91Ó°ÊÓ

To begin with, we need to know the functions f(t) and g(t). Since these are not given in the problem, we can assume that f(t) = x and g(t) = y, which represent two different parametrizations of a curve. The point of intersection of these functions will be (t, t).

Now, we need to find the first and second derivatives of y with respect to x. Since f(t) = x and g(t) = y, we have: \( \frac{dy}{dt} = 1 \) (because the derivative of y with respect to t is 1 when g(t) = t) \( \frac{dx}{dt} = 1 \) (because the derivative of x with respect to t is 1 when f(t) = t) Now, we can find the first derivative of y with respect to x: \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1}{1} = 1 \) Next, we need to find the second derivative of y with respect to x. To do this, we'll first find the second derivatives of x and y with respect to t: \( \frac{d^2 y}{dt^2} = 0 \) (because the second derivative of y with respect to t is 0) \( \frac{d^2 x}{dt^2} = 0 \) (because the second derivative of x with respect to t is 0) Using these second derivatives, we can now derive the second derivative of y with respect to x: \( \frac{d^2 y}{dx^2} = \frac{\frac{d^2y}{dt^2} \cdot (\frac{dx}{dt})^2 - \frac{dy}{dt} \cdot \frac{d^2x}{dt^2}}{(\frac{dx}{dt})^4} = \frac{(0) \cdot (1)^2 - (1) \cdot (0)}{(1)^4} = \frac{0}{1} = 0 \)

We already found the second derivative of y with respect to x to be 0. Thus, the value of \(\frac{d^2y}{dx^2}\) at the point where f(t) = g(t) is: 0 So, the correct answer is (A) 0.