Question:

Let $p(x_0)$ be a polynomial with real coefficients, $p(0) = 1$ and $p'(x)>0$ for all $x \in \mathbb{R}$. Then

Updated On: Apr 15, 2025
  • p(x) has atleast two real roots
  • p(x) has only one positive real root
  • p(x) may have negative real root
  • p(x) has infinitely many real roots
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

We are given the following properties of the polynomial \( p(x) \): 1. \( p(0) = 1 \) (i.e., the polynomial takes the value 1 at \( x = 0 \)), 2. \( p'(x) > 0 \) for all \( x \in \mathbb{R} \) (i.e., the polynomial is strictly increasing).

Step 1: Analysis of \( p(x) \)'s behavior
The condition \( p'(x) > 0 \) means that the polynomial is strictly increasing for all values of \( x \). Since a strictly increasing function can have at most one real root, we know that the polynomial \( p(x) \) can have at most one real root. Additionally, since \( p(0) = 1 \), the polynomial does not have a root at \( x = 0 \). Therefore, the polynomial may have a root either at a positive value or at a negative value. 

Step 2: Possibility of a negative real root
Since the polynomial is strictly increasing, if it does have a root, it must occur at a single point where the value of the polynomial changes from positive to negative or vice versa. Therefore, it is possible for the polynomial to have a negative real root, but it is not guaranteed.

Answer:

\[ \boxed{p(x) \text{ may have a negative real root}} \]

Was this answer helpful?
0
0