Given `f(x)=(ax+b)/(x+1), lim_(x to oo)f(x)=1 and lim_(x to 0)f(x)=2`, then f(-2) is

f(x)=(ax^(2)+1)/(x^(2)+1),lim_(x rarr0)f(x)=1 and lim_(x rarr oo)f(x)=1, then prove that f(-2)=f(2)=1

Let f(x) lt 0 AA x in (-oo, 0) and f (x) gt 0 AA x in (0,oo) also f (0)=0, Again f'(x) lt 0 AA x in (-oo, -1) and f '(x) gt AA x in (-1,oo) also f '(-1)=0 given lim _(x to oo) f (x)=0 and lim _(x to oo) f (x)=oo and function is twice differentiable. The minimum number of points where f'(x) is zero is:

Let f(x) lt 0 AA x in (-=oo, 0) and f (x) gt 0 AA x in (0,oo) also f (0)=o, Again f'(x) lt 0 AA x in (-oo, -1) and f '(x) gt AA x in (-1,oo) also f '(-1)=0 given lim _(x to oo) f (x)=0 and lim _(x to oo) f (x)=oo and function is twice differentiable. If f'(x) lt 0 AA x in (0,oo)and f'(0)=1 then number of solutions of equatin f (x)=x ^(2) is :

If f(x)=cos^(-1)(4x^3-3x)and lim_(xto1//2+)f'(x)=a and lim_(xto1//2-)f'(x)=b then a + b+ 3 is equal to

A function f(x) having the following properties, (i) f(x) is continuous except at x=3 (ii) f(x) is differentiable except at x=-2 and x=3 (iii) f(0) =0 lim_(x to 3) f(x) to - oo lim_(x to oo) f(x) =3 , lim_(x to oo) f(x)=0 (iv) f'(x) gt 0 AA in (-oo, -2) uu (3,oo) " and " f'(x) le 0 AA x in (-2,3) (v) f''(x) gt 0 AA x in (-oo,-2) uu (-2,0)" and "f''(x) lt 0 AA x in (0,3) uu(3,oo) Then answer the following questions Find the Maximum possible number of solutions of f(x)=|x|

Let f : R toR " be a real function. The function "f" is double differentiable. If there exists "ninN" and "p inR" such that "lim_(x to oo)x^(n)f(x)=p" and there exists "lim_(x to oo)x^(n+1)f(x), "then" lim_(x to oo)x^(n+1)f'(x) is equal to

For the function f(x) = 2 . Find lim_(x to 1) f(x)

Let f : R toR " be a real function. The function "f" is double differentiable. If there exists "ninN" and "p inR" such that "lim_(xto oo)x^(n)f(x)=p" and there exists "lim_(x to oo)x^(n+1)f(x), "then" lim_(x to oo) x^(n+2)f''(x) is equal to

f(x)=e^x then lim_(x rarr 0) f(f(x))^(1/{f(x)} is