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JEE Main 20254 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Application of Derivatives

Let a 0 . If the function f(x)=6 x^3-45 a x^2+108 a^2 x+1 attains...

Let a>0\mathrm{a} \gt 0. If the function f(x)=6x3−45ax2+108a2x+1\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 x1x_1 and x2x_2 respectively such that x1x2=54x_1 x_2=54, then a+x1+x2\mathrm{a}+\mathrm{x}_1+\mathrm{x}_2 is equal to :-

Options

  1. A1515
  2. B1818
  3. C2424
  4. D1313

Correct answer

B. 1818

Step-by-step solution

f′(x)=18x2−90ax+108a2=0x=2ax=3ax1=2ax2=3ax1x2=546a2=54a=3a+x1+x23+2×3+3×3=18 option (2)\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}
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