Start Learning Free

Before You Begin

Question #41648

Difficulty: 8/10 | 🟢 Free

The Question

Let a curve LL be the locus of points (x,y,z)(x, y, z) satisfying the equations: y=2x−1−xln⁡x+∫1xln⁡(t)dty = 2x - 1 - x \ln x + \int_1^x \ln(t) dt z=x+114−x22ln⁡x+x24+∫1xtln⁡(t)dtz = x + \frac{11}{4} - \frac{x^2}{2} \ln x + \frac{x^2}{4} + \int_1^x t \ln(t) dt Find the length of the segment of LL that lies inside the sphere S:x2+y2+z2−2x−4y−6z=0S: x^2 + y^2 + z^2 - 2x - 4y - 6z = 0.

Options

  • A.

    333\sqrt{3}

  • B.

    434\sqrt{3}

  • C.

    232\sqrt{3}

  • D.

    535\sqrt{3}

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

Line lying in a plane|Exact Differential Recognition|Integration by Parts (LIATE Rule)|Domain Restrictions

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize that simplifying integral equations via Integration by Parts can yield a linear trajectory in 3D space.

The primary trap lies in the domain restriction: the logarithmic terms implicitly restrict the line to a ray (x>0x > 0).

Students who blindly simplify the equations and treat the locus as a full infinite line will find two intersection points with the sphere (x=3x = 3 and x=−1x = -1) and incorrectly calculate the full chord length (434\sqrt{3}).

By enforcing the domain constraint, the intersection set is truncated, and the true segment length is measured from the ray's origin to the valid boundary point (333\sqrt{3}).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterThree-Dimensional Geometry - Class 12
ConceptLine lying in a plane
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree