ExamHabit
Practise free

1983

JEE Main 1983 Mathematics question on Sets Relations and Functions

Let f(x)=∣x−1∣f(x) = |x - 1|. Then,
  1. A.
    f(x2)=(f(x))2f(x^2) = (f(x))^2
  2. B.
    f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y)
  3. C.
    f(∣x∣)=∣f(x)∣f(|x|) = |f(x)|
  4. D.
    None of the above
Show answer and solution

Correct option: D

Solution

f(x²) = |x²-1| but (f(x))² = (|x-1|)² = (x-1)². These differ (e.g. x=0 gives 1 vs 1 — check x=2: |4-1|=3 vs (2-1)²=1). So option 1 fails.

f(x+y) = |x+y-1| but f(x)+f(y) = |x-1|+|y-1|. At x=2,y=2: f(4)=3 vs f(2)+f(2)=2. So option 2 fails.

f(|x|) = ||x|-1| and |f(x)| = ||x-1|| = |x-1|. At x=-2: f(|-2|)=f(2)=1 but |f(-2)|=|-3|=3. These are unequal, so option 3 fails.

Since all three are false, the answer is option 4.

[1983]

Practise Sets Relations and Functions the way it is asked

A timed set of real JEE Main Sets Relations and Functions questions, graded as the exam grades. No sign-up.

Start a Sets Relations and Functions test

More: all JEE Main Sets Relations and Functions questions · JEE Main 1983 · Mathematics chapter list