Let `f(x)=5-|x-2| and g(x)=|x+1|, x in R`. If f(x)n artains maximum value at `alpha` ang g(x) attains minimum value of `beta`, then `lim_(xto-alpha beta) ((x-1)(x^(2)-5x+6))/(x^(2)-6x+8)` is equal to

Let f(x)=5-|x-2| and g(x)=|x+1|, x in R . If f(x)n attains maximum value at alpha and g(x) attains minimum value of beta , then lim_(xto-alpha beta) ((x-1)(x^(2)-5x+6))/(x^(2)-6x+8) is equal to

Let f(x)=5-|x-2| and g(x)=|x+1|, x in R . If f(x) attains maximum value at alpha ang g(x) attains minimum value of beta , then underset(xto-alpha beta)lim((x-1)(x^(2)-5x+6))/(x^(2)-6x+8) is equal to (a) -1/2 (b) -3/2 (c) 1/2 (d) 3/2

Let f(x)=5-[x-2]g(x)=[x+1]+3 If maximum value of f(x) is alpha& minimum value of f(x) is beta then lim_(x rarr(alpha-beta))((x-3)(x^(2)-5x+6))/((x-1)(x^(2)-6x+8)) is

Let f(x)=5-[x-2] g(x)=[x+1]+3 If maximum value of f(x) is alpha & minimum value of f(x) is beta then underset(xrarr(alpha-beta))lim ((x-3)(x^(2)-5x+6))/((x-1)(x^(2)-6x+8)) is (A) -1/2 (B)1/2 (C)3/2 (D)-3/2

Let alpha (x) = f(x) -f (2x) and beta (x) =f (x) -f (4x) and alpha '(1) =5 alpha'(2) =7 then find the vlaue of beta'(1)-10

Let f(x)=|x-2| and g(x)=|3-x| and A be the number of real solutions of the equation f(x)=g(x),B be the minimum value of h(x)=f(x)+g(x),C be the area of triangle formed by f(x)=|x-2|,g(x)=|3-x| and x and x- axes and alpha

If minimum value of function f(x)=(1+alpha^(2))x^(2)-2 alpha x+1 is g(alpha) then range of g(alpha) is