Let a and B be non-zero vectors. If f (x) = (XA + b) (a-xb), X belongs to R and is even, A. A vertical B B. a parallel B C. | a | = | B | D. | a | unequal | B|

Let a and B be non-zero vectors. If f (x) = (XA + b) (a-xb), X belongs to R and is even, A. A vertical B B. a parallel B C. | a | = | B | D. | a | unequal | B|

Option C is correct!
Analysis:
f(x)=(xa+b)(a-xb)=x*|a|²-x²a*b+a*b-x*|b|²=-x²a*b+（|a|²-|b|²）x +a*b
If the function f (x) is even on R, then:
For any real number x, f (- x) = f (x)
That is - (- x) & #178; a * B + (|a | & #178; - |b | & #178;) (- x) + A * b = - X & #178; a * B + (|a | & #178; - |b | & #178;) x + A * B
Easy to get: 2 (| a | & # 178; - | B | & # 178;) x = 0
If the above formula is true for any real number x, it must be such that:
|A | a | B | B | a | = | B|