Question

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

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

Hint:

Only \(t=0\) is a possible problem point; compare the right and left derivatives at 0.

Answer 1

The Correct Option is A

For \(t\neq0\), \(f\) is a product of differentiable functions, so only \(t=0\) needs checking; \(f(0)=0\).
Right derivative: \(f'(0^+)=\lim_{t\to0^+}\frac{(|\lambda|e^{t}-\mu)\sin 2t}{t}=2(|\lambda|-\mu)\).
Left derivative (\(t<0\), \(|t|=-t\)): \(f'(0^-)=\lim_{t\to0^-}\frac{(|\lambda|e^{-t}-\mu)\sin(-2t)}{t}=-2(|\lambda|-\mu)\).
Differentiable at 0 requires \(2(|\lambda|-\mu)=-2(|\lambda|-\mu)\Rightarrow\mu=|\lambda|\).
So \(S=\{(\lambda,|\lambda|)\}\), where \(\mu=|\lambda|\ge0\) and \(\lambda\) can be any real number. Hence \(S\subseteq\mathbf{R}\times[0,\infty)\). It is not inside \([0,\infty)\times\mathbf{R}\) since \(\lambda\) can be negative.
Answer: (1) \(\mathbf{R}\times[0,\infty)\)