Let $\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}$ be defined as
$f(x)= \begin{cases}\frac{\lambda\left|x^{2}-5 x+6\right|}{\mu\left(5 x-x^{2}-6\right)}, & x<2 \\ e^{\frac{\tan (x-2)}{x-[x]}} & , x>2 \\ \mu & , x=2\end{cases}$
where $[\mathrm{x}]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x=2$, then $\lambda+\mu$ is equal to:
Correct Option: 1
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} e^{\frac{\tan (x-2)}{x-2}}=e^{1}$
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} \frac{-\lambda(x-2)(x-3)}{\mu(x-2)(x-3)}=-\frac{\lambda}{\mu}$
For continuity $\mu=\mathrm{e}=-\frac{\lambda}{\mu} \Rightarrow \mu=\mathrm{e}, \lambda=-\mathrm{e}^{2}$
$\lambda+\mu=\mathrm{e}(-\mathrm{e}+1)$
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