Units digit of x^4 - 5(X^2) + 1
St1: Units digit of x^2 = 4
X^2 can be 4 (x = 2) or x^2 can be 64 (x=8)
If x = 2, x^4 ends in (4*4 = 16) 6 + 1 = 7 and 5*(2^2) ends in 0 (as 5 * even number ends in 0) thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
If x = 8, x^4 ends in (64*64 = 4096) 6 + 1 = 7 and 5*(2^2) ends in 0 thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
Hence sufficient
St2: Units digit of x^4 = 6
X can be 2, 4, 6 and 8
If x = 2, x^4 ends in (4*4) 6 + 1 = 7 and 5*(even) ends in 0 thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
If x = 4, x^4 ends in (16*16) 6 + 1 = 7 and 5*(even) ends in 0 thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
If x = 6, x^4 ends in (36*36) 6 + 1 = 7 and 5*(even) ends in 0 thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
If x = 8, x^4 ends in (64*64) 6 + 1 = 7 and 5*(even) ends in 0 thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
Hence sufficient
Ans D
