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.

If f(x)={:{ (x-2 ,x 0) :} ,then number of points, where y=f(f(x)) is discontinuous is

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

If f(u) = 1/(u^(2)+u-2) , where u = 1/(x-1) , then the points of discontinuity of f are x = …..

The number of points where f(x) is discontinuous in [0,2] where f(x)={[cos pi x]:x 1}

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

Find the points of discontinuity of f(x)=(1)/(2sin x-1)

If f(x)=(1+x)/([x]),x in[1,3] , then the number of points of discontinuity are (where [.] is G.I.F.)

Let x=t^(2)+t+1,y=t^(2)-t+1, where t>0 The total number of tangents at (1,1) will be:

The number of points where f(x) is discontinuous in [0,2] where f(x) = { [cos pix ;x<1 , [x-2] ; x>1} where [.] represents G.I.F. is :