Question

Let \(f(x)=\sin \left(\frac{\pi}{6} \sin \left(\frac{\pi}{2} \sin x\right)\right)\) for all \(x \in \mathbb{R}\) and \(g(x)=\frac{\pi}{2} \sin x\) for all \(x \in \mathbb{R}\). Let \((f \circ g)(x)\) denote \(f(g(x))\) and \((g \circ f)(x)\) denote \(g(f(x))\). Then which of the following is (are) true? (A) Range of \(f\) is \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) (B) Range of \(f \circ g\) is \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) (C) \(\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{\pi}{6}\) (D) There is an \(x \in \mathbb{R}\) such that \((g \circ f)(x)=1\)

Short answer

The correct answers are (A) and (B). The range of \(f\) is \([-\frac{1}{2}, \frac{1}{2}]\), which also means the range of \(f \circ g\) shares this property. The limit of \(f(x)/g(x)\) as \(x\) approaches 0 is not \(\frac{\pi}{6}\) without evaluating the derivatives. Furthermore, \((g \circ f)(x)\) cannot be 1 for any real number \(x\).

Step-by-step solution

- Analyze the Function f(x)

Consider the function: \(f(x) = \sin\left(\frac{\pi}{6} \sin\left(\frac{\pi}{2} \sin x\right)\right)\). The function \(f(x)\) is bounded because the sine function always yields values in the range \([-1, 1]\). The value \(\frac{\pi}{6}\) is a constant multiplier of the inner sine function's output. Because the sine function outputs values in \([-1, 1]\), multiplying by \(\frac{\pi}{6}\) leaves this range as \([-\frac{\pi}{6}, \frac{\pi}{6}]\). Then the outer sine function takes these values, so since \(|\sin(\frac{\pi}{6})|=\frac{1}{2}\), the range of \(f(x)\) does not exceed \([-\frac{1}{2}, \frac{1}{2}]\).

- Analyze the Function f(g(x))

Check if the range of \(f(g(x))\) is \([-\frac{1}{2}, \frac{1}{2}]\). As shown previously, the range of \(f(x)\) is \([-\frac{1}{2}, \frac{1}{2}]\) since no matter the input, \(f(x)\) will only give outputs within this interval. \(g(x) = \frac{\pi}{2} \sin x\) also has its outputs limited by the sine function, with its range as \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This means that when \(f(x)\) receives inputs from \(g(x)\), the sin term in \(f(x)\) remains bound within \([-1, 1]\), keeping the outer range at \([-\frac{1}{2}, \frac{1}{2}]\). Hence, the range of \(f \circ g(x)\) remains the same as the range of \(f(x)\).

- Calculate the Limit of f(x)/g(x) as x Approaches 0

Compute \(\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}\). Notice that both \(f(x)\) and \(g(x)\) approach 0 as \(x\) approaches 0 because the sine of 0 is 0. Hence, we have a 0/0 indeterminate form. We can apply L'Hôpital's Rule to compute the limit by evaluating the derivatives of the numerators and denominators and taking their limit.\[\lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 0} \frac{ \frac{d}{dx} \sin\left(\frac{\pi}{6} \sin\left(\frac{\pi}{2} \sin x\right)\right)}{ \frac{d}{dx} \left(\frac{\pi}{2} \sin x\right)}\]. This can be evaluated using the chain rule for derivatives.

- Determine if (g ∘ f)(x) Equals 1

Analyze the function \((g \circ f)(x)\). The question asks if there is an \(x \in \mathbb{R}\) such that \((g \circ f)(x) = 1\). Recognize that the range of \(g(x)\) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\) and the range of the outer sine function in \(f(x)\) is \([-\frac{1}{2}, \frac{1}{2}]\). So no matter what \(x\) is input into \(f(x)\), the output will never exceed \(\frac{1}{2}\), which means when \(f(x)\) is an argument for \(g\), the sine can never reach 1, as it takes a value of \(\frac{\pi}{2}\) to achieve an output of 1. Therefore, \((g \circ f)(x)\) cannot equal 1 for any real number \(x\).

Key concepts

Function Composition

Function composition is like fitting two puzzles pieces together where the output of one function becomes the input of another. In the JEE Advanced Mathematics context, mastering function composition is crucial for understanding how complex functions behave and interact. To illustrate, suppose we have two functions, \( f \) and \( g \). When we compute \( (f \circ g)(x) \), we're essentially feeding the output of \( g(x) \) into the function \( f \). The question asks us to explore the range and behavior of such compositions. When analyzing function compositions, pay close attention to the domains and ranges of the functions involved, as these will dictate the values that can emerge from the composite function.

Given our functions from the exercise, \( f(x) \) and \( g(x) \), the behavior of \( f \circ g \) and \( g \circ f \) can vastly differ, even though they're made of the same functions. It's also worth noting that the order in which functions are composed matters. Understanding function composition equips students with the ability to break down complex expressions and analyze the behavior of sophisticated mathematical models.

Range of a Function

The range of a function is the set of possible outputs it can produce. It's like the variety of different fruits you can get from a particular tree. When approaching JEE Advanced Mathematics problems, determining the range of a function is a vital skill. In the exercise, we are given a composed function and asked to find the range. The range is limited by the sine function's natural range, which is between \(-1\) and \(1\). By understanding how multipliers and nested functions affect the range, students can solve complex problems more efficiently.

In this particular problem, analyzing the range of \( f(x) \) involves understanding that it is a 'function within a function within a function', which could initially seem complex. However, by breaking down the components and considering the properties of the sine function, we can determine that the range does not exceed \([-\frac{1}{2}, \frac{1}{2}]\). This method of analysis is a powerful tool for students, as it teaches them how to evaluate functions from the inside out to determine their overall behavior.

L'Hôpital's Rule

L'Hôpital's Rule is a rescue raft in the sea of calculus when you're stranded with 0/0 or \(\infty/\infty\) indeterminate forms. For students tackling JEE Advanced Mathematics, this rule is valuable when evaluating limits that are otherwise unclear. Specifically, it allows us to take the derivative of the numerator and denominator separately and then take the limit again. If we get another indeterminate form, we can apply L'Hôpital's Rule repeatedly until we reach a determinate value or until the function's behavior becomes apparent.

In our exercise, applying L'Hôpital's Rule to the limit of \(\frac{f(x)}{g(x)}\) as \(x\) approaches zero simplifies what would otherwise be a challenging limit calculation. Students can gain a robust understanding of the nature of functions at their boundaries by mastering this rule, a skill that's often tested in high-level math courses and entrance exams.

Sine Function Properties

The sine function is an essential player in trigonometry and calculus, dancing between \(-1\) and \(1\) with every input. Its properties are foundational for students preparing for the JEE Advanced Mathematics. Within the provided exercise, recognizing that the sine function's maximum and minimum values are \(1\) and \(-1\), respectively, is the key to determining the range of the composite functions and evaluating the limits involved.

When the sine function is multiplied by a constant or nested within another sine, as seen in \(f(x)\), these properties guide us in narrowing down the possible range of the function. It's like having a ladder; no matter how the sine function is scaled or shifted, it can't climb above or drop below its inherent bounds. This understanding opens the door for students to solve more complex trigonometric equations with confidence.