Question

Let $f(x) = x^2 + x \sin x - \cos x$. Then

• A. f(x) = 0 has at least one real root
• B. f(x) =0 has no real root
• C. f(x) =0 has at least one positive root
• D. f(x) =0 has at least one negative root

Answer 1

The Correct Option is A, C, D

We are given the function \( f(x) = x^2 + x \sin x - \cos x \), and we need to analyze whether it has real roots.

Step 1: Behavior of \( f(x) \) at some key points
- At \( x = 0 \): \[ f(0) = 0^2 + 0 \sin(0) - \cos(0) = 0 - 1 = -1 \] Thus, \( f(0) = -1 \). - At \( x = 1 \): \[ f(1) = 1^2 + 1 \sin(1) - \cos(1) \] Approximating \( \sin(1) \approx 0.841 \) and \( \cos(1) \approx 0.540 \): \[ f(1) \approx 1 + 0.841 - 0.540 = 1.301 \] Thus, \( f(1) \approx 1.301 \). 

Step 2: Intermediate Value Theorem
From the calculations, we know that: - \( f(0) = -1 \) (negative), - \( f(1) \approx 1.301 \) (positive). By the Intermediate Value Theorem, since \( f(x) \) is continuous, and \( f(0) \) and \( f(1) \) have opposite signs, there must be at least one root in the interval \( (0, 1) \). 

Step 3: Conclusion
Since the function changes sign between 0 and 1, we can conclude that \( f(x) = 0 \) has at least one real root in this interval.

Answer:

\[ \boxed{f(x) = 0 \text{ has at least one real root}} \]