The function y = a(1 - cos x) is maximum when | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the function $y = a(1 - \cos x)$.First,find the derivative $y'$ with respect to $x$:$y' = \frac{d}{dx}[a(1 - \cos x)] = a(0 - (-\sin x)) = a

Vedclass · Accepted Answer

$\pi $

Vedclass · Answer

(A) Given the function $y = a(1 - \cos x)$.First,find the derivative $y'$ with respect to $x$:$y' = \frac{d}{dx}[a(1 - \cos x)] = a(0 - (-\sin x)) = a \sin x$.To find the critical points,set $y' = 0$:$a \sin x = 0 \Rightarrow \sin x = 0 \Rightarrow x = n\pi$ for any integer $n$.Within the standard interval $[0, 2\pi]$,the critical points are $x = 0$ and $x = \pi$.Now,find the second derivative $y''$:$y'' = \frac{d}{dx}(a \sin x) = a \cos x$.Evaluate $y''$ at the critical points:For $x = 0$,$y''(0) = a \cos(0) = a$.For $x = \pi$,$y''(\pi) = a \cos(\pi) = -a$.Assuming $a > 0$,the function has a local maximum when $y'' < 0$,which occurs at $x = \pi$.

The function $y = a(1 - \cos x)$ is maximum when $x = $

Similar Questions

If $f(x) = x + \frac{1}{x}$,$x \neq 0$,then the local maximum and minimum values of the function $f$ are respectively....

Consider a cuboid of sides $2x$,$4x$,and $5x$ and a closed hemisphere of radius $r$. If the sum of their surface areas is a constant $k$,then the ratio $x:r$,for which the sum of their volumes is maximum,is

The minimum value of the function $f(x) = 2 x^3 - 15 x^2 + 36 x - 48$ on the set $A = \{x \mid x^2 + 20 \le 9 x\}$ is

The vertices of a triangle are $(0, 0)$,$(x, \cos x)$,and $(\sin^3 x, 0)$ where $0 < x < \frac{\pi}{2}$. The maximum area for such a triangle in sq. units is:

If $x$ is real,then the minimum value of $\frac{x^2-x+1}{x^2+x+1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module