Question

The function which is neither decreasing nor increasing in $\left(\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}\right)$ is:

• A. $\cos ecx$
• B. $\tan x$
• C. $x^2$
• D. $\left|x-1\right|$

Answer 1

The correct option is A

$\cos ecx$

Explanation for the correct option:

We can identify the function by drawing the graph of the given options between the given range.

The given range is $\left(\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}\right)$

Option (A): Draw the graph of $\cos ecx$

Here, in the graph we can clearly observe that the graph of $\cos ecx$ increasing in$\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)\cup \left(\mathrm{\pi },\frac{3\mathrm{\pi }}{2}\right)$.

But around $x=\mathrm{\pi }$

$f\left({\mathrm{\pi }}^{-}\right)>f\left({\mathrm{\pi }}^{+}\right)$

So, $\cos ecx$ is neither increasing nor decreasing in $\left(\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}\right)$.

Thus, option (A) $\cos ecx$ is the correct answer.

Explanation for incorrect options:

Option (B): Graph of $\tan x$ between $\left(\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}\right)$

As the graph shows, it is increasing between $\left(\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}\right)$.

So, the given option is incorrect.

Option (C): The given option is $x^2$

Now for positive value of $x$,$x^2$ is always increasing function.

As, $f\text{'}\left(x\right)=2x>0\forall x>0$

Thus, the function will be increasing. So, the given option is incorrect.

Option (D): For the given option, $\left|x-1\right|$,

For $x>1$,the function is increasing as $f\text{'}\left(x\right)=1$ and $x=\frac{\mathrm{\pi }}{2}>1$.Therefore fucntion will increase in the given interval.

Thus, option (D) is incorrect as well.

Therefore, the correct option is (A).