Problem 17 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f: R \rightarrow R, g: R \rightarrow R\) be two functions given by \(f(x)=2 x-3, g(x)=x^{3}+5\). Then \((f o g)^{-1}(x)\) is equal to (A) \(\left(\frac{x-7}{2}\right)^{1 / z}\) (B) \(\left(\frac{x+7}{2}\right)^{1 /}\) (C) \(\left(x-\frac{7}{2}\right)^{1 / 3}\) (D) \(\left(\frac{x-2}{7}\right)^{1 / z}\)

Step-by-Step Solution

Answer

Option (A): \(\left(\frac{x-7}{2}\right)^{1/3}\)

Step 1: Find the Composition of Functions

First, find the composition of the two functions, \( (f \circ g)(x) \). \( f(g(x)) = f(x^3 + 5) \). Applying the definition of function \( f \), we get: \[f(x^3 + 5) = 2(x^3 + 5) - 3 \= 2x^3 + 10 - 3 \= 2x^3 + 7.\]

Step 2: Set up the Inverse Problem

To find the inverse of \( (f \circ g)(x) = 2x^3 + 7 \), set \( y = 2x^3 + 7 \). We want \( x \) in terms of \( y \). Therefore, solve for \( x \): \[ y = 2x^3 + 7. \]

Step 3: Solve for x

To solve for \( x \), rearrange \( y = 2x^3 + 7 \):Subtract 7 from both sides: \[ y - 7 = 2x^3. \]Divide both sides by 2: \[ \frac{y - 7}{2} = x^3. \]Take the cube root of both sides: \[ x = \left(\frac{y - 7}{2}\right)^{1/3}. \]

Step 4: Write the Inverse Function

Now that we have \( x = \left(\frac{y - 7}{2}\right)^{1/3} \), the inverse function \( (f \circ g)^{-1}(x) \) is: \[ (f \circ g)^{-1}(x) = \left(\frac{x - 7}{2}\right)^{1/3}. \]

Step 5: Choose the Correct Option

Among the given options, identify which matches the expression of the inverse function \( (f \circ g)^{-1}(x) \). The correct option is:- (A) \(\left(\frac{x-7}{2}\right)^{1/3}\)

Key Concepts

Inverse Function

Inverse functions are like unwrapping a gift and then rewrapping it to see if it's the same gift. In mathematical terms, each step you take should find you back where you started, mirroring the operations of the original function but in reverse order. If you have a function \(f\) that maps elements from set \(X\) to set \(Y\), then the inverse function \(f^{-1}\) maps elements back from set \(Y\) to set \(X\).

An inverse function exists only if the original function is one-to-one, meaning no two different inputs have the same output. To find the inverse of a composed function like \((f \circ g)\), you're essentially peeling back each function in reverse order. This involves switching the roles of the input and output, then solving for the original variable.

• Start by writing the function in terms of \(y\).
• Solve the equation to express \(x\) in terms of \(y\).
• Ensure that all steps of function transformations are reversed.

By solving these equations, you unravel each function to return to the beginning, defining the inverse.

Cube Root

Taking the cube root is like asking “what number, when used as a factor three times, gives me this number?”. The cube root helps to inverse the exponentiation of three.

Given \(x^3 = y\), the cube root operation is \(x = y^{1/3}\). It answers the question of what base number, multiplied by itself twice more, equals your target (i.e., \(y\)).
Cube roots allow us to reverse cubic transformations and simplify expressions in equations, especially when dealing with cubic functions.

• A cube root looks for a factor of three identical terms.
• It effectively "undoes" cubing a number.

When applying cube roots, both sides of an equation are transformed, often simplifying complex algebraic problems by breaking them down into manageable parts.

Polynomial Function

Polynomial functions are like the most flexible scribbling system for mathematicians, expressed in the form \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\). These functions consist of variables and coefficients, making them very versatile for different equations and operations.

Incorporating terms with non-negative integer exponents, polynomial functions can become more complex with higher degree terms. For example, a cubic polynomial \(2x^3 + 7\) was utilized in the task at hand, which indicates there are cubic transformations involved.

• Each term in a polynomial can be manipulated within equations, which are often solved using inverse operations.
• The degree of the polynomial dictates the potential complexity and the behavior of the function’s graph.
• Solutions often involve factoring, dividing, and handling cubic or quadratic expressions.

Understanding how to manipulate and solve polynomial equations is essential for breaking down complex problems, leading to meaningful mathematical interpretations.