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Quadratic in x with definite-integral coefficients | JEE

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Question
If xx satisfies (∫01dtt2+2tcos⁡α+1)x2−(∫−33t2sin⁡2tt2+1 dt)x−2=0\left(\displaystyle\int_{0}^{1}\dfrac{dt}{t^{2}+2t\cos\alpha+1}\right)x^{2}-\left(\displaystyle\int_{-3}^{3}\dfrac{t^{2}\sin 2t}{t^{2}+1}\,dt\right)x-2=0 for 0<α<π0<\alpha<\pi, then the value of xx is
A±α2sin⁡α\pm\sqrt{\dfrac{\alpha}{2\sin\alpha}}
B±2sin⁡αα\pm\sqrt{\dfrac{2\sin\alpha}{\alpha}}
C±αsin⁡α\pm\sqrt{\dfrac{\alpha}{\sin\alpha}}
D±2sin⁡αα\pm 2\sqrt{\dfrac{\sin\alpha}{\alpha}}correct
Solution
Step 1: The middle coefficient vanishes. The integrand t2sin⁡2tt2+1\dfrac{t^{2}\sin 2t}{t^{2}+1} is odd in tt, so over the symmetric interval [−3,3][-3,3],
∫−33t2sin⁡2tt2+1 dt=0.\int_{-3}^{3}\dfrac{t^{2}\sin 2t}{t^{2}+1}\,dt=0.
 Step 2: Evaluate the leading coefficient. Complete the square: t2+2tcos⁡α+1=(t+cos⁡α)2+sin⁡2αt^{2}+2t\cos\alpha+1=(t+\cos\alpha)^{2}+\sin^{2}\alpha, hence
∫01dt(t+cos⁡α)2+sin⁡2α=1sin⁡α[tan⁡−1t+cos⁡αsin⁡α]01.\int_{0}^{1}\dfrac{dt}{(t+\cos\alpha)^{2}+\sin^{2}\alpha}=\dfrac{1}{\sin\alpha}\Big[\tan^{-1}\dfrac{t+\cos\alpha}{\sin\alpha}\Big]_{0}^{1}.
 Step 3: Simplify the bracket. Using 1+cos⁡αsin⁡α=cot⁡α2\dfrac{1+\cos\alpha}{\sin\alpha}=\cot\dfrac{\alpha}{2} and cos⁡αsin⁡α=cot⁡α\dfrac{\cos\alpha}{\sin\alpha}=\cot\alpha,
tan⁡−1 ⁣(cot⁡α2)−tan⁡−1 ⁣(cot⁡α)=(π2−α2)−(π2−α)=α2.\tan^{-1}\!\big(\cot\tfrac{\alpha}{2}\big)-\tan^{-1}\!\big(\cot\alpha\big)=\left(\dfrac{\pi}{2}-\dfrac{\alpha}{2}\right)-\left(\dfrac{\pi}{2}-\alpha\right)=\dfrac{\alpha}{2}.
 Therefore the leading coefficient is 1sin⁡α⋅α2=α2sin⁡α.\dfrac{1}{\sin\alpha}\cdot\dfrac{\alpha}{2}=\dfrac{\alpha}{2\sin\alpha}. Step 4: The equation reduces to α2sin⁡α x2−2=0\dfrac{\alpha}{2\sin\alpha}\,x^{2}-2=0, so
x2=4sin⁡αα ⇒ x=±2sin⁡αα.x^{2}=\dfrac{4\sin\alpha}{\alpha}\ \Rightarrow\ x=\pm 2\sqrt{\dfrac{\sin\alpha}{\alpha}}.
 Correct answer: (4)
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