If `f(x) = { 3 + |x-k|, x leq k; a^2 -2 + sin(x-k)/(x-k), xgtk}` has minimum at x =k, then show that `|a| >2`.

f (x)= {{:(3+|x-k|"," , x le k),(a ^(2) -2 + ( sin (x -k))/((x-k))"," , x gtk):} has minimum at x =k, then:

If f(x)={3+|x-k|,xlt=k a^2-2+(s n(x-k))/(x-k),x > k has minimum at x=k , then a in R b. |a| 2 d. 1<|a|<2

If f(x)={3+|x-k|,xlt=k a^2-2+(s n(x-k))/(x-k),x > k has minimum at x=k , then a in R b. |a| 2 d. 1<|a|<2

If f(x)={([3+|x-k| ,x k]) has minimum at x=k,then (a )|a| 2 (d) |a|>=2

If f(x)={([3+|x-k| ,x k]) has minimum at x=k,then (a )|a| 2 (d) |a|>=2

If f(x) =x-k/x has a maximum value at x=-2 , then k=

if f(x) = {underset(a^(2) -2+ (sin (x-K))/(x-K))(3+|x-k|)," "underset(x" "gt" "k)(x" "le" "k) has minimum at x=k , then show that |a| gt2

If f(x)=|x-7|+|x-10|+|x-12| has a minimum at x=k, then the value of k is

If f(x)=|x-2|-|x-4|, then f(x)=k has