Integrals · Question 3

Let the function f:[1, ) R be defined by f(t)= cases(-1)n+12, & if t=2n-1, n N,\ (2n+1-t)2f(2n-1)+ (t-(2n-1))2f(2n+1), & if 2n-1<t<2n+1, n N. cases Define g(x)= 1xf(t) dt, x (1, ). Let denote the number of solutions of the equation g(x)=0 in the interval (1,8] and = x 1+ g(x)x-1. Then the value of + is equal to .

Answer

5

Solution

f(1)=2, f(3)=-2, f(5)=2, f(7)=-2, f(9)=2 and f is linear in between: a zig-zag with zeros at t=2,4,6,8.

2n-12n+1f dt=0 (equal triangles of opposite sign), so g(3)=g(5)=g(7)=0.

g>0 on (1,3), g<0 on (3,5), g>0 on (5,7), and g<0 on (7,8] since f<0 on (7,8). Hence =3.

= x 1+ g(x)x-1=g'(1+)=f(1)=2.

+ =5. Answer: 5