Target Exam

CUET

Subject

Section B1

Chapter

Applications of Derivatives

Question:

Find the intervals in which the function $f$ given by $f(x) = \sin x + \cos x, 0 \leq x \leq 2\pi$ is increasing or decreasing.

Options:

Increasing in $[0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi]$ and decreasing in $(\frac{\pi}{4}, \frac{5\pi}{4})$

Increasing in $[\frac{\pi}{4}, \frac{3\pi}{4}]$ and decreasing in $[0, \frac{\pi}{4}] \cup [\frac{3\pi}{4}, 2\pi]$

Increasing in $[0, \frac{\pi}{2}]$ and decreasing in $[\frac{\pi}{2}, 2\pi]$

Increasing in $[\frac{\pi}{4}, \frac{5\pi}{4}]$ and decreasing in $[0, \frac{\pi}{4}] \cup [\frac{5\pi}{4}, 2\pi]$

Correct Answer:

Increasing in $[0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi]$ and decreasing in $(\frac{\pi}{4}, \frac{5\pi}{4})$

Explanation:

The correct answer is Option (1) → Increasing in $[0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi]$ and decreasing in $(\frac{\pi}{4}, \frac{5\pi}{4})$ ##

We have

$f(x) = \sin x + \cos x,$

$\text{or } \quad f'(x) = \cos x - \sin x$

Now $f'(x) = 0$ gives $\sin x = \cos x$ which gives that $x = \frac{\pi}{4}, \frac{5\pi}{4}$ as $0 \leq x \leq 2\pi$.

The points $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ divide the interval $[0, 2\pi]$ into three disjoint intervals, namely, $\left[0, \frac{\pi}{4}\right), \left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$ and $\left(\frac{5\pi}{4}, 2\pi\right]$.

Note that $f'(x) > 0$ if $x \in \left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right]$

or $f$ is increasing in the intervals $\left[0, \frac{\pi}{4}\right]$ and $\left[\frac{5\pi}{4}, 2\pi\right]$

Also $f'(x) < 0$ if $x \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$

or $f$ is decreasing in $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$