Question

The set of all values of x satisfying \[ \sqrt{x^2-2x+1}>x+2 \] is

• A. \((-2,\infty)\)
• B. \((-\infty,0)\)
• C. \((-\infty,-\frac12)\)
• D. \((-\frac12,\infty)\)

Hint:

Whenever square root contains perfect square expression, convert into modulus immediately.

Answer 1

The Correct Option is C

Concept: First simplify square root, then solve modulus inequality.

Step 1: Simplify expression.
Observe \[ x^2-2x+1=(x-1)^2 \] Thus \[ \sqrt{x^2-2x+1}=|x-1| \] Equation becomes \[ |x-1|& gt;x+2 \]

Step 2: Case 1: \(x\geq1\)
Then \[ x-1& gt;x+2 \] \[ -1& gt;2 \] Impossible. No solution.

Step 3: Case 2: \(x& lt;1\)
Then \[ -(x-1)& gt;x+2 \] \[ -x+1& gt;x+2 \] \[ -2x& gt;1 \] \[ x& lt;-\frac12 \] Thus solution set \[ (-\infty,-\frac12) \] Hence \[ \boxed{(-\infty,-\frac12)} \]