCET · Maths · Quadratic Equation

If \(\frac{32 x^2+186 x}{\left(x^2+1\right)(x+5)}=\frac{37 x+1}{x^2+1}+\frac{\lambda}{x+5}\), then \(\frac{\lambda}{2}\) is equal to

  1. A \(-5\)
  2. B \(\frac{-7}{2}\)
  3. C \(\frac{-3}{2}\)
  4. D \(\frac{-5}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{-5}{2}\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \frac{3 x^2+186 x}{\left(x^2+1\right)(x+5)}=\frac{37 x+1}{x^2+1}+\frac{\lambda}{x+5} \\ & \Rightarrow \frac{32 x^2+186 x}{\left(x^2+1\right)(x+5)}=\frac{(37 x+1)(x+5)+\lambda\left(x^2+1\right)}{\left(x^2+1\right)(x+5)} \\ & \Rightarrow \quad 32 x^2+186…

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