हिंदी

Find the Intervals in Which the Function F ( X ) = 4 Sin X 2 + Cos X − X ; 0 ≤ X ≤ 2 π is Strictly Increasing Or Strictly Decreasing.

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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. 

योग
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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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2015-2016 (March) All India Set 1 E

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