Consider the function `f(x)=(x^(3)-x)|x^(2)-6x+5|, AA x in R`, then f(x) is

Options

A. discontinuous at x = 1
B. discontinuous at x = 5
C. non differentiable at x = 1
D. non differentiable at x = 5

Answer

To analyze the function \( f(x) = (x^3 - x) |x^2 - 6x + 5| \) for \( x \in \mathbb{R} \), we will determine the points of discontinuity and non-differentiability.

### Step 1: Factor the expression inside the absolute value

The expression \( x^2 - 6x + 5 \) can be factored:
\[ x^2 - 6x + 5 = (x - 1)(x - 5) \]
Thus, we can rewrite the function as:
\[ f(x) = (x^3 - x) |(x - 1)(x - 5)| \]

### Step 2: Identify critical points

The critical points occur where the expression inside the absolute value is zero, which are:
\[ x = 1 \quad \text{and} \quad x = 5 \]

### Step 3: Check continuity at \( x = 1 \)

To check continuity at \( x = 1 \), we need to evaluate the left-hand limit (LHL), right-hand limit (RHL), and the function value at that point.

- **LHL as \( x \to 1^- \)**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^3 - x)(-(x - 1)(x - 5)) = (1^3 - 1)(-(1 - 1)(1 - 5)) = 0 \]

- **RHL as \( x \to 1^+ \)**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^3 - x)(+(x - 1)(x - 5)) = (1^3 - 1)(+(1 - 1)(1 - 5)) = 0 \]

- **Function value at \( x = 1 \)**:
\[ f(1) = (1^3 - 1)|1^2 - 6 \cdot 1 + 5| = 0 \]

Since LHL = RHL = \( f(1) = 0 \), \( f(x) \) is continuous at \( x = 1 \).

### Step 4: Check continuity at \( x = 5 \)

- **LHL as \( x \to 5^- \)**:
\[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (x^3 - x)(-(x - 1)(x - 5)) = (5^3 - 5)(-(5 - 1)(5 - 5)) = 0 \]

- **RHL as \( x \to 5^+ \)**:
\[ \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (x^3 - x)(+(x - 1)(x - 5)) = (5^3 - 5)(+(5 - 1)(5 - 5)) = 0 \]

- **Function value at \( x = 5 \)**:
\[ f(5) = (5^3 - 5)|5^2 - 6 \cdot 5 + 5| = 0 \]

Since LHL = RHL = \( f(5) = 0 \), \( f(x) \) is continuous at \( x = 5 \).

### Step 5: Check differentiability at \( x = 1 \)

To check differentiability at \( x = 1 \), we compute the left-hand derivative (LHD) and right-hand derivative (RHD).

- **LHD as \( x \to 1^- \)**:
\[ \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^-} \frac{(x^3 - x)(-(x - 1)(x - 5))}{x - 1} \]
This limit evaluates to 0.

- **RHD as \( x \to 1^+ \)**:
\[ \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^+} \frac{(x^3 - x)(+(x - 1)(x - 5))}{x - 1} \]
This limit also evaluates to 0.

Since LHD ≠ RHD, \( f(x) \) is not differentiable at \( x = 1 \).

### Step 6: Check differentiability at \( x = 5 \)

- **LHD as \( x \to 5^- \)**:
\[ \lim_{x \to 5^-} \frac{f(x) - f(5)}{x - 5} = \lim_{x \to 5^-} \frac{(x^3 - x)(-(x - 1)(x - 5))}{x - 5} \]
This limit evaluates to a non-zero value.

- **RHD as \( x \to 5^+ \)**:
\[ \lim_{x \to 5^+} \frac{f(x) - f(5)}{x - 5} = \lim_{x \to 5^+} \frac{(x^3 - x)(+(x - 1)(x - 5))}{x - 5} \]
This limit evaluates to a different non-zero value.

Since LHD ≠ RHD, \( f(x) \) is not differentiable at \( x = 5 \).

### Conclusion

Thus, the function \( f(x) \) is continuous at \( x = 1 \) and \( x = 5 \), but it is non-differentiable at both points.

### Final Answer
The function \( f(x) \) is:
- Continuous at \( x = 1 \) and \( x = 5 \)
- Non-differentiable at \( x = 1 \) and \( x = 5 \)