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Question #75811

Difficulty: 8/10 | 🟢 Free

The Question

Let AA be an n×nn \times n matrix (n≥2n \ge 2) with det⁡(A)=c2>0\det(A) = c^2 > 0. Let D(t)=det⁡(ktA)D(t) = \det(k t A) where kk is a non-zero scalar and t>0t > 0 is a parameter. The normal to the rectangular hyperbola xy=c2xy = c^2 at the point P(ct,ct)P\left(ct, \frac{c}{t}\right) intersects the y-axis at Y0Y_0. Which of the following correctly expresses Y0Y_0 in terms of the derivative D′(t)D'(t)?

Options

  • A.

    ct−t4−nD′(t)nknc\frac{c}{t} - \frac{t^{4-n} D'(t)}{n k^n c}

  • B.

    ct−t3D′(t)kc\frac{c}{t} - \frac{t^3 D'(t)}{k c}

  • C.

    ct−t3D′(t)knc\frac{c}{t} - \frac{t^3 D'(t)}{k^n c}

  • D.

    ct−t4−nD′(t)nkc\frac{c}{t} - \frac{t^{4-n} D'(t)}{n k c}

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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

Scalar Multiplication Geometric Scaling|Rectangular Hyperbola|Differentiation of Determinants|Elementary Row Operation Invariance

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the multilinear property of determinants and its integration with coordinate geometry.

A common trap (Concept 215 error) is treating the determinant as a linear operator, incorrectly assuming det⁡(ktA)=ktdet⁡(A)\det(ktA) = kt\det(A) or failing to scale both kk and tt by the matrix order nn.

This leads to a dimensional mismatch when substituting the derivative back into the geometric equation of the normal's y-intercept.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterVector Algebra - Class 12
ConceptScalar Multiplication Geometric Scaling
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree