Problem 91 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } f(x)=x \cos x \text { and } g(x)=\frac{x}{1+x^{2}} \text { , then find } f \circ g(x) \text { and } g o f(x) $$

Step-by-Step Solution

Answer

f(g(x)) = \(\frac{x}{1+x^{2}} \cdot \cos\left(\frac{x}{1+x^{2}}\right)\) and g(f(x)) = \(\frac{x \cos(x)}{1 + x^2 \cos^2(x)}\)

Step 1: Find f(g(x))

To find f(g(x)), we will substitute g(x) as the input to f(x). So we will replace x in f(x) with g(x):\[ f(g(x)) = g(x) \cdot \cos(g(x)) = \left( \frac{x}{1+x^{2}}\right) \cdot \cos\left(\frac{x}{1+x^{2}}\right) \]This is the desired composition f(g(x)).

Step 2: Find g(f(x))

To find g(f(x)), we will substitute f(x) as the input to g(x). So we will replace x in g(x) with f(x):\[ g(f(x)) = \frac{f(x)}{1+f^{2}(x)} = \frac{x \cos(x)}{1+(x \cos(x))^2} = \frac{x \cos(x)}{1 + x^2 \cos^2(x)} \]This is the desired composition g(f(x)). The final answer is: f(g(x)) = \(\frac{x}{1+x^{2}} \cdot \cos\left(\frac{x}{1+x^{2}}\right)\) and g(f(x)) = \(\frac{x \cos(x)}{1 + x^2 \cos^2(x)}\).

Key Concepts

Understanding Trigonometric Functions

Trigonometric functions are mathematical functions that relate angles to the ratios of the sides of right triangles. They are essential in many areas of mathematics, including calculus and geometry. In this exercise, we encounter the cosine function, one of the basic trigonometric functions.

Here's a quick rundown of some key aspects of the cosine function:

• Definition: For an angle \(\theta\), the cosine is the adjacent side over the hypotenuse in a right triangle.
• Range: The cosine of any real angle ranges between -1 and 1.
• Graph: The graph of \(\cos(x)\) is a wave-like pattern and has a period of \(2\pi\).

Understanding these properties will help you tackle problems involving trigonometric functions, especially in competitive exams like the IIT JEE.

Exploring Differential Calculus

Differential calculus is a branch of mathematics that studies how functions change. It's essential for understanding rates of change and slopes of curves. In our exercise, function composition involves derivatives, although it's not directly calculated here.

Some important concepts in differential calculus:

• Derivative: Represents the rate of change or slope of a function. If \(f(x)\) is differentiable, the derivative \(f'(x)\) gives us this rate.
• Product Rule: Used when differentiating products of two functions. If \(f(x) \cdot g(x)\) is a function, its derivative is \(f'(x)g(x) + f(x)g'(x)\).
• Chain Rule: Necessary for finding derivatives of composite functions like \(f(g(x))\). Write \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).

Mastering these rules can significantly improve your problem-solving skills in IIT JEE preparation.

Navigating IIT JEE Preparation with Composition of Functions

Preparing for the IIT JEE exams requires a strong understanding of function composition, among other topics. Composition of functions like \(f(g(x))\) and \(g(f(x))\) can be challenging but are crucial for scoring well in mathematics.

Some tips for mastering this concept in the context of IIT JEE include:

• Understand the Basics: Have a strong foundation in both the domain and range of functions before moving on to composition.
• Practice: Regular practice is vital. Solve various problems to solidify your understanding.
• Visualize: Diagramming functions and their compositions can provide a clearer picture of how they interact.
• Time Management: During the exam, manage your time efficiently. Spend the right amount of time on complex problems.

With a strategic approach, you can excel in the function composition problems in your IIT JEE exams.