about to only mathematics

Answer

`y - 2 =0`

Explanation

As `|f(x_1) - f(x_2) | le (x_1 - x_2)^(2), AA x _1, x_2 in R`
`rArr |f(x_1) - f(x_2) | le |x_1 - x _2|^(2) " " ` [as `x^(2) = |x|^(2)` ]
` therefore " " |(f(x_1) - f(x_2))/( x_1 - x_2)| le |x_1 - x _2|`
`rArr underset(x_1 to x_2) (lim) |(f(x_1 ) - f (x _2))/( x_1 - x _2) | le underset (x_1 to x_2)( lim) |x_1 - x_2|`
`rArr |f'(x_1 ) | le 0 , AA x_1 in R`
`therefore |f'(x)| le 0, ` which shows`|f' (x)| = 0`
`" " ` [ as modulus is non negative or `|f' (x) | ge 0]`
` therefore f'(x) =0 or f(x)` is constant function.
`rArr ` Equation of tangent at `(1, 2)` is
`" " ( y - 2)/( x - 1) = f ' (x)`
or ` " " y - 2 = 0" " [ because ` as` f' (x) = 0 ] `
`rArr y - 2 =0` is required equation of tangent.