Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $S=\{(\lambda ,\mu )\in R\times R:f(t)=(|\lambda |e^{|t|}-\mu )\sin (2|t|),t\in R,\text{ is a differentiable function}\}$. Then S is a subset of :

We are investigating the differentiability of the function $f(t)=(|\lambda |e^{|t|}-\mu )\sin (2|t|)$ for all $t\in \mathbb{R}$. The presence of the absolute value $|t|$ introduces a potential point of non-differentiability at $t=0$.

For the function to be differentiable everywhere, it must specifically be differentiable at $t=0$. We must ensure that the left-hand derivative and the right-hand derivative at the origin are equal.

Observe that the function $f(t)$ is an even function because $f(-t)=f(t)$. For any even function that is differentiable at the origin, the derivative must satisfy $f^{\prime }(0)=0$.

If $f'(0) eq 0$, the graph would exhibit a sharp corner (a cusp or a kink) at the origin due to the reflectional symmetry. Therefore, the condition $f^{\prime }(0)=0$ is a necessary requirement for smoothness.

We apply the formal definition of the derivative at $t=0$:

Given that $f(0)=(|\lambda |e^0-\mu )\sin (0)=0$, the expression simplifies to:

Using the standard limit ${\lim }_{h\to 0}\frac{\sin (2h)}{h}=2$ and noting that ${\lim }_{h\to 0}{e}^h=1$, we evaluate the limit:

For the function to be differentiable at $t=0$, we must set the derivative equal to zero:

Since $|\lambda |\ge 0$ for any real $\lambda$, it follows that $\mu$ must be non-negative. The set $S$ of parameters $(\lambda ,\mu )$ that satisfy this condition is defined by:

This result confirms that for any chosen $\lambda$, there exists a unique $\mu$ that ensures the function remains smooth across the entire real line.