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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES3 Marks
Question
Show that the function given by f(x) = sin x is:
  1. strictly increasing $\bigg(0,\frac{\pi}{2}\bigg),$
  2. strictly decreasing in $\bigg(\frac{\pi}{2},\pi\bigg),$
  3. neither increasing nor decreasing in $(0,\pi).$
✓

Answer

Given: f(x) = sin x
$\therefore$ f(x) = cos x
  1. Since, f'(x) = cos x > 0, i.e., positive in first quadrant, i.e., in $\bigg(0,\frac{\pi}{2}\bigg).$
Therefore, f(x) is strictly increasing in $\bigg(0,\frac{\pi}{2}\bigg).$
  1. Since, f'(x) = cos x < 0, i.e., negative in second quadrant, i.e., in $\bigg(\frac{\pi}{2},\pi\bigg).$
Therefore, f(x) is strictly decreasing in $\bigg(\frac{\pi}{2},\pi\bigg).$
  1. Since f'(x) = cos x > 0, i.e., positive in first quadrant, i.e., in $\bigg(0,\frac{\pi}{2}\bigg)\text{and }\text{f}' \text{(x)}=\cos \text{x} <0,$
i.e., negative in second quadrant, i.e., in $\bigg(\frac{\pi}{2},\pi\bigg)\text{and } \text{f'}\bigg(\frac{\pi}{2}\bigg) = \cos\frac{\pi }{2} = 0.$

$\therefore$ f'(x) does not have the same sign in the interval $(0,\ \pi).$

Therefore, f(x) is neither increasing nor decreasing in $(0,\ \pi).$

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