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Question #40843

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x)$ be the volume of the tetrahedron formed by the origin $O(0,0,0)$ and the points $A, B,$ and $C$ whose position vectors are given by: $\vec{a} = [x] \hat{i} + \cos(\pi [x]) \hat{j} + \hat{k}$ $\vec{b} = \hat{i} + ([x]-1) \hat{j} + \cos(\pi [x]) \hat{k}$ $\vec{c} = \cos(\pi [x]) \hat{i} + \hat{j} + ([x]-2) \hat{k}$ where $[x]$ denotes the greatest integer less than or equal to $x$ . The value of the integral $\int_0^4 6 f(x) \, dx$ is:

Options

A.
18
B.
16
C.
17
D.
20

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

Octant Sign Convention|Determinant as Geometric Area/Volume|Greatest Integer Function (GIF)

Why This Question Is Tricky

Common Mistakes

This question tests the geometric interpretation of the scalar triple product alongside piecewise evaluation using the Greatest Integer Function (GIF).

The volume of a tetrahedron is strictly positive, requiring the absolute value of the determinant.

As $x$ varies, the vertices shift across different octants, causing the determinant's sign to flip (specifically at $k=2$ ).

Students often blindly integrate the determinant function without applying the absolute value piecewise, leading to a 'signed volume' cancellation (yielding 16 instead of 18).

This traps those who lack rigor in handling absolute values over discontinuous intervals.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Octant Sign Convention
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free