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Solve |x-pi|+|x^2-pi^2| le 0...

Solve `|x-pi|+|x^2-pi^2| le 0 `

Text Solution

Verified by Experts

The correct Answer is:
`x=pi`
`|x- pi |+|x^2 - pi ^2 | le o `
or ` |x-pi|(1+|x+pi|)le 0`
or `|x-pi|=0`
or `x=pi`
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