Answer:
C = 59.17 nF 
Q = 2.6 
Step-by-step explanation:
given data 
frequencies = 40k Hz
frequencies = 90k Hz
solution
we take here R, L C take in series 
so cut off frequency is express as
Wc1 = 
 = 40000
wc2 = 
 = 90000
so here 
wc2 - wc1 will be 
wc2 - wc1 = 90000 - 40000 = 50000
so 
 = 50000 
we consider here R is 500 
so L = 
 
L = 10 m H 
and here total cut off frequency is 
total cut off frequency = 40000 + 90000 = 130000 
so capacitance will be 
capacitance C is = 
 
so C = 59.17 nF 
quality factor Q will be 
Q = 
 
Q = 2.6