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Question #40728

Difficulty: 8/10 | 🟢 Free

The Question

Let a point $P(x, y)$ move such that the following system of linear equations in variables $U, V, W$ : $(t^2+1)U + e^t V + \ln(t) W = 0$ $(t^3+t)U + (t e^t + 5x + 12y - 39)V + (t \ln(t) + \sin(t))W = 0$ $\left(t+\frac{1}{t}\right)U + \frac{e^t}{t} V + \left(\frac{\ln(t)}{t} + \sqrt{t-1}\right)W = 0$ has a non-trivial solution for all $t \in (1, 2)$ . If the locus of $P$ is a curve $L$ , then the shortest distance from the origin to $L$ is:

Options

A.
$1$
B.
$2$
C.
$3$
D.
$4$

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

Locus of a Moving Point|System of Linear Equations (Cramer/Inverse)|Triangular/Diagonal Matrix Shortcut

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to evaluate the determinant of a seemingly complex parameterized matrix by recognizing it can be reduced to an upper triangular form.

The primary trap is the 'Lethal Collision' of concepts: students often attempt to expand the determinant using Sarrus' Rule or standard cofactor expansion, which leads to an intractable algebraic mess involving exponential, logarithmic, and trigonometric functions.

By applying elementary row operations, the matrix simplifies drastically, revealing that the determinant is identically zero for a range of 't' only if the locus-dependent factor vanishes.

This elegantly collapses the locus into a simple straight line, penalizing brute-force calculation and rewarding structural observation of singular matrices.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Locus of a Moving Point
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free