Question

satisfying
The number of real numbers
the equation.
|x^2-1|+|x-1|^2+√x^2-3x+2=0is​

Answer 1

Answer:

Step-by-step explanation:

Answer with explanation:

The given equation is

  \rightarrow 2\sin x=x+\frac{1}{x}\\\\\rightarrow \sin x=\frac{x+\frac{1}{x}}{2}\\\\\rightarrow -1 \leq \sin x \leq 1\\\\\rightarrow -1 \leq \frac{x+\frac{1}{x}}{2}\leq 1\\\\\rightarrow \frac{x+\frac{1}{x}}{2}+1\geq 0\\\\\rightarrow \frac{x^2+2 x+1}{2 x}\geq 0\\\\\rightarrow \frac{(x+1)^2}{2 x}\geq 0\\\\ (x+1)^2\geq 0 \text{and} 2 x\geq 0\\\\x\geq 0\\\\ \rightarrow x \geq 0

\frac{x+\frac{1}{x}}{2}\leq 1\\\\\rightarrow \frac{x+\frac{1}{x}}{2}-1\leq 0\\\\\rightarrow \frac{x^2-2 x+1}{2 x}\leq 0\\\\\rightarrow \frac{(x-1)^2}{2 x}\leq 0\\\\ (x-1)^2\geq 0 \text{and} 2 x< 0\\\\x < 0

The number of solution satisfying the above equation

x∈ (0,∝) and x∈(-∝,0)

There will be infinite real values of x satisfying the equation.