Question

If $P(x)=(2013)x^{2012}-(2012)x^{2011}-16x+8$, then $P(x)=0$ for $x\in \left[0,8^{\frac{1}{2011}}\right]$ has

• A. exactly one real root.
• B. no real root.
• C. atleast one and at most two real roots.
• D. atleast two real roots

Answer 1

Answer: (D)

$F(x)=\int P(x)dx=x^{2013}-x^{2012}-8x^2+8x+C$, where $C$ is constant of integration.
$F(x)=x(x-1)\left(x^{2011}-8\right)+C$
$F(0)=F(1)=F\left(8^{1/2011}\right)=C$
$\Rightarrow F^{\prime }(x)=0\text{ has atleast two real roots. (Using Rolle’s Theorem) }$
$\text{ Note that }P(x)=0\text{ has exactly two real roots in }x\in \left[0,8^{\frac{1}{2011}}\right]$