Problem 128 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the function \(f:[1, \infty) \rightarrow[1, \infty)\) is defined by \(f(x)=\) \(2^{x(x-1)}\), then (A) \(f\) is one-one (B) \(f\) is onto (C) \(f^{-1}(x)=\frac{1+\sqrt{1+4 \log _{2} x}}{2}\) (D) \(f^{-1}(x)=\frac{1-\sqrt{1+4 \log _{2} x}}{2}\)

Step-by-Step Solution

Answer

(A), (B), and (C) are correct.

Step 1: Evaluate One-One Property

To determine if the function is one-one, check if the derivative is always positive. Calculate the derivative:\[ f(x) = 2^{x(x-1)} = e^{(x^2 - x) \ln 2}\]Using the chain rule, differentiate:\[ f'(x) = \frac{d}{dx}[e^{(x^2 - x) \ln 2}] = e^{(x^2-x) \ln 2} \times 2x - 1 \times \ln 2\]Since \(e^{(x^2 - x) \ln 2}\) and \(\ln 2\) are always positive, check the sign of \(2x - 1\). As \(x \geq 1\), \(2x - 1 \geq 1\). Thus, \(f'(x) > 0\), confirming \(f\) is one-one.

Step 2: Evaluate Onto Property

To check if \(f\) is onto, verify whether \(f(x)\) can reach every point in the range \([1, \infty)\).Since \(f(x) = 2^{x(x-1)}\), as \(x\) grows, \(f(x) = 2^{0} = 1\) when \(x = 1\) and increases without bound as \(x\) increases. Thus, \(f(x)\) covers every value in \([1, \infty)\), meaning \(f\) is onto.

Step 3: Find the Inverse Function

We need to determine whether choices (C) or (D) are correct expressions for the inverse. Start by expressing \(f(x)\) in terms of \(y\): Given \(y = 2^{x(x-1)}\), take the logarithm:\[\log_2 y = x(x-1)\]\[x^2 - x - \log_2 y = 0\]This is a quadratic equation: \(x^2 - x - \log_2 y = 0\). Solve for \(x\):\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-\log_2 y)}}{2} = \frac{1 \pm \sqrt{1 + 4 \log_2 y}}{2}\]Since \(x \geq 1\), we choose the positive solution:\[f^{-1}(y) = \frac{1 + \sqrt{1 + 4 \log_2 y}}{2}\]This matches option (C).

Key Concepts

One-One Function

A one-one function, or injective function, has a special characteristic where each element of the function's domain maps to a unique element in the range. This means that no two different inputs can produce the same output. In other words, if we have two values, say \(a\) and \(b\), and if \(f(a) = f(b)\), then it must follow that \(a = b\).
This unique property is often checked by examining the derivative of the function. For the function given in the exercise \(f(x) = 2^{x(x-1)}\):

- The derivative \(f'(x)\) was calculated, and it employs the exponential form \(e^{(x^2 - x) \ln 2}\), which remains positive.
- Paying attention to the component \(2x - 1\), which is derived from the original function, shows it is non-negative when \(x \geq 1\).

This confirms that \(f'(x)\) is always positive over its domain \([1, \infty)\), meaning the function does not have turning points and must therefore be one-one.

Onto Function

An onto function, or surjective function, ensures that every element in the function's codomain is the image of some element from its domain. An onto function effectively "covers" the entire range.

Let's look at the given function from the exercise, \(f(x) = 2^{x(x-1)}\). To determine if it is onto, we need to check whether it can produce every possible output value in its specified range \([1, \infty)\).

- At \(x = 1\), the function outputs \(2^{0} = 1\).
- As \(x\) increases beyond 1, \(x(x-1)\) and hence \(f(x)\) increases as well since exponential functions grow rapidly.

Thus, the function moves from 1 to infinity without skipping any value in between, ensuring it is onto.

Inverse Function

An inverse function essentially "reverses" the effect of a function, taking outputs back to their original inputs. For a function \(f(x)\) to have an inverse \(f^{-1}(x)\), it must first be one-one and onto (bijective).

In this exercise, we are provided with the function \(f(x) = 2^{x(x-1)}\) and need to find its inverse. To start, we express \(f(x) = y\) and attempt to solve for \(x\) in terms of \(y\):

- Taking the logarithm gives us \(\log_2 y = x(x-1)\). Reorganizing this into a quadratic equation, we get \(x^2 - x - \log_2 y = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1, b=-1, c=-\log_2 y\), we find:

• \(x = \frac{1 \pm \sqrt{1 + 4 \log_2 y}}{2}\)

Because the domain is constrained to \(x \geq 1\), the positive solution is chosen: \(f^{-1}(y) = \frac{1 + \sqrt{1 + 4 \log_2 y}}{2}\). This confirms that choice (C) is the correct inverse of \(f(x)\).