Differentiability Analysis of Absolute Value Trigonometric Function

To analyze the differentiability of the function $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = a \cos(|x^3 - x|) + b |x| \sin(|x^3 + x|)$ we simplify each term individually.

Step 1: Simplifying the First Term

Let $T_1(x) = a \cos(|x^3 - x|)$.

Since the cosine function is an even function, $\cos(-\theta) = \cos(\theta)$ for all $\theta \in \mathbb{R}$. Thus, for any real value $u$, we have: $\cos(|u|) = \cos(u)$

Substituting $u = x^3 - x$: $T_1(x) = a \cos(x^3 - x)$

Since $x^3 - x$ is a polynomial (smooth on $\mathbb{R}$) and $\cos(y)$ is infinitely differentiable on $\mathbb{R}$, the composite function $T_1(x)$ is differentiable on all of $\mathbb{R}$ for any $a \in \mathbb{R}$.

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Step 2: Simplifying the Second Term

Let $T_2(x) = b |x| \sin(|x^3 + x|)$.

Notice that $x^3 + x = x(x^2 + 1)$. Since $x^2 + 1 > 0$ for all $x \in \mathbb{R}$, the sign of $x^3 + x$ is always identical to the sign of $x$.

We consider two cases:

1. Case 1: $x \ge 0$ Here, $|x| = x$ and $|x^3 + x| = x^3 + x$. $T_2(x) = b x \sin(x^3 + x)$

2. Case 2: $x < 0$ Here, $|x| = -x$ and $|x^3 + x| = -(x^3 + x)$. Using the odd property of sine, $\sin(-u) = -\sin(u)$, we get: $\sin(|x^3 + x|) = \sin(-(x^3 + x)) = -\sin(x^3 + x)$ Therefore: $T_2(x) = b (-x) \left(-\sin(x^3 + x)\right) = b x \sin(x^3 + x)$

Since $T_2(x) = b x \sin(x^3 + x)$ holds for both $x \ge 0$ and $x < 0$, it is valid for all $x \in \mathbb{R}$: $T_2(x) = b x \sin(x^3 + x)$

This function is a product and composition of infinitely differentiable functions on $\mathbb{R}$, so $T_2(x)$ is differentiable on all of $\mathbb{R}$ for any $b \in \mathbb{R}$.

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Step 3: Analyzing the Differentiability of $f(x)$

Combining both terms, the function simplifies to: $f(x) = a \cos(x^3 - x) + b x \sin(x^3 + x)$

Since both terms are differentiable for all $x \in \mathbb{R}$, $f(x)$ is differentiable everywhere on $\mathbb{R}$ for any values of $a, b \in \mathbb{R}$.

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Step 4: Evaluating the Options

• Option (A): "$f$ is differentiable at $x = 0$ provided $a = 0$ and $b = 1$." Since $f$ is differentiable everywhere for all $a, b \in \mathbb{R}$, it is in particular differentiable at $x = 0$ when $a = 0$ and $b = 1$. (True)

• Option (B): "$f$ is differentiable at $x = 1$ provided $a = 1$ and $b = 0$." Since $f$ is differentiable everywhere for all $a, b \in \mathbb{R}$, it is in particular differentiable at $x = 1$ when $a = 1$ and $b = 0$. (True)

• Option (C): "NOT differentiable at $x = 0$ provided $a = 1$ and $b = 0$." (False)

• Option (D): "NOT differentiable at $x = 1$ provided $a = 1$ and $b = 1$." (False)

Conclusion

The correct options are A and B.