The given curve is y=\(\frac{x-1}{x-2}\).
\(\frac{dy}{dx}\)=\(\frac {(x-2)(1)-(x-1)(1)}{(x-2)^2}\)
=\(\frac {x-2-x+1}{(x-2)^2}\)=-\(\frac{-1}{(x-2)^2}\)
Thus, the slope of the tangent at x = 10 is given by,
\((\frac{\text{dy}}{\text{dx}}) \bigg]_{x=10}\)\(= \frac{-1}{(x-2)^2}\bigg] _{ x=10}\)=-\(\frac{-1}{(10-2)^2}\)=\(\frac{-1}{64}\)
Hence, the slope of the tangent at x = 10 is \(\frac{-1}{64}\)
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
m×n = -1
