Let `f:R->R` defined by `f(x)=(x^2+2k x+16)/(x^2-2k x+16),-2 , K , 0` and `g(x)=int_0^(x^2) f(t) dt,` then

Let f:R rarr R be defined by f(x)={x+2x^(2)sin((1)/(x)) for x!=0,0 for x=0 then

Let f:R-{2}rarr R be defined as f(x)=(x^(2)-4)/(x-2) and g:R rarr R be defined by g(x)=x+2. Find whether f=g or not.

Let f:R rarr R be defined by f(x)=int_(1)^(3)(dt)/(1+|t-x|)* If int_(1)^(3)f(t)dt=(6log_(e)-k)-(k+1) then k=

f(x) : [0, 5] → R, F(x) = int_(0)^(x) x^2 g(x) , f(1) = 3 g(x) = int_(1)^(x) f(t) dt then correct choice is

f(x) : [0, 5] → R, F(x) = int_(0)^(x) x^2 g(x) , f(1) = 3 g(x) = int_(1)^(x) f(t) dt then correct choice is

Let f(x)=1/x^2 int_0^x (4t^2-2f'(t))dt then find f'(4)

Consider the function f (x) and g (x), both defined from R to R f (x) = (x ^(3))/(2 )+1 -x int _(0)^(x) g (t) dt and g (x) =x - int _(0) ^(1) f (t) dt, then The number of points of intersection of f (x) and g (x) is/are:

Consider the function f (x) and g (x), both defined from R to R f (x) = (x ^(3))/(2 )+1 -x int _(0)^(x) g (t) dt and g (x) =x - int _(0) ^(1) f (t) dt, then The number of points of intersection of f (x) and g (x) is/are: