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

We have the function

$$f(t)=\bigl(|\lambda|\,e^{t}-\mu\bigr)\;\sin\!\bigl(2|t|\bigr), \qquad t\in\mathbb R.$$

For differentiability we must look carefully at the point where the expression $$|t|$$ changes its definition, namely at $$t=0.$$ Everywhere else, every symbol involved (the exponential, the absolute value already chosen as $$\pm t,$$ and the sine) is smooth, so no further problem occurs. Thus the whole problem reduces to checking the derivative at $$t=0.$$

First we split the function according to the sign of $$t$$ because $$|t|=\begin{cases}t,& t>0,\\ -t,& t<0.\end{cases}$$

For $$t>0$$ we get

$$f(t)=\bigl(|\lambda|\,e^{t}-\mu\bigr)\,\sin(2t). \quad -(1)$$

For $$t<0$$ we get

$$f(t)=\bigl(|\lambda|\,e^{t}-\mu\bigr)\,\sin\!\bigl(2(-t)\bigr)=\bigl(|\lambda|\,e^{t}-\mu\bigr)\,\sin(-2t).$$

Because $$\sin(-x)=-\sin x,$$ this becomes

$$f(t)=-\bigl(|\lambda|\,e^{t}-\mu\bigr)\,\sin(2t). \quad -(2)$$

Exactly at $$t=0$$ we have $$|t|=0,\;e^{0}=1,\;\sin 0 =0,$$ so

$$f(0)=\bigl(|\lambda|\cdot 1-\mu\bigr)\cdot 0=0.$$

Thus the function is already continuous at the origin for every pair $$(\lambda,\mu).$$ We only have to match the two one-sided derivatives there.

Derivative from the right (using (1)): First recall the product rule
$$\dfrac{d}{dt}\{u(t)\,v(t)\}=u'(t)\,v(t)+u(t)\,v'(t).$$

Here $$u(t)=|\lambda|e^{t}\!,\;\;u'(t)=|\lambda|e^{t},$$ and $$v(t)=\sin(2t),\;\;v'(t)=2\cos(2t).$$

So for $$t>0$$

$$f'(t)=|\lambda|e^{t}\sin(2t)+\bigl(|\lambda|e^{t}-\mu\bigr)\;2\cos(2t).$$

Evaluating at $$t=0$$ gives

$$f'_{+}(0)=|\lambda|e^{0}\sin 0+\bigl(|\lambda|e^{0}-\mu\bigr)\;2\cos 0 =|\lambda|\cdot 0+\bigl(|\lambda|-\mu\bigr)\;2\cdot 1 =2\bigl(|\lambda|-\mu\bigr).$$

Derivative from the left (using (2)): We again apply the product rule, remembering the extra minus sign.

Starting with $$g(t)= -\bigl(|\lambda|e^{t}-\mu\bigr)\sin(2t),$$ we have $$g'(t)= -\Bigl[\,|\lambda|e^{t}\sin(2t)+\bigl(|\lambda|e^{t}-\mu\bigr)\;2\cos(2t)\Bigr].$$

At $$t=0$$ this becomes

$$f'_{-}(0)=-\Bigl[\,|\lambda|e^{0}\sin 0+\bigl(|\lambda|e^{0}-\mu\bigr)\;2\cos 0\Bigr] =-\Bigl[\,|\lambda|\cdot 0+\bigl(|\lambda|-\mu\bigr)\;2\Bigr] =-2\bigl(|\lambda|-\mu\bigr).$$

For differentiability at $$t=0$$ we require the two one-sided derivatives to coincide:

$$f'_{+}(0)=f'_{-}(0)\quad\Longrightarrow\quad 2\bigl(|\lambda|-\mu\bigr)=-2\bigl(|\lambda|-\mu\bigr).$$

Adding the right-hand side to the left gives

$$4\bigl(|\lambda|-\mu\bigr)=0,$$ so

$$|\lambda|-\mu=0\quad\Longrightarrow\quad\mu=|\lambda|.$$

Because $$|\lambda|\ge 0$$ for every real $$\lambda,$$ the equality $$\mu=|\lambda|$$ forces $$\mu$$ to be a non-negative real number, while $$\lambda$$ itself can still be any real number.

Therefore the admissible pairs $$(\lambda,\mu)$$ satisfy $$\lambda\in\mathbb R,\quad\mu\in[0,\infty).$$ Symbolically, $$S\subset \mathbb R\times[0,\infty).$$

Among the given options, this set matches exactly with Option A.

Hence, the correct answer is Option A.