Let $\mathrm{a} \gt 0$. If the function $\mathrm{f}(\mathrm{x})=6 \mathrm{x}^3-45 \mathrm{a}…

Let $\mathrm{a} \gt 0$. If the function $\mathrm{f}(\mathrm{x})=6 \mathrm{x}^3-45 \mathrm{a} \mathrm{x}^2+108 \mathrm{a}^2 \mathrm{x}+1$ attains its local maximum and minimum values at the points $x_1$ and $x_2$ respectively such that $x_1 x_2=54$, then $\mathrm{a}+\mathrm{x}_1+\mathrm{x}_2$ is equal to :-
  1. $15$
  2. $18$
  3. $24$
  4. $13$

Solution

$\begin{aligned} & f^{\prime}(x)=18 x^2-90 a x+108 a^2=0 \\ & x=2 a \& x=3 a \\ & x_1=2 a \quad x_2=3 a \\ & x_1 x_2=54 \\ & 6 a^2=54 \\ & a=3 \\ & a+x_1+x_2 \\ & 3+2 \times 3+3 \times 3=18 \\ & \text { option }(2)\end{aligned}$

Asked in: JEE Main 2025 (04 Apr Shift 2)

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