91Ó°ÊÓ

We need to examine whether \(\sin^2 x < x\sin(\sin x)\) for \(0<x<\frac{\pi}{2}\). To do that, let's divide both sides by \(\sin^2 x\): \[\frac{\sin^2 x}{\sin^2 x} < \frac{x\sin (\sin x)}{\sin^2 x}\] \[1 < \frac{x}{\sin x}\sin(\sin x)\] Now let's examine the function \(g(x) = \frac{x}{\sin x}\). We will determine its behavior in the given interval and use this information to analyze the inequality.

To find the behavior of \(g(x)\) in the given interval, we can use the derivative. Find the derivative of \(g(x)\) using the quotient rule: \[\frac{d}{dx}g(x) = \frac{d}{dx}\frac{x}{\sin x} = \frac{\sin x(1) - x\cos x}{(\sin x)^2} = \frac{\sin x - x\cos x}{\sin^2 x}\] Now we will check the sign of the first derivative to determine the increasing or decreasing behavior of \(g(x)\): 1. For \(x = 0\), the first derivative is undefined. We cannot determine the behavior at this point. 2. For \(0 < x < \frac{\pi}{2}\), notice that \(\cos x > 0\), and according to the sine function, \(\sin x < x\). Therefore, \(\sin x < x\cos x\), and the first derivative is positive. 3. For \(x = \frac{\pi}{2}\), both the numerator and denominator of the first derivative are zero, resulting in an indeterminate form. Thus, for the given interval, we can conclude that \(g(x) = \frac{x}{\sin x}\) is an increasing function.

Since \(g(x) = \frac{x}{\sin x}\) is an increasing function for \(0 \frac{0}{\sin 0} = 0\) Now we will use this fact to analyze the inequality from Step 1: \[1 < \frac{x}{\sin x} \sin(\sin x)\] Since the inequality holds true for \(0<x<\frac{\pi}{2}\), we know that (a) is correct.

Since we have found option (a) to be true, we now need to verify whether (b) or (c) can also be true. (b) \(\sin^2 x > x\sin(\sin x)\): Since we have proved that \(\sin^2 x < x\sin(\sin x)\) for the interval \(0 1 + x\sin(\sin x)\): Since \(\sin^2 x < x\sin(\sin x)\), then adding 1 to the right side of the inequality, we would also have \(\sin^2 x < 1 + x\sin (\sin x)\), making this option false as well. In conclusion, since we have proven that option (a) holds true for the interval \(0<x<\frac{\pi}{2}\), and both options (b) and (c) are false, we have found the correct answer: The correct choice is (a) \(\sin^2 x < x\sin(\sin x)\) for \(0<x<\frac{\pi}{2}\).