Question

Let $f: R \rightarrow R$ 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 $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x=2$, then $\lambda+\mu$ is equal to:

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

Answer 1

Answer: (A)

$\displaystyle\lim _{x \rightarrow 2^{+}} f(x)=\displaystyle\lim _{x \rightarrow 2^{+}} e^{\frac{\tan (x-2)}{x-2}}=e^{1}$
$\displaystyle\lim _{x \rightarrow 2^{-}} f(x)=\displaystyle\lim _{x \rightarrow 2^{-}} \frac{-\lambda(x-2)(x-3)}{\mu(x-2)(x-3)}=-\frac{\lambda}{\mu}$
For continuity $\mu=e=-\frac{\lambda}{\mu}$
$\Rightarrow \mu=e, \lambda=-e^{2}$
$\lambda+\mu=e(-e+1)$