Trigonometric EquationHard
Question
Let 
(the set of all real numbers) be a positive, non-constant and differentiable function such that f′(x) < 2f(x) and 
Then the value of 
lies in the interval
(the set of all real numbers) be a positive, non-constant and differentiable function such that f′(x) < 2f(x) and Then the value of lies in the intervalOptions
A.(2e - 1, 2e)
B.(e - 1, 2e - 1)
C.
D.
Solution
f′(x) - 2 f(x) < 0
Multiply both side by e-2x
e-2x f′(x) - 2e-2x f(x) < 0
(e-2x f(x)) < 0
Now, g(x) = e-2x f(x)
∴ g(x) is a decreasing function.
⇒ e-2x f(x)<
⇒ f(x) < e2x-1
⇒
⇒
obviously f(x) is positive
∴
Multiply both side by e-2x
e-2x f′(x) - 2e-2x f(x) < 0
(e-2x f(x)) < 0 Now, g(x) = e-2x f(x)
∴ g(x) is a decreasing function.
⇒ e-2x f(x)<
⇒ f(x) < e2x-1
⇒
⇒
obviously f(x) is positive
∴
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