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MathMedAdv 2017 • Paper 2

Derivative Critical Points and Extrema of Determinant Expression

If f(x)=cos⁡(2x)cos⁡(2x)sin⁡(2x)−cos⁡xcos⁡x−sin⁡xsin⁡xsin⁡xcos⁡xf(x) = \begin{array}{|ccc|} \cos(2x) & \cos(2x) & \sin(2x) \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{array}, then choose the true statement(s):

Options

A

f′(x)=0f'(x) = 0 at exactly three points in (−π,π)(-\pi, \pi)

B

f′(x)=0f'(x) = 0 at more than three points in (−π,π)(-\pi, \pi)

Correct
C

f(x)f(x) attains its maximum at x=0x = 0

Correct
D

f(x)f(x) attains its minimum at x=0x = 0

Topics & Concepts

DeterminantsDeterminantsMaxima and Minima

Step-by-Step Solution

To determine the correct statements, we first simplify the given determinant f(x)f(x):

f(x)=∣cos⁡(2x)cos⁡(2x)sin⁡(2x)−cos⁡xcos⁡x−sin⁡xsin⁡xsin⁡xcos⁡x∣f(x) = \begin{vmatrix} \cos(2x) & \cos(2x) & \sin(2x) \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{vmatrix}

Applying the column operation C2→C2−C1C_2 \to C_2 - C_1:

f(x)=∣cos⁡(2x)0sin⁡(2x)−cos⁡x2cos⁡x−sin⁡xsin⁡x0cos⁡x∣f(x) = \begin{vmatrix} \cos(2x) & 0 & \sin(2x) \\ -\cos x & 2\cos x & -\sin x \\ \sin x & 0 & \cos x \end{vmatrix}

Expanding along the second column (C2C_2):

f(x)=2cos⁡x∣cos⁡(2x)sin⁡(2x)sin⁡xcos⁡x∣f(x) = 2\cos x \begin{vmatrix} \cos(2x) & \sin(2x) \\ \sin x & \cos x \end{vmatrix}

f(x)=2cos⁡x(cos⁡(2x)cos⁡x−sin⁡(2x)sin⁡x)f(x) = 2\cos x \left(\cos(2x)\cos x - \sin(2x)\sin x\right)

Using the identity cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A \cos B - \sin A \sin B, we get:

f(x)=2cos⁡xcos⁡(3x)f(x) = 2\cos x \cos(3x)

Using the product-to-sum formula 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A \cos B = \cos(A+B) + \cos(A-B):

f(x)=cos⁡(4x)+cos⁡(2x)f(x) = \cos(4x) + \cos(2x)

Step 1: Analyze critical points where f′(x)=0f'(x) = 0

Differentiating f(x)f(x) with respect to xx:

f′(x)=−4sin⁡(4x)−2sin⁡(2x)f'(x) = -4\sin(4x) - 2\sin(2x)

Using the double angle formula sin⁡(4x)=2sin⁡(2x)cos⁡(2x)\sin(4x) = 2\sin(2x)\cos(2x):

f′(x)=−4(2sin⁡(2x)cos⁡(2x))−2sin⁡(2x)=−2sin⁡(2x)(4cos⁡(2x)+1)f'(x) = -4(2\sin(2x)\cos(2x)) - 2\sin(2x) = -2\sin(2x)(4\cos(2x) + 1)

For f′(x)=0f'(x) = 0 in the interval x∈(−π,π)x \in (-\pi, \pi), we have two cases:

  1. sin⁡(2x)=0\sin(2x) = 0: Since x∈(−π,π)x \in (-\pi, \pi), 2x∈(−2π,2π)2x \in (-2\pi, 2\pi). 2x=−π,0,π  ⟹  x=−π2,0,π22x = -\pi, 0, \pi \implies x = -\frac{\pi}{2}, 0, \frac{\pi}{2} This yields 33 distinct solutions.

  2. 4cos⁡(2x)+1=0  ⟹  cos⁡(2x)=−144\cos(2x) + 1 = 0 \implies \cos(2x) = -\frac{1}{4}: Since −14∈(−1,1)-\frac{1}{4} \in (-1, 1), cos⁡(2x)=−14\cos(2x) = -\frac{1}{4} has 22 solutions in (0,2π)(0, 2\pi) and 22 solutions in (−2π,0)(-2\pi, 0). This yields 44 distinct solutions for x∈(−π,π)x \in (-\pi, \pi).

Total number of points in (−π,π)(-\pi, \pi) where f′(x)=0f'(x) = 0 is 3+4=73 + 4 = 7.

Therefore, f′(x)=0f'(x) = 0 at more than three points in (−π,π)(-\pi, \pi).

  • This confirms Option B is correct and Option A is incorrect.

Step 2: Analyze nature of extrema at x=0x = 0

Evaluating the first and second derivatives at x=0x = 0:

f′(0)=−2sin⁡(0)(4cos⁡(0)+1)=0f'(0) = -2\sin(0)(4\cos(0) + 1) = 0

Differentiating f′(x)f'(x) to find f′′(x)f''(x):

f′′(x)=−16cos⁡(4x)−4cos⁡(2x)f''(x) = -16\cos(4x) - 4\cos(2x)

At x=0x = 0:

f′′(0)=−16cos⁡(0)−4cos⁡(0)=−16−4=−20<0f''(0) = -16\cos(0) - 4\cos(0) = -16 - 4 = -20 < 0

Since f′(0)=0f'(0) = 0 and f′′(0)<0f''(0) < 0, f(x)f(x) attains a local maximum at x=0x = 0.

Furthermore, since cos⁡(4x)≤1\cos(4x) \le 1 and cos⁡(2x)≤1\cos(2x) \le 1 for all real xx: f(x)=cos⁡(4x)+cos⁡(2x)≤1+1=2=f(0)f(x) = \cos(4x) + \cos(2x) \le 1 + 1 = 2 = f(0)

Thus, f(x)f(x) attains its maximum at x=0x = 0.

  • This confirms Option C is correct and Option D is incorrect.

Final Conclusion:

The true statements are B and C.

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