If `y=1/(t^2+t-2),where \ t=1/(x-1),` then find the number of points where `f(x)` is discontinuous.

If y = frac(1)(t^2-t-6 and t = frac(1)(x-2) then the values of x which make the function y discontinuous, are

Let f(x) = [x^3 - 3] where [.] denotes the greatest integer function Then the number of points in the interval ( 1, 2) where the function is discontinuous, is

If x=3t " and " y=(2t)/3-1 , then find the value of t for which x = 3y.

If x=(2t)/(1+t^2) , y=(1-t^2)/(1+t^2) , then find (dy)/(dx) .

Find the points of minima for f(x)=int_0^x t(t-1)(t-2)dt

Find the locus of the point (t^2+t+1,t^2-t+1),t in Rdot

Find the set of points where the following functions are discontinuous f(x)=1/(1-e^((x-1)/(x-2))

If x=t+(1/t), y=t-(1/t) and t=parameter then the locus of the point (x, y) is __

Find a continuous function f, where (x^(4)-4x^(2))lef (x) le (2x^(2)-x^(3)) such that the area bounded by y=f(x), y=x^(4)-4x^(2), the y-axis, and the line x=t," where "(0letle2) is k times the area bounded by y=f(x),y=2x^(2)-x^(3), y-axis, and line x=t("where "0le t le 2).

If int_(0)^(x) f(t)dt=x+int_(x)^(1)t f(t)dt , find the value of f(1).