Set, Relation and FunctionHard
Question
Let f be a real-valued function defined on the interval (0, ∞) by f(x) = ln x + 
. Then which ofthe following statement(s) is (are) true?
. Then which ofthe following statement(s) is (are) true?Options
A.f″(x) exists for all x ∈ (0, ∞)
B.f′(x) exists for all x ∈ (0, ∞) and f′ is continuous on (0, ∞) but not differentiable on (0, ∞)
C.there exists α > 1 such that |f′(x)| < |f(x)| for all x ∈ (α, ∞)
D.there exists β > 0 such that |f(x)| + |f′(x)| ≤ β for all x ∈ (0, ∞)
Solution
f′(x) = 
f′(x) is not differentiable at sinx = −1 or x = 2nπ -
n ∈ N
In x ∈ (1, ∞) f(x) > 0, f′(x) > 0
Consider f(x) - f′(x)
= ln x +
Consider g(x) =
It can be proved that g(x) ≥ 2√2 - √10 ∀ x ∈ (0, ∞)
Now there exists some α > 1 such that
- ln x ≤ 2√2 - √10 for all x ∈ (α, ∞) as - ln x is strictly decreasing function.
⇒ g(x) ≥ - ln x.
f′(x) is not differentiable at sinx = −1 or x = 2nπ -
n ∈ NIn x ∈ (1, ∞) f(x) > 0, f′(x) > 0
Consider f(x) - f′(x)
= ln x +
Consider g(x) =
It can be proved that g(x) ≥ 2√2 - √10 ∀ x ∈ (0, ∞)
Now there exists some α > 1 such that
- ln x ≤ 2√2 - √10 for all x ∈ (α, ∞) as - ln x is strictly decreasing function.⇒ g(x) ≥
- ln x.Create a free account to view solution
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