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Determinant of aₙ=∫(1−cos2nx)/(1−cos2x)dx | JEE

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Question
Let an=∫0π/21−cos⁡2nx1−cos⁡2x dxa_{n}=\displaystyle\int_{0}^{\pi/2}\dfrac{1-\cos 2nx}{1-\cos 2x}\,dx. Then the value of ∣a1a2a3a4a5a6a7a8a9∣\begin{vmatrix}a_{1} & a_{2} & a_{3}\\[2pt] a_{4} & a_{5} & a_{6}\\[2pt] a_{7} & a_{8} & a_{9}\end{vmatrix} equals
A11
B22
C33
D00correct
Solution
Step 1: Show the sequence is arithmetic. Consider the combination
an+an+2−2an+1=∫0π/2(1−cos⁡2nx)+(1−cos⁡(2n+4)x)−2(1−cos⁡(2n+2)x)1−cos⁡2x dx.a_{n}+a_{n+2}-2a_{n+1}=\int_{0}^{\pi/2}\dfrac{\big(1-\cos 2nx\big)+\big(1-\cos(2n+4)x\big)-2\big(1-\cos(2n+2)x\big)}{1-\cos 2x}\,dx.
 The numerator collapses to 2cos⁡(2n+2)x−cos⁡2nx−cos⁡(2n+4)x2\cos(2n+2)x-\cos 2nx-\cos(2n+4)x. Step 2: Apply the product-to-sum identity cos⁡2nx+cos⁡(2n+4)x=2cos⁡(2n+2)xcos⁡2x\cos 2nx+\cos(2n+4)x=2\cos(2n+2)x\cos 2x:
2cos⁡(2n+2)x−2cos⁡(2n+2)xcos⁡2x=2cos⁡(2n+2)x(1−cos⁡2x).2\cos(2n+2)x-2\cos(2n+2)x\cos 2x=2\cos(2n+2)x\big(1-\cos 2x\big).
 Therefore the integrand reduces, and the denominator cancels:
an+an+2−2an+1=∫0π/22cos⁡(2n+2)x dx.a_{n}+a_{n+2}-2a_{n+1}=\int_{0}^{\pi/2}2\cos(2n+2)x\,dx.
 Step 3: Evaluate this integral:
∫0π/22cos⁡(2n+2)x dx=2[sin⁡(2n+2)x2n+2]0π/2=2sin⁡((n+1)π)2n+2=0.\int_{0}^{\pi/2}2\cos(2n+2)x\,dx=2\left[\dfrac{\sin(2n+2)x}{2n+2}\right]_{0}^{\pi/2}=\dfrac{2\sin\big((n+1)\pi\big)}{2n+2}=0.
 Hence an+an+2−2an+1=0a_{n}+a_{n+2}-2a_{n+1}=0, so {an}\{a_{n}\} is an arithmetic progression (with a1=∫0π/21 dx=π2a_{1}=\displaystyle\int_{0}^{\pi/2}1\,dx=\dfrac{\pi}{2}). Step 4: Use the A.P. property in the determinant. Apply the column operation C1→C1+C3−2C2C_{1}\to C_{1}+C_{3}-2C_{2}. Each entry of the new first column has the form ak+ak+2−2ak+1=0a_{k}+a_{k+2}-2a_{k+1}=0, producing a zero column. A determinant with a zero column is 00. Correct answer: (4)
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