Q. 115.0( 1 Vote )
Show that f(x) =
Given:- Function f(x) = cos2 x
Theorem:- Let f be a differentiable real function defined on an open interval (a,b).
(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)
(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)
(i) Obtain the function and put it equal to f(x)
(ii) Find f’(x)
(iii) Put f’(x) > 0 and solve this inequation.
For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.
Here we have,
f(x) = cos2 x
⇒ f’(x) = 3cosx(–sinx)
⇒ f’(x) = –2sin(x)cos(x)
⇒ f’(x) = –sin2x ; as sin2A = 2sinA cosA
Now, as given
⇒ 2x ∈ (0,π)
⇒ Sin(2x)> 0
⇒ –Sin(2x)< 0
⇒ f’(x) < 0
hence, Condition for f(x) to be decreasing
Thus f(x) is decreasing on interval
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