nhi samjh aa rha..

The main thing to notice here is that (x^2+(1/x)^2)= ((x+(1/x))^2)-2. Now we can take (x+(1/x)) as 'a'. This will let us handle the sum more easily.

(x^2)+(x)+(1/x^2)+(1/x)<0

((a^2)-2)+a<0.... assuming (x+(1/x)) as 'a' and using the above mentioned identity. Now this becomes a simple inequality involving a quadratic.

a^2+a-2<0

(a+2)(a-1)<0

This implies that a is between -2 and 1.

Now, what was 'a'? This was the sum of x and its reciprocal. That means that the sum of x and its reciprocal is between -2 and 1. Hence, the correct answer must be option A. Is that right?