Let `f(x)` be defined as `f(x)={tan^(-1)alpha-5x^2,0ltxlt1 and -6x ,xgeq1` if `f(x)` has a maximum at `x=1,` then find the values of `alpha` .

Let f:R rarr R be defined by, f(x)={[k-x;x le-1],[x^(2)+3,,xgt-1] .If f(x) has least value at x=-1 , then the possible value of k is

If f(x)={{:(,5 tan^(-1)alpha-3x^(2),0 lt x lt 1),(,5x-3,x ge 1):} then find the least integer value of alpha for which f(x) has minima at x=1.

Let f(x)={((alpha cotx)/x+beta/x^2,0<|x|lt=1),(1/3,x=0)) If f(x) is continuous at x=0 then the value of alpha^2+beta^2 is 1 b. 2 c. 5 d. 9

Let f : R to [0, pi/2) be defined by f ( x) = tan^(-1) ( 3x^(2) + 6x + a)". If " f(x) is an onto function . then the value of a si

Let f(x)={{:((alphacotx)/(x)+(beta)/(x^(2))",",0ltxle1),((1)/(3)",",x=0):} If f(x) is continuous at x =0, then the value of alpha^(2)+beta^(2) is

Let f(x)=(alpha x)/(x+1) Then the value of alpha for which f(f(x)=x is

Let f(x) be a function defined by f(x)=int_(1)^(x)t(t^(2)-3t+2)dt,1ltxlt3 then the maximum value of f(x) is

If f(x)=(tan x cot alpha)^((1)/(x-a)),x!=alpha is continuous at x=alpha, then the value of f(alpha) is