UPSC NDA I 2015 Mathematics Solved Paper

The function is said to have a minimum value at some x if the first derivative of the function at x is 0 and the second derivative of the function at x is greater than 0.
Hence,
f(x)=

x2−1

x2+1

f′(x)=

(x2+1)2x−(x2−1)2x

(x2+1)2

=0

4x

(x2+1)2

=0
x=0
Thus zero is a critical point for the function.
Now finding the second derivative of the function at 0, we get
f′′(x)=

(x2+1)24−16x2(x2+1)

(x2+1)4

f′′(0)=4
Thus we can see that the second derivative of the function at zero is greater than zero.
Hence the function attains the minimum value at 0.
∴f(0)=−1
The minimum value of the function is −1.