Question

If f(x) is invertible polynomial function of degree 'n' defined on $x \in R$ (where n > 2) and f''(x) = 0 has (n - 2) distinct real roots if

• A. f'(x) = 0 has $\frac{n-1}{2}$ distinct real root
• B. f'(x) = 0 has n - 1 distinct real root
• C. all the distinct root of f'(x) = 0 are repeated at least three times
• D. None of these

Answer 1

Answer: (A)

f'(x) = 0 has $\frac{n-1}{2}$ distinct real root

Answer 2

The problem states that $f(x)$ is an invertible polynomial function of degree 'n' (where n > 2) defined on $x \in R$. It also states that $f''(x) = 0$ has (n - 2) distinct real roots. We need to find which condition on $f'(x)$ is true.

Step 1: Analyze the condition for $f(x)$ to be an invertible polynomial function on R.

For a polynomial function to be invertible on $R$, it must be strictly monotonic (either strictly increasing or strictly decreasing) over its entire domain. This implies that its derivative, $f'(x)$, must maintain a constant sign (either $f'(x) \ge 0$ for all $x \in R$ or $f'(x) \le 0$ for all $x \in R$).

If $f'(x)$ has any real roots, these roots must be of even multiplicity. If a root had odd multiplicity, $f'(x)$ would change sign at that root, making $f(x)$ non-monotonic.

The degree of $f(x)$ is $n$. So, the degree of $f'(x)$ is $n-1$.

Step 2: Determine the parity of 'n'.

If $n-1$ is odd (i.e., $n$ is even), then $f'(x)$ is an odd-degree polynomial. An odd-degree polynomial must have at least one real root. As established in Step 1, for $f(x)$ to be invertible, all real roots of $f'(x)$ must have even multiplicity. However, if an odd-degree polynomial has all its real roots with even multiplicity, this is a contradiction, because the sum of even multiplicities would be an even number, but the degree of the polynomial is odd.

Therefore, for $f(x)$ to be an invertible polynomial function on $R$, the degree of $f'(x)$ (which is $n-1$) must be even. This implies that $n$ must be an odd number.

Step 3: Relate the number of distinct real roots of $f'(x)$ to its degree.

Since $n$ is odd, $n-1$ is even.

Let $k$ be the number of distinct real roots of $f'(x) = 0$. Let their multiplicities be $m_1, m_2, ..., m_k$.

From Step 1, each $m_i$ must be an even integer, so $m_i \ge 2$.

The sum of the multiplicities equals the degree of $f'(x)$: $\sum_{i=1}^k m_i = n-1$.

Since $m_i \ge 2$, we have $\sum_{i=1}^k m_i \ge 2k$.

Therefore, $n-1 \ge 2k \implies k \le \frac{n-1}{2}$.

This means the number of distinct real roots of $f'(x)=0$ must be less than or equal to $\frac{n-1}{2}$.

Step 4: Evaluate the given options based on the derived conditions.

(A) f'(x) = 0 has $\frac{n-1}{2}$ distinct real root

This option implies $k = \frac{n-1}{2}$. Since $n$ is odd, $\frac{n-1}{2}$ is an integer.

If $k = \frac{n-1}{2}$, and $n-1 \ge 2k$ (which is $n-1 \ge 2 \times \frac{n-1}{2} = n-1$), this implies that each of the $k$ distinct real roots must have a multiplicity of exactly 2.

So, $f'(x)$ would be of the form $C \prod_{i=1}^{(n-1)/2} (x-a_i)^2$, where $a_i$ are the distinct real roots and $C$ is a non-zero constant.

Let $P(x) = \prod_{i=1}^{(n-1)/2} (x-a_i)$. Then $f'(x) = C [P(x)]^2$.

Now, let's find $f''(x)$:

$f''(x) = C \cdot 2 P(x) P'(x)$.

The roots of $f''(x)=0$ are the roots of $P(x)=0$ and the roots of $P'(x)=0$.

$P(x)$ has $\frac{n-1}{2}$ distinct real roots ($a_1, a_2, ..., a_{(n-1)/2}$).

Since $P(x)$ has $\frac{n-1}{2}$ distinct real roots, by Rolle's Theorem, $P'(x)$ must have exactly $\left(\frac{n-1}{2} - 1\right)$ distinct real roots, and these roots interlace with the roots of $P(x)$. Therefore, the roots of $P(x)$ and $P'(x)$ are distinct from each other.

The total number of distinct real roots of $f''(x)=0$ is the sum of the number of distinct roots of $P(x)$ and $P'(x)$:

Number of distinct roots = $\frac{n-1}{2} + \left(\frac{n-1}{2} - 1\right) = \frac{n-1+n-1-2}{2} = \frac{2n-4}{2} = n-2$.

This matches the given condition that $f''(x)=0$ has (n-2) distinct real roots.

Thus, option (A) is consistent with all conditions.

(B) f'(x) = 0 has n - 1 distinct real root

This implies $k = n-1$. From Step 3, we know $k \le \frac{n-1}{2}$.

So, $n-1 \le \frac{n-1}{2}$. Since $n>2$, $n-1 > 1$, so $\frac{n-1}{2} > 0$. This inequality implies $1 \le \frac{1}{2}$, which is false.

Also, if $f'(x)$ has $n-1$ distinct real roots, each must have multiplicity 1 (since the degree of $f'(x)$ is $n-1$). Roots of odd multiplicity cause $f'(x)$ to change sign, making $f(x)$ non-monotonic and thus not invertible. So, this option is incorrect.

(C) all the distinct root of f'(x) = 0 are repeated at least three times

For $f(x)$ to be invertible, all real roots of $f'(x)$ must have even multiplicity. If a root is repeated at least three times AND has even multiplicity, its multiplicity must be at least 4.

Let $k$ be the number of distinct real roots of $f'(x)=0$. The sum of multiplicities is $n-1$.

If each of the $k$ roots has multiplicity at least 4, then $n-1 \ge 4k \implies k \le \frac{n-1}{4}$.

If this were the case, the number of distinct roots of $f'(x)$ would be smaller than or equal to $\frac{n-1}{4}$.

Let's check if this leads to $n-2$ distinct roots for $f''(x)=0$.

If $f'(x)$ has $k$ distinct roots, each with multiplicity $m_i \ge 4$ (and $m_i$ is even).

This condition is not specific enough to determine the exact number of distinct roots of $f''(x)$. For example, if $f'(x) = (x-a)^4$, then $f''(x) = 4(x-a)^3$, which has only one distinct root. Here $n-1=4 \implies n=5$. $n-2=3$. But $f''(x)$ has only 1 root. This contradicts the condition. So, option (C) is incorrect.

(D) None of these

Since option (A) is consistent with all the given conditions, (D) is incorrect.

Conclusion:

The invertibility of $f(x)$ on $R$ implies that $n$ must be an odd integer and all real roots of $f'(x)$ must have even multiplicity. The condition that $f''(x)=0$ has (n-2) distinct real roots forces $f'(x)$ to have exactly $\frac{n-1}{2}$ distinct real roots, each with multiplicity 2.