If `f(x) = {[x] +sqrt{x}, x lt 1;1/([x]+{x}^2) xgeq1}` then

If f(x)={[x]+sqrt({x}),x<1 1/([x]+{x}^2),xgeq1 , then [where [.] and {.] represent the greatest integer and fractional part functions respectively] f(x) is continuous at x=1 f(x) is not continuous at x=1 f(x) is differentiable at x=1 (lim)_(xvec1)f(x) does not exist

If f(x)={[x]+sqrt({x}),x<1 1/([x]+{x}^2),xgeq1 , then [where [.] and {.] represent the greatest integer and fractional part functions respectively] f(x) is continuous at x=1 f(x) is not continuous at x=1 f(x) is differentiable at x=1 (lim)_(xvec1)f(x) does not exist

If f(x)= {(|1-4x^2|,; 0 lt= x lt 1), ([x^2-2x],; 1 lt= x lt 2):} , where [.] denotes the greatest integer function, then f(x) is

The function f(x) = min {|x|, sqrt(1-x^(2))}, -1lt x lt 1 possesses

If f (x) = {{:((sqrt(1 + ax )- sqrt(1- ax ))/(x)"," -1 le x lt 0), ((x ^(2) +2)/(x-2)"," 0 le x le 1):} continuous on [-1,1], then a =

f(x)=x+1, x<0, f(x)=x^2, xgeq0 and g(x)={x^3, x<1, 2x-1, xgeq1 Then find f(g(x)) and find its domain and range.

If f (x) ={{:(x",", "when " 0 lt x lt 1),(2-x",", "when" 1 lt x le 2):} then f'(1) is equal to-

f(x)={x+1, x<0,x^2, xgeq0 and g(x)={x^3, x<1, 2x-1, xgeq1 Then find f(g(x)) and find its domain and range.

If f(x) = sqrt(x-1) + sqrt(x+24 -10sqrt(x-1)), 1 lt x lt 26 be real valued function, then 'f' (x) for 1 lt x lt 26 is: