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APPLICATION OF INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF INTEGRALS5 Marks
Question
Find the local maximum and local minima, of the function $\text{f(x)} = \sin x - \cos x, 0< x < 2\pi.$ Also find the local maximum and local minimum values.
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Answer

$\text{f(x)} = \sin x - \cos x, 0< x < 2\pi.$
$\text{f' (x) = 0} \Rightarrow \cos\text{x} + \sin \text{x} = \text{0 or} \tan \text{x} = -1, $
$\therefore \text{x} = 3\pi/4, \frac{7\pi}{4}$
$\text{f"(x)} = \cos\text{x}- \sin\text{x}$
$\text{f"} (3\pi/4) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \text{i.e. - ve so, x} = 3\pi/ \text{4 is Local Maxima}$
$\text{and f"} (7\pi/4) = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \text{i.e + ve so, x} = 7\pi/ \text{4 is Local Minima }$
Local Maximum value = $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$
Local Minimum value = $-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \sqrt{2}$

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