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If $F( - 1,1) \to B$ is a function defined by $f(x) = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$, then f is both one-one and onto when B is the intervalA. $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$B. $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$C. $\left[ {0,\dfrac{\pi }{2}} \right]$D. $\left( {0,\dfrac{\pi }{2}} \right)$

Last updated date: 08th Sep 2024
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Hint: Given, $F( - 1,1) \to B$ is a function defined by $f(x) = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$, then f is both one-one and onto. We have to find the interval of B. First, we will assume $x = \tan \theta$, then simplify the value of $f(x)$. After simplification we put the value of $\theta$ in $f(x)$ and find where the function is strictly increasing then we find the interval of B.

Complete step by step Solution:
A function is a process or a relation that associates each element $x$ of a non-empty set $X$, at least to a single element $y$ of another non-empty set $Y$. A relation $f$ from a set $X$ (the domain of the function) to another set $Y$ (the co-domain of the function) is called a function in math.
A one-to-one function is a function that maps distinct elements of its domain to distinct elements of its codomain.
If for every element of $Y$, there is at least one or more than one element mapped with $X$, then the function is said to be onto function.
Given, $F( - 1,1) \to B$
$f(x) = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$
Let $x = \tan \theta$
Putting in $f(x)$
$f(x) = {\tan ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)$
We know that $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
$f(x) = {\tan ^{ - 1}}\left( {\tan 2\theta } \right)$
$f(x) = 2\theta$ (1)
$x = \tan \theta$
$\theta = {\tan ^{ - 1}}x$
Putting in (1)
$f(x) = 2{\tan ^{ - 1}}x$
$f(x)$ is strictly increasing in $( - 1,1)$
A strictly increasing function is a function that always increases over its entire domain. It isn’t allowed any dips and the function cannot stay constant for even a short time. For example, if a function might increase everywhere then at infinity, it might drop off into an abyss.
$f( - 1) = 2\tan \left( {{{\tan }^{ - 1}}( - 1)} \right)$
$= 2\left( { - \dfrac{\pi }{4}} \right)$
$= - \dfrac{\pi }{2}$
$f(1) = 2\tan \left( {{{\tan }^{ - 1}}(1)} \right)$
$= 2\left( {\dfrac{\pi }{4}} \right)$
$= \dfrac{\pi }{2}$
So, $B = \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$

Hence, the correct option is B.

Note: Students should pay attention while solving the solution to get the correct interval. They should assume the correct function and the wrong function can lead to giving incorrect answers.