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(D) Let $f(x) = 5x^3 - 15x$. The equation $5x^3 - 15x - a = 0$ can be written as $f(x) = a$.To have three distinct real roots,the horizontal line $y =

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$30$

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(D) Let $f(x) = 5x^3 - 15x$. The equation $5x^3 - 15x - a = 0$ can be written as $f(x) = a$.To have three distinct real roots,the horizontal line $y = a$ must intersect the graph of $f(x)$ at three distinct points.First,find the critical points of $f(x)$ by setting $f'(x) = 0$:$f'(x) = 15x^2 - 15 = 15(x^2 - 1) = 15(x - 1)(x + 1)$.The critical points are $x = 1$ and $x = -1$.The local maximum value is $f(-1) = 5(-1)^3 - 15(-1) = -5 + 15 = 10$.The local minimum value is $f(1) = 5(1)^3 - 15(1) = 5 - 15 = -10$.For the equation $f(x) = a$ to have three distinct real roots,$a$ must lie strictly between the local minimum and local maximum values:$-10 Thus,the interval $(\alpha, \beta)$ is $(-10, 10)$,so $\alpha = -10$ and $\beta = 10$.We need to calculate $\beta - 2\alpha$:$\beta - 2\alpha = 10 - 2(-10) = 10 + 20 = 30$.

If the set of all values of $a$,for which the equation $5x^3 - 15x - a = 0$ has three distinct real roots,is the interval $(\alpha, \beta)$,then $\beta - 2\alpha$ is equal to . . . . . . .

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