Let `f(x) ={x^2-2|x|+a,x leq 1 and 6+x,xlt 1` number of positive integral value(s) of 'a' forwhich `f(x)` has local minima at `x=1` is/are

Given f(x)={(x^(2)-4|x|+a ,if, x le 2),(6-x, if, x gt 2):} ,then number of positive integral values of a for which f(x) has local minima at x=2 .

Let f(x)={(|x-2|+a^2-9a-9 ,ifx lt 2, 2x-3 ,ifx ge 2:} Then find value of 'a' for which f(x) has local minima at x= 2

The number of integral values of x satisfying |x^2-1|leq3 is

Let f(x) ={underset(x^(2) +xb " " x ge1)(3-x " "0le x lt1). Find the set of values of b such that f(x) has a local minima at x=1.

Let f(x) ={underset(x^(2) +xb " " x ge1)(3-x " "0le x lt1). Find the set of values of b such that f(x) has a local minima at x=1.

Let f(x)={1x-2l+a^(2)-9a-9 if x =2 if f(x) has local minimum at x=2, then

let f(x)=(x^2-1)^n (x^2+x-1) then f(x) has local minimum at x=1 when

Number of integral value of x, satisfying |x-1|+|x-2|+|x-3|leq6 is equal to :