Question

Let $f:R\to R$ be defined as
$f\left(x\right)=\{\begin{array}{cc}\frac{\lambda |x^2-5x+6|}{\mu (5x-x^2-6)},& x<2\\ & \\ e^{\frac{\tan (x-2)}{x-\left[x\right]}},& x>2\\ \mu ,& x=2\end{array}$
where $\left[x\right]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x=2,$ then $\lambda +\mu$ is equal to

• A. $1$
• B. $e(e-2)$
• C. $e(-e+1)$
• D. $2e-1$

Answer 1

The correct option is C $e(-e+1)$

For continuity $\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)=f\left(2\right)$
$\Rightarrow \underset{x\to 2^{-}}{lim}\frac{\lambda |(x-2)(x-3)|}{\mu (x-2)(3-x)}=\underset{x\to 2^{+}}{lim}{e}^{}\frac{\tan (x-2)}{x-2}=\mu$
$\Rightarrow \frac{\lambda \left(1\right)}{\mu (-1)}=e=\mu$
$\Rightarrow \mu =e$ and $\lambda =-e^2$
$\Rightarrow \lambda +\mu =e-e^2=e(1-e)$