Question

Let f(x) = |x – 1|. Then, 

A. f(x²) = [f(x)]² 

B. f(x + y) = f(x) f(y) 

C. f(|x|) = |f(x)| 

D. None of these

Answer 1

Option : (D)

f(x) = |x-1| 

f(x²) = |x²-1| 

f(x)² = (x-1)² 

= x²+1-2x 

So, 

f(x²) ≠ [f(x)]² 

f(x + y) = |x+y-1| 

f(x)f(y) = (x-1)(y-1) 

So, 

f(x + y) ≠ f(x)f(y) 

f(|x|) = ||x|-1| 

Therefore, 

Option D is correct.