2024/029) Consider the equation $x^{2021} + x^{2020} + \cdots+x - 1=0$

 Then

a) exactly one real root is -ve

b) all real roots are positive

c) exactly one real root is positive

d) no real root is positive.

Solution

we have 

 $x^{2021} + x^{2020} + \cdots+x - 1=0$

or   $x^{2021} + x^{2020} + \cdots+x + 1=2$

Note that 1 is not a root of this equation.

 Multiplying both sides by $x-1$ we get

$x^{2022} - 1  = 2(x-1)$

or  $f(x) = x^{2022} - 2x  + 1 = 0$

 Note that as we have multiplied by x-1 so x = 1 so there is at least one positive root that is 1

As there is change of sign two times as per Descarte rule there are two or zero positive roots root but as it has at least one positive root so there are two positive roots so original equation has one positive root.

Now  $f(-x) = x^{2022} + 2x  + 1 = 0$

As there is no change of sign so there is no -ve root

 So the equation has has one root and it is positive.

So answer is (c)-

Descartes