If `A=[[1,1],[0,-1]]` and `f(x) =1-x^2,` then `f(A)` is

If f'(x)=(1)/((1+x^(2))^(3//2)) and f(0)=0, then f(1) is equal to :

For x in RR - {0, 1}, let f_1(x) =1/x, f_2(x) = 1-x and f_3(x) = 1/(1-x) be three given functions. If a function, J(x) satisfies (f_2oJ_of_1)(x) = f_3(x) then J(x) is equal to :

If f^2(x)*f((1-x)/(1+x))=x^3, [x!=-1,1 and f(x)!=0], then find |[f(-2)]| (where [] is the greatest integer function).

Suppose |[f'(x),f(x)],[f''(x),f'(x)]|=0 is continuously differentiable function with f^(prime)(x)!=0 and satisfies f(0)=1 and f'(0)=2 then (lim)_(x->0)(f(x)-1)/x is 1//2 b. 1 c. 2 d. 0

If f(x) = {:{(px^(2)-q, x in [0,1)), ( x+1 , x in (1,2]):} and f(1) =2 , then the value of the pair ( p,q) for which f(x) cannot be continuous at x =1 is:

Let 'f is a function, continuous on [0,1] such that f(x) leq sqrt5 AA x in [0,1/2] and f(x) leq 2/x AA x in [1/2,1] then smallest 'a' for which int_0^1 f(x)dx leq a holds for all 'f' is

If f(x) is continuous in [0,2] and f(0)=f(2). Then the equation f(x)=f(x+1) has

If the function f:[1,∞)→[1,∞) is defined by f(x)=2 ^(x(x-1)) then f^-1(x) is

Let f(x)=int x^2/((1+x^2)(1+sqrt(1+x^2)))dx and f(0)=0 then f(1) is

A function y=f(x) satisfies (x+1)f^(prime)(x)-2(x^2+x)f(x)=((e^x)^2)/((x+1)),AAx >-1. If f(0)=5, then f(x) is