Question ID: #262
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
Solution:
Let $f(x) = 5x^3 – 15x$.
The equation is $f(x) = a$. For three distinct real roots, the line $y=a$ must cut the graph of $f(x)$ at three distinct points.
Find local maxima and minima:
$f'(x) = 15x^2 – 15 = 15(x-1)(x+1)$.
Critical points at $x=1$ and $x=-1$.
Values at critical points:
$f(1) = 5(1)^3 – 15(1) = 5 – 15 = -10$ (Local Minimum)
$f(-1) = 5(-1)^3 – 15(-1) = -5 + 15 = 10$ (Local Maximum)
For 3 distinct roots, $a$ must lie strictly between the local minimum and local maximum values.
$-10 < a < 10$.
Thus, interval $(\alpha, \beta)$ is $(-10, 10)$.
Here, $\alpha = -10$ and $\beta = 10$.
Value of $\beta – 2\alpha$:
$10 – 2(-10) = 10 + 20 = 30$
Ans. 30
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