`f:R rarr R,f(x)` is differentiable such that `f(x)=k(x^(5)+x).(kne0)` Then f(x) is increasing .
In the above question k can take value,

Options

A. -1
B. -2
C. 1
D. none of these

Answer

To determine the values that \( k \) can take such that the function \( f(x) = k(x^5 + x) \) is increasing, we need to analyze the derivative of the function.

### Step-by-Step Solution:

1. **Find the Derivative of \( f(x) \)**:
\[ f(x) = k(x^5 + x) \]
To find \( f'(x) \), we apply the power rule:
\[ f'(x) = k \left( \frac{d}{dx}(x^5) + \frac{d}{dx}(x) \right) = k(5x^4 + 1) \]

2. **Condition for Increasing Function**:
A function is increasing if its derivative is greater than or equal to zero:
\[ f'(x) > 0 \]
Therefore, we need:
\[ k(5x^4 + 1) > 0 \]

3. **Analyze \( 5x^4 + 1 \)**:
The term \( 5x^4 + 1 \) is always positive for all real \( x \) because:
- \( x^4 \) is non-negative for all \( x \).
- Thus, \( 5x^4 \geq 0 \) and \( 5x^4 + 1 \geq 1 > 0 \).

4. **Determine the Sign of \( k \)**:
Since \( 5x^4 + 1 > 0 \), the sign of the product \( k(5x^4 + 1) \) depends solely on \( k \):
\[ k(5x^4 + 1) > 0 \implies k > 0 \]

5. **Conclusion**:
Since \( k \) must be greater than zero and \( k \neq 0 \) (as given in the problem), the possible values for \( k \) are:
\[ k > 0 \]