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प्रश्न
The maximum value of \[\sin^2 \left( \frac{2\pi}{3} + x \right) + \sin^2 \left( \frac{2\pi}{3} - x \right)\] is
पर्याय
1/2
- \[\frac{3}{2}\]
1/4
3/4
उत्तर
\[ = \left[ \cos(30 + x) \right]^2 + \left[ \cos(30 - x) \right]^2 \left[\text{ Using }\sin(90 + A) = \cos A \right]\]
\[ = \left[ \frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x \right]^2 + \left[ \frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x \right]^2 \]
\[ = \frac{3}{4} \cos^2 x + \frac{1}{4} \sin^2 x - \frac{\sqrt{3}}{2}\cos x \sin x + \frac{3}{4} \cos^2 x + \frac{1}{4} \sin^2 x + \frac{\sqrt{3}}{2}\cos x \sin x\]
\[ = \frac{3}{2} \cos^2 x + \frac{1}{2} \sin^2 x\]
\[ = \frac{3}{2}\left( 1 - \sin^2 x \right) + \frac{1}{2} \sin^2 x\]
\[ = \frac{3}{2} - \frac{3}{2} \sin^2 x + \frac{1}{2} \sin^2 x\]
\[ = \frac{3}{2} - \sin^2 x\]
\[\text{ For }f(x)\text{ to be maximum, }\sin^2 x \text{ must have minimum value, which is 0. }\]
\[ \therefore \frac{3}{2}\text{ is the maximum value of }f\left( x \right) .\]
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