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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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The maximum distance of a point on the graph of the function $y=\sqrt 3\sin x+\cos x$ from $x$-axis is ________

$\begin{array}{1 1}(A)\;2&(B)\;3\\(C)\;6&(D)\;4\end{array} $

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  • $a\cos \theta+b\sin \theta=R\cos (\theta-\alpha)$
$y=\sqrt 3\sin x+\cos x$
$y=\sqrt{(\sqrt 3)^2+1^2}\big(\large\frac{\sqrt 3}{\sqrt{(\sqrt 3)^2+1^2}}$$\sin x+\large\frac{1}{\sqrt{(\sqrt 3)^2+1^2}}$$\cos x\big)$
$y=\sqrt{(\sqrt 3)^2+1^2}\big(\sin \alpha\sin x+\cos \alpha \cos x\big)$
$y=2\cos (\alpha-x)$
$y_{max}$ when $\cos (\alpha-x)=1$
$\therefore y=2$
Hence (A) is the correct answer.
answered May 20, 2014 by sreemathi.v
 

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