AP EAPCET 23-Aug-2021 Shift 2 Paper

Given equation,

(x2+1)3

x3

+

x2+1

3x

=0(x≠0)
⇒

3(x2+1)3+x2(x2+1)

3x3

=0
⇒(x2+1)[3(x2+1)2+x2]=0 ...(i)
∵x2≥0
⇒(x2+1)≥1 and 3(x2+1)2+x2≥3
By using this result in Eq. (i), we can observe that for no real value of x will satisfy Eq. (i).
Hence, the number of real roots is zero.