The function `f: RvecR` is defined by `f(x)=cos^2x+sin^4xdotT h e n ,f(R)=` `[3//4,1)` (b) `(3//4,1]` (c) `[3//4,1]` (d) (`3//4,1)`

The function f:Rvec R is defined by f(x)=cos^(2)x+sin^(4)x Then ,f(R)=[3/4,1)(b)(3/4,1](c)[3/4,1](d)(3/4,1)

The function f: R ⇒ R is defined by f(x)=cos^2x+sin^4x , r a n g e o f f(x)= (a) [3//4,1) (b) (3//4,1] (c) [3//4,1] (d) ( 3//4,1)

The function f: R-R is defined by f(x)=cos^2x+sin^4xforx in Rdot Then the range of f(x) is (3/4,1] (b) [3/4,1) (c) [3/4,1] (d) (3/4,1)

The function f: R-R is defined by f(x)=cos^2x+sin^4xforx in Rdot Then the range of f(x) is (3/4,1] (b) [3/4,1) (c) [3/4,1] (d) (3/4,1)

The function f: R-R is defined by f(x)=cos^2x+sin^4xforx in Rdot Then the range of f(x) is (3/4,1] (b) [3/4,1) (c) [3/4,1] (d) (3/4,1)

The function f:R-R is defined by f(x)=cos^(2)x+sin^(4)xf or x in R. Then the range of f(x) is ((3)/(4),1] (b) [(3)/(4),1) (c) [(3)/(4),1] (d) ((3)/(4),1)

If the function f:R rarr R defined by f(x) = 3x – 4 is invertible, find f^(-1)

The function f:R rarr R defined as f(x)=(3x^2+3x-4)/(3+3x-4x^2) is :

Let A={x in R : x!=0,\ -4lt=xlt=4} and f: A->R be defined by f(x)=(|x|)/x for x in Adot Then \ f is a. {-1, 1} b. (3//4,1] c. [3//4,1] d. (3//4,1)

If f(x)=sin^6x+cos^6x , then range of f(x) is [1/4,1] (b) [1/4,3/4] (c) [3/4,1] (d) none of these