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  • Let <img alt="\left(x^3+\alpha x^2+2 x-5\right)^{19}\left(x^2+\beta x-41\right)^8\left(x^4-x^3+x-7\right)^6=" src="https://

Let \left(x^3+\alpha x^2+2 x-5\right)^{19}\left(x^2+\beta x-41\right)^8\left(x^4-x^3+x-7\right)^6=x^{97}+391 x^{96}+a_{95} x^{95}+a_{94} x^{94}+\ldots+a_1 x+a_0 be an identity, where  \alpha, \beta, a_{95}, a_{94}, \ldots, a_1, a_0 are integers. If \alpha+\beta<10, then the smallest possible value of \alpha is

Option: 1

7


Option: 2

8


Option: 3

31


Option: 4

23


Answers (1)

best_answer

It will be an identity even if we replace x by  \frac{1}{y} and considering numerator alone. Differentiating on both sides with respect to y, at y=0 we get  19 \alpha+8 \beta=397, \alpha+\beta=10-\mathrm{k} where k  is positive integer. Put  \beta=10-\alpha-\mathrm{k} in first equation we get 11\alpha-8 \mathrm{k}=317 \therefore \alpha=31 is the smallest possible value

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Shailly goel

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