Prove that the function f given by f(x) = log cos x is strictly decreasing on \((0,\frac{\pi}{2})\) and strictly increasing on \((\frac{\pi}{2},\pi)\).
Solution and Explanation
We have,
f(x)=log cos x
\(fx=\frac {1}{cosx} (-sinx)=-tan x\)
In interval \((0,\frac{\pi}{2})\), tan x>0=- tanx<0
f'(x)<0 on \((0,\frac{\pi}{2})\)
∴ f is strictly decreasing in \((0,\frac{\pi}{2})\)
In interval \((\frac{\pi}{2},\pi)\),tan x<0=- tanx>0
f'(x)>0 on \((\frac{\pi}{2},\pi)\)
∴f is strictly increasing in \((\frac{\pi}{2},\pi)\).
Top Questions on Applications of Derivatives
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Consider the following two statements :
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(II) \( g \) is decreasing in \( (\frac{\pi}{4}, \frac{\pi}{2}) \)
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Concepts Used:
Increasing and Decreasing Functions
Increasing Function:
On an interval I, a function f(x) is said to be increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≤ f(y)
Decreasing Function:
On an interval I, a function f(x) is said to be decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≥ f(y)
Strictly Increasing Function:
On an interval I, a function f(x) is said to be strictly increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) < f(y)
Strictly Decreasing Function:
On an interval I, a function f(x) is said to be strictly decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) > f(y)
Graphical Representation of Increasing and Decreasing Functions
