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Chapter 1: Problem 100

There are three function given in column-A and its inverse in column-B \begin{tabular}{l|l} \(\frac{\text { Column-A }}{\text { (1) } f(x)=1-2^{-x}}\) & (a) \(\left.f^{-1}(x)=\left[x / \sqrt{(}-x^{2}\right)\right]\) \\ (2) \(f(x)=\sin \left(\tan ^{-1} x\right)\) & (b) \(f^{-1}(x)=-\log _{2}(1-x)\) \\\ (3) \(f(x)=2 x+3\) & (c) \(f^{-1}(x)=[(x-3) / 2]\) \end{tabular} which one of the following matches is correct? (a) (1) \(\mathrm{a},(2) \mathrm{b},(3) \mathrm{c}\) (b) (1) b, (2) c, (3) a (c) (1) b, (2) a, (3) c (d) (1) c, (2) b, (3) a

Short Answer

Expert verified
The correct match is: (a) (1) c, (2) b, (3) a.

Step by step solution

01

Testing Option (a)

Option (a) states (1) a, (2) b, (3) c. Let's test it. For (1): \(f(x)=1-2^{-x}\) and \(f^{-1}(x)=\frac{x}{\sqrt{-x^2}}\) Substitute \(f^{-1}(x)\) in \(f(x)\): \[f(f^{-1}(x)) = 1-2^{-\frac{x}{\sqrt{-x^2}}}\] Since f(f^{-1}(x)) ≠ x, option (a) is not correct.
02

Testing Option (b)

Option (b) states (1) b, (2) c, (3) a. Let's test it. For (1): \(f(x)=1-2^{-x}\) and \(f^{-1}(x)=-\log_2(1-x)\) Substitute \(f^{-1}(x)\) in \(f(x)\): \[f(f^{-1}(x)) = 1-2^{-(-\log_2(1-x))}\] \[f(f^{-1}(x)) = 1-(1-x)\] \[f(f^{-1}(x)) = x\] For (2): \(f(x)=\sin(\tan^{-1}(x))\) and \(f^{-1}(x)=\frac{(x-3)}{2}\) It can be seen that the function f(x) and its inverse are not the same form. Thus, option (b) is not correct.
03

Testing Option (c)

Option (c) states (1) b, (2) a, (3) c. Let's test it. We already showed that pair (1) b is correct. For (2): \(f(x)=\sin(\tan^{-1}(x))\) and \(f^{-1}(x)=x/\sqrt{-x^2}\) It can also be seen here that the function f(x) and its inverse are not the same form. Thus, option (c) is not correct.
04

Testing Option (d)

Option (d) states (1) c, (2) b, (3) a. Since option (d) has only one new pair other than (1) b, it must be the correct answer. We check it anyway. For (3): \(f(x)=2x+3\) and \(f^{-1}(x)=\frac{(x-3)}{2}\) Substitute \(f^{-1}(x)\) in \(f(x)\): \[f(f^{-1}(x)) = 2\left(\frac{(x-3)}{2}\right)+3\] \[f(f^{-1}(x)) = x-3+3\] \[f(f^{-1}(x)) = x\] Since f(f^{-1}(x)) = x, option (d) is correct. The correct match is: (a) (1) c, (2) b, (3) a.

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Most popular questions from this chapter

Part of the domain of the function \(\left.\mathrm{f}(\mathrm{x})=\sqrt{[}\\{\cos \mathrm{x}-(1 / 2)\\} /\left(6+35 \mathrm{x}-6 \mathrm{x}^{2}\right)\right]\) lying in the interval \([-1,6]\) is (a) \([-(1 / 6),(\pi / 3)]\) (b) \([(5 \pi / 3), 6]\) (c) \([-(1 / 5),(\pi / 3)] \cup[(5 \pi / 3), 6]\) (d) \((-(1 / 6),(\pi / 3)] \cup[(5 \pi / 3), 6)\)

Which of the following relation is one-one (a) \(R_{1}=\left\\{(x, y) / x^{2}+y^{2}=1, x, y \in R\right\\}\) (b) \(\mathrm{R}_{2}=\left\\{(\mathrm{x}, \mathrm{y}) / \mathrm{y}=\mathrm{e}^{(\mathrm{x}) 2} / \mathrm{x}, \mathrm{y} \in \mathrm{R}\right\\}\) (c) \(R_{3}=\left\\{(x, y) / y=x^{2}-3 x+3, x, y \in R\right\\}\) (d) None of these

If \(f: R \rightarrow R\) defined by \(f(x)=x^{4}+2\) then the value of \(f^{-1}(83)\) and \(\mathrm{f}^{-1}(-2)\) respectively are. (a) \(\Phi,\\{3,-3\\}\) (b) \(\\{3,-3\\}, \Phi\) (c) \(\\{4,-4\\}, \Phi\) (d) \(\\{4,-4\\},\\{2,-2\\}\)

The range of \(\mathrm{f}(\mathrm{x})=6^{\mathrm{x}}+3^{\mathrm{x}}+6^{-\mathrm{x}}+3^{-\mathrm{x}}+2\) is subset of \(\ldots\) (a) \([-\infty,-2)\) (b) \((-2,3)\) (c) \((6, \infty)\) (d) \([6, \infty)\)

Given the relation on \(\mathrm{R}=\\{(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c})\\}\) in the set \(\mathrm{A}=\\{\mathrm{a}, \mathrm{b}, \mathrm{c}\\}\) Then the minimum number of ordered pairs which added to \(R\) make it an equivalence relation is (a) 5 (b) 6 (c) 7 (d) 8
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