Let $ f:\left[ {0,\sqrt 3 } \right] \to \left[ {0,\dfrac{\pi }{3} + {{\log }_e}2} \right] $ defined by $ f\left( x \right) = {\log _e}\sqrt {{x^2} + 1} + {\tan ^{ - 1}}x $ , then $ f\left( x \right) $ is
A) One-one and onto
B) One-one but not onto
C) Onto but not one-one
D) Neither one-one nor onto

Hint: First, to find whether the function is one-one, we have to see whether $ f\left( {{x_1}} \right) = f\left( {{x_2}} \right) \Rightarrow {x_1} = {x_2} $
If this property is satisfied, then the function is one-one. Then, we are to find whether the range of $ f\left( x \right) $ lies within the codomain of the function. If so, then the function is onto as well.

Complete step-by-step answer:
Now, given, $ f\left( x \right) = {\log _e}\sqrt {{x^2} + 1} + {\tan ^{ - 1}}x $ .
Differentiating $ f\left( x \right) $ with respect to $ x $ , we get,
 $ f'\left( x \right) = \dfrac{1}{{\sqrt {{x^2} + 1} }}\dfrac{d}{{dx}}\left( {\sqrt {{x^2} + 1} } \right) + \dfrac{1}{{1 + {x^2}}} $
 $ \Rightarrow f'\left( x \right) = \dfrac{1}{{\sqrt {{x^2} + 1} }} \times \dfrac{1}{{2\sqrt {{x^2} + 1} }}\dfrac{d}{{dx}}\left( {{x^2} + 1} \right) + \dfrac{1}{{1 + {x^2}}} $
Using chain rule of differentiation, we get,
 $ \Rightarrow f'\left( x \right) = \dfrac{1}{{\sqrt {{x^2} + 1} }} \times \dfrac{1}{{2\sqrt {{x^2} + 1} }} \times \dfrac{d}{{dx}}\left( {{x^2} + 1} \right) + \dfrac{1}{{1 + {x^2}}} $
Now, simplifying, we get,
 $ \Rightarrow f'\left( x \right) = \dfrac{1}{{\sqrt {{x^2} + 1} }} \times \dfrac{1}{{2\sqrt {{x^2} + 1} }} \times 2x + \dfrac{1}{{1 + {x^2}}} $
Multiplying terms in denominator, we get,
\[ \Rightarrow f'\left( x \right) = \dfrac{1}{{2\left( {{x^2} + 1} \right)}} \times 2x + \dfrac{1}{{1 + {x^2}}}\]
Cancelling common factors in numerator and denominator, we get,
\[ \Rightarrow f'\left( x \right) = \dfrac{{\left( {x + 1} \right)}}{{\left( {{x^2} + 1} \right)}}\]
Now, to find the critical points, solving, we have,
 $ f'\left( x \right) = 0 $
\[ \Rightarrow \dfrac{{\left( {x + 1} \right)}}{{\left( {{x^2} + 1} \right)}} = 0\]
Now, cross multiplying the terms of equation and shifting all the constant terms to the right side of the equation, we get,
\[ \Rightarrow x = - 1\]
So, we have a critical point of function for \[x = - 1\].
But the domain of the function is $ \left[ {0,\sqrt 3 } \right] $ .
The first derivative of the function is positive for all values of x greater than \[x = - 1\].
So, the function is strictly increasing in the domain $ \left[ {0,\sqrt 3 } \right] $ .
Now, $ f\left( 0 \right) = {\log _e}\sqrt {{0^2} + 1} + {\tan ^{ - 1}}\left( 0 \right) $
 $ \Rightarrow f\left( 0 \right) = {\log _e}\sqrt 1 + 0 $
 $ \Rightarrow f\left( 0 \right) = 0 $
Also, $ f\left( {\sqrt 3 } \right) = {\log _e}\sqrt {{{\left( {\sqrt 3 } \right)}^2} + 1} + {\tan ^{ - 1}}\left( {\sqrt 3 } \right) $
 $ \Rightarrow f\left( {\sqrt 3 } \right) = {\log _e}\sqrt {3 + 1} + \dfrac{\pi }{3} $
 $ \Rightarrow f\left( {\sqrt 3 } \right) = {\log _e}2 + \dfrac{\pi }{3} $
So, the range of the function $ f\left( x \right) = {\log _e}\sqrt {{x^2} + 1} + {\tan ^{ - 1}}x $ for x varying in the interval $ \left[ {0,\sqrt 3 } \right] $ is $ \left[ {0,\dfrac{\pi }{3} + {{\log }_e}2} \right] $ .
Therefore, the range and codomain of the function are the same because the greatest and smallest values of the function lie within the codomain of the function.
So, we can say that every term in the co-domain of the function has a preimage in the domain of the function.
Thus, we can say that the function is onto.
Also, as the function is strictly increasing in the domain of the function, that is $ \left[ {0,\sqrt 3 } \right] $ . So, the values of the function keep on increasing with the increasing values of x. Hence, there are no two values of x for which the function attains the same value.
Hence, the function is one-one.
Therefore, the function is both one-to-one and onto. Hence, option (A) is the correct answer.
So, the correct answer is “Option A”.

Note: If a function is strictly increasing or decreasing within a certain range of values or the domain of the function, then the function is surely one-one as each preimage in the domain has only one unique image in the co-domain. We should be clear with the concepts of one-one and onto functions before attempting such problems. We should also have a clear understanding of the monotonicity of a function.