For the function `f(x)=x cos ""1/x, x ge 1 `which one of the following is incorrect ?

Options

A. (a)for atleast one x in the interval `[1, oo), f(x + 2) - f(x) lt 2`
B. (b)`underset(x rarr oo)(lim) f'(x) = 1`
C. (c)for all x in the interval `[1, oo), f(x+2) - f(x) gt 2`
D. (d)f'(x) is strictly decreasing in the interval `[1, oo)`

Answer

To solve the problem regarding the function \( f(x) = x \cos\left(\frac{1}{x}\right) \) for \( x \geq 1 \), we need to analyze the function and its derivatives to determine which statement is incorrect among the provided options.

### Step 1: Analyze the function
The function is defined as:
\[ f(x) = x \cos\left(\frac{1}{x}\right) \quad \text{for } x \geq 1 \]

### Step 2: Find the first derivative \( f'(x) \)
Using the product rule and chain rule:
\[ f'(x) = \frac{d}{dx}\left(x\right) \cdot \cos\left(\frac{1}{x}\right) + x \cdot \frac{d}{dx}\left(\cos\left(\frac{1}{x}\right)\right) \]
Calculating the derivative of \( \cos\left(\frac{1}{x}\right) \):
\[ \frac{d}{dx}\left(\cos\left(\frac{1}{x}\right)\right) = -\sin\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = \frac{\sin\left(\frac{1}{x}\right)}{x^2} \]
Thus, we have:
\[ f'(x) = \cos\left(\frac{1}{x}\right) + x \cdot \frac{\sin\left(\frac{1}{x}\right)}{x^2} \]
Simplifying this gives:
\[ f'(x) = \cos\left(\frac{1}{x}\right) + \frac{\sin\left(\frac{1}{x}\right)}{x} \]

### Step 3: Evaluate the limit as \( x \to \infty \)
We need to find:
\[ \lim_{x \to \infty} f'(x) \]
As \( x \to \infty \), \( \frac{1}{x} \to 0 \). Therefore:
\[ \cos\left(\frac{1}{x}\right) \to \cos(0) = 1 \quad \text{and} \quad \sin\left(\frac{1}{x}\right) \to \sin(0) = 0 \]
Thus:
\[ \lim_{x \to \infty} f'(x) = 1 + 0 = 1 \]

### Step 4: Find the second derivative \( f''(x) \)
To find \( f''(x) \), we differentiate \( f'(x) \):
\[ f''(x) = \frac{d}{dx}\left(\cos\left(\frac{1}{x}\right)\right) + \frac{d}{dx}\left(\frac{\sin\left(\frac{1}{x}\right)}{x}\right) \]
Calculating \( \frac{d}{dx}\left(\cos\left(\frac{1}{x}\right)\right) \) gives:
\[ -\sin\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = \frac{\sin\left(\frac{1}{x}\right)}{x^2} \]
For \( \frac{d}{dx}\left(\frac{\sin\left(\frac{1}{x}\right)}{x}\right) \), we use the quotient rule:
\[ \frac{d}{dx}\left(\frac{\sin\left(\frac{1}{x}\right)}{x}\right) = \frac{x \cdot \frac{\cos\left(\frac{1}{x}\right)}{x^2} - \sin\left(\frac{1}{x}\right)}{x^2} \]
Combining these results gives:
\[ f''(x) = \frac{\sin\left(\frac{1}{x}\right)}{x^2} + \frac{\cos\left(\frac{1}{x}\right)}{x^2} - \frac{\sin\left(\frac{1}{x}\right)}{x^2} \]
Thus:
\[ f''(x) = \frac{\cos\left(\frac{1}{x}\right)}{x^2} \]

### Step 5: Analyze the sign of \( f''(x) \)
Since \( \cos\left(\frac{1}{x}\right) \) is positive for \( x \geq 1 \), we conclude that:
\[ f''(x) > 0 \quad \text{for } x \geq 1 \]
This indicates that \( f'(x) \) is increasing.

### Step 6: Conclusion
Now, we check the options provided in the question to identify which statement is incorrect. Based on our analysis:
- \( f'(x) \) is strictly increasing.
- The limit of \( f'(x) \) as \( x \to \infty \) is 1.
- \( f(x) \) is continuous and differentiable for \( x \geq 1 \).

### Final Answer
The incorrect option can be determined based on the analysis above.