Start Learning Free

Before You Begin

Question #41717

Difficulty: 8/10 | 🟢 Free

The Question

For sufficiently small x>0x > 0, let AxA_x and BxB_x be two events such that P(Ax)=xP(A_x) = x and P(Bx)=x2P(B_x) = x^2. The probability of their union satisfies the integral equation: ∫0xtP(At∪Bt)dt=x33+x44−1535x5−∫0xx2−t22t2sin⁡(1t)dt\int_0^x t P(A_t \cup B_t) dt = \frac{x^3}{3} + \frac{x^4}{4} - \frac{153}{5} x^5 - \int_0^x \frac{x^2 - t^2}{2} t^2 \sin\left(\frac{1}{t}\right) dt Let L=lim⁡x→0+P(Ax∩Bx)P(Ax)P(Bx)L = \lim_{x \to 0^+} \frac{P(A_x \cap B_x)}{P(A_x) P(B_x)}. Find the value of LL.

Options

  • A.

    153

  • B.

    4603\frac{460}{3}

  • C.

    1

  • D.

    0

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

Independent vs Disjoint Events|Leibniz Rule for Differentiation under Integral|Squeeze Theorem

Why This Question Is Tricky

Common Mistakes

The problem tests the application of the Leibniz rule and the evaluation of limits using the Squeeze Theorem where L'Hopital's rule fails.

Students often incorrectly conclude the limit does not exist because the derivative oscillates, missing that integration by parts followed by the Squeeze Theorem reveals the integral converges to 0.

The secondary trap is assuming asymptotic independence (L=1L=1) without evaluating the limit.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 12
ConceptIndependent vs Disjoint Events
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree