Question

12. Find the maximum point in the first cycle for the
equation : Y-3 = 4 sin (π×x/2))
​

Answer 1

Given : y-3 = 4 sin (π×x/2))

To Find : Maximum point

​

Solution:

y-3 = 4 sin (π×x/2))

=> y = 3 + 4 sin (π×x/2))

dy/dx = 4 Cos (π×x/2))  / (π/2)

dy/dx = 0

=>  4 Cos (π×x/2))  / (π/2) = 0

=> Cos (π×x/2)) = 0

=> Cos (π×x/2) = Cosπ/2

=> x = 1

d²y/dx² = -4 sin (π×x/2))  / (π/2)²

x = 1

=> d²y/dx² = -4 sin (π/2))  / (π/2)²

- 4/(π/2)² < 0

Hence maximum value at x = 1

y = 3 + 4 sin (π×x/2))

=  3 + 4 sin (π /2))

= 3 + 4

= 7

Maximum value is 7

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