JEE Main 2021 (Online) 25th July Morning ShiftQuestion 1
Let $f:\mathbf{R}\to\mathbf{R}$ be defined as $f(x)=\begin{cases}\frac{\lambda|x^{2}-5x+6|}{\mu(5x-x^{2}-6)}, & x<2\\ e^{\frac{\tan(x-2)}{x-[x]}}, & x>2\\ \mu, & x=2\end{cases}$
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 :
Limits & Continuity
Solution
Concept. Remove the modulus using the sign of $(x-2)(x-3)$ for $x<2$, and use $x-[x]=x-2$ just right of $2$. Formulas used. (1) $x^2-5x+6=(x-2)(x-3)$(2) $\lim_{t\to0}\frac{\tan t}{t}=1$(3) continuity: left limit $=$ right limit $=f(2)$Step 1 - Left of 2. For $x<2$, both $x-2<0$ and $x-3<0$, so $x^2-5x+6>0$ and $|x^2-5x+6|=x^2-5x+6=-(5x-x^2-6)$. So $f(x)=\frac{-\lambda(5x-x^2-6)}{\mu(5x-x^2-6)}=-\frac{\lambda}{\mu}$ (constant). Step 2 - Right of 2. For $2<x<3$, $[x]=2$, so $x-[x]=x-2$ and by (2): $$\frac{\tan(x-2)}{x-2}\to1,\quad f(x)\to e$$ Step 3 - Continuity (3). $f(2)=\mu=e$ and $-\frac{\lambda}{\mu}=e\Rightarrow\lambda=-e^2$. Step 4 - Compute. $\lambda+\mu=e-e^2=e(1-e)=e(-e+1)$. Answer: D