JEE Main 2025 (Online) 4th April Evening ShiftQuestion 1
Let $a>0$. If the function $f(x)=6x^{3}-45ax^{2}+108a^{2}x+1$ attains its local maximum and minimum values at the points $x_1$ and $x_2$ respectively such that $x_1x_2=54$, then $a+x_1+x_2$ is equal to :
Differentiation and its Application
Solution
Concept. Factorise $f'$; for a cubic with positive leading coefficient the smaller root is the maximum. Formulas used. (1) $f'(x)=18x^2-90ax+108a^2=18(x-2a)(x-3a)$Step 1 - Max and min. By (1), with $a>0$: $f'$ changes $+\to-$ at $2a$ and $-\to+$ at $3a$. So $x_1=2a$, $x_2=3a$. Step 2 - Use $x_1x_2=54$. $6a^2=54$, so $a=3$, $x_1=6$, $x_2=9$. Step 3 - Value. $a+x_1+x_2=3+6+9=18$. Answer: (C)