Question:

The function $ f(x) = cos^2x $ is strictly decreasing on

Updated On: Jun 14, 2022
  • $ \left[0,\frac{\pi}{2}\right] $
  • $ \left[0,\frac{\pi}{2}\right) $
  • $ \left(0,\frac{\pi}{2}\right) $
  • $ \left(0,\frac{\pi}{2}\right] $
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

We have, $f(x) = cos^2\, x$
On differentiating both sides w.r.t. ?$x$?, we get
$f'(x) = -2 \,cosx \, sinx = -sin2x$
$\because -1 \le sin\,2x \le 1$
$\therefore f'(x) > 0 $ for $x \in (\frac{-\pi}{2} , 0)$
and $f'(x) < 0 $ for $ x\in (0, \frac{\pi}{2})$
$\therefore f(x) = cos^2\,x$ is strictly decreasing on $( 0, \frac{\pi}{2})$.
Was this answer helpful?
0
0

Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives