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प्रश्न
Find the intervals in which the function `f("x") = (4sin"x")/(2+cos"x") -"x";0≤"x"≤2pi` is strictly increasing or strictly decreasing.
उत्तर
Given `f("x") = (4sin"x")/(2+cos"x") -"x";0≤"x"≤2pi`
⇒ `f("x") = ((2+cos"x")4 cos "x" + 4 sin^2 "x")/(2+ cos"x")^2 - 1`
⇒ `f("x") = (8 cos "x" + 4(sin^2 "x" + cos^2 "x")-4-cos^2 "x"-4cos"x")/((2+ cos"x"))`
⇒ `f("x") = (4cos "x"- cos^2 "x")/((2 + cos "x")^2`
For critical points, `f("x") = [(4- cos"x")/((2+ cos "x")^2]] cos "x" = 0`
f(x) is strictly increasing for f'(x) > 0
i.e., cos x > 0 ⇒ x ∈ `[0, pi/2), ∪ (3pi/2, 2pi ]`
and f(x) is strictly decreasing for f'(x) < 0
i.e., cos x < 0 ⇒ x ∈
Interval | Sign of f (x) | f (x) is strictly |
(0,π/2) | Positive | Increasing |
(π/2, 3π/2) | Negative | Decreasing |
(3π/2,2π) | Positive | Increasing |
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