Problem 126 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f: R \rightarrow R\) be defined by \(f(x)=\frac{e^{x}-e^{-x}}{2}\), then (A) \(f\) is one-one (B) \(f\) is onto (C) \(f^{-1}(x)=\log \left(x-\sqrt{x^{2}+1}\right)\) (D) \(f^{-1}(x)=\log \left(x+\sqrt{x^{2}+1}\right)\)

Step-by-Step Solution

Answer

Options (A), (B), and (D) are correct.

Step 1: Understanding the Function

The function given is \(f(x) = \frac{e^x - e^{-x}}{2}\), which is a hyperbolic sine function, denoted as \(\sinh(x)\). The function \(\sinh(x)\) is defined as \(\sinh(x) = \frac{e^x - e^{-x}}{2}\).

Step 2: Determine if the Function is One-One

To determine if \(f(x)\) is one-one, check if it is strictly increasing or decreasing. Calculate \(f'(x)\): \[ f'(x) = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x + e^{-x}}{2} = \cosh(x) \]. Since \(\cosh(x) \geq 0\) and is strictly positive except at \(x = 0\), \(f(x)\) is strictly increasing for all \(x\). Thus, \(f\) is one-one.

Step 3: Determine if the Function is Onto

For \(f(x)\) to be onto, it must cover all real numbers. Since \(\sinh(x)\) has a range of all real numbers (\(\mathbb{R}\)), \(f(x)\) is onto. Therefore, option B is correct.

Step 4: Find the Inverse Function

To find \(f^{-1}(x)\), set \(\sinh(x) = \frac{e^u - e^{-u}}{2} = x\) and solve for \(u\). This equation becomes \( e^u - e^{-u} = 2x \). Now, squaring both sides: \[ \left(e^u - e^{-u}\right)^2 = (2x)^2 \] which simplifies to:\[ e^{2u} - 2 + e^{-2u} = 4x^2 \]. Rearrange to form a quadratic in \(e^u\): \[ e^{2u} + 1 = 4x^2 + 2 \]. Solve for \(e^{2u}\):\[ (e^u)^2 - 2x(e^u) - 1 = 0 \].

Step 5: Solve the Quadratic Equation

Solving the quadratic \(t^2 - 2xt - 1 = 0\) where \(t = e^u\), using the quadratic formula:\[ t = \frac{2x \pm \sqrt{4x^2 + 4}}{2} \].This becomes:\[ t = x \pm \sqrt{x^2 + 1} \].We choose the positive solution since \(e^u > 0\), giving us:\[ e^u = x + \sqrt{x^2 + 1} \].

Step 6: Express the Inverse Function

Taking natural logarithms to solve for \(u\):\[ u = \ln(x + \sqrt{x^2 + 1}) \]. Thus, the inverse function is \(\log(x + \sqrt{x^2 + 1})\). Therefore, option D is the inverse function.

Key Concepts

One-One Functions

A function is labeled as one-one (or injective) if every element in the domain maps to a distinct and unique element in the codomain. In simpler terms, no two different inputs produce the same output. To determine if a function is one-one, we often look to see if the function is strictly increasing or decreasing.

For the hyperbolic sine function, given by \(f(x) = \frac{e^x - e^{-x}}{2}\), we investigate its derivative. The derivative, \(f'(x)\), is obtained by differentiating the function, resulting in \(f'(x) = \cosh(x) = \frac{e^x + e^{-x}}{2}\). This derivative is always non-negative, with \(\cosh(x)\) being positive for all \(x\) except at the point \(x = 0\). Thus, \(f(x)\) is strictly increasing, which implies that the function is one-one. This characteristic guarantees that each output value is obtained by exactly one input from the domain.

Recognizing a function as one-one is crucial because it allows us to define its inverse function.

Onto Functions

An onto function, or surjective function, is one where every element in the codomain is the image of at least one element from the domain. In other words, the function's range is equal to the codomain. For a function to be onto, its output should cover all possible values within the codomain.

In the case of the hyperbolic sine function \(\sinh(x) = \frac{e^x - e^{-x}}{2}\), the range is all real numbers \(\mathbb{R}\). This means for every real number \(y\), there exists an \(x\) such that \(f(x) = y\). Thus, every possible value of the codomain \(\mathbb{R}\) is hit by at least one element from the domain. As a result, \(\sinh(x)\) is considered onto.

Being onto is a vital property for a function to possess when we want to determine if a function has an inverse on the entire real line.

Inverse Functions

An inverse function is one that reverses the mapping of the original function. Essentially, if a function \(f\) maps \(x\) to \(y\), the inverse function \(f^{-1}\) maps \(y\) back to \(x\). In order to have an inverse, a function has to be both one-one and onto, ensuring that every output is paired with a unique input.

For our hyperbolic sine function, we determined earlier that it is both one-one and onto. Now, we can find the inverse function. We set \(\sinh(x) = \frac{e^u - e^{-u}}{2} = x\) and solve for \(u\) in terms of \(x\). By converting the hyperbolic relationship, we derived \(e^u = x + \sqrt{x^2 + 1}\). Taking the natural log of both sides, we get the inverse function:

• \(f^{-1}(x) = \ln(x + \sqrt{x^2 + 1})\)

This way, the inverse of \(\sinh(x)\) is expressed in terms of logarithmic functions, providing a way to convert outputs back to inputs uniquely based on the given conditions.

Hyperbolic Sine Function (\sinh)

The hyperbolic sine function, denoted as \(\sinh(x)\), is an analog of the sine function from trigonometry. However, it is based on hyperbolic angles rather than circular angles. It is defined by the expression \(\sinh(x) = \frac{e^x - e^{-x}}{2}\). This function is part of the suite of hyperbolic functions, which are related to the hyperbola in much the same way that trigonometric functions relate to the circle.

Key properties of \(\sinh(x)\) include:

• The range of \(\sinh(x)\) is all real numbers \(\mathbb{R}\).
• \(\sinh(x)\) is an odd function, meaning \(\sinh(-x) = -\sinh(x)\).
• The derivative of \(\sinh(x)\) is \(\cosh(x)\).
• \(\sinh(x)\) grows exponentially for large positive and negative values of \(x\).

Understanding the hyperbolic sine function and its properties can be crucial for solving problems in calculus, differential equations, and mathematical physics, where these functions frequently appear.