JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 5)
Let $\mathrm{a}>0$. If the function $f(x)=6 x^3-45 \mathrm{a} x^2+108 \mathrm{a}^2 x+1$ attains its local maximum and minimum values at the points $x_1$ and $x_2$ respectively such that $x_1 x_2=54$, then $\mathrm{a}+x_1+x_2$ is equal to :
15
13
24
18
Explanation
$$\begin{aligned} &f(x)=6 x^3-45 a x^2+108 a^2 x+1\\ &\text { For maxima or minima } f^{\prime}(x)=0 \end{aligned}$$
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$$\begin{aligned} & x_1 x_2=\frac{108 a^2}{18}=54 \\ & \Rightarrow a^2=9 \Rightarrow a=3 \\ & \text { Now, } a+x_1+x_2=3+\frac{90}{6}=3+15=18 \end{aligned}$$
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