Question

Let $n$ be a natural number and let $a$ be a real number. The number of zeros of $x^{2n+1}-(2n+1)x+a=0$ in the interval $[-1,1]$ is

• A. $2$ if $a>0$
• B. $2$ if $a<0$
• C. At most one for every value of $a$
• D. At least three for every value of $a$

Answer 1

The correct option is C At most one for every value of $a$

We will approach the question by checking its monotonicity

$f\left(x\right)=x^{2n+1}-(2n+1)x+a$

$f^{\prime }\left(x\right)=(2n+1)x^{2n}-(2n+1)$

to find the maxima and minima

$f^{\prime }\left(x\right)=0(2n+1)x^{2n}-(2n+1)=0x^{2n}-1=0x=1,-1$

using the second deivative test

$f^{\prime \prime }\left(x\right)=(2n+1)(2n{)x}^{2n-1}$

$f^{\prime \prime }\left(1\right)=(2n+1)\left(2n\right)>0$

which gives point of minima

$f^{\prime \prime }(-1)=(2n+1)\left(2n\right)<0$

which gives point of maxima

The graph of the polynomial function is strictly decreasing, So it can atmost cross the x axis once

$\therefore$option $C$ is correct