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Question #41372

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 2x & 4x^2 & 8x^3 \\ 3x & 9x^2 & 27x^3 \end{vmatrix}$ . Define a function $g(x) = f'(x) + 90x^4 - 240x^3$ . If $x = \alpha$ is the unique point of local minimum for $g(x)$ , find the value of $\alpha$ .

Options

A.
1
B.
$\frac{7}{5}$
C.
-2
D.
0

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

First Derivative Test Trap|Factor Theorem in Determinants|Determinant Scaling Property

Why This Question Is Tricky

Common Mistakes

This question tests the 'Derivative of a Determinant' trap combined with the First Derivative Test.

A common lethal mistake among students is assuming that the derivative of a determinant is the determinant of the derivative matrix (i.e., differentiating all elements simultaneously), which incorrectly yields $f'(x) = \det(A'(x)) = 72x^3$ .

This leads to a flawed function $g_{trap}(x) = 90x^4 - 168x^3$ and a false local minimum at $x = 7/5$ .

The correct approach requires applying the determinant scaling property first to extract $x^6$ , simplifying the function to $f(x) = 12x^6$ before differentiation, thereby revealing the true critical points.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	First Derivative Test Trap
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free