The number of solution(s) of the equation e∫0∞tdx(x+x2025)(x+1x)=∫0txet2−x2dx is R, then the value of R2−42R+1000 is equal to _______

et∫0∞dx(x2+1)(1+x2024)=et2∫0txe−x2dx eπ4t=12(et2−1)By graph, we get 2 solutions R=2 R2−42R+1000=4−84+1000=1000−80=920

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The number of solution(s) of the equation $e^{\int_{0}^{\infty} \frac{t d x}{(x + x^{2025}) (x + \frac{1}{x})}} = \int_{0}^{t} x e^{t^{2} - x^{2}} d x$ is R, then the value of $R^{2} - 42 R + 1000$ is equal to _______

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answer is 920.

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Detailed Solution

$e^{t \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (1 + x^{2024})}} = e^{t^{2}} \int_{0}^{t} x e^{- x^{2} d x} e^{\frac{π}{4} t} = \frac{1}{2} (e^{t^{2}} - 1)$
By graph, we get 2 solutions
Question Image
$R = 2$
$R^{2} - 42 R + 1000 = 4 - 84 + 1000 = 1000 - 80 = 920$