Problem:
The number of real solutions to the equation
100xβ=sinx
Answer Choices:
A. 61
B. 62
C. 63
D. 64
E. 65
Solution:
Since 100βxβ=sin(βx), the equation has an equal number of positive and negative solutions. Also x=0 is a solution. Furthermore, all positive solutions are less than or equal to 100, since β£xβ£=100β£sinxβ£β©½100.
Since 15.5<2Ο100β<16, the graphs of 100xβ and sinx are as shown in the adjoining figure. Thus there is one solution to the given equation between 0 and Ο and two solutions in each of the intervals from (2kβ1)Ο, to (2k+1)Ο, 1β©½kβ©½15.
The total number of solutions is, therefore,
1+2(1+2β
15)=63