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JEE Main 20252 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Application of Derivatives

If the function f(x)=2 x^3-9 ax^2+12 a^2 x+1 , where a 0 , attains...

If the function f(x)=2x3−9ax2+12a2x+1f(x)=2 x^3-9 \mathrm{ax}^2+12 \mathrm{a}^2 \mathrm{x}+1, where a>0\mathrm{a} \gt 0, attains its local maximum and local minimum values at pp and qq, respectively, such that p2=q\mathrm{p}^2=\mathrm{q}, then f(3)f(3) is equal to:

Options

  1. A55
  2. B10
  3. C23
  4. D37

Correct answer

D. 37

Step-by-step solution

f′(x)=6x2−18ax+12a2f^{\prime}(x)=6 x^2-18 a x+12 a^2
f′(x)=6(x2−3ax+2a2)f^{\prime}(x)=6\left(x^2-3 a x+2 a^2\right)
roots are a,2aa, 2 a
p2=q⇒a2=2a\mathrm{p}^2=\mathrm{q} \Rightarrow \mathrm{a}^2=2 \mathrm{a}
a=2a=2
f(x)=2x3−18x2+48x+1f(x)=2 x^3-18 x^2+48 x+1
f(3)=37f(3)=37
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