T=2，

v=2×2-2/2？ =4- 1/2=7/2

2 Solution: Let the coordinates of the tangent point be (x0, x0? )

Y'|x=x0=2x0, so the tangent equation is y-x0? =2x0(x-x0)

∵ parabola y=x? Intersection p (5/2,2)

∴2-x0？ =2x0(5/2-x0)

X0=2 or 3

So the coordinates of the tangent point are (2,4) or (3,9).

The tangent passes through point p (5/2,2).

The tangent equation is 4x-y-4=0 or 6x-y-9=0.

3.A

Solution: ∵ The premise is: "For differentiable function f(x), if f'(x0)=0, then x=x0 is the extreme point of function f(x)", which is not a true proposition.

Because for the differentiable function f(x), if f'(x0)=0 and the derivative function values are different when x > x0 and x < x0 are satisfied, then x=x0 is the extreme point of the function f(x).

∴ major premise error,