The absolute maximum value of f(x) = sin x + | vedclass

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

Question

(D) To find the absolute maximum value of $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$,we first find the critical points by setting the derivat

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(D) To find the absolute maximum value of $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$,we first find the critical points by setting the derivative to zero.$f'(x) = \cos x - \sin x$.Setting $f'(x) = 0$,we get $\cos x = \sin x$,which implies $	an x = 1$.For $x \in [0, \pi]$,the solution is $x = \pi/4$.Now,we evaluate $f(x)$ at the critical point and the endpoints of the interval:$1$. At $x = 0$: $f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$.$2$. At $x = \pi/4$: $f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = 1/\sqrt{2} + 1/\sqrt{2} = 2/\sqrt{2} = \sqrt{2}$.$3$. At $x = \pi$: $f(\pi) = \sin(\pi) + \cos(\pi) = 0 - 1 = -1$.Comparing these values,the absolute maximum value is $\sqrt{2}$.

The absolute maximum value of $f(x) = \sin x + \cos x$ for $x \in [0, \pi]$ is . . . . . . .

Similar Questions

An extreme value of $f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$ in $(0, \pi / 2)$ is

$A$ particle moving in a straight line starts from rest and the acceleration at any time $t$ is $a - kt^2$, where $a$ and $k$ are positive constants. The maximum velocity attained by the particle is

The greatest value of the function $f(x) = -1 + \frac{2}{2^{x^2} + 1}$ is:

The maximum area of a rectangle that can be inscribed in a circle of radius $2 \text{ unit}$ is (in square unit)

Vedclass Test Series

Exam Paper Generator

Online Exam Module