Verified Solution • 2025–26 Session moderate

QJEEM24Ma6046: JEE Main 2024 Mathematics - Class 12 JEEM24Ma6046 Solution: If the function has a local maximum at and a local minimum at , …

If the function f(x)=2 x^3-9 \mathrm{ax}^2+12 \mathrm{a}^2 x+1, \mathrm{a}> 0 has a local maximum at x=αx=\alpha and a local minimum at x=α2x=\alpha^2 , then α\alpha and α2\alpha^2 are the roots of the equation :

Multiple Choice Options

  • A x2−6x+8=0x^2-6 x+8=0
  • B 8x2−6x+1=08 x^2-6 x+1=0
  • C 8x2+6x−1=08 x^2+6 x-1=0
  • D x2+6x+8=0x^2+6 x+8=0

Step-by-Step Verified Solution

amp;f(x)=6x2−18ax+12a2amp;=6(x2−3a+2a2)amp;=6(x−a)(x−2a)=0amp;x=a,2a\begin{aligned}& f(x)=6 x^2-18 a x+12 a^2 \\& =6\left(x^2-3 a+2 a^2\right) \\& =6(x-a)(x-2 a)=0 \\& x=a, 2 a\end{aligned}

a=α,2a=α2⇒α=0,2a=\alpha, \quad 2 a=\alpha^2 \quad \Rightarrow \alpha=0,2

agt;0amp;∴amp;α=2amp;amp;α2=4\begin{array}{lll}a>0 & \therefore & \alpha=2 \\& & \alpha^2=4\end{array}

∴x2−6x+8=0\therefore x^2-6 x+8=0 is the required quadratic equation.

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