
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 interval
A. $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$
B. $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$
C. $\left[ {0,\dfrac{\pi }{2}} \right]$
D. $\left( {0,\dfrac{\pi }{2}} \right)$
Answer
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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.
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.
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