Question

Let S = \\{ \left( {\lambda ,\mu } \right) \in R \times R:f\left( t \right) = \left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right) \cdot \sin \left( {2\left| t \right|} \right),t \in R, is a differentiable function \\}
Then $S$ is a subset of ?
A. $R \times [0,\infty )$
B. $( - \infty ,0) \times R$
C. $[0,\infty ) \times R$
D. $R \times ( - \infty ,0)$

Answer 1

In the above question, we have given a function $f\left( t \right) = \left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right) \cdot \sin \left( {2\left| t \right|} \right)$. Now, we will see the value of function at $t > 0$ and $t < 0$. Now, we have given that this function is differentiable, so, we will differentiate the function and we will see the value of the function at $t > 0$ and $t < 0$. Now, we will see that at $t = 0$, the Right hand derivative is equal to the left hand derivative. Now, we will take the value of the differentiated function at $t = 0$. Then on simplifying, we will get the range of $\left| \lambda \right|,\mu$ and then checking the options will provide us the correct answer.

Complete step-by-step answer:
In the above question, we have given an equation. That is S = \\{ \left( {\lambda ,\mu } \right) \in R \times R:f\left( t \right) = \left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right) \cdot \sin \left( {2\left| t \right|} \right),t \in R,
Now, we have also given that $f\left( t \right) = \left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right) \cdot \sin \left( {2\left| t \right|} \right)$ is differentiable.
Now, we can say that
$f\left( t \right) = \left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right) \cdot \sin \left( {e\left| t \right|} \right)$ at $t > 0$
$f\left( t \right) = \left( {\left| \lambda \right|{e^{\left| { - t} \right|}} - \mu } \right) \cdot \left( { - \sin \left( {2\left| t \right|} \right)} \right)$ at $t < 0$
Now, differentiating the function as it is given that the function is differentiable.
$f'\left( t \right) = \left( {\left( {\left| \lambda \right|{e^{\left| t \right|}}} \right)\sin 2t + \left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\left( {2\cos et} \right)} \right)$ at $t > 0$
$f'\left( t \right) = \left( {\left( {\left| \lambda \right|{e^{ - t}}} \right)\sin 2t + \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - \cos 2t} \right)} \right)$ at $t < 0$
Now, we now that $f\left( t \right)$ is differentiable,
Therefore, Left Hand Derivative is equal to Right Hand Derivative.

$$\Rightarrow \left( {\left( {\left| \lambda \right|} \right)\sin 2\left( 0 \right) + \left( {\left| \lambda \right|{e^0} - \mu } \right)\left( {2\cos \infty } \right)} \right) = \left( {\left( {\left| \lambda \right|{e^{ - 0}}} \right)\sin \infty + \left( {\left| \lambda \right|{e^{ - 0}} - \mu } \right)\left( { - \cos 0} \right)} \right) \\\ \Rightarrow 0 + \left( {\left| \lambda \right| - \mu } \right)2 = 0 - 2\left( {\left| \lambda \right|e - \mu } \right) \\\ \Rightarrow 4\left( {\left| \lambda \right| - \mu } \right) = 0 \\\ \Rightarrow \left| \lambda \right| = \mu \\\$$

Now, we know that,
S = \left( {\lambda ,\mu } \right) = \left\\{ {\lambda \in R,\mu \in \left( {0,\infty } \right)} \right\\}
Now, $0 \leqslant \lambda < \infty$ and $0 < \mu < \infty$
Now, by checking the options, we will get a correct option,
So, we get that Set $S$ is subset of $R \times [0,\infty )$

Hence, the correct option for this question is A.

Note:
We can solve the above question with another approach. As we know that $\sin \left( {2\left| t \right|} \right)$ is not differentiable at $\infty$ points. But in the problem the given function is differentiable. So, to have this function is differentiable, we need to make $\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)$ equal to zero. Now, putting this equal to zero, will provide us the range of the ${e^{\left| t \right|}}$. Hence, we will get a correct option for our answer.