Question:
Let $P_n(x) = 1 + 2x + 3x^2$ + ..... + $(n + 1)x^n$ be a polynomial such that $n$ is even. Then the number of real roots of $P_n(x) = 0$ is
Let $P_n(x) = 1 + 2x + 3x^2$ + ..... + $(n + 1)x^n$ be a polynomial such that $n$ is even. Then the number of real roots of $P_n(x) = 0$ is
Updated On: Jul 6, 2022
- $0$
- $n$
- $1$
- none of these
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The Correct Option is A
Solution and Explanation
If $n$ is even, then there is no change of sign in this expression
$\therefore$ there is no negative real root of $f(x)$
Hence there is no real root
When $x > 0, P_n(x) > 0$ and
$\therefore \, P_n(x) = 0$ can have no $+ve$ real root.
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
