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Question #41888

Difficulty: 8/10 | 🟢 Free

The Question

Let $E_1$ and $E_2$ be two mutually exclusive and exhaustive events associated with a random experiment, with probabilities $p_1$ and $p_2$ respectively. Let $X = \begin{pmatrix} 1 & 2p_1 \\ 0 & 1 \end{pmatrix}$ and $Y = \begin{pmatrix} 1 & 0 \\ 2p_2 & 1 \end{pmatrix}$ . A matrix-valued function is defined as $M(t) = X^2 + 2tXY + t^2Y^2$ . If $f(t) = \det(M(t))$ , then the minimum possible value of $f'(1)$ is:

Options

A.
16
B.
18
C.
24
D.
32

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

Exhaustive Events Concept|Differentiation of Determinants|Matrix Multiplication Non-Commutativity

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of probability constraints and matrix algebra, specifically targeting the non-commutativity of matrix multiplication.

A common trap for students is to blindly apply the algebraic identity $X^2 + 2tXY + t^2Y^2 = (X+tY)^2$ .

Because $X$ and $Y$ do not commute, this simplification is mathematically invalid.

If a student falls for this trap, they will compute $f(t) = (\det(X+tY))^2$ , leading to an incorrect derivative $f'(1) = 32(1-p_1 p_2)^2$ and an erroneous minimum value of 18.

The problem forces a rigorous, term-by-term expansion and differentiation, rewarding students who respect matrix properties.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Exhaustive Events Concept
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free