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: Rewrite the problem using the gaps between A, B and C.
Let $x = B - A$ and $y = C - B$; both are positive since $A < B < C$. Then $C - A = x + y$.
A useful identity connects the sum of squares of the pairwise differences to the given sums: $(A-B)^2+(B-C)^2+(C-A)^2 = 2(A^2+B^2+C^2) - 2(AB+BC+CA)$.

Step 2: Find $A^2+B^2+C^2$ from the given data.
Since $(A+B+C)^2 = A^2+B^2+C^2+2(AB+BC+CA)$, we get $36 = A^2+B^2+C^2 + 2(99)$, so $A^2+B^2+C^2 = 36 - 18 = 18$.

Step 3: Compute the sum of squared gaps and simplify.
$x^2+y^2+(x+y)^2 = 2(18) - 2(99) = 18$.
Expanding $(x+y)^2 = x^2+2xy+y^2$, this becomes $2x^2+2xy+2y^2 = 18$, so $x^2+xy+y^2 = 9$, with $x>0$ and $y>0$.

Step 4: Express B in terms of x and y, then find its range.
From $A+B+C=6$ with $A=B-x$ and $C=B+y$: $(B-x)+B+(B+y)=6$, so $3B = 6+x-y$, giving $B = \frac{6+(x-y)}{3}$.
Let $u = x - y$. From $x^2+xy+y^2=9$ and $(x-y)^2 = x^2-2xy+y^2$, we get $xy = \frac{9-u^2}{3}$. For $x$ and $y$ to be positive real numbers we need $xy>0$, so $u^2 < 9$, that is $-3 < u < 3$.
Checking that positive $x$ and $y$ actually exist for every such $u$ confirms the full open range $u \in (-3,3)$ is reachable, with $x$ or $y$ shrinking to $0$ only at the excluded endpoints $u = \pm 3$.
So $B = \frac{6+u}{3}$ ranges over $\left(\frac{6-3}{3}, \frac{6+3}{3}\right) = (1,3)$, which proves Conclusion I.

Step 5: Show Conclusions II and III can fail using a sample case.
Take $u=1$: then $xy = \frac{9-1}{3} = \frac{8}{3}$ and $x+y = \sqrt{u^2+4xy} = \sqrt{1 + \frac{32}{3}} \approx 3.42$, giving $x \approx 2.21$ and $y \approx 1.21$.
Then $B = \frac{6+1}{3} \approx 2.33$, $A = B - x \approx 0.12$, and $C = B+y \approx 3.54$, and indeed $A < B < C$ holds.
Here $A \approx 0.12$ is not between 2 and 3, so Conclusion II fails, and $C \approx 3.54$ is not between 0 and 1, so Conclusion III fails as well.
Since this is one valid case where both II and III break down, neither of them can be a guaranteed consequence of the statement.

Final Answer:
Only Conclusion I is always true, so the statement supports option (AA) and no other.