JEE Main 2025 — Application Of Derivatives Question with Solution

To determine the value of $f(3)$ for the function $f(x)=2x^3-9a{x}^2+12a^{2}x+1$, where $a>0$, we follow these steps:

First, find the critical points by setting the derivative equal to zero:

$f^{\prime }(x)=6x^2-18ax+12a^2=0$

Factoring gives:

$6(x-a)(x-2a)=0$

Thus, the critical points are $x=a$ and $x=2a$.

$x=a$ corresponds to a local maximum.

$x=2a$ corresponds to a local minimum.

According to the problem, $p^2=q$. Substituting $p=a$ and $q=2a$ gives:

$a^2=2a$

Solving for $a$ gives:

$a(a-2)=0$

Since $a>0$, we have $a=2$.

Now, substitute $a=2$ back into the function:

$f(x)=2x^3-18x^2+48x+1$

To find $f(3)$:

$f(3)=2(3{)}^3-18(3{)}^2+48(3)+1$

Calculate each term:

$2(3{)}^3=54$

$18(3{)}^2=162$

$48(3)=144$

Thus,

$f(3)=54-162+144+1=37$

So, $f(3)=37$.