f and g differentiable functions of x such that `f(g(x))=x," If "g'(a)ne0" and "g(a)=b," then "f'(b)=`

If f and g are derivable function of x such that g'(a)ne0,g(a)=b" and "f(g(x))=x," then 'f'(b)=

Let f :R to R be a continous and differentiable function such that f (x+y ) =f (x). F(y)AA x, y, f(x) ne 0 and f (0 )=1 and f '(0) =2. Let g (xy) =g (x). g (y) AA x, y and g'(1)=2.g (1) ne =0 The number of values of x, where f (x) g(x):

Let f :R to R be a continous and differentiable function such that f (x+y ) =f (x). F(y)AA x, y, f(x) ne 0 and f (0 )=1 and f '(0) =2. Let f (xy) =g (x). G (y) AA x, y and g'(1)=2.g (1) ne =0 The number of values of x, where f (x) g(x):

If f and g are differentiable at a in R such that f(a)=g(a)=0 and g'(a)!=0 then show that lim_(x rarr a)(f(x))/(g(x))=(f'(a))/(g'(a))

Let f and g be two differentiable functions on R such that f'(x)>0 and g′(x) g(f(x-1)) (b) f(g(x))>f(g(x+1)) (c) g(f(x+1))

If f(x),g(x) be twice differentiable functions on [0,2] satisfying f''(x)=g''(x)f'(1)=2g'(1)=4 and f(2)=3g(2)=9 then f(x)-g(x) at x=4 equals (A) 0 (B) 10 (C) 8 (D) 2

If f and g are continuous functions on [0,a] such that f(x)=f(a-x) and g(x)+g(a-x)=2 then show that int_(0)^(a)f(x)g(x)dx=int_(0)^(a)f(x)dx

Let f and g be differentiable functions such that: xg(f(x))f\'(g(x))g\'(x)=f(g(x))g\'(f(x))f\'(x) AA x in R Also, f(x)gt0 and g(x)gt0 AA x in R int_0^xf(g(t))dt=1-e^(-2x)/2, AA x in R and g(f(0))=1, h(x)=g(f(x))/f(g(x)) AA x in R Now answer the question: f(g(0))+g(f(0))= (A) 1 (B) 2 (C) 3 (D) 4

Let f and g be differentiable functions such that: xg(f(x))f\'(g(x))g\'(x)=f(g(x))g\'(f(x))f\'(x) AA x in R Also, f(x)gt0 and g(x)gt0 AA x in R int_0^xf(g(t))dt=1-e^(-2x)/2, AA x in R and g(f(0))=1, h(x)=g(f(x))/f(g(x)) AA x in R Now answer the question: f(g(0))+g(f(0))= (A) 1 (B) 2 (C) 3 (D) 4