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For what value of x does 6 sin (2x - 25)° attain its maximum value in the range...


Question

For what value of x does 6 sin (2x - 25)° attain its maximum value in the range \(0^{\circ} \leq x \leq 180^{\circ}\)?

Options

A) 12

B) 32

C) 57

D) 14

The correct answer is C.

Explanation:

\(y = 6\sin (2x - 25)\);
\(\frac{\delta y}{\delta x} = 2 \cos (2x - 25)\)
Equating to zero gives:
\(12\cos (2x - 25) = 0\)
\(\cos(2x - 25) = 0\)
\(2x - 25 = \cos^{-1}0\)
\(2x - 25 = 90^{\circ};\)
\(2x = 90^{\circ} + 25^{\circ} = 115^{\circ}\)
\(2x = 115^{\circ};\)
\(x = \cfrac{115^{\circ}}{2}\)
\(x = 57\frac{1}{2}\)

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Dicussion (1)

  • \(y = 6\sin (2x - 25)\);
    \(\frac{\delta y}{\delta x} = 2 \cos (2x - 25)\)
    Equating to zero gives:
    \(12\cos (2x - 25) = 0\)
    \(\cos(2x - 25) = 0\)
    \(2x - 25 = \cos^{-1}0\)
    \(2x - 25 = 90^{\circ};\)
    \(2x = 90^{\circ} + 25^{\circ} = 115^{\circ}\)
    \(2x = 115^{\circ};\)
    \(x = \cfrac{115^{\circ}}{2}\)
    \(x = 57\frac{1}{2}\)

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