NDA 2018 Maths · Mathematics · Derivatives · medium

The maximum value of f(x)=(x)/(1+x^2 for x>0 is

Single correct option

The maximum value of

f(x)=\frac{x}{1+x^2}

for

x>0

Show answer and solution

Answer: (D) $\frac12$

f'(x)=\frac{1-x^2}{(1+x^2)^2}=0

gives

x=1

, with the sign changing from positive to negative, so

x=1

is a maximum and

f(1)=\frac12

. Equivalently,

1+x^2\ge2x

gives

\frac{x}{1+x^2}\le\frac12

. The value 1 is the critical point itself, not the function value.

Original question written in the NDA 2018 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake