Solve for the number of real solutions of the equation: √(x^2 − x + 1) + 1/√(x^2 − x + 1) = 2 − x^2.

Difficulty: Medium

Correct Answer: 1

Explanation:

Introduction / Context:
This equation mixes a radical expression and its reciprocal on the left with a quadratic on the right. An inequality insight about t + 1/t helps bound the left-hand side and identify feasible x-values without brute force algebra.

Given Data / Assumptions:

A(x) = x^2 − x + 1.
LHS: √A + 1/√A, domain requires A > 0.
RHS: 2 − x^2.

Concept / Approach:
Let t = √A > 0. Then LHS = t + 1/t ≥ 2 by AM ≥ GM, with equality iff t = 1. Also, A = x^2 − x + 1 has minimum 3/4 (at x = 1/2), so t ≥ √(3/4) = √3/2 > 0, ensuring the expression is defined for all real x.

Step-by-Step Solution:

Since t + 1/t ≥ 2, LHS ≥ 2, with equality when t = 1 → A = 1.A = 1 ⇒ x^2 − x + 1 = 1 ⇒ x(x − 1) = 0 ⇒ x = 0 or x = 1.Check RHS at these x: for x = 0, RHS = 2 − 0 = 2; LHS = 1 + 1 = 2 → valid.For x = 1, RHS = 2 − 1 = 1; LHS = 2 → not valid.For any other x, LHS > 2 while RHS ≤ 2, so no equality.

Verification / Alternative check:
Numerical sampling near x = 0 and x = 1 confirms only x = 0 satisfies the equality.

Why Other Options Are Wrong:
0, 2, 3, 4 solutions contradict the inequality structure; only one real solution exists.

Common Pitfalls:
Forgetting the equality condition t = 1 for t + 1/t = 2; mishandling the domain by allowing A ≤ 0 (which never occurs here).

Solve for the number of real solutions of the equation: √(x^2 − x + 1) + 1/√(x^2 − x + 1) = 2 − x^2.

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