If alpha and beta (where alpha beta) are tw | vedclass

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(D) Given the equation: $3\cos^{-1}\left(x^2 - 5x - \frac{11}{2}\right) = \pi$Dividing by $3$,we get: $\cos^{-1}\left(x^2 - 5x - \frac{11}{2}\right) =

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$35$

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(D) Given the equation: $3\cos^{-1}\left(x^2 - 5x - \frac{11}{2}\right) = \pi$Dividing by $3$,we get: $\cos^{-1}\left(x^2 - 5x - \frac{11}{2}\right) = \frac{\pi}{3}$Taking cosine on both sides: $x^2 - 5x - \frac{11}{2} = \cos\left(\frac{\pi}{3}\right)$Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$,we have: $x^2 - 5x - \frac{11}{2} = \frac{1}{2}$Rearranging the terms: $x^2 - 5x - 6 = 0$Factoring the quadratic equation: $(x - 6)(x + 1) = 0$Thus,the roots are $x = 6$ and $x = -1$.Given $\alpha > \beta$,we have $\alpha = 6$ and $\beta = -1$.Now,calculate $(\alpha^2 + \beta^3) = (6)^2 + (-1)^3 = 36 - 1 = 35$.

If $\alpha$ and $\beta$ (where $\alpha > \beta$) are two zeroes of the equation $3\cos^{-1}\left(x^2 - 5x - \frac{11}{2}\right) = \pi$,then $(\alpha^2 + \beta^3)$ is -

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