FunctionHard
Question
If [2 cos x] + [sin x] = - 3, then the range of the function, f(x) = sin x + √3 cos x in [0, 2 π] is:
(where [. ] denotes greatest integer function)
(where [. ] denotes greatest integer function)
Options
A.[- 2, -1)
B.(- 2, - 1]
C.(- 2, -1)
D.[-2, -√3)
Solution
[2 cos x] + [sin x] = - 3
-2 ≤ 2 cos x ≤ 2 ⇒ [2 cos x] = 2, 1, - 1, - 2
- 1 ≤ sin x ≤ 1 ⇒ [sin x] = 1, 0, - 1
Equation holds true for [2 cos x] = - 2 and [sin x] = -1
⇒ - 1 ≤ cos x < -
and - 1 ≤ sin x < 0
⇒ x ∈
and x ∈ (π, 2π)
⇒ x ∈
f(x) = sin x + √3 cos x
f(x) = 2 sin
⇒ - 1 ≤ sin
⇒ - 2 ≤ 2 sin
< - √3
Hence range is [-2, -√3]
-2 ≤ 2 cos x ≤ 2 ⇒ [2 cos x] = 2, 1, - 1, - 2
- 1 ≤ sin x ≤ 1 ⇒ [sin x] = 1, 0, - 1
Equation holds true for [2 cos x] = - 2 and [sin x] = -1
⇒ - 1 ≤ cos x < -
and - 1 ≤ sin x < 0⇒ x ∈
and x ∈ (π, 2π)⇒ x ∈
f(x) = sin x + √3 cos x
f(x) = 2 sin
⇒ - 1 ≤ sin
⇒ - 2 ≤ 2 sin
< - √3Hence range is [-2, -√3]
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