Answer
Verified
430.2k+ views
Hint: First simplify the given equation by substituting $ x = \tan \theta $ . After simplification, use the range of $ \sin \theta $ to find the range of the given expression. You can use the fact that $ {\tan ^{ - 1}}x $ is an increasing function. So the inequality will not change.
Complete step-by-step answer:
The given equation is
$ \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) $
To simplify this equation, put $ x = \tan \theta $ . Then
$ \dfrac{{2x}}{{1 + {x^2}}} = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} $
We have a formula,
$ \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \sin 2\theta $
By using this formula, we can write
$ \dfrac{{2x}}{{1 + {x^2}}} = \sin 2\theta $
Now, we know that the range of $ \sin 2\theta $ is $ \left[ { - 1,1} \right] $ . Because the maximum value of $ \sin 2\theta $ is 1 and its minimum value is -1. Also it is a continuous function. So it takes all the values between -1 and 1.
Thus, $ \dfrac{{2x}}{{1 + {x^2}}} \in [ - 1,1] $
Now, by applying $ {\tan ^{ - 1}} $ to both the sides. And knowing that $ {\tan ^{ - 1}} $ is an increasing function. So the inequality in intervals will not change.
$ \therefore {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) \in \left[ {{{\tan }^{ - 1}}( - 1),{{\tan }^{ - 1}}(1)} \right] $
By using the property, $ {\tan ^{ - 1}}( - x) = {-\tan ^{ - 1}}x $ , we can write
$ = \left[ { - {{\tan }^{ - 1}}1,{{\tan }^{ - 1}}1} \right] $
We know that, $ {\tan ^{ - 1}}1 = \dfrac{\pi }{4} $
Thus, we get the range as
$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) \in \left[ { - \dfrac{\pi }{4},\dfrac{\pi }{4}} \right] $
Therefore, from the above explanation, the correct answer is, option (A) $ \left[ { - \dfrac{\pi }{4},\dfrac{\pi }{4}} \right] $
So, the correct answer is “Option A”.
Note: To solve this question, you need to know the trigonometric formulae. Then only it would click you that you can simplify the equation in terms of $ \sin \theta $ . The key point here is to know the range of the sine function and know that the inequality does not change when you apply an increasing function to it. If you check the graph of $ {\tan ^{ - 1}}x $ . You will observe that, it is an increasing function.
Complete step-by-step answer:
The given equation is
$ \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) $
To simplify this equation, put $ x = \tan \theta $ . Then
$ \dfrac{{2x}}{{1 + {x^2}}} = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} $
We have a formula,
$ \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \sin 2\theta $
By using this formula, we can write
$ \dfrac{{2x}}{{1 + {x^2}}} = \sin 2\theta $
Now, we know that the range of $ \sin 2\theta $ is $ \left[ { - 1,1} \right] $ . Because the maximum value of $ \sin 2\theta $ is 1 and its minimum value is -1. Also it is a continuous function. So it takes all the values between -1 and 1.
Thus, $ \dfrac{{2x}}{{1 + {x^2}}} \in [ - 1,1] $
Now, by applying $ {\tan ^{ - 1}} $ to both the sides. And knowing that $ {\tan ^{ - 1}} $ is an increasing function. So the inequality in intervals will not change.
$ \therefore {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) \in \left[ {{{\tan }^{ - 1}}( - 1),{{\tan }^{ - 1}}(1)} \right] $
By using the property, $ {\tan ^{ - 1}}( - x) = {-\tan ^{ - 1}}x $ , we can write
$ = \left[ { - {{\tan }^{ - 1}}1,{{\tan }^{ - 1}}1} \right] $
We know that, $ {\tan ^{ - 1}}1 = \dfrac{\pi }{4} $
Thus, we get the range as
$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) \in \left[ { - \dfrac{\pi }{4},\dfrac{\pi }{4}} \right] $
Therefore, from the above explanation, the correct answer is, option (A) $ \left[ { - \dfrac{\pi }{4},\dfrac{\pi }{4}} \right] $
So, the correct answer is “Option A”.
Note: To solve this question, you need to know the trigonometric formulae. Then only it would click you that you can simplify the equation in terms of $ \sin \theta $ . The key point here is to know the range of the sine function and know that the inequality does not change when you apply an increasing function to it. If you check the graph of $ {\tan ^{ - 1}}x $ . You will observe that, it is an increasing function.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE