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AP EAPCET 23-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 13 of 160

Marks: +1, -0

\frac{(x^{2}+1)^{3}}{x^{3}}+\frac{x^{2}+1}{3x}=0\ (x \neq 0)

Solution:

Given equation,

\frac{(x^{2}+1)^{3}}{x^{3}}+\frac{x^{2}+1}{3x}=0\ (x \neq 0)

\Rightarrow \frac{3(x^{2}+1)^{3}+x^{2}(x^{2}+1)}{3x^{3}}=0

\Rightarrow (x^{2}+1)[3(x^{2}+1)^{2}+x^{2}]=0

...(i)

\because x^{2} \geq 0

\Rightarrow (x^{2}+1) \geq 1

and

3(x^{2}+1)^{2}+x^{2} ≥ 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.

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