The range of the function f ( x ) = c o s 2 x 4 + s i n x 4 , x ϵ R is Options: A . &nbsp [ 0 , 5 4 ] B . &nbsp [ 1 , 5 4 ] C . &nbsp ( − 1 , 5 4 ) D . &nbsp [ − 1 , 5 4 ]

Answer: Option D : D f ( x ) = 1 − s i n 2 x 4 + s i n x 4 = − { s i n 2 x 4 − s i n x 4 } + 1 = − { ( s i n x 4 − 1 2 ) 2 − 1 4 } + 1 = 5 4 − ( s i n x 4 − 1 2 ) 2 Maximun f ( x ) = 5 4 Minimum f ( x ) = 5 4 − ( − 1 − 1 2 ) 2 = 5 4 − 9 4 = − 1 Range of f ( x ) = [ − 1 , 5 4 ]

The range of the function f(x)=cos2x4+sinx4,x ϵ R is

Discussion Forum : Relations And Functions Ii

Question -

The range of the function $f (x) = c o s^{2} \frac{x}{4} + s i n \frac{x}{4}, x ϵ R$ is

Options:

A .

[0, \frac{5}{4}]

B .

[1, \frac{5}{4}]

C .

(- 1, \frac{5}{4})

D .

[- 1, \frac{5}{4}]

Answer: Option D
:
D

f (x) = 1 - s i n^{2} \frac{x}{4} + s i n \frac{x}{4} = - {s i n^{2} \frac{x}{4} - s i n \frac{x}{4}} + 1 = - {{(s i n \frac{x}{4} - \frac{1}{2})}^{2} - \frac{1}{4}} + 1

= \frac{5}{4} - {(s i n \frac{x}{4} - \frac{1}{2})}^{2}

Maximun

f (x) = \frac{5}{4}

Minimum

f (x) = \frac{5}{4} - {(- 1 - \frac{1}{2})}^{2} = \frac{5}{4} - \frac{9}{4} = - 1

Range of

f (x) = [- 1, \frac{5}{4}]

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