Question

Let \(\mathrm{f}:(-1,1) \rightarrow \mathrm{B}\) be a functions defined by \(\mathrm{f}(\mathrm{x})=\sin ^{-1}\left[2 \mathrm{x} /\left(1+\mathrm{x}^{2}\right)\right]\) then \(\mathrm{f}\) is both one-one and onto then \(\mathrm{B}\) is in the (a) \([-(\pi / 4),(\pi / 4)]\) (b) \([-(\pi / 2),(\pi / 2)]\) (c) \([-(\pi / 4),(\pi / 4)]\) (d) \([-(\pi / 2),(\pi / 2)]\)

Short answer

The range of the function f(x) is \(-\frac{\pi}{2}\leq f(x)\leq\frac{\pi}{2}\). Therefore, option (b) \([-\frac{\pi}{2},\frac{\pi}{2}]\) is the correct answer.

Step-by-step solution

Find the range of the function

To find the range of the function, first, let's analyse the expression inside the sine inverse function. We have: \[\frac{2x}{1+x^2}\] From the domain of the function, we know that \(x\) can take any value between \(-1\) and \(1\). By plugging in the minimum and maximum possible values of x, we can find the range of this expression: Minimum value of x: \(x = -1\) \[\frac{2(-1)}{1+(-1)^2}=-1\] Maximum value of x: \(x = 1\) \[\frac{2(1)}{1+1^2}=1\] So the range of the expression inside the sine inverse function is \[-1\leq\frac{2x}{1+x^2}\leq1\]

Apply sine inverse function

Now we will apply the sine inverse function to the expression's range and find the range of the function \(f(x)\): Minimum value: \(\sin^{-1}(-1)=-\frac{\pi}{2}\) Maximum value: \(\sin^{-1}(1)=\frac{\pi}{2}\) So the range of the function is \(-\frac{\pi}{2}\leq f(x)\leq\frac{\pi}{2}\).

Choose the appropriate option

With the calculated range, we can determine the correct option for the function's range B. Since the range of the function is \(-\frac{\pi}{2}\leq f(x)\leq\frac{\pi}{2}\), the correct option is: (b) \([-\frac{\pi}{2},\frac{\pi}{2}]\)

Key concepts

Arcsine Function

The arcsine function, denoted as \( \sin^{-1}(x) \), is an inverse trigonometric function. It is used to determine the angle whose sine is \( x \). The arcsine function is essential when reversing sine operations to find an angle. When you input a value into \( \sin^{-1}(x) \), it outputs an angle in radians. This angle is within a defined [range](#ranges-for-inverse-functions) to ensure it remains a function. For the arcsine function, this range is \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \).

**Characteristics of the Arcsine Function:**

• Only takes inputs between \(-1\) and \(1\), because those are the limits of the sine function's output.
• Outputs an angle in radians between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
• Is strictly increasing, meaning as input \(x\) increases, the output \(\sin^{-1}(x)\) also increases.

This makes the arcsine function particularly useful when dealing with problems requiring you to backtrack from a known sine value to the probable angle.

Range of a Function

The concept of a function's range is crucial in understanding how a function behaves in terms of output values. The **range of a function** is the complete set of all possible output values (y-values) it can produce when it takes inputs from its domain.

To determine the range effectively:

• Compute the expression within the function, especially if it contains inverse operations or trigonometric elements.
• Evaluate the possible outputs for the boundary values of the domain.

In our specific function \(f(x) = \sin^{-1}( \frac{2x}{1+x^2} )\), the expression \(\frac{2x}{1+x^2}\) has a range of \([-1, 1]\) when evaluated over its domain of \((-1, 1)\). Applying the arcsine to this range extends the range of the function \(f(x)\) to \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

Thus, understanding a function's range helps us accurately describe the scope of output values a function can achieve, making it a vital aspect in solving trigonometric and calculus-related problems.

One-one and Onto Functions

A function is described as **one-one** (or injective) if each output is uniquely connected to a single input. In simpler terms, no two different inputs produce the same output. This property ensures that each element in the domain maps to a distinct element in the co-domain.**Onto** (or surjective) functions, on the other hand, cover every element of the co-domain at least once. This implies the range of the function is precisely the entire co-domain.

• **Example of One-one:** The function \(y = 2x + 1\) is one-one because different values of \(x\) yield different values of \(y\).
• **Example of Onto:** The same function \(y = 2x + 1\) can be considered onto from \(\mathbb{R}\) to \(\mathbb{R}\), since any real number \(y\) can be achieved by some real number \(x\).

For the function in the original exercise, \(f(x) = \sin^{-1}( \frac{2x}{1+x^2} )\), our task is to check these properties for identifying the range. By covering the outputs from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) without repetition, we establish that it is both one-one and onto, hence the function maps uniquely and covers the entire co-domain in its defined limits.