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

Answer: (D)

None of these

Answer 2

f(x) = |x – 1| = $\left{ \begin{matrix}

• x + 1, & x < 1 \ x - 1, & x \geq 1 \end{matrix} \right.\ $

Consider f(x²) = (f(x))²

If it is true it should be true \overset{̶}{V} x ∴ Put x = 2

LHS = f(2²) = | 4 − 1| = 3

RHS = f((2)²) = 1

∴ (1) is not correct

Consider f(x + y) = f(x) + f(y)

Put x = 2, y = 5 we get

F(7) = 6; f(2) + f(5) = 1 + 4 = 5 ∴ (2) is not correct.

Consider f(|x|) = |f(x)|

Put x = − 5 then f(|−5|) = f(5) = 4

|f(−5)| = |−5 − 1| = 6

∴ (3) is not correct.

Hence (4) is the correct alternative