Verified Solution • 2025–26 Session moderate

QJEEM03Maa2c2: JEE Main 2003 Mathematics - Class 12 JEEM03Maa2c2 Solution: If the function where attains its maximum and minimum at and…

If the function f(x)=2x3−9ax2+12a2x+1,f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1, where a>0, attains its maximum and minimum at pp and qq respectively such that p2=q{p^2} = q , then aa equals

Multiple Choice Options

  • A 12{1 \over 2}
  • B 33
  • C 11
  • D 22

Step-by-Step Verified Solution

f(x)=2x3−9ax2+12a2x+1f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1

f′(x)=6x2−18ax+12a2;f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};

          f′′(x)=12x−18a\,\,\,\,\,\,\,\,\,\,f''\left( x \right) = 12x - 18a

For max. or min.

6x2−18ax+12a2=06{x^2} - 18ax + 12{a^2} = 0

⇒x2−3ax+2a2=0\Rightarrow {x^2} - 3ax + 2{a^2} = 0

⇒x=a\Rightarrow x = a or x=2a.x=2a.

At x=ax=a

f''(a) = 12a - 18a = -6a < 0

At x=ax=a maximum As f''( aa ) < 0.

At x=2ax=2a

f''(a) = 24a - 18a = 6a &gt; 0

At x=2ax=2a minimum As f''(2 aa ) > 0.

∴\therefore p=ap=a and q=2aq=2a

As per question p2=q{p^2} = q

∴\therefore a2=2a{a^2} = 2a

⇒\Rightarrow a(a−2)=0a(a - 2) = 0

∴\therefore aa = 0, 2

but a &gt; 0, therefore, a=2.a=2.

Related Learning Resources & Exam Preparation