Question

Questions 43 and 44 are based on the following instructions: each question gives a statement followed by three conclusions, I, II and III. Pick the option that tells you which of the three conclusions can be derived from the statement alone.

Statement: Let \(A\), \(B\), \(C\) be real numbers satisfying \(A < B < C\), \(A + B + C = 6\), and \(AB + BC + CA = 9\).

Conclusion I: \(1 < B < 3\)
Conclusion II: \(2 < A < 3\)
Conclusion III: \(0 < C < 1\)

• A. Using the given statement, only conclusion I can be derived.
• B. Using the given statement, only conclusion II can be derived.
• C. Using the given statement, only conclusion III can be derived.
• D. Using the given statement, all conclusions can be derived.

Hint:

Treat A and C as the two roots of a quadratic equation whose coefficients depend on B, then use the condition that B must lie strictly between those two roots.

Answer 1

The Correct Option is A

Step 1: Express A and C as the roots of a quadratic in terms of B.
From \(A + B + C = 6\), we get \(A + C = 6 - B\).
From \(AB + BC + CA = 9\), factor B out of the first two terms: \(B(A + C) + AC = 9\), so \(AC = 9 - B(6-B) = B^2 - 6B + 9\).
So A and C are the two roots of the quadratic \(x^2 - (6-B)x + (B^2 - 6B + 9) = 0\).

Step 2: Use the discriminant to find when A and C are real and distinct.
For real, distinct roots the discriminant must be positive:
\[ (6-B)^2 - 4(B^2-6B+9) > 0 \]
Expanding: \(36 - 12B + B^2 - 4B^2 + 24B - 36 > 0\), which simplifies to \(-3B^2 + 12B > 0\), or \(3B(4-B) > 0\).
This holds when \(0 < B < 4\).

Step 3: Use the fact that B must lie strictly between A and C.
Since the quadratic \(f(x) = x^2 - (6-B)x + (B^2-6B+9)\) opens upward, its two roots A and C straddle B (that is, \(A < B < C\)) exactly when \(f(B) < 0\).
Substitute \(x = B\):
\[ f(BB) = B^2 - (6-B)B + B^2 - 6B + 9 = 3B^2 - 12B + 9 = 3(B-1)(B-3) \]
So \(f(B) < 0\) exactly when \(1 < B < 3\), which already sits inside the wider range \(0 < B < 4\) found in Step 2.
This proves Conclusion I, \(1 < B < 3\), always holds.

Step 4: Test Conclusions II and III with one valid example.
Take \(B = 2\), a value allowed by Step 3. Then \(A + C = 4\) and \(AC = 2^2 - 6(2) + 9 = 1\), so A and C solve \(x^2 - 4x + 1 = 0\), giving \(x = 2 \pm \sqrt{3}\).
So \(A = 2 - \sqrt{3} \approx 0.27\) and \(C = 2 + \sqrt{3} \approx 3.73\), and indeed \(A < B < C\) holds since \(0.27 < 2 < 3.73\).
Here \(A \approx 0.27\), which does not lie between 2 and 3, so Conclusion II (\(2 < A < 3\)) fails for this valid case.
Also \(C \approx 3.73\), which does not lie between 0 and 1, so Conclusion III (\(0 < C < 1\)) fails too.
Since one valid case is enough to break Conclusions II and III, neither of them can be guaranteed by the statement.

Final Answer:
Only Conclusion I can always be derived from the statement, so the correct choice is option (AA). \[ \boxed{\text{Option (AA)}} \]