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Question #73578

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \int_0^x e^{-t^2} (t^2 - 4)(t - 3)^2 dt$ and $\lambda$ be the number of points where $f(x)$ attains a local extremum. Consider two skew lines $L_1$ and $L_2$ given by: $L_1: \frac{x-1}{2} = \frac{y+1}{1} = \frac{z-2}{-1}$ $L_2: \frac{x+2}{1} = \frac{y-1}{2} = \frac{z-\mu}{1}$ If the shortest distance between $L_1$ and $L_2$ is at most $\lambda \sqrt{3}$ , then the sum of all possible integer values of $\mu$ is:

Options

A.
91
B.
133
C.
49
D.
112

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

Geometric Interpretation of Cross Product|Shortest distance between skew lines|Modulus Inequality Logic|First Derivative Test Trap

Why This Question Is Tricky

Examiner Levers

Logical Assertion

Tests the causal relationship between two true statements.

Difficulty Boost : +4

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the intersection of vector geometry and differential calculus.

The primary trap lies in the First Derivative Test: students frequently assume that all roots of $f'(x)=0$ (critical points) are points of local extrema.

Here, $x=3$ is a stationary point but not an extremum because the derivative does not change sign (it is a point of inflection).

Failing to verify this yields an incorrect parameter $\lambda=3$ instead of $\lambda=2$ , which subsequently corrupts the modulus inequality bound for the shortest distance between the skew lines, leading directly to the trap answer 133.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Geometric Interpretation of Cross Product
Difficulty	8
Cognitive Load	8
Target Solve Time	180 sec
Access	Free