Sin2600-x-sin2600-x-k-2-k-2- k-

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Standard XII

Mathematics

Definition of Functions

sin2600+x-sin...

Question

${sin}^{2} (60^{0} + x) - {sin}^{2} (60^{0} - x) \in [- \frac{k}{2}, \frac{k}{2}]$ , k=

3

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2

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\sqrt{3}

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\sqrt{2}

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Solution

The correct option is C $\sqrt{3}$
$f (x) = s i n^{2} (60 + x) - s i n^{2} (60 - x)$
$= [s i n (60 + x) + s i n (60 - x)] [s i n (60 + x) - s i n (60 - x)]$
$= (2 s i n (60) c o s x) (2 c o s (60) s i n x)$
$f (x) = \frac{\sqrt{3}}{2} s i n 2 x$
$∴$ range of $f (x) ϵ [- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}] = [- \frac{k}{2}, \frac{k}{2}]$
$k = \sqrt{3}$

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sin2(600+x)−sin2(600−x)∈[−k2,k2], k=

The correct option is C √3f(x)=sin2(60+x)−sin2(60−x)=[sin(60+x)+sin(60−x)][sin(60+x)−sin(60−x)]=(2sin(60)cosx)(2cos(60)sinx)f(x)=√32sin2x∴ range of f(x)ϵ[−√32,√32]=[−k2,k2]k=√3

${sin}^{2} (60^{0} + x) - {sin}^{2} (60^{0} - x) \in [- \frac{k}{2}, \frac{k}{2}]$ , k=