If f(x)= ^2 x+ ^2 2 x+ ^2 3 x , then the number of values of x [0,2 ] for which f(x)=1 is

Step-by-step solution

aligned & f(x)= ^2 x+ ^2 2 x+ ^2 3 x=1 & ^2 x+ (2 ^2 x-1 )^2+ (4 ^3 x-3 x )^2=1 & ^2 x+4 ^4 x+1-4 ^2 x+16 ^6 x & +9 ^2 x-24 ^4 x=1 & 6 ^2 x-20 ^4 x+16 ^6 x=0 & 2 ^2 x (3-10 ^2 x+8 ^4 x )=0 & ^2 x=0 or 8 ^4 x-10 ^2 x+3=0 & ^2 x=0 or (4 ^2 x-3 ) (2 ^2 x-1 )=0 & ^2 x=0 or x= 3 4 or ^2 x= 1 2 & ^2 x=0 or x= 3 2 or x= 1 2 & ^2 x=(2 n+1) 2 , & x=2 n 6 & aligned or array ll or & x=2 n + 4 & x= 2 , 3 2 , 4 , 6 , 7 4 , 11 6 array Hence, there are six values of x for which x [0,2 ] .