Question

Let $S=\left\{\right(\lambda ,\mu )\in R\times R:f(t)=(|\lambda |e^{|t|}-\mu )\cdot \sin (2|t|),t\in R,\text{is a differentiable function}\}.$
Then $S$ is a subset of :

• A. $R\times [0,\infty )$
• B. $[0,\infty )\times R$
• C. $R\times (-\infty ,0)$
• D. $(-\infty ,0)\times R$

Answer 1

The correct option is A $R\times [0,\infty )$

Point of non-differentiability can be at $t=0$
Checking for continuity at $t=0$
$L.H.L.=\underset{t\to 0^{-}}{lim}(|\lambda |e^{|t|}-\mu )\cdot \sin \left(2|t|\right)=0$
$R.H.L.=\underset{t\to 0^{+}}{lim}(|\lambda |e^{|t|}-\mu )\cdot \sin \left(2|t|\right)=0$
$f\left(0\right)=0$
$\therefore L.H.L.=R.H.L.=f\left(0\right)$
So the function is continuous at $t=0$

Now, check for differentiablity at $t=0$,
$R.H.D.=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h-0}\Rightarrow R.H.D.=\underset{h\to 0}{lim}\frac{(|\lambda |e^h-\mu )\cdot \sin \left(2h\right)}{h}\Rightarrow R.H.D.=2(|\lambda |-\mu )L.H.D.=\underset{h\to 0}{lim}\frac{f(-h)-f\left(0\right)}{-h-0}\Rightarrow L.H.D.=\underset{h\to 0}{lim}\frac{(|\lambda |e^h-\mu )\cdot \sin \left(2h\right)}{-h}\Rightarrow L.H.D.=-2(|\lambda |-\mu )$
So, for the function to be differentiable,
$R.H.D.=L.H.D.\Rightarrow 2(|\lambda |-\mu )=-2(|\lambda |-\mu )\Rightarrow |\lambda |=\mu \Rightarrow \lambda \in R;\mu \in [0,\infty )$
Hence, $S$ is a subset of $R\times [0,\infty )$