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Question #40432

Difficulty: 8/10 | 🟢 Free

The Question

Let $\Delta(x) = \begin{vmatrix} \sin\left(x+\frac{\pi}{4}\right) & \cos\left(x+\frac{\pi}{4}\right) & \ln(1+\sin^2 x) \\ \sin\left(x-\frac{\pi}{4}\right) & \cos\left(x-\frac{\pi}{4}\right) & \sin 2x \\ \sin\left(x+\frac{3\pi}{4}\right) & \cos\left(x+\frac{3\pi}{4}\right) & \cos 2x \end{vmatrix}$ and $f(x) = [\Delta(x)] \left( \Delta^2(x) - 1 \right)$ , where $[ \cdot ]$ denotes the greatest integer function. If $L$ and $R$ are the left-hand and right-hand derivatives of $f(x)$ at $x = \frac{\pi}{4}$ respectively, then the value of $L^2 + R^2$ is

Options

A.
4
B.
8
C.
16
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

Allied Angle Periodicity|Differentiation of Determinants|Greatest Integer Function Traps

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to simplify a determinant with allied angles and recognize the jump discontinuity introduced by the Greatest Integer Function (GIF).

The trap lies in blindly applying the product rule to $f(x) = [\Delta(x)](\Delta^2(x)-1)$ without analyzing the neighborhood of $x=\pi/4$ .

Because $\Delta(x)$ crosses an integer value at the phase shift, the GIF causes the function to split into two different definitions on the left and right.

Students who ignore the piecewise nature will incorrectly compute the derivative or assume it is undefined, missing the fact that the zero from $(\Delta^2(x)-1)$ ensures continuity, allowing distinct but valid LHD and RHD.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Allied Angle Periodicity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free