Using Rolle's theorem, prove that there is at least one root in (45^{1/100}, 46) of the polynomial P

Q: Wait, why can't we just use energy conservation or some other method? Why Rolle's Theorem?

That is a fantastic question! In JEE Advanced, the choice of tool is half the battle. We cannot use energy conservation because this is a pure polynomial root problem, not a physics system. We use Rolle's Theorem because it is the most elegant bridge between a function's values at boundaries and the existence of a root for its derivative. Trying to solve a degree-101 polynomial algebraically is a dead end; Rolle's Theorem turns an impossible algebraic problem into a simple evaluation problem.

Q: Why can we just set the constant of integration $C$ to zero? Doesn't that change the function?

You are absolutely right that $C$ changes the function, but it shifts the entire graph vertically. Rolle's Theorem requires $F(a) = F(b)$. If we add a constant $C$ to both sides, the condition becomes $F(a) + C = F(b) + C$, which is still $F(a) = F(b)$. The constant cancels out! Setting $C=0$ is just a strategic choice to make our algebra cleaner.

Q: How did you know to group the terms like that in the factorization step?

Pattern recognition is a skill you build with practice! When I saw the coefficients $51$ and $102$, and $2323$ and $101$, I immediately saw the ratio $1/2$ and $23$. When you see such clean ratios, it is a massive hint that the expression is designed to be factored. Always look for common binomials like $(x-46)$ when you have high-degree polynomials in these types of problems.

Question

Using Rolle's theorem, prove that there is at least one root in $(45^{1/100}, 46)$ of the polynomial $P(x) = 51x^{101} - 2323x^{100} - 45x + 1035$.

Sigma Percentile · Accepted Answer

### The Mountain of Degree 101: A Journey into Rolle's Theorem

Imagine you are standing at the base of a massive, jagged mountain. This mountain is our polynomial, $P(x) = 51x^{101} - 2323x^{100} - 45x + 1035$. It is a beast of degree 101.

If you try to climb it by solving for its roots directly, you will be lost in a forest of algebra, never to return. But as a JEE aspirant, you know that we do not fight these mountains head-on; we find the path of least resistance. Today, that path is **Rolle's Theorem**.

### The Strategy: The Auxiliary Function

We are asked to prove that there is a root in the interval $(45^{1/100}, 46)$. Direct calculation is impossible. However, Rolle's Theorem gives us a beautiful shortcut.

It tells us that if we have a function $F(x)$ that is continuous and differentiable, and if $F(a) = F(b)$, then there must be a point $c$ where $F'(c) = 0$. If we can construct an $F(x)$ such that its derivative is our original polynomial $P(x)$, then finding a root for $P(x)$ becomes as simple as finding where the tangent of $F(x)$ is horizontal.

So, let us build our auxiliary function. We define $F(x)$ as the integral of $P(x)$:

$$F(x) = \int (51x^{101} - 2323x^{100} - 45x + 1035) dx$$

Applying the power rule, $\int x^n dx = \frac{x^{n+1}}{n+1}$, we get:

$$F(x) = 51 \cdot \frac{x^{102}}{102} - 2323 \cdot \frac{x^{101}}{101} - 45 \cdot \frac{x^2}{2} + 1035x + C$$

We set $C = 0$ for simplicity. Now, look at the coefficients. This is where the magic happens: $51/102$ is exactly $1/2$, and $2323/101$ is exactly $23$. Our function simplifies to:

$$F(x) = \frac{1}{2}x^{102} - 23x^{101} - \frac{45}{2}x^2 + 1035x$$

### The Art of Factorization

Now, we need to see if $F(a) = F(b)$. To do this, we must factorize $F(x)$. Let us group the terms:

$$F(x) = \left(\frac{1}{2}x^{102} - 23x^{101}\right) - \left(\frac{45}{2}x^2 - 1035x\right)$$

Factoring out common terms from each group gives us:

$$F(x) = \frac{1}{2}x^{101}(x - 46) - \frac{45}{2}x(x - 46)$$

And finally, pulling out the common binomial $(x - 46)$ and the term $\frac{1}{2}x$, we arrive at the elegant form:

$$F(x) = \frac{1}{2}x(x - 46)(x^{100} - 45)$$

### The Climax: Boundary Evaluation

This is the moment of truth. We must check the boundaries of our interval $(45^{1/100}, 46)$.

First, let $x = 45^{1/100}$. Substituting this into our factored $F(x)$:

$$F(45^{1/100}) = \frac{1}{2}(45^{1/100})(45^{1/100} - 46)((45^{1/100})^{100} - 45)$$

Since $(45^{1/100})^{100} = 45$, the last term becomes $(45 - 45) = 0$. Thus, $F(45^{1/100}) = 0$.

Next, let $x = 46$. Substituting this into our factored $F(x)$:

$$F(46) = \frac{1}{2}(46)(46 - 46)(46^{100} - 45)$$

Since $(46 - 46) = 0$, the entire expression becomes zero. Thus, $F(46) = 0$.

### Conclusion: Q.E.D.

We have shown that $F(45^{1/100}) = 0$ and $F(46) = 0$. Because $F(x)$ is a polynomial, it is continuous and differentiable everywhere.

By Rolle's Theorem, there must exist at least one $c$ in the interval $(45^{1/100}, 46)$ such that $F'(c) = 0$. Since $F'(x) = P(x)$, this means $P(c) = 0$.

We have conquered the mountain! You have successfully proven the existence of the root without ever having to solve for it directly. This is the power of calculus—the ability to see the truth without needing to see every detail.

Animated Solution for Mathematics - Differentiation: Using Rolle's theorem, prove that there is at least one root in $(4 5^{1/100}, 46)$ of the polynomial $P (x) = 51 x^{101} - 2323 x^{100} - 45 x + 1035$ .

Visualized Solution

Analyzing the Polynomial $P (x)$

Recalling Rolle's Theorem

Constructing the Auxiliary Function $F (x)$

Integrating the Polynomial

Simplifying the Coefficients

Grouping Terms for Factorization

The Factored Form of $F (x)$

Evaluating $F (x)$ at $x = 4 5^{\frac{1}{100}}$

Evaluating $F (x)$ at $x = 46$

Applying Rolle's Theorem

Connecting Back to $P (x)$

The Sigma Insight: Mean Value Theorems

The Mountain of Degree 101

The Strategy

The Art of Factorization

The Climax

Conclusion

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Visualized Solution

Analyzing the Polynomial .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x)

Recalling Rolle's Theorem

Constructing the Auxiliary Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F(x)