Application of Derivatives — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives2 Marks
Question
What is the maximum value of the function sin x + cos x?
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Answer
Let f(x) = sin x + cos x, $\Rightarrow$ f'(x) = cos x - sin x Now, f'(x) = 0 $\Rightarrow$ cos x - sin x = 0 $\Rightarrow$ cos x = sin x $\Rightarrow$ tan x = 1 $\Rightarrow \mathrm{x}=\frac{\pi}{4}, \frac{5 \pi}{4}, \ldots$ Now, If f''(x) will be negative when (sin x + cos x) > 0, means both sin x and cos x are positive. And, we know that sin x and cos x both are positive in the first quadrant. Then, f''(x) will be negative when $\mathrm{x} \in\left(0, \frac{\pi}{2}\right)$. f''(x) = -sin x - cos x = -(sin x + cos x) Now, let us take x = $\frac{\pi}{4}$ $f^{\prime \prime}\left(\frac{\pi}{4}\right)=-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0$ Then, by second derivative test, f will be maximum at $x=\frac{\pi}{4}$ And, the maximum value of f is $f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$
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