The range of the function f x -cos 2 x -4-sin x -4- x - R isA- -0- 5-4-1- 5-4-C- -1- 5-4D- -1- 5-4-

The correct option is D f(x)=1 sin^2 x/4+sinx/4= { sin^2x/4 sinx/4}+1 = { ( sin x/4 1/2 )^2 1/4}+1= 5/4 ( sin x/4 1/2 )^2 Maximum f(x)=5/4 Minimum f(x) ...

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Byju's Answer

Standard XI

Mathematics

Sum of n Terms in GP

The range of ...

Question

$T h e r a n g e o f t h e f u n c t i o n f (x) = c o s^{2} \frac{x}{4} + s i n \frac{x}{4}, x ϵ R i s$

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Solution

The correct option is D

$f (x) = 1 - s i n^{2} \frac{x}{4} + s i n \frac{x}{4} = - {s i n^{2} \frac{x}{4} - s i n \frac{x}{4}} + 1 = - {{(s i n \frac{x}{4} - \frac{1}{2})}^{2} - \frac{1}{4}} + 1 = \frac{5}{4} - {(s i n \frac{x}{4} - \frac{1}{2})}^{2} M a x i m u m f (x) = \frac{5}{4} M i n i m u m f (x) = \frac{5}{4} = {(- 1 - \frac{1}{2})}^{2} = \frac{5}{4} - \frac{9}{4} = - 1 R a n g e o f f (x) = [- 1, \frac{5}{4}]$

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Similar questions

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

Q. The range of the function

f (x) = c o s^{2} \frac{x}{4} + s i n \frac{x}{4}, x ϵ R

Q. The range of

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

x \in R - {- 1, 5}

Q. If both sum of roots and product of roots of

x^{2} - (λ^{2} - 5 λ + 5) x + (2 λ^{2} - 3 λ - 4) = 0

are less than

1,

then the range of

λ

Q. The values of

k

for which the equation

∣ ∣ | x - 2 | - 5 ∣ ∣ = k

has

4

distinct solutions is

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

The correct option is D f(x)=1−sin2x4+sinx4=−{sin2x4−sinx4}+1=−{(sinx4−12)2−14}+1=54−(sinx4−12)2Maximum f(x)=54Minimum f(x)=54=(−1−12)2=54−94=−1Range of f(x)=[−1,54]

$T h e r a n g e o f t h e f u n c t i o n f (x) = c o s^{2} \frac{x}{4} + s i n \frac{x}{4}, x ϵ R i s$