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# Range of ${\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$ is(A) $\left[ { - \dfrac{\pi }{4},\dfrac{\pi }{4}} \right]$ (B) $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ (C) $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{4}} \right]$ (D) $\left[ {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right]$

Last updated date: 20th Jun 2024
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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.

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.