FunctionHard
Question
The range of the function f(x) = log√2 (2 - log2(16 sin2x + 1)) is
Options
A.(- ∞, 1)
B.(- ∞, 2)
C.(- ∞, 1]
D.(- ∞, 2]
Solution
Here (2 - log2 (16 sin2x + 1) > 0
⇒ 0 < 16 sin2x + 1 < 4
⇒ 0 ≤ sin2x < 3/16
⇒ 1 ≤ 16 sin2x + 1 ≤ 4
⇒ 0 ≤ log2 (16 sin2x + 1) < 2
⇒ 2 ≥ 2 - log2 (16 sin2x + 1) > 0
⇒ log√2 2 ≥ log√2 (2 - log2 (16 sin2x + 1)) > - ∞
⇒ 2 ≥ y > - ∞
Hence range is y ∈ (- ∞, 2]
⇒ 0 < 16 sin2x + 1 < 4
⇒ 0 ≤ sin2x < 3/16
⇒ 1 ≤ 16 sin2x + 1 ≤ 4
⇒ 0 ≤ log2 (16 sin2x + 1) < 2
⇒ 2 ≥ 2 - log2 (16 sin2x + 1) > 0
⇒ log√2 2 ≥ log√2 (2 - log2 (16 sin2x + 1)) > - ∞
⇒ 2 ≥ y > - ∞
Hence range is y ∈ (- ∞, 2]
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