f(x)=(1)/(1+(1)/(x));g(x)=(1)/(1+(1)/(f(x)))rArr g'(2)=

f (x)= (1)/(1+ (1)/(x))g (x) =(1)/(1 + (1)/(f(x)))implies g'(2)=

If f(1) =g(1)=2 , then lim_(xrarr1) (f(1)g(x)-f(x)g(1)-f(1)+g(1))/(f(x)-g(x)) is equal to

If f(1) =g(1)=2 , then lim_(xrarr1) (f(1)g(x)-f(x)g(1)-f(1)+g(1))/(f(x)-g(x)) is equal to

Suppose that the function f,g,f', and g' are continuous over [0,1],g(x)!=0 for x in[0,1],f(0)=0,g(0)=pi,f(1)=(2015)/(2),g(1)=1 The value of int_(0)^(1)(f(x)g'(x)(g^(2)(x)-1)+f'(x)-g(x)(g^(2)(x)+1))/(g^(2)(x))dx is equal to

Let f(x), g(x) be differentiable functions and f(1) = g(1) = 2, then : lim_(x rarr 1) (f(1) g(x) - f(x) g(1) - f(1) + g(1))/(g(x) - f(x)) equals :

If f(x)=x+tan x and g(x) is inverse of f(x) then g'(x) is equal to (1)(1)/(1+(g(x)-x)^(2))(2)(1)/(1-(g(x)-x)^(2))(1)/(2+(g(x)-x)^(2))(4)(1)/(2-(g(x)-x)^(2))

Statement-1:lim_(x rarr0)((g(2-x^(2)))/(g(2)))^((4)/(x^(2)))=e^(-(1)/(2)) where g(2)=-40 and g(2)=-5 Statement-2: If lim_(x rarr a)f(x)=1,lim_(x rarr a)g(x)rarr oo then lim_(x rarr a){f(x)}^(g(x))=e^(lim_(a))(f(x)-1)g(x)

If f(x)=x+tan x and f(x) is inverse of g(x), then g'(x) is equal to (1)/(1+(g(x)-x)^(2)) (b) (1)/(1+(g(x)+x)^(2))(1)/(2-(g(x)-x)^(2)) (d) (1)/(2+(g(x)-x)^(2))

If f(x)=(1)/((1-x)) and g(x)=f[f{f(x))} then g(x) is discontinuous at