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find range: y= 1/2-sin3x

Note by Akash Sinha
3 years, 10 months ago

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\( y = \frac {1} {2 - sin 3x} \) or \( y = \frac {1}{2} - sin 3x \) ? Lokesh Sharma · 3 years, 10 months ago

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@Lokesh Sharma Anyway, for the first one, Range is \( (\frac {1}{3}, 1) \)

For second one, Range is \((-\frac{1}{2}, \frac{3}{2}) \) Lokesh Sharma · 3 years, 10 months ago

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@Lokesh Sharma Reason: The min and max value of \(sin x\) is \(-1\) and \(+1\) respectively. Lokesh Sharma · 3 years, 10 months ago

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@Lokesh Sharma How do we find min. and max. values ? Priyansh Sangule · 3 years, 10 months ago

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@Lokesh Sharma thanks Akash Sinha · 3 years, 10 months ago

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@Lokesh Sharma 1st one Akash Sinha · 3 years, 10 months ago

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