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
- A \(-5\)
- B \(\frac{-7}{2}\)
- C \(\frac{-3}{2}\)
- D \(\frac{-5}{2}\)
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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