If sin ,θ + √ 3 cos ,θ = 6x - {x^2} - 11,x ∈ R , 0 le θ le 2π

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Trigonometrical Equations

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If $\sin \,\theta + \sqrt 3 \cos \,\theta = 6x - {x^2} - 11,x \in R$ , $0 \le \theta \le 2\pi $ , then the equation has solution for

A

one value of $x$

B

two value of $x$

C

infinite value of $x$

D

no value of $x$

Solution

$\sin \theta+\sqrt{3} \cos \theta=-2-(x-3)^{2}$

minimum vlaue of $\sin \theta+\sqrt{3} \cos \theta$ is $-2$

Hence solution of equation is possible only if $x=3$

Standard 11

Mathematics

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