On which of the following intervals is the function 'f' given by `f(x)=x^(100)+sinx-1` strictly increasing?

Options

A. `(-1, 1)`
B. `(0, 1)`
C. `((pi)/(2), pi)`
D. `(0, (pi )/(2))`.

Answer

To determine the intervals on which the function \( f(x) = x^{100} + \sin x - 1 \) is strictly increasing, we need to analyze its derivative.

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^{100}) + \frac{d}{dx}(\sin x) - \frac{d}{dx}(1) \]

Calculating the derivatives, we get:

\[ f'(x) = 100x^{99} + \cos x \]

### Step 2: Set the derivative greater than zero
To find where the function is strictly increasing, we need to solve the inequality:

\[ f'(x) > 0 \]

This translates to:

\[ 100x^{99} + \cos x > 0 \]

### Step 3: Analyze the inequality
We can rearrange the inequality:

\[ \cos x > -100x^{99} \]

### Step 4: Consider the behavior of \( \cos x \) and \( -100x^{99} \)
1. The function \( \cos x \) oscillates between -1 and 1 for all \( x \).
2. The term \( -100x^{99} \) approaches 0 as \( x \) approaches 0 and becomes very negative as \( |x| \) increases.

### Step 5: Identify intervals
- For \( x = 0 \):
\[ f'(0) = 100(0)^{99} + \cos(0) = 1 > 0 \]
Hence, \( f(x) \) is increasing at \( x = 0 \).

- For \( x > 0 \):
As \( x \) increases, \( 100x^{99} \) grows much faster than \( \cos x \) can oscillate. Therefore, for sufficiently large \( x \), \( f'(x) > 0 \).

- For \( x < 0 \):
As \( x \) becomes more negative, \( -100x^{99} \) becomes very large and positive, while \( \cos x \) remains bounded between -1 and 1. Thus, \( f'(x) > 0 \) for negative \( x \) as well.

### Step 6: Conclusion
From the analysis, we find that \( f'(x) > 0 \) for all \( x \). Therefore, the function \( f(x) \) is strictly increasing on the entire real line.

### Summary of Intervals
- The function \( f(x) = x^{100} + \sin x - 1 \) is strictly increasing on the interval \( (-\infty, \infty) \).