is The function `f(x)=(x^2-1)|x^2-3x+2|+cos(|x|)` is differentiable not differentiable at (a)-1 (b)0 (c)1 (d)2

Options

A. -1
B. 0
C. 1
D. 2

Answer

To determine whether the function \( f(x) = (x^2 - 1)|x^2 - 3x + 2| + \cos(|x|) \) is differentiable at the points \( -1, 0, 1, \) and \( 2 \), we will analyze each point step by step.

### Step 1: Identify Non-Differentiable Points
The function contains absolute values, which can lead to points of non-differentiability.

1. **For \( |x| \)**: This is not differentiable at \( x = 0 \).
2. **For \( |x^2 - 3x + 2| \)**: We need to factor \( x^2 - 3x + 2 \):
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]
This expression is zero at \( x = 1 \) and \( x = 2 \). Therefore, \( f(x) \) is not differentiable at \( x = 1 \) and \( x = 2 \) as well.

### Step 2: Analyze Differentiability at Each Point
Now we will check the differentiability at each of the points \( -1, 0, 1, \) and \( 2 \).

#### At \( x = -1 \)
- The function is continuous and differentiable at \( x = -1 \) since neither absolute value affects the function here.
- **Conclusion**: \( f(x) \) is differentiable at \( x = -1 \).

#### At \( x = 0 \)
- The term \( \cos(|x|) \) is differentiable everywhere, but \( |x| \) is not differentiable at \( x = 0 \).
- **Conclusion**: \( f(x) \) is not differentiable at \( x = 0 \).

#### At \( x = 1 \)
- The term \( |x^2 - 3x + 2| \) is not differentiable at \( x = 1 \) (as shown in Step 1).
- **Conclusion**: \( f(x) \) is not differentiable at \( x = 1 \).

#### At \( x = 2 \)
- The term \( |x^2 - 3x + 2| \) is also not differentiable at \( x = 2 \).
- **Conclusion**: \( f(x) \) is not differentiable at \( x = 2 \).

### Final Conclusion
The function \( f(x) \) is not differentiable at the points \( 0, 1, \) and \( 2 \). Therefore, the answer is:
- (a) -1: Differentiable
- (b) 0: Not differentiable
- (c) 1: Not differentiable
- (d) 2: Not differentiable

The correct answer is **(b) 0, (c) 1, and (d) 2**.