Ars Mathematica: Solución al problema 76

(\frac{π}{2} + 2 k π, \frac{3 π}{2} + 2 k π), k \in Z,

s e n (x)

\frac{x}{100}

f (x) = s e n (x) - \frac{x}{100}

f (\frac{π}{2} + 2 k π) = s e n (\frac{π}{2} + 2 k π) - \frac{\frac{π}{2} + 2 k π}{100} = 1 - \frac{π + 4 k π}{200}

1 - \frac{π + 4 k π}{200} > 0 \Rightarrow 1 > \frac{π + 4 k π}{200} \Rightarrow \frac{200 - π}{4 k} > π \Rightarrow 15.66 > π

k = 0, \dots, 15, f (\frac{π}{2} + 2 k π) > 0

f (\frac{3 π}{2} + 2 k π) = s e n (\frac{3 π}{2} + 2 k π) - \frac{\frac{3 π}{2} + 2 k π}{100} = - 1 - \frac{3 π + 4 k π}{200}

k > 0

(\frac{π}{2} + 2 k π, \frac{3 π}{2} + 2 k π), k = 0, \dots, 15

\frac{\frac{π}{2} + 2 \cdot 16 π}{100} = 1.02

s e n (\frac{π}{2} + 2 k π) - \frac{\frac{π}{2} + 2 k π}{100} < 0

k > 15

f (x) < 0

100

(\frac{3 π}{2} + 2 k π, \frac{5 π}{2} + 2 k π)

k > 0

f (\frac{3 π}{2} + 2 k π) = s e n (\frac{3 π}{2} + 2 k π) - \frac{\frac{3 π}{2} + 2 k π}{100} = - 1 - \frac{3 π + 4 k π}{200} < 0

f (\frac{5 π}{2} + 2 k π) = s e n (\frac{5 π}{2} + 2 k π) - \frac{\frac{5 π}{2} + 2 k π}{100} = 1 - \frac{5 π + 4 k π}{200}

1 - \frac{5 π + 4 k π}{200} > 0 \Rightarrow 1 > \frac{5 π + 4 k π}{200} \Rightarrow \frac{200 - 5 π}{4 π} > k \Rightarrow k < 14.6

(\frac{3 π}{2} + 2 k π, \frac{5 π}{2} + 2 k π), k = 0, \dots, 14

15 + 16

(0, 0)

31 + 31 + 1 = 63

Ars Mathematica