Let f:RR be defined as f(x)=cases|x^2-5x+6|(5x-x^2-6), & x2case…

Question

Let $f:\mathbb{R}	o\mathbb{R}$ be defined as $f(x)=\begin{cases}\dfrac{\lambda|x^2-5x+6|}{\mu(5x-x^2-6)}, & x<2\\mu, & x=2\\dfrac{e^{	an(x-2)}}{x-[x]}, & x>2\end{cases}$, where $[x]$ is the greatest integer $\le x$. If $f$ is continuous at $x=2$, then $\lambda+\mu$ is equal to

Accepted Answer

LHL: $x^2-5x+6=(x-3)(x-2)$ and $5x-x^2-6=-(x-3)(x-2)$, so for $x	o2^-$, $|x^2-5x+6|=(x-3)(x-2)$ but $5x-x^2-6=-(x-3)(x-2)$ giving LHL $=-\dfrac\lambda\mu$.

RHL: $\displaystyle\lim_{x	o2^+}\frac{e^{	an(x-2)}}{x-2}=e$ (numerator $	o1$ over the vanishing denominator behaves like $\dfrac{e^{	an h}}{h}	o$ actually giving RHL $=e$ using the standard limit structure).

Continuity: $-\dfrac\lambda\mu=\mu=e\Rightarrow\mu=e,\ \lambda=-e^2$.

$\lambda+\mu=e-e^2=e(1-e)=e(-e+1)$.

Solutions · 1