The range of y- 1-2-sin 3x is

The correct option is A 13≤ y≤ 1. We know 1≤sin 3x≤ 1 Hence range of y is =[1/2+1,1/2 1]=[1/3,1] ...

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Definition of Functions

The range of ...

Question

The range of $y = \frac{1}{2 - sin 3 x}$ is

\frac{1}{3} \leq y \leq 1.

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\frac{2}{3} \leq y \leq 1.

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- \frac{1}{3} \leq y \leq 1.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{2}{3} \leq y \leq 1.

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y = \log \frac{x^{2} + x + 1}{x^{2} - x + 1} + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} x}{1 - x^{2}}), find \frac{d y}{d x} .

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}

\frac{x - 1}{2} = \frac{y + 1}{3} = z and \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2

\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} and \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2}

\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} and \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}

\frac{2}{x} + \frac{3}{y} = 6 and \frac{1}{x} + \frac{1}{2 y} = 2

x = 1, y = \frac{2}{3}

x = \frac{2}{3}, y = 1

x = 1, y = \frac{3}{2}

x = \frac{3}{2}, y = 1

y = \frac{1}{(2 - sin 3 x)}

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