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Question #41542

Difficulty: 8/10 | 🟢 Free

The Question

Let $D(x) = \begin{vmatrix} \log_2 2 & \log_2 4 & \log_2 8 \\\\ x^3 & x^2 & x \\\\ \sec^2\left(\frac{\pi}{4}\right) & \tan^2\left(\frac{\pi}{4}\right) & 3 \end{vmatrix}$ and $f(x) = D'(x)$ .

Find the area of the region bounded by the curves $y = f(x)$ and $y = -3x^2 + 6x + 3$ .

Options

A.
8
B.
$4\sqrt{2}$
C.
40
D.
$8\sqrt{2}$

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

Horizontal vs Vertical Strips|Differentiation of Determinants|Row/Column Exchange Parity|Linear Dependency of Rows

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply the derivative to a determinant and compute the area between two curves.

The primary trap lies in the evaluation of the determinant $D(x)$ before differentiation.

A common reflex is to swap the first and second rows to place the variable terms at the top, making the expansion feel more standard.

However, students frequently forget that swapping two adjacent rows multiplies the determinant's value by $-1$ .

Failing to account for this parity change leads to the incorrect derivative $f_{wrong}(x) = 9x^2 + 6x - 3$ , which intersects the given parabola at $x = \pm 1/\sqrt{2}$ , resulting in the trap answer $4\sqrt{2}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Horizontal vs Vertical Strips
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free