Question

Prove that for every $x > 0\,,\,\dfrac{x}{{1 + {x^2}}} < {\tan ^{ - 1}}\left( x \right) < x$ ?

Answer 1

Hint: In this question, we have to use the definition of increasing function. First we have to prove that the function ${\tan ^{ - 1}}x - \dfrac{x}{{1 + {x^2}}}$ and $x - {\tan ^{ - 1}}x$ are increasing and then we will use the definition that if a function f(x) is an increasing function and ${x_1} > {x_2}$ then, $f\left( {{x_1}} \right) > f\left( {{x_2}} \right)$.

Complete step by step answer:
To do the above question, first we should know about some properties of increasing function.
$\left( 1 \right)\,If\,,\,{f^1}\left( x \right) > 0\,,\,x\, \in \left( {a,b} \right)\,$then f is strictly increasing function on (a, b).
$\left( 2 \right)$f is strictly increasing function on (a, b), then ${x_1} > {x_2}\, \Rightarrow \,f\left( {{x_1}} \right) > f\left( {{x_2}} \right)$.
Now, we have, $x > 0 \Rightarrow x \in \left( {0,\infty } \right)$ given to prove $\dfrac{x}{{1 + {x^2}}} < {\tan ^{ - 1}}x < x$.Let,
$f\left( x \right) = {\tan ^{ - 1}}x - \dfrac{x}{{1 + {x^2}}}................\left( 1 \right)$

Now differentiate the following equation. In the above function, first is the standard derivative of the inverse of the tangent function and in the second we have to use the quotient method.
$ \Rightarrow {f^1}\left( x \right) = \dfrac{1}{{1 + {x^2}}} - \dfrac{{\left( {1 + {x^2}} \right).1 - x\left( {2x} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}$
$ \Rightarrow {f^1}\left( x \right) = \dfrac{1}{{1 + {x^2}}} - \dfrac{{\left( {1 - {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}$
Now, taking LCM and on further simplification, we get
$ \Rightarrow {f^1}\left( x \right) = \dfrac{{2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} > 0$
$ \Rightarrow {f^1}\left( x \right) > 0$
Therefore, the above function is strictly increasing.

Now, we know that if$x > 0$, then
$f\left( x \right) > f\left( 0 \right)$
Also,
$ \Rightarrow f\left( 0 \right) = {\tan ^{ - 1}}0 - \dfrac{0}{1} = 0$
Therefore,
$ \Rightarrow f\left( x \right) > 0$
$ \Rightarrow {\tan ^{ - 1}}x - \dfrac{x}{{1 + {x^2}}} > 0$
$ \Rightarrow {\tan ^{ - 1}}x > \dfrac{x}{{1 + {x^2}}}................\left( 2 \right)$
Again let,
$g\left( x \right) = x - {\tan ^{ - 1}}x.............\left( 3 \right)$
$ \Rightarrow g\left( x \right) = 1 - \dfrac{1}{{1 + {x^2}}} = \dfrac{{{x^2}}}{{1 + {x^2}}} > 0$
$i.e.\,x > 0\, \Rightarrow \,\,g'\left( x \right) > 0$
Therefore, g(x) is a strictly increasing function.
Now, we know that if $x > 0$, then
$g\left( x \right) > g\left( 0 \right)$
Also,
\[ \Rightarrow g\left( 0 \right) = 0 - {\tan ^{ - 1}}0 = 0\]
Therefore,
$ \Rightarrow g\left( 0 \right) > 0$
$ \Rightarrow x - {\tan ^{ - 1}}x > 0$
$ \Rightarrow x > {\tan ^{ - 1}}x................\left( 4 \right)$
Now from $\left( 3 \right)\,and\,\left( 4 \right)$, we get
$ \Rightarrow x > {\tan ^{ - 1}}x > \dfrac{x}{{1 + {x^2}}}$
$\therefore \dfrac{x}{{1 + {x^2}}} < {\tan ^{ - 1}}x < x$.
Hence proved.

Note: In the above question, we have taken the two functions according to the requirement or what we have to prove in the question. we should also understand the if a function is strictly decreasing and ${x_1} > {x_2}$, then $f\left( {{x_1}} \right) < f\left( {{x_2}} \right)$ for all values of $x$.