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
Create a free account to view solution
View Solution FreeMore Trigonometric Equation Questions
(cos α + cosβ)2 + (sin α + sin β)2 = .......Let f(n) = tan-1 + tan-1 + tan-1 + .... n terms, then f(n) is -...In ᐃABC, a = 3, b = 4 and c = 5, then value of sin A + sin2B + sin3C is -...If tan (π/4 + θ) + tan(π/4 - θ) = λ sec 2θ, then λ =...If A = cos2θ + sin4θ, then for all value of θ-...