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=1/(t^2+t-2),w h e r et=1/(x-1), then find the number of points where f(x) is discontinuous.

Let f be a composite function.of x defined by f(u)=(1)/(u^(3)-6u^(2)+11u-6) where u(x)=(1)/(x) .Then the number of points x where f is discontinuous is

Let f(x) ={{:( [x],-2le xle -(1)/(2) ),( 2x^(2)-1,-(1)/(2)lt xle2):}and g(x) =f(|x|)+|f(x)|, where [.] represents the greatest integer function . the number of point where g(x) is discontinuous is

Let f(x)=int_(0)^(x) (sint-cost)(e^t-2)(t-1)^3(t-1)^3(t-2)^5 dt , 0lt xle4 Then , the number of points where f(x) assumes local maximum value , is

If a function y=f(x) is defined as y=(1)/(t^(2)-t-6)and t=(1)/(x-2), t in R . Then f(x) is discontinuous at

If function y=f(x) has only two points of discontinuity say x_1,x_2, where x_1,x_2<0, then the number of points of discontinuity of y=f(-|x+1|) is (a) 0 (b) 2 (c) 6 (d) 4

Consider the curve represented parametrically by the equation x = t^3-4t^2-3t and y = 2t^2 + 3t-5 where t in R .If H denotes the number of point on the curve where the tangent is horizontal and V the number of point where the tangent is vertical then

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