Question

Between any two real roots of the equation \( e^x \sin x = 1 \), the equation \( e^x \cos x = -1 \) has:

• A. At least one root
• B. Exactly one root
• C. No root
• D. None of these

Hint:

The Intermediate Value Theorem guarantees that between two roots of a continuous function, there must be a root of any intermediate value. This applies to oscillating functions like \( \sin x \) and \( \cos x \) combined with exponential terms.

Answer 1

The Correct Option is A

We are given the two equations: 1. \( e^x \sin x = 1 \) 2. \( e^x \cos x = -1 \) 

Step 1: Behavior of \( e^x \sin x = 1 \) The equation \( e^x \sin x = 1 \) implies that the product of \( e^x \) (an exponentially increasing function) and \( \sin x \) (a periodic function oscillating between -1 and 1) equals 1. Since \( e^x \) is always positive, we can infer that this equation will have real roots at certain intervals where \( \sin x \) reaches values such that the product equals 1. 

Step 2: Behavior of \( e^x \cos x = -1 \) Similarly, the equation \( e^x \cos x = -1 \) suggests that for certain values of \( x \), the function \( e^x \) and \( \cos x \) (which oscillates between -1 and 1) will multiply to give -1. For real solutions to exist, \( e^x \) needs to be large enough and the cosine function will take negative values (between -1 and 0). 

Step 3: Intermediate Roots By the Intermediate Value Theorem, since \( e^x \sin x = 1 \) has real roots, and the function \( e^x \cos x \) is continuous, there must be at least one root of \( e^x \cos x = -1 \) between any two consecutive real roots of \( e^x \sin x = 1 \), as the functions \( e^x \sin x \) and \( e^x \cos x \) oscillate between positive and negative values. Thus, between any two real roots of \( e^x \sin x = 1 \), there is at least one root of \( e^x \cos x = -1 \). 

Hence, the correct answer is \( \boxed{\text{(a)}} \).