Question

The function $f\left(x\right)=\sin x+\cos x$ is increasing in the interval

• A. $\left[\frac{3\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right]$
• B. $\left[0,\frac{3\mathrm{\pi }}{4}\right]$
• C. $\left[\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}\right]$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Step 1: Solve for the critical points

Given function is $f\left(x\right)=\sin x+\cos x$

We know that a function is increasing at $x$ if $f\text{'}\left(x\right)>0$

We have,
$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(\sin x+\cos x\right)\\ & =& \cos x-\sin x\end{array}$

The critical points are when $f\text{'}\left(x\right)=0$. So,
$\begin{array}{cc}& \cos x-\sin x=0\\ \Rightarrow & \cos x=\sin x\\ \Rightarrow & \cos x=\cos \left(\frac{\mathrm{\pi }}{2}-x\right)\\ \Rightarrow & x=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2}-\mathrm{x}\\ \Rightarrow & x=\mathrm{n\pi }+\frac{\mathrm{\pi }}{4},\mathrm{n}=0,\pm 1,\pm 2,\dots \end{array}$

Thus the critical values of $f\left(x\right)$ are $\frac{\mathrm{\pi }}{4}$, $\frac{5\mathrm{\pi }}{4}$, $\frac{9\mathrm{\pi }}{4}$,… (by substituting the values for $\mathrm{n}$)

Step 2: Solve for the interval in which the given function is increasing in nature

A function is increasing between two successive critical points if the function has local minima at the smaller critical point and local maxima at the larger one.
a function has local maxima at a point if $f\text{'}\text{'}\left(a\right)<0$ at point $a$

So we have,
$\begin{array}{rcl}f\text{'}\text{'}\left(x\right)& =& \frac{d}{dx}f\text{'}\left(x\right)\\ & =& \frac{d}{dx}\left(\cos x-\sin x\right)\\ & =& -\sin x-\cos x\end{array}$

At $x=\frac{\mathrm{\pi }}{4}$,
$\begin{array}{rcl}f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{4}\right)& =& -\sin \left(\frac{\mathrm{\pi }}{4}\right)-\cos \left(\frac{\mathrm{\pi }}{4}\right)\\ & =& -\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\\ & =& -\sqrt{2}<0\end{array}$

Thus the function has local maxima at $x=\frac{\mathrm{\pi }}{4}$.

None of the options have $\frac{\mathrm{\pi }}{4}$ as their upper bound. So $f\left(x\right)$ increases at none of these intervals.

Hence, option(D) is correct.