Let `a, b in RR` and `f : RR rarr RR` be defined by `f(x) = a cos(|x^3-x|) + b|x| sin(|x^3+x|).` Then `f` is

Options

A. differentiable at x=0 if a=0 and b=1
B. differentiable at x=1 if a=1 and b=0
C. NOT differentiable at x=0 if a=1 and b=0
D. Not differentiable at x=1 if a=1 and b=1

Answer

To determine the differentiability of the function \( f(x) = a \cos(|x^3 - x|) + b |x| \sin(|x^3 + x|) \), we will analyze the function based on the values of \( a \) and \( b \).

### Step 1: Analyze the function
The function \( f(x) \) is composed of two parts:
1. \( a \cos(|x^3 - x|) \)
2. \( b |x| \sin(|x^3 + x|) \)

Both parts involve absolute values and trigonometric functions, which are continuous everywhere in \( \mathbb{R} \).

### Step 2: Check continuity
Since \( \cos \) and \( \sin \) are continuous functions and the absolute value function is also continuous, the composition of these functions will be continuous. Therefore, \( f(x) \) is continuous for all \( x \in \mathbb{R} \).

### Step 3: Check differentiability
To check differentiability, we need to ensure that the derivatives of both components exist for all \( x \).

1. **For \( a \cos(|x^3 - x|) \)**:
- The function \( |x^3 - x| \) is differentiable everywhere except possibly at points where \( x^3 - x = 0 \) (i.e., \( x(x^2 - 1) = 0 \), which gives \( x = -1, 0, 1 \)).
- At these points, we need to check the left-hand and right-hand derivatives.

2. **For \( b |x| \sin(|x^3 + x|) \)**:
- The function \( |x| \) is not differentiable at \( x = 0 \).
- The function \( \sin(|x^3 + x|) \) is continuous and differentiable everywhere.

### Step 4: Evaluate specific cases
1. **Case when \( a = 0 \) and \( b = 1 \)**:
- The function simplifies to \( f(x) = |x| \sin(|x^3 + x|) \).
- At \( x = 0 \), we check the differentiability:
- Left-hand derivative: \( \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{|h| \sin(|h^3 + h|)}{h} = \lim_{h \to 0^-} -\sin(0) = 0 \).
- Right-hand derivative: \( \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h| \sin(|h^3 + h|)}{h} = \lim_{h \to 0^+} \sin(0) = 0 \).
- Since both derivatives equal, \( f(x) \) is differentiable at \( x = 0 \).

2. **Case when \( a = 1 \) and \( b = 0 \)**:
- The function simplifies to \( f(x) = \cos(|x^3 - x|) \).
- The analysis is similar; we check points \( -1, 0, 1 \) for differentiability.
- At these points, the left-hand and right-hand derivatives can be calculated and found to be equal.

### Conclusion
The function \( f(x) \) is differentiable for all \( x \in \mathbb{R} \) under the given conditions on \( a \) and \( b \).