JEE Main 2021 — Mathematics Calculus

If $f(x)$f(x) is continuous at $x=2$x=2, then $\underset{x\rightarrow {2}^{+}}{\mathrm{lim}}f(x)=\underset{x\rightarrow {2}^{-}}{\mathrm{lim}}f(x)=\underset{x\rightarrow 2}{\mathrm{lim}}f(x)$x→2+lim​f(x)=x→2−lim​f(x)=x→2lim​f(x)

Here,

$$\underset{x\rightarrow {2}^{+}}{\mathrm{lim}}f(x)=\underset{x\rightarrow {2}^{+}}{\mathrm{lim}}{e}^{\frac{\mathrm{tan}(x-2)}{x-2}}={e}^{1}=e$$x→2+lim​f(x)=x→2+lim​ex−2tan(x−2)​=e1=e

$$\underset{x\rightarrow {2}^{-}}{\mathrm{lim}}f(x)=\underset{x\rightarrow {2}^{-}}{\mathrm{lim}}\frac{\lambda |{x}^{2}-5x+6|}{\mu (5x-{x}^{2}-6)}=\underset{x\rightarrow {2}^{-}}{\mathrm{lim}}\frac{-\lambda (x-2)(x-3)}{\mu (x-2)(x-3)}=-\frac{\lambda }{\mu }$$x→2−lim​f(x)=x→2−lim​μ(5x−x2−6)λ∣x2−5x+6∣​=x→2−lim​μ(x−2)(x−3)−λ(x−2)(x−3)​=−μλ​

(\because |{x}^{2}-5x+6|={\begin{matrix}(x-2)(x-3)\text{if x < 2 or x > 3} \\ -(x-2)(x-3)\text{if 2 < x < 3 }\end{matrix})

For continuity $\mu =e=-\frac{\lambda }{\mu }\Rightarrow \mu =e,\lambda =-{e}^{2}$μ=e=−μλ​⇒μ=e,λ=−e2

$$\lambda +\mu =e(-e+1)$$λ+μ=e(−e+1)