A function `y=\ f(x)` is defined on [0,4] as `f(x)={2x\ \ \ \ if\ 0lt=xlt=1(x-2)^2\ \ \ \ \ if\ \ \ \ 1

The points of discontinuity of the function f(x)={2sqrt(x)\ \ \ ,\ \ \ 0lt=xlt=1 ,\ \ \4-2x\ \ \ ,\ \ \ 1< x

The function f(x) is defined by f(x)={x^2+a x+b\ \ \ ,\ \ \ 0lt=x 4 f is continuous them determine the value of a and b

Let f be an even function defined on R. Suppose f is defined on [0,4] as follows: f(x) = {{:(3x, if, 0 le x lt 1),(4-x, if, 1 le x le 4):} and f satisfies the condition. f(x-2) = f{x + [(6x^(2) + 272)/(x^(2) + 2)]) AA x in R Where [x] = greatest integer le x . Then f(3271) +f(-2052) + f(806) =..............

Discuss the continuity of the function f, where f is defined by : f(x)={{:(2x", if "xlt0),(0", if "0lexle1),(4x", if "xgt1):}

f:[0,\ 1]->R is defined as f(x)=[x^3(1-x)sin(1/(x^2))\ \ \ if\ 0

Consider function f(x)={x{x}+1,\ \ \ \ \ \ \ \ \ \ \ \ 0lt=x<1 2-{x},\ \ \ \ 1lt=xlt=2\ \ \ where {x}: fractional part function of x . Statement-1: Rolles Theorem is not applicable to f(x) in [0,2] Statement-2: f(0)\ !=f(2)

A function f is defined by f(x)=\ int_0^picostcos(x-t)dt ,\ 0lt=xlt=2pi then find its minimum value.

The function f : [0,oo)to[0,oo) defined by f(x)=(2x)/(1+2x) is

A function y=f(x) satisfying f'(x)=x^(-(3)/(2)),f'(4)=2 and f(0)=0 is

If a function f(x) is defined as f(x) = {{:(-x",",x lt 0),(x^(2)",",0 le x le 1),(x^(2)-x + 1",",x gt 1):} then