APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES3 Marks
Question
What is the maximum value of the function sinx + cosx?
✓
Answer
Let $\text{f}\text{(x)} =\sin\text{x}+\cos\text{x}\ \Rightarrow\ \ \text{f}'\text{(x)}=\cos\text{x}-\sin\text{x}$
Now $\text{f}'\text{(x)}=0\ \Rightarrow\ \cos\text{x}-\sin\text{x}=0\ \Rightarrow\ \ -\sin\text{x}=-\cos\text{x}$
$\Rightarrow\ \tan\text{x}=1=\tan\frac{\pi}{4}$ $\Rightarrow\ \text{x}=\text{n}\pi+\frac{\pi}{4}\ \ [\text{Turning point}]$
$\therefore\ \text{f} \bigg(\text{n}\pi+\frac{\pi}{4}\bigg)=\sin\bigg(\text{n}\pi+\frac{\pi}{4}\bigg)+\cos \bigg(\text{n}\pi+\frac{\pi} {4}\bigg)$
$=(-1)^\text{n}\sin\frac{\pi}{4}+(-1)^\text{n}\cos\frac{\pi}{4}$
$=(-1)^\text{n}\frac{1}{\sqrt{2}}+(-1)^\text{n}\frac{1}{\sqrt{2}}$ $=2(-1)^\text{n}\frac{1}{\sqrt{2}}=\sqrt{2}(-1)^\text{n}$
If n is even, then $\ \text{f}\bigg(\text{n}\pi+\frac{\pi}{4}\bigg)=\sqrt{2}$
If n is odd, then $\ \text{f}\bigg(\text{n}\pi+\frac{\pi}{4}\bigg)=-\sqrt{2}$
Therefore, maximum value of $\text{f}\text{(x)} \text{ is }\sqrt{2}$ and minimum value of $\text{f}\text{(x)} \text{ is }-\sqrt{2}.$
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