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Question #40563

Difficulty: 8/10 | 🟢 Free

The Question

Let $m$ be a non-zero real number such that the quadratic equations $x^2 - 11x + m = 0$ and $x^2 - 14x + 2m = 0$ have a common root. Let $\alpha$ and $\beta$ be the non-common roots of the first and second equations, respectively. Let $S$ be the set of all integers $r$ satisfying the inequality $|r^2 - (\beta - 4)r - 4\alpha| \le 2\beta$ Let $N(r)$ denote the number of ways to distribute $r$ identical coins into $3$ distinct boxes (if such a distribution is not possible, $N(r)=0$ ). Find the value of $\sum_{r \in S} N(r)$ .

Options

A.
81
B.
84
C.
85
D.
88

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Permutations of Identical Objects|Condition for Common Roots|Modulus Inequality Logic

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of quadratic common roots, modulus inequalities, and combinatorics.

The primary trap lies in the domain of the combinatorial function.

Students often solve the inequality to find $S = \{-4, -3, -2, -1, 0, 2, 3, 4, 5, 6\}$ and blindly apply the algebraic formula $\binom{r+2}{2} = \frac{(r+2)(r+1)}{2}$ to all elements, including negative ones.

This yields extraneous values (e.g., $r=-4 \implies 3$ , $r=-3 \implies 1$ ), leading to the trap answer 85.

A secondary trap catches students who only solve the upper bound of the modulus ( $r^2 - 2r - 24 \le 0$ ), incorrectly including $r=1$ to get 84 or 88.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	Permutations of Identical Objects
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free