JEE Mathematics 2025 Question with Solution

Given: $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ with $a > 0$.

Find: The value of $f(3)$ when the local maximum and minimum occur at $p$ and $q$ respectively, and $p^2 = q$.

Local extrema occur at critical points, so first compute the derivative:

$$f'(x) = \frac{d}{dx}(2x^3 - 9ax^2 + 12a^2x + 1) = 6x^2 - 18ax + 12a^2$$

Set $f'(x) = 0$:

$$6x^2 - 18ax + 12a^2 = 0$$

Dividing by $6$,

$$x^2 - 3ax + 2a^2 = 0$$

Factorizing,

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

So the critical points are $x=a$ and $x=2a$.

Now use the second derivative:

$$f''(x) = 12x - 18a$$

At $x=a$,

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

So the function has a local maximum at $x=a$. Hence, $p=a$.

At $x=2a$,

$$f''(2a) = 24a - 18a = 6a > 0$$

So the function has a local minimum at $x=2a$. Hence, $q=2a$.

Given $p^2 = q$, we get

$$a^2 = 2a$$ $$a(a-2)=0$$

Since $a>0$, it follows that

$$a=2$$

Now substitute $a=2$ into the function:

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

Then,

$$f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1$$ $$f(3) = 54 - 162 + 144 + 1$$ $$f(3) = 37$$

Therefore, the value of $f(3)$ is $37$, so the correct option is D.

Given: The local maximum and minimum occur at $p$ and $q$ for $f(x)=2x^3-9ax^2+12a^2x+1$.

Find: $f(3)$.

From

$$f'(x)=6x^2-18ax+12a^2=6(x-a)(x-2a)$$

the critical points are immediately $a$ and $2a$. Since for a cubic with positive leading coefficient, the smaller critical point is the local maximum and the larger one is the local minimum, we take

$$p=a, \qquad q=2a$$

Using $p^2=q$,

$$a^2=2a$$

Since $a>0$,

$$a=2$$

Now evaluate:

$$f(3)=2(27)-9(2)(9)+12(4)(3)+1=54-162+144+1=37$$

Therefore, the correct option is D.