The interval on which `f(x)=sqrt(1-x^2)` is continuous in

What is the interval in which the function f(x) = sqrt(9-x^(2)) is increasing ? (f(x)gt0)

If f(x) = {((1-cos 4x)/x^(2), x lt 0),(a, x =0),((sqrtx)/(sqrt(16+sqrt(x))-4), xgt 0):} then the value of a for which f(x) is continuous at x = 0 is

The value of f(0) so that f(x)=(sqrt(1+x)-1)/(x) is continuous at x=0 is

Find the value of f (0) for which f(x)=(64(sqrt(x+4)-2))/(sin2x) continuous .

f : R^(+)to R^(+), f(x)=(log x)/(sqrt(x)) . Find the intervals in which f(x) is increasing or decreasing.

f(x)= [x] + sqrt(x -[x]) , where [.] is a greatest integer function then …….. (a) f(x) is continuous in R+ (b) f(x) is continuous in R (C) f(x) is continuous in R - 1 (d) None of these

If f and g are continuous and real valued function on an interval, then - (i) f+g is continuous on the interval, (ii) fg is continuous on the interval, (iii) (f)/(g) is continuous on the interval. which of the statements given above are correct ?