Problem 83 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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

Step-by-Step Solution

Answer

The function \(f(x) = 1 + x^2\), which is option (A).

Step 1: Understand the Composition

We have two functions, \(g(x)=1+\sqrt{x}\) and \(f[g(x)]=3+2\sqrt{x}+x\). Our task is to find \(f(x)\) in terms of \(x\). By the definition, \(f[g(x)]\) means inserting \(g(x)\) into \(f(x)\).

Step 2: Substitute and Simplify

Since \(g(x) = 1 + \sqrt{x}\), we set \(y = g(x) = 1 + \sqrt{x}\). We know \(f(y) = 3 + 2\sqrt{x} + x\). From this substitution, solve for \(\sqrt{x}\): \(\sqrt{x} = y - 1\).

Step 3: Express \(x\) in terms of \(y\)

Continuing from Step 2, square both sides of \(\sqrt{x} = y - 1\) to find \(x = (y - 1)^2\).

Step 4: Substitute Back to Find \(f(y)\)

Now replace \(\sqrt{x}\) and \(x\) in \(f[g(x)] = 3 + 2\sqrt{x} + x\) with expressions in terms of \(y\): \(f(y) = 3 + 2(y - 1) + (y - 1)^2\).

Step 5: Simplify \(f(y)\)

Expand and simplify the expression for \(f(y)\):\[f(y) = 3 + 2(y - 1) + (y - 1)^2 = 3 + 2y - 2 + (y^2 - 2y + 1) = 1 + 2y + y^2 - 2y + 1 = y^2 + 1.\]

Step 6: Conclusion

After simplification, we find \(f(y) = 1 + y^2\). Since \(y = x\), the original \(f(x)\) is \(1 + x^2\), which corresponds to option (A).

Key Concepts

Function Composition

In mathematics, function composition is the process of applying one function to the results of another. This way, you can build more complex functions from simpler ones. For example, if you have a function \( g(x) \) and another function \( f(x) \), the composition \( f(g(x)) \) means that you first apply \( g \) to \( x \), and then apply \( f \) to the result of \( g(x) \). In this exercise, \( f[g(x)] = 3 + 2\sqrt{x} + x \), meaning that \( g(x) \) is substituted into \( f(x) \).

• \( g(x) = 1 + \sqrt{x} \)
• Substitute \( g(x) \) into \( f(x) \), resulting in \( f[g(x)] \)
• The goal is to find \( f(x) \) in simpler terms

As you can see, function composition allows us to identify how two functions can be combined in various ways by substitution.

Substitution Method

The substitution method is a vital technique in many mathematical contexts, including solving equations and function operations. The primary idea is to replace a complex part of an expression with a simpler variable to make solving more manageable. In our exercise, we make use of the substitution method to find \( f(x) \). We have:- \( g(x) = 1 + \sqrt{x} \)- Let \( y = g(x) = 1 + \sqrt{x} \), making it a single expression This turns our target equation into a simpler form: \( f(y) = 3 + 2 \sqrt{x} + x \).To progress further, we solve for \( \sqrt{x} \):- \( \sqrt{x} = y - 1 \)Next, solve for \( x \):- Square both sides: \( x = (y - 1)^2 \)Now, all parts needed for substitution into \( f(y) \) are expressed in terms of \( y \), simplifying the solution process.

Algebraic Manipulation

Algebraic manipulation involves a series of operations used to simplify or solve equations, expressions, and functions. This involves expanding, factoring, or rearranging terms in a way that reveals more straightforward expressions. In this exercise, algebraic manipulation is essential in expressing \( f(y) \) in simpler terms.Once we have:- \( f(y) = 3 + 2(y - 1) + (y - 1)^2 \)You use algebraic manipulation to expand and simplify step-by-step:1. Expand the expression: \( 3 + 2y - 2 + (y^2 - 2y + 1) \)2. Combine like terms: \( 1 + y^2 \)Now, everything is expressed in terms of \( y \), simplifying \( f(y) \) to \( y^2 + 1 \). Since \( y = x \), it means \( f(x) = x^2 + 1 \). This solution exemplifies how algebraic manipulation can transform a complex problem into a more approachable format.