Question

If $P(x)=2016\cdot x^{2015}-2015\cdot x^{2014}-10x+5$, then

• A. $P(x)=0$ will have at least two real roots in $\left(0,5^{\frac{1}{2014}}\right)$.
• B. $P^{\prime }(x)=0$ will have at least two real roots.
• C. $P^{\prime \prime }(x)=0$ will have at least one real root.
• D. $P(x)=0$ will have at least three real roots.

Answer 1

Answer: (A), (B), (C), (D)

Let $f(x)=\int P(x)dx$
$=x^{2016}-x^{2015}-5x^2+5x$
$=x\left(x^{2015}-x^{2014}-5x+5\right)$
$=x(x-1)\left(x^{2014}-5\right)$
$\Theta f(x)=0$ will have real roots $x=0,1,5^{\frac{1}{2014}}$
$\therefore$ According to Rolle's theorem $P(x)=0$ will have at least two real roots in $\left(0,5^{\frac{1}{2014}}\right)$.
$\Theta P(x)=0$ is odd degree polynomial continuous function
$\therefore P(x)=0$ will have at least three real roots.
$\therefore P^{\prime }(x)=0$ will have at least two real roots.
$\therefore P^I(x)=0$ will have at least one real root.