Key Concepts and Formulas

• Differentiability at a Point: A function $f(t)$ is differentiable at a point $t=a$ if the limit of the difference quotient exists. For a function involving absolute values and piecewise definitions, differentiability at the point where the definition changes (here, $t=0$) requires the left-hand derivative (LHD) to be equal to the right-hand derivative (RHD).
• Derivative of $|t|$: The derivative of $|t|$ is $\text{sgn}(t)$, which is 1 for $t>0$ and -1 for $t<0$. It is undefined at $t=0$.
• Chain Rule: If $y = g(u)$ and $u = h(t)$, then $\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt}$.
• Product Rule: If $f(t) = u(t)v(t)$, then $f'(t) = u'(t)v(t) + u(t)v'(t)$.

Step-by-Step Solution

Step 1: Analyze the function $f(t)$ and its form for $t>0$ and $t<0$. The given function is $f(t) = (|\lambda| e^{|t|} - \mu) \sin(2|t|)$. We need to consider the cases $t>0$ and $t<0$ separately because of the $|t|$ term.

For $t > 0$, $|t| = t$. So, $f(t) = (|\lambda| e^t - \mu) \sin(2t)$. For $t < 0$, $|t| = -t$. So, $f(t) = (|\lambda| e^{-t} - \mu) \sin(2(-t)) = (|\lambda| e^{-t} - \mu) (-\sin(2t))$.

We can write $f(t)$ as:

$$f(t) = \begin{cases} (|\lambda| e^t - \mu) \sin(2t) & \text{if } t > 0 \\ (|\lambda| e^{-t} - \mu) (-\sin(2t)) & \text{if } t < 0 \end{cases}$$

Note that $f(0) = (|\lambda| e^0 - \mu) \sin(0) = (|\lambda| - \mu) \cdot 0 = 0$.

Step 2: Calculate the derivative of $f(t)$ for $t>0$ and $t<0$. We will use the product rule and the chain rule.

For $t > 0$: Let $u(t) = |\lambda| e^t - \mu$ and $v(t) = \sin(2t)$. Then $u'(t) = |\lambda| e^t$ and $v'(t) = \cos(2t) \cdot 2 = 2\cos(2t)$. So, $f'(t) = u'(t)v(t) + u(t)v'(t) = (|\lambda| e^t) \sin(2t) + (|\lambda| e^t - \mu)(2\cos(2t))$.

For $t < 0$: Let $p(t) = |\lambda| e^{-t} - \mu$ and $q(t) = -\sin(2t)$. Then $p'(t) = |\lambda| e^{-t} \cdot (-1) = -|\lambda| e^{-t}$ and $q'(t) = -\cos(2t) \cdot 2 = -2\cos(2t)$. So, $f'(t) = p'(t)q(t) + p(t)q'(t) = (-|\lambda| e^{-t}) (-\sin(2t)) + (|\lambda| e^{-t} - \mu)(-2\cos(2t))$ $f'(t) = |\lambda| e^{-t} \sin(2t) - 2(|\lambda| e^{-t} - \mu)\cos(2t)$.

We can write $f'(t)$ as:

$$f'(t) = \begin{cases} |\lambda| e^t \sin(2t) + 2(|\lambda| e^t - \mu) \cos(2t) & \text{if } t > 0 \\ |\lambda| e^{-t} \sin(2t) - 2(|\lambda| e^{-t} - \mu) \cos(2t) & \text{if } t < 0 \end{cases}$$

Step 3: Apply the condition for differentiability at $t=0$. For $f(t)$ to be differentiable at $t=0$, the left-hand derivative (LHD) must equal the right-hand derivative (RHD) at $t=0$.

The RHD at $t=0$ is the limit of $f'(t)$ as $t \to 0^+$: $\text{RHD} = \lim_{t \to 0^+} [|\lambda| e^t \sin(2t) + 2(|\lambda| e^t - \mu) \cos(2t)]$ $\text{RHD} = |\lambda| e^0 \sin(0) + 2(|\lambda| e^0 - \mu) \cos(0)$ $\text{RHD} = |\lambda| \cdot 1 \cdot 0 + 2(|\lambda| - \mu) \cdot 1 = 0 + 2(|\lambda| - \mu) = 2(|\lambda| - \mu)$

The LHD at $t=0$ is the limit of $f'(t)$ as $t \to 0^-$: $\text{LHD} = \lim_{t \to 0^-} [|\lambda| e^{-t} \sin(2t) - 2(|\lambda| e^{-t} - \mu) \cos(2t)]$ $\text{LHD} = |\lambda| e^0 \sin(0) - 2(|\lambda| e^0 - \mu) \cos(0)$ $\text{LHD} = |\lambda| \cdot 1 \cdot 0 - 2(|\lambda| - \mu) \cdot 1 = 0 - 2(|\lambda| - \mu) = -2(|\lambda| - \mu)$

For $f(t)$ to be differentiable at $t=0$, we must have LHD = RHD: $2(|\lambda| - \mu) = -2(|\lambda| - \mu)$ $4(|\lambda| - \mu) = 0$ $|\lambda| - \mu = 0$ $|\lambda| = \mu$

Step 4: Determine the set S based on the condition found. The set S consists of pairs $(\lambda, \mu) \in \mathbb{R} \times \mathbb{R}$ such that $f(t)$ is differentiable. We found that the condition for differentiability is $|\lambda| = \mu$.

This means that $\mu$ must be non-negative, since $|\lambda| \ge 0$. So, $\mu \in [0, \infty)$. The value of $\lambda$ can be any real number, $\lambda \in \mathbb{R}$.

Therefore, the set S is given by: $S = \{(\lambda, \mu) \in \mathbb{R} \times \mathbb{R} \mid \mu = |\lambda|\}$ This set can be described as $\lambda \in \mathbb{R}$ and $\mu \in [0, \infty)$.

Step 5: Identify which option the set S is a subset of. We need to find the option that contains all pairs $(\lambda, \mu)$ where $\lambda \in \mathbb{R}$ and $\mu \in [0, \infty)$.

Option (A) is $\mathbb{R} \times [0, \infty)$. This means $\lambda \in \mathbb{R}$ and $\mu \in [0, \infty)$. This matches our findings for S. Option (B) is $[0, \infty) \times \mathbb{R}$. This means $\lambda \in [0, \infty)$ and $\mu \in \mathbb{R}$, which is incorrect as $\lambda$ can be negative. Option (C) is $\mathbb{R} \times (-\infty, 0)$. This means $\lambda \in \mathbb{R}$ and $\mu \in (-\infty, 0)$, which is incorrect as $\mu$ must be non-negative. Option (D) is $(-\infty, 0) \times \mathbb{R}$. This means $\lambda \in (-\infty, 0)$ and $\mu \in \mathbb{R}$, which is incorrect as $\lambda$ can be positive and $\mu$ has restrictions.

Thus, S is a subset of $\mathbb{R} \times [0, \infty)$.

Common Mistakes & Tips

• Incorrectly handling $|t|$ derivative: Remember that the derivative of $|t|$ is $\text{sgn}(t)$, not simply 1. Also, be careful when differentiating composite functions involving $|t|$.
• Forgetting continuity requirement: While the question asks for differentiability, which implies continuity, it's good practice to mentally check continuity at $t=0$. In this case, $f(0)=0$, and the limits of the piecewise functions as $t \to 0^+$ and $t \to 0^-$ also yield 0, so continuity holds.
• Algebraic errors in LHD=RHD: The calculation of LHD and RHD and setting them equal requires careful algebraic manipulation. Double-checking the signs and terms is crucial.

Summary

To determine the set S of pairs $(\lambda, \mu)$ for which $f(t) = (|\lambda| e^{|t|} - \mu) \sin(2|t|)$ is differentiable, we analyzed the function's behavior for $t>0$ and $t<0$. We then derived the expressions for the right-hand derivative (RHD) and the left-hand derivative (LHD) at $t=0$. For differentiability, these two must be equal. Setting RHD = LHD led to the condition $\mu = |\lambda|$. This implies that $\lambda$ can be any real number, while $\mu$ must be a non-negative real number. Therefore, the set S is a subset of $\mathbb{R} \times [0, \infty)$.

The final answer is \boxed{A}.