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At what point in the interval [0,2?] does the function Sin2x attain its maximum value?

Manvendra Singh chahar , 11 Years ago
Grade Upto college level
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
f(x) = sin(2x)
f^{'}(x) = 2cos(2x) = 0
2x = (n + \frac{1}{2})\pi
x = (n + \frac{1}{2})\frac{\pi }{2}
x = \frac{\pi }{4}, \frac{3\pi }{4}, \frac{5\pi }{4}, \frac{7\pi }{4}
f^{''}(x) = -4sin(2x)
f^{''}(\frac{\pi }{4}) = -4sin(2.\frac{\pi }{4}) = -4
f^{''}(\frac{3\pi }{4}) = -4sin(2.\frac{3\pi }{4}) = 4
f^{''}(\frac{5\pi }{4}) = -4sin(2.\frac{5\pi }{4}) = -4
f^{''}(\frac{7\pi }{4}) = -4sin(2.\frac{7\pi }{4}) = 4
x = \frac{\pi }{4}, \frac{5\pi }{4}
is the pont of maxima.
Thanks & Regards
Jitender Singh
IIT Delhi
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