Maths. Section B1Applications of Derivatives

The function f(x)=sin x+cos x;\ 0≤ x≤2π is(A) Increasing in ≤ft(0,(π)…

The function $ f(x)=\sin x+\cos x;\ 0\le x\le2\pi $ is

(A) Increasing in $ \left(0,\displaystyle\frac{\pi}{4}\right) $

(B) Decreasing in $ (0,\pi) $

(D) Decreasing in $ \left(\displaystyle\frac{\pi}{4},\displaystyle\frac{5\pi}{4}\right) $

(E) Neither increasing nor decreasing in $ \left(0,\displaystyle\frac{\pi}{4}\right) $

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 3

The correct answer is Option (3) → (A), (C) and (D) only

$ f'(x)=\cos x-\sin x $

In $ [0,2\pi] $, $ f'(x)=0 $ when $ \tan x=1 $:

$ \displaystyle x=\frac{\pi}{4},\ \frac{5\pi}{4} $

The sign of f′ at one point of each part:

$ \displaystyle f'\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}-1}{2}\gt0 $

$ \displaystyle f'\left(\frac{\pi}{2}\right)=-1\lt0 $

$ \displaystyle f'\left(\frac{3\pi}{2}\right)=1\gt0 $

So f is increasing on $ \left(0,\frac{\pi}{4}\right) $, decreasing on $ \left(\frac{\pi}{4},\frac{5\pi}{4}\right) $ and increasing on $ \left(\frac{5\pi}{4},2\pi\right) $.

Checking the statements

(A) Increasing in $ \left(0,\frac{\pi}{4}\right) $ — Correct. $ f'\gt0 $ there.

(B) Decreasing in $ (0,\pi) $ — Incorrect. f increases on $ \left(0,\frac{\pi}{4}\right) $.

(C) Increasing in $ \left(\frac{5\pi}{4},2\pi\right) $ — Correct. $ f'\gt0 $ there.

(D) Decreasing in $ \left(\frac{\pi}{4},\frac{5\pi}{4}\right) $ — Correct. $ f'\lt0 $ there.

(E) Neither increasing nor decreasing in $ \left(0,\frac{\pi}{4}\right) $ — Incorrect. f is increasing there.

NCERT (2026-27 edition):
Increasing and Decreasing Functions

$NCERT extract$