Question

Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) be two functions such that \(f(x)=2 x-3, g(x)=x^{3}+5\) The function \((f \circ g)^{-1}(x)\) is equal to (a) \([(\mathrm{x}+7) / 2]^{(1 / 3)}\) (b) \([x-(7 / 2)]^{(1 / 3)}\) (c) \([(\mathrm{x}-2) / 7]^{(1 / 3)}\) (d) \([(x-7) / 2]^{(1 / 3)}\)

Short answer

The function $(f \circ g)^{-1}(x)$ is equal to (d) $[(x-7) / 2]^{(1 / 3)}$.

Step-by-step solution

Find the function f◦g(x)

The function composition f(g(x)) is defined as f(g(x)) = f(x^3 + 5). Replace f(x) with the given function f(x) = 2x - 3: f(g(x)) = 2(x^3 + 5) - 3 Now we can simplify: f(g(x)) = 2x^3 + 10 - 3 f(g(x)) = 2x^3 + 7

Finding the inverse of f◦g(x)

Let the inverse function be denoted as (f◦g)^(-1)(y), such that f◦g((f◦g)^(-1)(y)) = y. We have f◦g(x) = 2x^3 + 7. To find the inverse, replace f◦g(x) with y and then solve for x. y = 2x^3 + 7 Now, isolate x^3: x^3 = (y - 7) / 2 Next, take the cube root both sides: x = [(y - 7) / 2]^(1/3) Thus, (f◦g)^(-1)(y) = [(y - 7) / 2]^(1/3).

Matching the form of the inverse function with the given options

Comparing our inverse function with the given options, we can see that it matches option (d): (f◦g)^(-1)(x) = [(x - 7) / 2]^(1/3) Therefore, the correct answer is (d).

Key concepts

Function Composition

Understanding function composition is a fundamental skill in mathematics, particularly when dealing with complex functions. In essence, function composition combines two functions into a single function by applying one function to the result of another.

Consider two functions, say \( f(x) \) and \( g(x) \). The composition of these two functions, denoted as \( f \circ g \) or \( f(g(x)) \), is the result of applying \( g \) to \( x \) and then taking that output and applying \( f \) to it. Visually, you can imagine this as a two-step journey where \( x \) first transforms under \( g \), and then this new value undergoes a second transformation under \( f \).

The great benefit of understanding function composition is that it helps to simplify complex problems and can also give insight into how changes in one function can affect the output of the composite function. When you're working on mathematics for competitive exams like IIT JEE, grasping the nuances of function composition can give you a significant advantage.

Finding Inverse Functions

The notion of an inverse function is a bit like a mathematical 'undo' button. If you have a function that performs a certain operation on a number, its inverse function reverses that operation. To find an inverse function, denoted as \( f^{-1}(x) \), you essentially express the function's equation in terms of \( x \).

In practice, you swap the output and the input and then solve the equation to express the original input in terms of the new output. This process typically involves isolating the input variable on one side of the equation.

However, it's important to note that not all functions have inverses. For a function to have an inverse, it must be bijective, meaning it's both injective (one-to-one) and surjective (onto). This concept is crucial in various mathematical fields and solving problems in JEE Maths requires understanding how to find and apply inverse functions effectively.

Polynomial Functions

Polynomial functions are algebraic expressions that involve only non-negative integer powers of the variable. These functions are written in the form \( f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_2 x^2 + a_1 x + a_0 \), where \( a_n, a_{n-1}, ... , a_0 \) are constants, and \( n \) is a non-negative integer.

Polynomials are a central theme in algebra and appear across various areas of mathematics and applications in engineering, physics, economics, and beyond. They are particularly valued for their simplicity and the insights they provide into the behavior of more complex functions. In preparing for JEE or any other competitive exam, mastering polynomials is crucial as they serve as the foundation for more advanced topics in calculus, such as finding limits, derivatives, and integrals.