The product of all possible values of alpha,f | vedclass

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(C) Let $L = \lim_{x 	o 0} \frac{1 - \cos(\alpha x) \cos((\alpha+1)x) \cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}$.Using the approximation $\cos 	heta \

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(C) Let $L = \lim_{x 	o 0} \frac{1 - \cos(\alpha x) \cos((\alpha+1)x) \cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}$.Using the approximation $\cos 	heta \approx 1 - \frac{	heta^2}{2}$ and $\sin 	heta \approx 	heta$ as $x 	o 0$:$L = \lim_{x 	o 0} \frac{1 - (1 - \frac{\alpha^2 x^2}{2})(1 - \frac{(\alpha+1)^2 x^2}{2})(1 - \frac{(\alpha+2)^2 x^2}{2})}{((\alpha+1)x)^2}$.Neglecting higher order terms of $x^4$ and above,the numerator becomes:$1 - (1 - \frac{x^2}{2}(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2)) = \frac{x^2}{2}(\alpha^2 + \alpha^2 + 2\alpha + 1 + \alpha^2 + 4\alpha + 4) = \frac{x^2}{2}(3\alpha^2 + 6\alpha + 5)$.Thus,$L = \frac{3\alpha^2 + 6\alpha + 5}{2(\alpha+1)^2} = 2$.$3\alpha^2 + 6\alpha + 5 = 4(\alpha^2 + 2\alpha + 1) = 4\alpha^2 + 8\alpha + 4$.Rearranging gives $\alpha^2 + 2\alpha - 1 = 0$.The product of the roots of this quadratic equation is given by $\frac{c}{a} = \frac{-1}{1} = -1$.

The product of all possible values of $\alpha$,for which $\lim_{x \to 0} \left( \frac{1 - \cos(\alpha x) \cos((\alpha + 1)x) \cos((\alpha + 2)x)}{\sin^2((\alpha + 1)x)} \right) = 2$,is:

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