JEE Main 2025 (Online) 23rd January Morning ShiftQuestion 1
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 $\_\_\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$30$
Solution
Concept. $5x^3-15x=a$ has three distinct roots when the line $y=a$ lies strictly between the local minimum and maximum values of $g(x)=5x^3-15x$. Formulas used. (1) $g'(x)=15x^2-15=15(x-1)(x+1)$Step 1 - Extreme values. By (1), $g$ has a local maximum at $x=-1$ and a local minimum at $x=1$: $$g(-1)=-5+15=10,\qquad g(1)=5-15=-10.$$ Step 2 - The interval. Three distinct roots iff $-10<a<10$, so $\alpha=-10$, $\beta=10$. Step 3 - Value. $\beta-2\alpha=10+20=30$. Answer: 30