Question

Let f(x) = cos(x) and g(x) =\(1-\frac{x^2}{2}\) for  \(x \in (-\frac{\pi}{2},\frac{\pi}{2})\). Then

• A. f(x) ≥ g(x) for all \(x \in (-\frac{\pi}{2},\frac{\pi}{2})\)
• B. f(x) ≤ g(x) for all \(x \in (-\frac{\pi}{2},\frac{\pi}{2})\)
• C. f(x) − g(x) changes sign exactly once on \((-\frac{\pi}{2},\frac{\pi}{2})\)
• D. f(x) − g(x) changes sign more than once on \((-\frac{\pi}{2},\frac{\pi}{2})\)

Answer 1

The Correct Option is A

To solve this problem, we need to compare the functions \( f(x) = \cos(x) \) and \( g(x) = 1-\frac{x^2}{2} \) over the interval \( x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

1. First, let's understand the nature of each function:
   • \( f(x) = \cos(x) \): The cosine function on \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) is a decreasing function that ranges from 0 to 1.
   • \( g(x) = 1 - \frac{x^2}{2} \): This function is a quadratic approximation of the cosine function, derived from the Taylor series expansion of \( \cos(x) \) around 0 up to the \( x^2 \) term.
2. Now, we compute the difference between the functions: \( f(x) - g(x) = \cos(x) - \left(1 - \frac{x^2}{2}\right) \). Simplifying, we have:
   \(\cos(x) - 1 + \frac{x^2}{2}\)
3. We analyze the behavior of the difference function \(d(x) = \cos(x) - 1 + \frac{x^2}{2}\):
   • At \( x = 0 \), \( d(0) = \cos(0) - 1 + \frac{0^2}{2} = 0 \).
   • For small values of \( x \), the cosine function is well-approximated by \( 1 - \frac{x^2}{2} \) because higher-order terms become negligible. Therefore, \( d(x) \approx 0 \) for small \( x \).
   • As \( x \) moves further from 0, \( \cos(x) \) remains greater than \( 1 - \frac{x^2}{2} \) because the next terms in the Taylor series expansion of \( \cos(x) \) are negative, making the approximation \( g(x) \) slightly less than \( f(x) \).
4. Overall, the difference \( d(x) = \cos(x) - 1 + \frac{x^2}{2} \geq 0 \) for \( x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), suggesting that \( f(x) \geq g(x) \) over the specified interval.
5. Therefore, the correct answer is: \( f(x) \geq g(x) \) for all \( x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).