Animated Solution for Mathematics - Differentiation: If $f(x)=x^3-x^2{f}^{\prime }(1)+x{f}^{\prime \prime }(2)-f^{\prime \prime \prime }(3),x\in \mathbb{R}$, then

Welcome, future engineer. Today, we are going to dismantle a problem that often intimidates students at first glance. It looks like a recursive nightmare—a function defined by its own derivatives.

But I want you to take a deep breath. In the world of JEE Advanced, the most complex-looking problems are often just simple concepts wearing a clever disguise. Let us peel back the mask.

The Art of Substitution

Look at the given function:

Your instinct might be to panic. How can we define a function using its own derivatives? But look closer at the terms $f^{\prime }(1)$, $f^{\prime \prime }(2)$, and $f^{\prime \prime \prime }(3)$.

These are not functions of $x$. They are values. They are constants. Let us stop treating them as scary symbols and start treating them as what they are: numbers. Let us define them as follows:

$a=f^{\prime }(1)$ $b=f^{\prime \prime }(2)$ $c=f^{\prime \prime \prime }(3)$

Now, our function transforms into something beautiful and familiar:

This is just a standard cubic polynomial. The entire problem has now shifted from a calculus mystery to a simple puzzle of finding three constants: $a$, $b$, and $c$.

The Calculus Engine

To find these constants, we need to use the definitions we just created. We need the derivatives of our new polynomial $f(x)=x^3-a{x}^2+bx-c$. Let us differentiate step-by-step:

First derivative:

Second derivative:

Third derivative:

Do you see the elegance here? The third derivative is a constant, $6$. This is the key that unlocks the entire problem.

Because $f^{\prime \prime \prime }(x)=6$ for all $x$, it must be true that $f^{\prime \prime \prime }(3)=6$. And since we defined $c=f^{\prime \prime \prime }(3)$, we have our first victory: $c=6$.

Solving the System

Now that we have $c$, we only need $a$ and $b$. We have two definitions left to exploit:

1. $a=f^{\prime }(1)$ 2. $b=f^{\prime \prime }(2)$

Using our derivative expressions, let us plug in the values:

For $a$: (Let us call this Equation 1).

For $b$: (Let us call this Equation 2).

We now have a classic system of linear equations:

Adding these two equations is a joy. The $b$ terms cancel out perfectly:

Now, substitute $a=3$ back into Equation 2:

The Final Verification

We have conquered the constants: $a=3$, $b=6$, and $c=6$. Our function is fully revealed:

To solve the problem, we simply evaluate the function at the necessary points to check the options:

$f(0)=-6$ $f(1)=1-3+6-6=-2$ $f(2)=8-12+12-6=2$ $f(3)=27-27+18-6=12$

Testing the option $2f(0)-f(1)+f(3)=f(2)$:

LHS: $2(-6)-(-2)+12=-12+2+12=2$ RHS: $f(2)=2$

It matches! We have successfully navigated the complexity and arrived at the truth. Remember, in JEE Advanced, never let the notation intimidate you. Break it down, define your constants, and let the calculus do the heavy lifting.