AP EAMCET · Maths · Functions
The number of real roots of the equation \(\frac{\left(x^2+1\right)^3}{x^3}+\frac{x^2+1}{3 x}=0,(x \neq 0)\) is
- A \(1\)
- B \(0\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(0\)
Step-by-step Solution
Detailed explanation
Given equation, \(\frac{\left(x^2+1\right)^3}{x^3}+\frac{x^2+1}{3 x}=0(x \neq 0)\) \(\Rightarrow \quad \frac{3\left(x^2+1\right)^3+x^2\left(x^2+1\right)}{3 x^3}=0\) \(\Rightarrow \quad\left(x^2+1\right)\left[3\left(x^2+1\right)^2+x^2\right]=0\) ...(i)…
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