The function `y = f(x)` is defined by `x=2t-|t|, y=t^2+|t|, t in R` in the interval `x in [-1,1]`, then

If y = f(x) defined parametrically by x = 2t - |t - 1| and y = 2t^(2) + t|t| , then

Let y = f(x) be defined parametrically as y = t^(2) + t|t|, x = 2t - |t|, t in R . Then, at x = 0,find f(x) and discuss continuity.

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

A function y=f(x) is given by x=1/(1+t^2) and y=1/(t(1+t^2)) for all tgt0 then f is

Let the function y = f(x) be given by x= t^(5) -5t^(3) -20t +7 and y= 4t^(3) -3t^(2) -18t + 3 where t in ( -2,2) then f'(x) at t = 1 is

Let function F be defined as f(x)= int_1^x e^t/t dt x > 0 then the value of the integral int_1^1 e^t/(t+a) dt where a > 0 is

Show that the function y=f(x) defined by the parametric equations x=e^(t)sin(t),y=e^(t).cos(t), satisfies the relation y''(x+y)^(2)=2(xy'-y)

The locus of the represented by x = t^ 2 + t + 1 , y = t ^ 2 - t + 1 is

If the function y=f(x) is represented as x=phi(t)=t^5-5t^3-20 t+7 , y=psi(t)=4t^3-3t^2-18 t+3(|t|<2), then find the maximum and minimum values of y=f(x)dot

Is the function f defined by f(x)={{:(x , ifxlt=1),( 5, ifx >1):} continuous at x = 0? A t x = 1? A t x = 2?