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Question #41721

Difficulty: 8/10 | 🟢 Free

The Question

A point zz is chosen uniformly at random from the region S={z∈C:4≤∣z∣≤5}S = \{z \in \mathbb{C} : 4 \le |z| \le 5\}. Let w=u+ivw = u + iv be a complex number defined by the transformation w=z+16zw = z + \frac{16}{z}. A Bernoulli trial is considered a success if ww satisfies both of the following conditions: (i) det⁡(uv1801−801)≥0\det \begin{pmatrix} u & v & 1 \\ 8 & 0 & 1 \\ -8 & 0 & 1 \end{pmatrix} \ge 0 (ii) ∣w−8∣+∣w+8∣≤12|w - 8| + |w + 8| \le 12 Let XX be the number of successes in n=2024n = 2024 independent trials. Find the value of E(X2+X+1)E(X^2 + X + 1), where E(⋅)E(\cdot) denotes the expected value.

Options

  • A.

    0

  • B.

    1

  • C.

    2024

  • D.

    2025

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

Bernoulli Trials Logic|Determinant as Geometric Area/Volume|Locus Problems via Transformation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate complex transformations, determinant-based geometric constraints, and probability distributions, while avoiding a critical feasibility trap.

The Joukowsky transformation w=z+16/zw = z + 16/z acts as a complex distractor, baiting students into mapping the annulus to the ww-plane to find the area of intersection with an assumed ellipse.

Students often blindly apply the area formula for an ellipse upon seeing the sum of distances to foci, calculating 2a=12  ⟹  a=62a=12 \implies a=6 and 2ae=16  ⟹  ae=82ae=16 \implies ae=8, leading to a negative b2=36−64=−28b^2 = 36 - 64 = -28.

If they ignore the negative sign or fail to check the domain, they will calculate a non-zero probability.

The trap is that the triangle inequality ∣w−8∣+∣w+8∣≥16|w - 8| + |w + 8| \ge 16 makes the condition ∣w−8∣+∣w+8∣≤12|w - 8| + |w + 8| \le 12 an empty set.

Consequently, the probability pp collapses to 0, nullifying the binomial distribution parameters and trivially reducing the expected value expression to 1.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 12
ConceptBernoulli Trials Logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree