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Question #40401

Difficulty: 8/10 | 🟢 Free

The Question

Let EE be the ellipse x23+y2=1\frac{x^2}{3} + y^2 = 1. For a point PP on EE, let θ∈[0,π]\theta \in [0, \pi] be its eccentric angle. If (xP,yP)(x_P, y_P) are the coordinates of PP, then the value of the integral I=∫0π[xP+yP] dθI = \int_{0}^{\pi} [x_P + y_P] \, d\theta where [⋅][ \cdot ] denotes the greatest integer function, is:

Options

  • A.

    0

  • B.

    π3\frac{\pi}{3}

  • C.

    2π3\frac{2\pi}{3}

  • D.

    2

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

Auxiliary Angle Method|Ellipse Auxiliary Circle|Integral Part and Fractional Part

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from the geometric locus of an ellipse using the auxiliary angle to a rigorous evaluation of a transcendental function within the Greatest Integer Function (GIF).

The primary trap lies in the domain partitioning of 2sin⁡(θ+π/3)2\sin(\theta + \pi/3).

Students frequently make two critical errors: first, they ignore the phase shift and integrate [2sin⁡θ][2\sin\theta] over [0,π][0, \pi], yielding 2π/32\pi/3; second, they mishandle the GIF for negative values, incorrectly assuming [−0.5]=0[-0.5] = 0 instead of −1-1, which leads to the distractor π/3\pi/3.

Correctly identifying the sub-intervals where the function crosses integers, especially in the negative range [−3,0)[-\sqrt{3}, 0), is essential for arriving at the exact cancellation that results in 0.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterTrigonometric Functions - Class 11
ConceptAuxiliary Angle Method
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree