Question

Given the function \(\mathrm{f}(\mathrm{x})=\left[\left(3^{\mathrm{x}}+3^{-\mathrm{x}}\right) / 2\right]\), then \(\mathrm{f}(\mathrm{x}+\mathrm{y})+\mathrm{f}(\mathrm{x}-\mathrm{y})=\ldots\) (a) \(\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})\) (b) \(\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})\) (c) \([\\{\mathrm{f}(\mathrm{x})\\} /\\{\mathrm{f}(\mathrm{y})\\}]\) (d) \(2 \mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y})\)

Short answer

f(x+y) + f(x-y) = \(\frac{3^{x+y} + 3^{-(x+y)} + 3^{x-y} + 3^{-(x-y)}}{2}\). None of the given options match this expression.

Step-by-step solution

Write down the given function

The given function is f(x) = \(\left[\left(3^{x}+3^{-x}\right) / 2\right]\).

Replace x with (x+y) and (x-y) in the given function

Firstly, replace x with (x+y) in the function f(x): f(x+y) = \(\left[\left(3^{x+y}+3^{-(x+y)}\right) / 2\right]\). Secondly, replace x with (x-y) in the function f(x): f(x-y) = \(\left[\left(3^{x-y}+3^{-(x-y)}\right) / 2\right]\).

Sum the expressions for f(x+y) and f(x-y)

Now, add f(x+y) and f(x-y): f(x+y) + f(x-y) = \(\left[\left(3^{x+y}+3^{-(x+y)}\right) / 2\right] + \left[\left(3^{x-y}+3^{-(x-y)}\right) / 2\right]\).

Simplify the expression

Combine the fractions in the expression: f(x+y) + f(x-y) = \(\frac{3^{x+y} + 3^{-(x+y)} + 3^{x-y} + 3^{-(x-y)}}{2}\).

Comparing the expression to the given options

Compare the simplified expression to the given options: Option (a): f(x) + f(y) = \(\frac{3^{x} + 3^{-x}}{2}\) + \(\frac{3^{y} + 3^{-y}}{2}\) ≠ f(x+y) + f(x-y) Option (b): f(x) * f(y) = \(\left[\left(3^{x}+3^{-x}\right) / 2\right] * \left[\left(3^{y}+3^{-y}\right) / 2\right]\) ≠ f(x+y) + f(x-y) Option (c): \([\frac{f(x)}{f(y)}]\) = \(\frac{\left[\left(3^{x}+3^{-x}\right) / 2\right]}{\left[\left(3^{y}+3^{-y}\right) / 2\right]}\) ≠ f(x+y) + f(x-y) Option (d): \(2 \cdot f(x) \cdot f(y)\) = \(2 \cdot \left[\left(3^{x}+3^{-x}\right) / 2\right] \cdot \left[\left(3^{y}+3^{-y}\right) / 2\right]\) ≠ f(x+y) + f(x-y) None of the given options match the simplified expression for f(x+y) + f(x-y).

Key concepts

Exponential Functions

Exponential functions, such as the one in the problem \( f(x) = \left[\left(3^{x}+3^{-x}\right) / 2\right] \), are mathematical expressions where a constant base is raised to a variable exponent. In this case, the constant base is 3, and the exponent is \( x \) or \( -x \).

Exponential growth and decay are fundamental concepts showcased by exponential functions. These functions are distinguished by their rapid increase or decrease as the value of the exponent changes. They are used extensively in fields such as finance to calculate compound interest, physics to represent radioactive decay, and in population modeling.

Key Characteristics of Exponential Functions

• Rate of Change: The rate at which an exponential function increases or decreases is proportional to its current value.
• Asymptotic Behavior: Exponential functions approach a horizontal asymptote as \( x \) approaches positive or negative infinity.
• Domain and Range: Exponential functions have a domain of all real numbers and, if the base is greater than one, have a range of positive real numbers.

Understanding these characteristics will help students to tackle a wide variety of problems involving exponential functions, which are a staple in advanced mathematics, including JEE maths problems.

Function Addition

Function addition is a fundamental operation which involves adding the outputs of two functions point by point. If you have two functions \( f(x) \) and \( g(x) \), their sum \( h(x) \) is defined as \( h(x) = f(x) + g(x) \). In our exercise, solving for \( f(x+y) + f(x-y) \) required this concept.

Adding functions is like combining their effects, and this process is used not only in pure mathematics but also in applications such as signal processing, where different waves are added to form a new signal.

Steps for Function Addition

• Determine the Functions: Identify the functions that need to be added.
• Point-by-Point Addition: At any given point \( x \), calculate both \( f(x) \) and \( g(x) \), then add them together to find \( h(x) \).
• Simplify the Result: Combine like terms and simplify the sum to its simplest form.

This same methodology was applied in the exercise, emphasizing the importance of step-by-step processing to avoid mistakes, and underpinning the 'JEE maths problems' style of questioning.

JEE Maths Problems

Joint Entrance Examination (JEE) is a highly competitive examination in India for students aspiring to enter various engineering colleges. JEE maths problems are designed to test a student’s understanding of a wide range of mathematical concepts and their ability to apply these concepts to solve complex problems.

The problems generally require a strong foundational understanding, quick thinking, and often an ability to connect different concepts, such as the given problem which is a mix of exponential functions and function addition.

JEE Maths Problem-Solving Strategies

• Analyze: Carefully read the problem to understand what is being asked and what information is provided.
• Conceptual Knowledge: Use your knowledge of the concept, like exponential functions, to begin formulating an approach.
• Application: Apply the appropriate mathematical operations, in this case, function addition, to arrive at a solution.
• Option Elimination: In objective-type questions, use the process of elimination to narrow down the correct answer.
• Validation: Check if the answer makes sense and matches any given options, or if none of the options fit, as was the case in the original exercise.

Preparing for such problems usually involves practicing a wide range of questions and mastering the underlying concepts, which will aid in efficiently tackling the unique challenges presented by the JEE exam.