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

Let a function y=y(x) be defined parametrically by x=2t-|t|y=t^(2)+t|t| then y'(x),x>0

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

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

A function y=f(x) is given by x=(1)/(1+t^(2)) and y=(1)/(t(1+t^(2))) for all t>0 then f is

Let f(x) be a function defined by f(x)=int_(1)^(x)t(t^2-3t+2)dt,x in [1,3] Then the range of f(x), is

Let f(x) be a function defined by f(x)=int_(1)^(x)t(t^(2)-3t+2t)dt,1lexle3 then the maximum value of f(x) is