Show that there exists no polynomial `f(x)` with integral coefficients which satisfy `f(a)=b ,f(b)=c ,f(c)=a ,` where `a , b , c ,` are distinct integers.

Let a,b be integers and f(x) be a polynomial with integer coefficients such that f(b)-f(a)=1. Then, the value of b-a, is

If f(x) is a polynomial of degree at least two with integral co-efficients then the remainder when it is divided by(x-a) (x-b) is , where a ne b (1) x[(f(a)-f(b))/(b-a)] + (af(b) -bf(a))/(a-b) (2) x[(f(a) -f(b))/(a-b)] + (af(b) -bf(a))/(a-b) (3) x[(f(b) -f(a))/(a-b)] + (af(b) -bf(a))/(a-b) (4) x [ (f(b) -f(a))/(a-b)] + (bf(a) -af(a))/(a-b)

If f'(c)=(f (b)-f(a))/(b-a) , where f(x) =e^(x), a=0 and b=1 , then: c = ….

If f(x) is a polynomial then the remainder of f(x) when divided by (x-a) is A) f(0), B) f(a), C) f(1) ,D) f(-a)

Let f(x)=a+b|x|+c|x|^(2) , where a,b,c are real constants. The, f'(0) exists if

Let fx) be a polynomial function of second degree.If f(1)=f(-1) and a ,b,c are in A.P. the f'(a),f'(b) and f'(c) are in