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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
On which of the following intervals is the function f given by $\text{f(x)} = \text{x}^{100} +\sin \text{x} - 1$ is strictly decreasing:
  1. $(0,1)$
  2. $\Big(\frac{\pi}{2},\pi\Big)$
  3. $\Big(0,\frac{\pi}{2}\Big)$
  4. $\text{None of there.}$
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Answer

Given: $\text{f}\text{(x)} = \text{x}^{100 }+ \sin \text{x} - 1$ $\Rightarrow \ \text{f}'\text{(x)} = 100\text{x}^{99 } + \cos \text{x}$
  1. $\text{ On } (0,10),\ \text{x} > 0 \text{ threfor } 100\text{x}^{99} > 0$
$\text{And for } \cos \text{x}\ \Rightarrow\ ( 0,1 \text{ radian}) =(0,57^{\circ} \text{nearly)} >0$

Therefore, f(x) is strictly increasing on $(0,1).$
  1. $\text{For } 100\text{x}^{99} \text{x}\in \Big(\frac{\pi}{2},\pi\Big)$ $= \Big(\frac{11}{7},\frac{22}{7}\Big)=(1.5,3.1)>1\text{ and hence}\ 100^{99} >100$
For $\cos\text{x}$ $ \Big(\frac {\pi}{2}, \pi\ \Big)$ is in second quadrant and hence cos x is negative and between -1 and 0.

Therefore, f(x) is strictly increasing on $\Big(\frac {\pi}{2},\pi\Big).$
  1. On $\Big(0,\ \frac{\pi}{2}\Big)=(0,\ 1.5)$ both terms of given function are positive.
Therefore, f(x) is strictly increasing on $\Big(0,\ \frac{\pi}{2}\Big).$
  1. Option (d) is the correct answer.

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