The range of the function fx-log-5 3-cos 3-4 -x -cos -4-x -cos -4-x -cos 3-4 -x is

F(x)=log_√(5) ( 3+2sin ( 3π4 )sin( x)+2cos ( π4 ) ·cos (x) ) =log_√(5)(3+√(2)(cos x sin x)) ∵ Range of cos x – sin x is nbsp;[ √(2),√(2)] Then range of f(x) i ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

The range of ...

Question

The range of the function
$f (x) = {log}_{\sqrt{5}} (3 + cos (\frac{3 π}{4} + x) + cos (\frac{π}{4} + x) + cos (\frac{π}{4} - x) - cos (\frac{3 π}{4} - x))$
is

[0, 2]

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[- 2, 2]

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(0, \sqrt{5})

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

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Solution

The correct option is A $[0, 2]$
$f (x) = {log}_{\sqrt{5}} (3 + 2 sin (\frac{3 π}{4}) sin (- x) + 2 cos (\frac{π}{4}) \cdot cos (x))$
$= {log}_{\sqrt{5}} (3 + \sqrt{2} (cos x - sin x))$
$∵$ Range of $cos x - sin x$ is $[- \sqrt{2}, \sqrt{2}]$
Then range of $f (x)$ is $[0, 2]$

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Similar questions

Q. Find

(cos (\frac{π}{4} + x) + cos (\frac{π}{4} - x) - \sqrt{2} cos x)

(cos (\frac{3 π}{4} + x) - cos (\frac{3 π}{4} - x) + \sqrt{2} sin x)

Prove that

$(i) cos (\frac{π}{4} + x) + cos (\frac{π}{4} - x) = \sqrt{2} cos x$

(ii) $cos (\frac{3 π}{4} + x) - cos (\frac{3 π}{4} - x) = - \sqrt{2} sin x$

cos (\frac{3 π}{4} + x) - cos (\frac{3 π}{4} - x) = \sqrt{2} . sin x

Q. Prove the following:

c o s (\frac{3 π}{4} + x) - c o s (\frac{3 π}{4} - x) = - \sqrt{2} s i n x

Q. Prove that:

cos (\frac{3 π}{4} + x) - cos (\frac{3 π}{4} - x) = - \sqrt{2} sin x

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The range of the function f(x)=log√5(3+cos(3π4+x)+cos(π4+x)+cos(π4−x)−cos(3π4−x)) is

The correct option is A [0,2]f(x)=log√5(3+2sin(3π4)sin(−x)+2cos(π4)⋅cos(x)) =log√5(3+√2(cosx−sinx)) ∵ Range of cosx–sinx is [−√2,√2] Then range of f(x) is [0,2]

The range of the function
$f (x) = {log}_{\sqrt{5}} (3 + cos (\frac{3 π}{4} + x) + cos (\frac{π}{4} + x) + cos (\frac{π}{4} - x) - cos (\frac{3 π}{4} - x))$
is

The correct option is A $[0, 2]$
$f (x) = {log}_{\sqrt{5}} (3 + 2 sin (\frac{3 π}{4}) sin (- x) + 2 cos (\frac{π}{4}) \cdot cos (x))$
$= {log}_{\sqrt{5}} (3 + \sqrt{2} (cos x - sin x))$
$∵$ Range of $cos x - sin x$ is $[- \sqrt{2}, \sqrt{2}]$
Then range of $f (x)$ is $[0, 2]$