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` c. `|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|,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

f(x)={3+|x-k|,x k} has minimum at x=k, then show that |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

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) = {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

Let f:""R vec R be defined by f(x)={k-2x , if""xlt=-1 (-2x+3),x >-1} . If f has a local minimum at x=-1 , then a possible value of k is (1) 0 (2) -1/2 (3) -1 (4) 1

Let f:""R to R be defined by f(x)""={k-2x , if""xlt=-1," "2x+3,if""x > -1} . If f has a local minimum at x""=""1 , then a possible value of k is