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Question #41890

Difficulty: 8/10 | 🟢 Free

The Question

Let $E_1, E_2,$ and $E_3$ be mutually exclusive and exhaustive events in a sample space. The probabilities of events $E_1$ and $E_2$ are given by: $P(E_1) = \lim_{x \to 0} \frac{1}{x^3} \int_0^{\sin x - x \cos x} e^{t^2} dt$ $P(E_2) = \lim_{x \to 0} \frac{1}{x^3} \int_0^{\sin^{-1}x - x} \frac{\ln(1+t)}{t} dt$ If $P(E_3)$ is the probability of the event $E_3$ , then the value of $L = \lim_{x \to 0} \frac{1}{x^3} \int_0^{P(E_3) \cdot x} \sin(t^2) dt$ is equal to:

Options

A.
$\frac{1}{24}$
B.
$\frac{1}{12}$
C.
$\frac{1}{8}$
D.
$\frac{1}{6}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Exhaustive Events Concept|Newton-Leibniz Formula with variable limits|Trigonometric Standard Limits

Why This Question Is Tricky

Common Mistakes

This question fuses probability axioms with advanced limit evaluation using the Newton-Leibniz formula.

The primary trap lies in the differentiation of the variable upper limits: students frequently forget the chain rule component (multiplying by the derivative of the limit itself) or mishandle the standard limit $\lim_{t \to 0} \frac{\ln(1+t)}{t} = 1$ embedded within the integral.

The 'exhaustive events' condition acts as a rigorous self-check mechanism; if the limits are evaluated incorrectly, the probabilities will not sum to 1, leading to an erroneous $P(E_3)$ and a cascading failure in the final integral limit.

This effectively punishes rote application of L'Hôpital's rule without careful algebraic expansion.

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