Question

Let $f:R\to R$ be defined as $f(x)=\{\begin{array}{l}\frac{\lambda |x^2-5x+6|}{\mu \left(5x-x^2-6\right)},x<2\\ e^{\frac{\tan (x-2)}{x-[x]}},x>2\\ \mu ,x=2\end{array}$
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. $2e-1$

Answer 1

Answer: (A)

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