Problem 27 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(g(x)=1+\sqrt{x}\) and \(f[g(x)]=3+2 \sqrt{x}+x\), then \(f(x)=\) (A) \(1+2 x^{2}\) (B) \(2+x^{2}\) (C) \(1+x\) (D) \(2+x\)

Step-by-Step Solution

Answer

(D) \(2+x\)

Step 1: Understand the problem

We are given two functions, \(g(x) = 1 + \sqrt{x}\) and \(f[g(x)] = 3 + 2\sqrt{x} + x\), and we need to find an expression for \(f(x)\).

Step 2: Substitute and simplify

Given that \(f[g(x)] = 3 + 2 \sqrt{x} + x\), we substitute \(g(x)\) for \(x\) in \(f(x)\), which means replacing \(x\) in \(f(x)\) with \(1+\sqrt{x}\). Thus, \(f(1 + \sqrt{x}) = 3 + 2\sqrt{x} + x \).

Step 3: Express \(f(x)\) based on substitution

To find \(f(x)\), let \(u = \sqrt{x}\) so that \(x = u^2\). Substituting \(x = u^2\) into the function \(f[g(x)] = 3 + 2u + u^2\), we see that \(x = 1 + u\), so \(f(1 + u) = 3 + 2u + u^2\). Therefore, \(f(x) = x + 2\).

Step 4: Match the result with given options

The expression \(f(x) = x + 2\) matches option (D) \(2+x\). So the correct answer is option (D).

Key Concepts

Functions

A function is an essential concept in mathematics, which represents a relationship between a set of inputs and a set of possible outputs. Every element in the domain, which is the input, is linked to exactly one element in the codomain, which is the output. Functions can be represented using equations, such as \( f(x) = x + 2 \), tables, graphs, or words.
Understanding functions is crucial due to several reasons:

• Functions help in visualizing relationships between quantities.
• They are building blocks for more advanced mathematical concepts.
• Functions are ubiquitous in modeling real-world situations.

Functions can vary in complexity, from simple linear functions, like \( f(x) = x + 2 \), to complex polynomial or trigonometric functions. In the context of the given problem, we need to decode the functions provided: \( g(x) = 1 + \sqrt{x} \) and \( f[g(x)] = 3 + 2\sqrt{x} + x \).
Here, the output of \( g(x) \) becomes the input for another function \( f \), showcasing how functions can interact through composition.

Composite Functions

Composite functions result when one function is applied to the result of another function. This process, known as composition, creates a new function that combines the operations of both functions. It’s denoted as \( f(g(x)) \) where \( g(x) \) is applied first, and then \( f \) to the result.
This is useful in various areas of math and science, allowing complex operations to be simplified into sequences of simpler operations.
Composite functions are important because they:

• Facilitate the simplification of complex problems.
• Allow the chaining of functions to model multi-step processes.

In the problem, we explore composite functions by substituting \( g(x) = 1 + \sqrt{x} \) into another function \( f \), revealing the relationship \( f[g(x)] = 3 + 2\sqrt{x} + x \). This highlights how composite functions transform inputs in a multi-step fashion and underscores the importance of understanding each function involved.

Algebraic Expressions

Algebraic expressions involve variables, numbers, and arithmetic operations like addition, subtraction, multiplication, and division. They are used to represent mathematical ideas and problems concisely.
Algebraic expressions can include monomials (single terms) like \(2x\), binomials (two terms) like \(x + 2\), or polynomials, which include two or more terms.
In the given problem, we deal with the algebraic expression \( f(g(x)) = 3 + 2\sqrt{x} + x \). This expression results from substituting \( g(x) \) into \( f \) and demonstrates the power of algebra to neatly encapsulate multi-step processes.
Here are some key points about algebraic expressions:

• They provide a way to generalize arithmetic.
• They can express patterns and relationships in a succinct form.
• Algebraic expressions are foundational for solving equations and inequalities.

Understanding how to manipulate and simplify these expressions is critical for solving mathematical problems, like determining the function \( f(x) \) as done in this exercise.