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Matrices & Determinants: Vmatrix Vmatrix Equal

JEE Maths question with a full step-by-step solution.

Question
If f(x)=∣x5∣sin⁡x∣2x4tan⁡3x1sec⁡2xsin⁡3xx45∣f(x) = \begin{vmatrix} x^5 & |\sin x| & 2x^4 \\ \tan^3 x & 1 & \sec 2x \\ \sin^3 x & x^4 & 5 \end{vmatrix}, then ∫−π/2π/2f(x) dx\displaystyle\int_{-\pi/2}^{\pi/2} f(x)\,dx is equal to:
A22
B−2-2
C00correct
DNone of these
Solution
Step 1: Determine the symmetry of ff Substituting x→−xx \to -x:
f(−x)=∣−x5∣sin⁡x∣2x4−tan⁡3x1sec⁡2x−sin⁡3xx45∣f(-x) = \begin{vmatrix} -x^5 & |\sin x| & 2x^4 \\ -\tan^3 x & 1 & \sec 2x \\ -\sin^3 x & x^4 & 5 \end{vmatrix}
 Step 2: Factor −1-1 from column 1
f(−x)=−1⋅∣x5∣sin⁡x∣2x4tan⁡3x1sec⁡2xsin⁡3xx45∣=−f(x)f(-x) = -1 \cdot \begin{vmatrix} x^5 & |\sin x| & 2x^4 \\ \tan^3 x & 1 & \sec 2x \\ \sin^3 x & x^4 & 5 \end{vmatrix} = -f(x)
 Step 3: Apply the property of odd functions over a symmetric interval Since f(−x)=−f(x)f(-x) = -f(x), ff is an odd function. Therefore:
∫−π/2π/2f(x) dx=0\int_{-\pi/2}^{\pi/2} f(x)\,dx = 0
 Answer: (3)
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