" (ii) "f(x)={[(x-1)tan(pi x)/(2),," if "x!=1],[k,," if "x=1]" at "x=1

Find the value of the constant k so that the given function is continuous at the indicated point: f(x)={(x-1)(tan(pi x))/(2),quad if quad x!=1k,quad if x=1 at x=1

Let f(x)={(|x+1|)/(tan^(- 1)(x+1)), x!=-1 ,1, x!=-1 Then f(x) is

Let f(x)={(|x+1|)/(tan^(-1)(x+1)),x!=-1,1,x!=-1 Then f(x) is

If f(x) = (1 - x) tan (pi x)/(2) then "limit"_(x->1) f(x) is equal to :

Let f(X) = {:{((x-1) tan (pi/2x), x ne 1),(k, x =1):} . Find k if f is continuous at x = 1.

If the function f(x)={((sqrt(2+cosx)-1)/(pi-x)^2 ,; x != pi), (k, ; x = pi):} is continuous at x = 1, then k equals:

If f(x)=[tan((pi)/(4)+x)]^((1)/(x)),xne0" " f(x)=k , "x=0 , " is continuous at x=0 " Then k= . . . .

If f(x)=tan^(-1)x,g(x)=tan^(-1)((1+x)/(1-x))" for "|x|lt1 , show that f'(x)=g'(x)andg(x)-f(x)=pi/4 .