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Question #41812

Difficulty: 8/10 | 🟢 Free

The Question

Let u⃗=sec⁡2θi^+cot⁡2θj^+csc⁡2θk^\vec{u} = \sec^2 \theta \hat{i} + \cot^2 \theta \hat{j} + \csc^2 \theta \hat{k}, v⃗=tan⁡2θi^+csc⁡2θj^+sec⁡2θk^\vec{v} = \tan^2 \theta \hat{i} + \csc^2 \theta \hat{j} + \sec^2 \theta \hat{k}, and w⃗=i^−j^+(cot⁡2θ−tan⁡2θ)k^\vec{w} = \hat{i} - \hat{j} + (\cot^2 \theta - \tan^2 \theta)\hat{k}, where θ∈(0,π4)\theta \in \left(0, \frac{\pi}{4}\right). Let pp be the volume of the parallelepiped whose coterminous edges are u⃗,v⃗,w⃗\vec{u}, \vec{v}, \vec{w}. A biased coin, which shows heads with probability pp, is tossed 10 times. Let XX denote the number of heads obtained. Find the value of ∑k=010(k2−3k+2)P(X=k)+2024p\sum_{k=0}^{10} (k^2 - 3k + 2) P(X=k) + 2024^p

Options

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

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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|Scalar Triple Product Volume Logic|Linear Independence vs Dependence

Why This Question Is Tricky

Common Mistakes

The question employs the 'Degeneracy Trap' by linking the probability of a Bernoulli trial to the volume of a parallelepiped formed by three complex-looking trigonometric vectors.

Students often rush to expand the 3×33 \times 3 determinant, which is algebraically tedious and highly prone to errors.

The trap tests whether the student checks for linear dependence first; recognizing that u⃗−v⃗=w⃗\vec{u} - \vec{v} = \vec{w} immediately yields a volume (and thus probability) of 0.

Consequently, the binomial distribution collapses to a trivial deterministic case where X=0X=0 with certainty, simplifying the expectation sum instantly.

Quick Facts About This Question

Question Information

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