Question

If p(x) = 51x^{101} - 2323x^{100} - 45x + 1035, using Rolle's theorem, prove that at least one root of p(x) lies between (45^{(1/100)}, 46). [JEE 2004]

Answer 1

The question is a proof based question and doesn't have options.

Answer 2

To prove that at least one root of $p(x) = 51x^{101} - 2323x^{100} - 45x + 1035$ lies between $(45^{(1/100)}, 46)$ using Rolle's theorem, we need to find an auxiliary function $F(x)$ such that $F'(x) = p(x)$ and then find two points $a, b$ such that $F(a) = F(b)$. If $F(a) = F(b)$, Rolle's theorem guarantees that there is at least one root of $F'(x) = p(x)$ in the interval $(a, b)$.

Step 1: Find the anti-derivative of $p(x)$.

Let $F(x) = \int p(x) dx$.

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

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

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

Step 2: Find two values $a$ and $b$ such that $F(a) = F(b)$.

Let's choose $C=0$ for simplicity, as the constant of integration will cancel out when comparing $F(a)$ and $F(b)$.

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

We can factor out $x$:

$F(x) = x \left( \frac{1}{2}x^{101} - 23x^{100} - \frac{45}{2}x + 1035 \right)$

Immediately, we see that $F(0) = 0$. So, let $a=0$.

Now we need to find another value $b$ such that $F(b)=0$. This means we need to find a root of the expression inside the parenthesis:

$G(x) = \frac{1}{2}x^{101} - 23x^{100} - \frac{45}{2}x + 1035$

Let's test $x=46$:

$G(46) = \frac{1}{2}(46)^{101} - 23(46)^{100} - \frac{45}{2}(46) + 1035$

$G(46) = (46)^{100} \left( \frac{1}{2} \cdot 46 - 23 \right) - 45 \cdot 23 + 1035$

$G(46) = (46)^{100} (23 - 23) - 1035 + 1035$

$G(46) = (46)^{100} (0) - 0 = 0$

So, $x=46$ is a root of $G(x)$, which means $F(46)=0$. Let $b=46$.

Step 3: Apply Rolle's Theorem.

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

1. $F(x)$ is a polynomial, so it is continuous on the closed interval $[0, 46]$.
2. $F(x)$ is a polynomial, so it is differentiable on the open interval $(0, 46)$.
3. We found $F(0) = 0$ and $F(46) = 0$, so $F(0) = F(46)$.

According to Rolle's Theorem, there exists at least one value $c \in (0, 46)$ such that $F'(c) = 0$.

Since $F'(x) = p(x)$, this implies that there exists at least one root of $p(x)$ in the interval $(0, 46)$.

Step 4: Refine the interval to $(45^{(1/100)}, 46)$.

We have established that a root $c$ exists in $(0, 46)$. We need to show that this root (or another one) lies in $(45^{(1/100)}, 46)$. This means we need to show that there are no roots of $p(x)$ in the interval $(0, 45^{(1/100)}]$.

Let $x_0 = 45^{(1/100)}$. We evaluate $p(x_0)$:

$p(x) = 51x^{101} - 2323x^{100} - 45x + 1035$

$p(x_0) = 51x_0^{101} - 2323x_0^{100} - 45x_0 + 1035$

Since $x_0^{100} = 45$:

$p(x_0) = 51x_0 \cdot (x_0^{100}) - 2323(x_0^{100}) - 45x_0 + 1035$

$p(x_0) = 51x_0 \cdot 45 - 2323 \cdot 45 - 45x_0 + 1035$

Factor out 45 from the first two terms and notice $1035 = 23 \times 45$:

$p(x_0) = 45(51x_0 - 2323 - x_0 + 23)$

$p(x_0) = 45(50x_0 - 2300)$

$p(x_0) = 45 \times 50 (x_0 - 46)$

$p(x_0) = 2250 (x_0 - 46)$

Since $x_0 = 45^{(1/100)}$, and $45 < 46$, it implies $45^{(1/100)} < 46^{(1/100)}$. Also, $46^{(1/100)} < 46$ (because $46^{100} > 46$). Thus, $x_0 < 46$.

Therefore, $(x_0 - 46)$ is negative.

So, $p(x_0) = 2250 (x_0 - 46) < 0$.

Now consider the interval $[0, x_0]$.

$F(0) = 0$.

$F(x_0) = x_0 \left( \frac{1}{2}x_0^{101} - 23x_0^{100} - \frac{45}{2}x_0 + 1035 \right)$

$F(x_0) = x_0 \left( \frac{1}{2}x_0 \cdot 45 - 23 \cdot 45 - \frac{45}{2}x_0 + 1035 \right)$

$F(x_0) = x_0 \left( \frac{45}{2}x_0 - 1035 - \frac{45}{2}x_0 + 1035 \right)$

$F(x_0) = x_0 (0) = 0$.

So, we have $F(0) = 0$ and $F(45^{(1/100)}) = 0$.

Applying Rolle's Theorem to $F(x)$ on the interval $[0, 45^{(1/100)}]$:

1. $F(x)$ is continuous on $[0, 45^{(1/100)}]$.
2. $F(x)$ is differentiable on $(0, 45^{(1/100)})$.
3. $F(0) = F(45^{(1/100)}) = 0$.

Therefore, there exists at least one $c_1 \in (0, 45^{(1/100)})$ such that $F'(c_1) = p(c_1) = 0$.

This means $p(x)$ has a root in $(0, 45^{(1/100)})$.

However, the question asks to prove a root lies in $(45^{(1/100)}, 46)$. Let's re-examine.

We have $F(0)=0$ and $F(46)=0$. This implies there is a root $c \in (0, 46)$ for $p(x)$.

We also found that $p(45^{(1/100)}) < 0$.

And we previously calculated $p(46) = 23(46)^{100} - 1035$. Since $46^{100}$ is a very large positive number, $p(46) > 0$.

Since $p(x)$ is a polynomial, it is continuous.

We have $p(45^{(1/100)}) < 0$ and $p(46) > 0$.

By the Intermediate Value Theorem, there must be at least one root of $p(x)$ in the interval $(45^{(1/100)}, 46)$.

The problem explicitly asks to use Rolle's theorem. Let's reconsider the definition of Rolle's theorem.

If $f(a)=f(b)$, then $f'(c)=0$ for some $c \in (a,b)$.

We established that $F(0)=0$ and $F(46)=0$.

By Rolle's theorem, there exists $c \in (0, 46)$ such that $p(c)=0$.

Now, to prove that this root lies in $(45^{(1/100)}, 46)$, we need to show that $p(x)$ does not have a root in $(0, 45^{(1/100)}]$. This is where the initial calculation of $p(x_0)$ comes in.

Let's assume there is a root $c_0$ in $(0, 45^{(1/100)})$.

Consider the interval $[c_0, 46]$.

We have $p(c_0)=0$ and $p(46)>0$. This does not directly help with Rolle's theorem for $p(x)$.

The standard application of Rolle's theorem for proving a root of $p(x)$ in $(a,b)$ is to find $F(x)$ such that $F'(x)=p(x)$ and $F(a)=F(b)$.

We found $F(0)=0$ and $F(46)=0$. This guarantees a root $c \in (0, 46)$.

We also need to use the point $45^{(1/100)}$. Let $x_0 = 45^{(1/100)}$.

We found $F(x_0) = 0$.

So we have three roots for $F(x)$: $0$, $x_0$, and $46$.

Applying Rolle's Theorem:

1. On the interval $[0, x_0]$: Since $F(0) = F(x_0) = 0$, there exists at least one root $c_1 \in (0, x_0)$ such that $p(c_1) = 0$.
2. On the interval $[x_0, 46]$: Since $F(x_0) = F(46) = 0$, there exists at least one root $c_2 \in (x_0, 46)$ such that $p(c_2) = 0$.

The question asks to prove that at least one root of $p(x)$ lies between $(45^{(1/100)}, 46)$.

The second application of Rolle's Theorem directly provides this: there exists at least one root $c_2 \in (x_0, 46)$, i.e., $c_2 \in (45^{(1/100)}, 46)$.

Thus, by applying Rolle's theorem to the function $F(x)$ on the interval $[45^{(1/100)}, 46]$, we can prove the statement.