Question

The smallest and the largest values of ${\tan ^{ - 1}}\left[ {\dfrac{{\left( {1 - x} \right)}}{{\left( {1 + x} \right)}}} \right]$, $0 \leqslant x \leqslant 1$ are
A.$0,\pi$
B.$0,\dfrac{\pi }{4}$
C.$- \dfrac{\pi }{4},\dfrac{\pi }{4}$
D.$\dfrac{\pi }{4},\dfrac{\pi }{2}$

Answer 1

Hint : In the question, we are given the inverse function of $\tan gent$ and we have to find the smallest and largest values of it. We will consider the given expression as a function. We will substitute $x = \tan \theta$ and try to convert the function into the form of identity and simplify it. Then, we will substitute the value of the range given in the question to find the smallest and largest values of the given expression.
Formula used
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
${\tan ^{ - 1}}\left( {\tan A} \right) = A$

Complete step-by-step answer :
We have, ${\tan ^{ - 1}}\left[ {\dfrac{{\left( {1 - x} \right)}}{{\left( {1 + x} \right)}}} \right]$
Let $f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)$
Let us substitute $x = \tan \theta$ in the above written function
$\Rightarrow f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right)$
As we know $\tan \dfrac{\pi }{4} = 1$. So now we will replace $1$ by $\tan \dfrac{\pi }{4}$.
$\Rightarrow f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{\tan \dfrac{\pi }{4} - \tan \theta }}{{\tan \dfrac{\pi }{4} + \tan \theta }}} \right)$
It can also be written as
$\Rightarrow f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{\tan \dfrac{\pi }{4} - \tan \theta }}{{\tan \dfrac{\pi }{4} + \tan \dfrac{\pi }{4}.\tan \theta }}} \right)$. Because $\tan \dfrac{\pi }{4} = 1$.
Now, we can see that above written function is in the form of an identity $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
Here, $\tan A = \tan \dfrac{\pi }{4}$ and $\tan B = \tan \theta$. Let us simplify the function using this identity.
$\Rightarrow f\left( x \right) = {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} - \theta } \right)} \right)$
As we know ${\tan ^{ - 1}}\left( {\tan A} \right) = A$, we get
$\Rightarrow f\left( x \right) = \dfrac{\pi }{4} - \theta$
If $x = \tan \theta$, then $\theta = {\tan ^{ - 1}}x$.
$\Rightarrow f\left( x \right) = \dfrac{\pi }{4} - {\tan ^{ - 1}}x$
In the range we have, $0 \leqslant x \leqslant 1$.
Now, let us find out the minimum value by substituting $1$ in the function
As we know
$\Rightarrow f\left( 1 \right) = \dfrac{\pi }{4} - {\tan ^{ - 1}}\left( 1 \right)$
As we know $\tan \dfrac{\pi }{4} = 1$ . Then,
$\Rightarrow f\left( 1 \right) = \dfrac{\pi }{4} - {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right)$
As we know ${\tan ^{ - 1}}\left( {\tan A} \right) = A$. Therefore, we get
$\Rightarrow f\left( 1 \right) = \dfrac{\pi }{4} - \dfrac{\pi }{4}$
$\Rightarrow f\left( 1 \right) = 0$
This is the minimum value.
Now, let us find out the maximum value by substituting $0$ in the function
$\Rightarrow f\left( 0 \right) = \dfrac{\pi }{4} - {\tan ^{ - 1}}\left( 0 \right)$
$\Rightarrow f\left( 0 \right) = \dfrac{\pi }{4} - {\tan ^{ - 1}}\left( {\tan {0^\circ }} \right)$
As we know, the value of $\tan gent$ is zero at $0^\circ$.
$\Rightarrow f\left( 0 \right) = \dfrac{\pi }{4} - 0$
$\Rightarrow f\left( 0 \right) = \dfrac{\pi }{4}$
This is the maximum value.
Therefore, the smallest and largest values are $0,\dfrac{\pi }{4}$.
So, the correct answer is “Option B”.

Note : Whenever we come across such problems, we try to replace $x$, to make the function in the form an identity and simplify it. The given expression is in tangent function, make sure that it says in tan function only, do not simplify it as sin and cosine. Remember that range determines the smallest and largest values of any function.