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,$f(x)=\frac{x^2-1}{x^2+1}$ $f'(x)=\frac{(x^2+1)2x-(x^2-1)2x}{(x^2+1)^2}=0$$\frac{4x}{(x^2+1)^2}=0$$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)=\frac{(x^2+1)^{2}4-16x^2(x^2+1)}{(x^2+1)^4}$$f''(0)=4$Thus we can see that the second derivative of the function at zero is greater than zero.Hence the function attains the minimum value at 0.$\therefore f(0)=-1$The minimum value of the function is $-1$.