If `f(x) = |x + 1| - 2|x - 1|` then (A) maximum value of f(x) is 2 (C) there is `f(x)=2`. (B) there are two solutions of `f(x) = 1` (D) there are two solutions of `f(x) = 3.`

If f(x)=|x-1|+|x-2|-|x-3|, then (a) maximum value of f(x) is 3 if x in[2,3] (b) maximum value of f(x) is -3 if x in[1,3] (c) maximum value of f(x) is -3 if x in[1,2] (d) maximum value of f(x) is 3 if x in[1,3]

Let f(x)=2x+1. AA x , then the solution of the equation f(x)=f^(-1)(x) is

If f(x)=(x+1)/(x-1) , then the value of f(f(2)) is

If f(x)={0,x<1 and 2x-2,x<=1 then the number of solutions of the equation f(f(f(x)))=x is

If f(x)=ax^(2)+bx+c and f(x+1)=f(x)+x+1 , then the value of (a+b) is __

if f(x)=3|x|+|x-2| then f'(1) is equal to a)2 b)1 c)3 d)0

If f(x)=|x|-2|-2 for x in[-4,4] then the number of real solutions of f(f(x))=-1 is

If f(x) = | x | + [ x ] , then f( -3/2 ) + f( 3/2 ) is equal to : ( a ) 1 ( b ) -1/2 ( c ) 2 ( d ) 1/2

Let f (x) =(2 |x| -1)/(x-3) Range of the values of 'k' for which f (x) = k has exactly two distinct solutions: