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Chapter 1: Problem 67

If \(\mathrm{f}(\mathrm{x})=[(1-\mathrm{x}) /(1+\mathrm{x})]\) then \(\mathrm{f}(\mathrm{f}(\cos 2 \theta))=\ldots\) (a) \(\tan 2 \theta\) (b) \(\sec 2 \theta\) (c) \(\cos 2 \theta\) (d) \(\cot 2 \theta\)

Short Answer

Expert verified
(c) \(\cos 2 \theta\)

Step by step solution

01

Find the value of \(\mathrm{f}(\cos 2 \theta)\)

Substitute \(\mathrm{x} = \cos 2 \theta\) in the function \(\mathrm{f}(\mathrm{x})=\dfrac{1 - \mathrm{x}}{1 + \mathrm{x}}\): \(\mathrm{f}(\cos 2 \theta) = \dfrac{1 - \cos 2 \theta}{1 + \cos 2 \theta}\)
02

Find the value of \(\mathrm{f}(\mathrm{f}(\cos 2 \theta))\)

Now, replace \(\mathrm{x}\) with the result of step 1 in the function \(\mathrm{f}(\mathrm{x})=\dfrac{1 - \mathrm{x}}{1 + \mathrm{x}}\): \(\mathrm{f}(\mathrm{f}(\cos 2 \theta)) =\mathrm{f} \left( \dfrac{1 - \cos 2 \theta}{1 + \cos 2 \theta} \right)\) Using the existing function notation, we have: \(\mathrm{f}\left(\dfrac{1 - \cos 2 \theta}{1 + \cos 2 \theta}\right) = \dfrac{1 - \dfrac{1 - \cos 2 \theta}{1 + \cos 2 \theta}}{1 + \dfrac{1 - \cos 2 \theta}{1 + \cos 2 \theta}}\)
03

Simplify and match the result with the given options

To simplify the expression, let's use the fact that \(a - \dfrac{b}{c} = \dfrac{ac - b}{c}\) and \(a + \dfrac{b}{c} = \dfrac{ac + b}{c}\). Thus, we get: \(\dfrac{1 - \dfrac{1 - \cos 2 \theta}{1 + \cos 2 \theta}}{1 + \dfrac{1 - \cos 2 \theta}{1 + \cos 2 \theta}} = \dfrac{\dfrac{1 + \cos 2 \theta - 1 + \cos 2 \theta}{1 + \cos 2 \theta}}{\dfrac{1 + \cos 2 \theta + 1 - \cos 2 \theta}{1 + \cos 2 \theta}}\) Simplify the numerator and denominator: \(\dfrac{2 \cos 2 \theta}{2} = \cos 2 \theta\) Therefore, the final result is: \(\mathrm{f}(\mathrm{f}(\cos 2 \theta))=\cos 2 \theta\) Comparing this result to the given options, our answer is:
04

Answer

(c) \(\cos 2 \theta\)

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Most popular questions from this chapter

Suppose sets \(\mathrm{A}_{\mathrm{i}}(\mathrm{i}=1,2, \ldots 60)\) each set having 12 elements and set \(\mathrm{B}_{\mathrm{j}}(\mathrm{j}=1,2,3 \ldots . \mathrm{n})\) each set having 4 elements let \({ }^{60} \mathrm{U}_{\mathrm{i}=1} \mathrm{~A}_{1}={ }^{\mathrm{n}} \mathrm{U}_{\mathrm{j}=1} \mathrm{~B}_{\mathrm{j}}=\mathrm{C}\) and each element of \(\mathrm{C}\) belongs to exactly 20 of \(\mathrm{A}_{i}\) 's exactly 18 of \(\mathrm{B}_{j}\) 's then \(\mathrm{n}\) is equal to (a) 162 (b) 36 (c) 60 (d) 120

If \(\mathrm{aN}=\\{\mathrm{ax} / \mathrm{x} \in \mathrm{N}\\}\) and \(\mathrm{bN} \cap \mathrm{cN}=\mathrm{d} \mathrm{N}\) Where \(\mathrm{b}, \mathrm{c} \in \mathrm{N}\) are relatively prime then (a) \(\mathrm{d}=\mathrm{bc}\) (b) \(c=b d\) (c) \(b=c d\) (d) \(a=b d\)

The inverse of \(\left[\left(7^{x}-7^{-x}\right) /\left(7^{x}+7^{-x}\right)\right]\) is (a) \((1 / 2) \log _{7}[(1+\mathrm{x}) /(1-\mathrm{x})]\) (b) \(\log _{7}[(1-x) /(1+x)]\) (c) \(\log _{(1 / 2)}[(1-x) /(1+x)]\) (d) \((1 / 2) \log _{e}[(1+x) /(1-x)]\)

Two finite sets have \(\mathrm{m}\) and \(\mathrm{n}\) element respectively. The total number of subsets of first set is 112 more than the total number of sub sets of the second set. The value of \(\mathrm{m}\) and \(\mathrm{n}\) respectively are (a) 5,2 (b) 4,7 (c) 7,4 (d) 2,5

The domain of the function \(\mathrm{f}(\mathrm{x})={ }^{21-\mathrm{x}} \mathrm{C}_{3 \mathrm{x}-1}+{ }_{25-3 \mathrm{x}} \mathrm{P}_{5 \mathrm{x}-3}\) is (a) \(\\{1,2,3\\}\) (b) \(\\{2,3\\}\) (c) \(\\{2,3,4\\}\) (d) \(\\{2,3,4,5\\}\)
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