Animated Solution for Mathematics - Differentiation: Let $f(x)$ be a polynomial function such that $f(x)+f^{\prime }(x)+f^{\prime \prime }(x)=x^5+64$. Then, the value of ${\lim }_{x\to 1}\frac{f(x)}{x-1}$ is

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the elegant architecture of calculus.

When you look at a problem like this—a polynomial hidden behind a differential equation—it is easy to feel overwhelmed. But I want you to take a deep breath. We are going to peel back the layers of this function, $f(x)$, one by one, until its secrets are laid bare.

The Gateway of Limits

We begin with the expression ${\lim }_{x\to 1}\frac{f(x)}{x-1}$. This is our compass.

In the world of JEE Advanced, whenever you see a limit where the denominator vanishes as $x\to 1$, your intuition should immediately scream: "Indeterminate form!" If the numerator $f(x)$ did not also vanish at $x=1$, the limit would diverge to infinity.

But we are looking for a specific, finite value. Therefore, we must have $f(1)=0$.

With this realization, we invoke the power of L'Hopital's Rule. Since we have a $\frac{0}{0}$ form, we differentiate the numerator and denominator with respect to $x$:

Suddenly, the problem transforms. We no longer care about the entire function $f(x)$; we only need to find the value of its derivative at $x=1$. The hunt is on!

The Algebraic Dance

We are given the master equation:

Look at the right-hand side. It is a polynomial of degree $5$. This tells us everything we need to know about $f(x)$.

Since the derivatives $f^{\prime }(x)$ and $f^{\prime \prime }(x)$ have lower degrees than $f(x)$, the leading term $x^5$ must have originated from $f(x)$ itself. This confirms that $f(x)$ is a fifth-degree polynomial.

Now, here is where the magic happens. Instead of solving for every coefficient, let us differentiate this equation repeatedly. This is a technique of "successive reduction."

1. Original: $f(x)+f^{\prime }(x)+f^{\prime \prime }(x)=x^5+64$ 2. First Derivative: $f^{\prime }(x)+f^{\prime \prime }(x)+f^{\prime \prime \prime }(x)=5x^4$ 3. Second Derivative: $f^{\prime \prime }(x)+f^{\prime \prime \prime }(x)+f^{(4)}(x)=20x^3$ 4. Third Derivative: $f^{\prime \prime \prime }(x)+f^{(4)}(x)+f^{(5)}(x)=60x^2$

The Grand Cancellation

We are closing in on our target. We need $f^{\prime }(1)$. From our first equation, if we plug in $x=1$, we get:

Since we already established that $f(1)=0$, this simplifies beautifully to:

To find $f^{\prime }(1)$, we now only need $f^{\prime \prime }(1)$. Look at our system of derivatives. If we subtract the second derivative equation from the third, watch what happens:

Wait, let us refine our approach to isolate $f^{\prime \prime }(1)$ directly. If we look at the third derivative equation: $f^{\prime \prime \prime }(x)+f^{(4)}(x)+f^{(5)}(x)=60x^2$.

The Final Victory

We know that for a polynomial of degree $5$, the fifth derivative is a constant: $f^{(5)}(x)=5!=120$.

Using the relation $f^{\prime \prime }(x)+f^{\prime \prime \prime }(x)+f^{(4)}(x)=20x^3$, and evaluating the derivatives at $x=1$, we can deduce the values. However, a more direct path is to use the system:

By substituting $x=1$ into the derivative chain:

Solving for $f^{\prime \prime }(1)$, we get:

Finally, we return to our Relation A: $f^{\prime }(1)+f^{\prime \prime }(1)=65$. Substituting our value for $f^{\prime \prime }(1)$:

And there it is. The limit is $-15$.

Do you see the beauty in this? We didn't need to find the full polynomial. We simply used the structure of the calculus to isolate the information we needed. Keep this mindset—look for the path of least resistance, trust the math, and you will conquer any problem JEE throws your way.