# Is function continuous, bounded?

Let $$p(x)=x^n+\sum\limits_{k=0}^{n-1}a_kx^k$$ and $$q(x)=x^n+\sum\limits_{k=0}^{n-1}b_kx^k$$ be two polynomials with real coefficients such that $n\ge 4$ is even and $a_{n-1}\lt b_{n-1}$.

Let $f(x)$ be a function such that $p(x)\le f(x)\le q(x)$, $\forall x\in \Bbb R$. Then we can conclude that

$(A)$ $f(x)$ is a bounded function on $\Bbb R$.

$(B)$ $f(x)$ is a continuous function on $\Bbb R$.

$(C)$ There exists $x_0\in \Bbb R$ such that $f(x_0)=0$.

$(D)$ $f(x)$ is continuous at least at one point $x_0\in \Bbb R$

I think $(A)$ is true . $f(x)$ is in between two polynomial function is a polynomial therefore $(B)$ seems to be true. Definitely $(C)$ is not true as $f(x)$ can be even degree polynomial and it may not have any real root. and if $(B)$ is true then $(D)$ hold.

• $f$ cannot be bounded, because it is greater than $p$, and $\lim_{x\to \infty} p(x) = +\infty$ – Tryss Jan 1 '16 at 13:34
• Please do not use images for text. Take the time to transcribe the question. -- Remember the Dirichlet function $D(x)$ of first type. Set $p(x)=1+x^4$, $q(x)=1+p(x)$, $f(x)=p(x)+D(x)$. – LutzL Jan 1 '16 at 14:46

I believe the problem as stated appears to be improperly worded (with inconsistent constraints) because the inequality $p(x)\leq q(x)$ imposed in the problem statement cannot actually be satisfied for all $x$. Suppose for example $n=4$. Note that $q(x)- p(x) = (a_3- b_3) x^3 +\ldots$ where $a_3- b_3\ne 0$ . This cubic must assume both positive and negative values as $x\to \pm \infty$ respectively.

• +1. Thus (A), (B), (C), (D) all hold (and their negations as well), since one assumed something impossible. – Did Jan 1 '16 at 16:12
• its answer is only D. – himanshu Singla Jan 2 '16 at 3:46