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Question #40507
The Question
Let be the minimum value of the function and be the maximum value of the function . Consider the function defined on by Let be the set of all points such that . Find the sum of all elements in .
Options
- A.
- B.
- C.
- D.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Monster
Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.
Concepts Behind This Question
Maximum/Minimum of Quadratic Expressions|Newton-Leibniz Formula with variable limits|Logarithmic Integration Modulus
Why This Question Is Tricky
Common Mistakes
This question tests the application of the Newton-Leibniz formula combined with the 'Logarithmic Integration Modulus' trap.
Students must first determine the integration interval by finding the extrema of two quadratics.
The core trap lies in solving .
A common mistake is to ignore the absolute value, solving only and missing the root derived from .
Because the vertex of the quadratic inside the logarithm drops below zero, the modulus is activated, making a valid critical point where the derivative is zero.
Quick Facts About This Question
Question Information
| Subject | Mathematics |
| Chapter | Complex Numbers and Quadratic Equations - Class 11 |
| Concept | Maximum/Minimum of Quadratic Expressions |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |