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Question #72131

Difficulty: 5/10 | 🟢 Free

The Question

Let $f(x) = x^3 - 6x^2 + cx + 2$ . Given that $f(x)$ is strictly increasing on $\mathbb{R}$ , let $S$ be the set of all possible values of $c$ . Let $m \in S$ .

Line $L_1$ passes through the origin and is parallel to the vector $\hat{i} + m\hat{j} + 2\hat{k}$ . Line $L_2$ passes through the point $(-1, 0, -2)$ and is parallel to the vector $2\hat{i} + \hat{j} - \hat{k}$ .

If $d$ is the shortest distance between $L_1$ and $L_2$ , find the minimum possible value of $d$ over all valid $m$ .

Options

A.
$\frac{2\sqrt{30}}{5}$
B.
0
C.
$\sqrt{5}$
D.
$2\sqrt{6}$

Check Your Answer

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

Shortest distance between skew lines|Slope Sensitivity|Increasing/Decreasing Strictness

Why This Question Is Tricky

Common Mistakes

This question fuses differential calculus with 3D vector geometry to create a constrained optimization problem.

The primary trap lies in the parameter $m$ .

Students often assume skew lines can be made to intersect by varying a parameter, incorrectly concluding the minimum distance is 0.

However, the parameter $m$ is constrained by the condition that $f(x)$ is strictly increasing, which requires its derivative's discriminant to be non-positive ( $c \ge 12$ ).

By forcing students to first determine this domain $S = [12, \infty)$ and then optimize the shortest distance function $d(m)$ , the problem tests their ability to handle slope sensitivity and boundary extrema rather than just applying standard formulas.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Shortest distance between skew lines
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free