What is the maximum value of a x+b x ? - Vidyadip

Question

What is the maximum value of $a \sin x+b \cos x$ ?

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Answer

Suppose
$y =a \sin x+b \sin x$
$\frac{d y}{d x} =a \cos x-b \sin x$
For, maxima and minima $\frac{d y}{d x}=0$
$\therefore a \cos x-b \sin x=0$
$\tan x=\frac{a}{b}$
$\frac{d^2 y}{d x^2}=-a \sin x-b \cos x$
$=-(a \sin x+b \cos x)=-ve$
$\therefore$maximum value $=\frac{a \times a}{\sqrt{a^2+b^2}}+\frac{b \times b}{\sqrt{a^2+b^2}}$
$=\frac{a^2+b^2}{\sqrt{a^2+b^2}}=\sqrt{a^2+b^2}$

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