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Increasing and Decreasing Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIncreasing and Decreasing Functions5 Marks
Question
Show that $\text{f}(\text{x})=\cos\text{x}$ is a decreasing function on $(0,\pi),$ increasing in $(-\pi,0)$ and neither increasing nor decreasing in $(-\pi,\pi).$
✓

Answer

$\text{f}(\text{x})=\cos\text{x}$
Domain of $\cos\text{x}$ is $(-\pi,\pi).$
$\Rightarrow\text{f}'(\text{x})=-\sin\text{x}$
For $\text{x}\in(-\pi,0),\sin\text{x}<0$
$[\because$ Sine function is negative in third and fourth quadrant$]$
$\Rightarrow-\sin\text{x}>0$
$\Rightarrow\text{f}'(\text{x})>0$
So, $\cos\text{x}$ is increasing in $(-\pi,0).$
For $\text{x}\in(0,\pi),\sin\text{x}>0$
$[\because$ Sine function is positive in first and second quadrant$]$
$\Rightarrow-\sin\text{x}<0$
$\Rightarrow\text{f}'(\text{x})<0$
So, f(x) is decreasing on $(0,\pi).$
Thus, f(x) is neither increasing nor decreasing in $(-\pi,\pi).$

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