Question

The numerator of the fraction $\frac{4x+1}{x^2+3x+2}$, can be expressed as
$\lambda \times \frac{d}{dx}\left(x^2+3x+2\right)+\mu$, the values of $\lambda \mathrm{and}\mu$ are?

• A.
  $\lambda =-2,\mu =-5$
• B.
  $\lambda =-2,\mu =5$
• C.
  $\lambda =2,\mu =-5$
• D.
  $\lambda =2,\mu =5$

Answer 1

The correct option is C

$\lambda =2,\mu =-5$
$\mathrm{Here},\mathrm{we}\mathrm{represent}(4x+1)=\lambda \frac{d}{dx}\left(x^2+3x+2\right)+\mu$
$\Rightarrow \left(4x+1\right)=\lambda (2x+3)+\mu$
$\Rightarrow (4x+1)=2x\lambda +(3\lambda +\mu )$
Now, equating the coefficients of $x$ and constant term we get,
$\lambda =2\mathrm{and}\mu =-5$.