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Chapter 1: Problem 50

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

Expert verified
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

01

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 \[-1\leq\frac{2x}{1+x^2}\leq1\]
02

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 \(-\frac{\pi}{2}\leq f(x)\leq\frac{\pi}{2}\).
03

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) \([-\frac{\pi}{2},\frac{\pi}{2}]\)

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