Trigonometric EquationHard
Question
For function f(x) = x cos 
, x ≥ 1,
, x ≥ 1,Options
A.for atleast one x in interval [1, ∞), f(x + 2) - f(x) < 2
B.
f′(x) = 1
f′(x) = 1C.for all x in the interval [1, ∞), f(x + 2) - f(x) > 2
D.f′(x) is strictly decreasing in the interval [1, ∞)
Solution
For f(x) = x cos 
, x ≥ 1
f′(x) = cos
→ 1 for x → ∞
also f″(x) =
=
< 0 for x ≥ 1
⇒ f′(x) is decreasing for [1, ∞)
⇒ f′(x + 2) < f′(x). Also,
f(x + 2)- f(x) = 
∴ f(x + 2) - f(x) > 2 ∀ x ≥ 1
, x ≥ 1f′(x) = cos
→ 1 for x → ∞ also f″(x) =
=
< 0 for x ≥ 1 ⇒ f′(x) is decreasing for [1, ∞)
⇒ f′(x + 2) < f′(x). Also,
f(x + 2)- f(x) = ∴ f(x + 2) - f(x) > 2 ∀ x ≥ 1
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