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Question #40798

Difficulty: 8/10 | 🟢 Free

The Question

Let EE be the set of all values attained by the function f(x)=x22−4∣x∣f(x) = \frac{x^2}{2} - 4|x| at its points of local extrema, and let k=n(E)k = n(E) be the number of elements in EE. For each m∈Em \in E, let LmL_m be the line with slope mm which is a valid tangent to the parabola y2=4xy^2 = 4x. If LmL_m intersects the latus rectum of the parabola at Qm(1,ym)Q_m(1, y_m), find the sum of the values of (m2ym2+k)(m^2 y_m^2 + k) over all valid lines LmL_m.

Options

  • A.

    4227

  • B.

    4226

  • C.

    4230

  • D.

    4228

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

Latus Rectum Geometry|Tangent/Normal Slope Interplay|The Modulus Trap|Non-Differentiable Extrema

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of calculus (non-differentiable extrema) and coordinate geometry (tangent conditions).

Students often differentiate f(x)f(x) ignoring the modulus, missing the local maximum at x=0x=0, leading to an incorrect count k=1k=1 (yielding 4226).

Furthermore, those who do find m=0m=0 often blindly substitute it into the algebraic simplification m2ym2=m2(m+1/m)2=(m2+1)2=1m^2 y_m^2 = m^2(m + 1/m)^2 = (m^2+1)^2 = 1, failing to realize that a parabola y2=4axy^2=4ax has no valid tangent with slope m=0m=0.

This 'Lethal Collision' of blindly applying formulas without geometric validation leads to the trap answer 4230, effectively separating top candidates.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterConic Sections - Class 11
ConceptLatus Rectum Geometry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree