Question

If $P(x)=(2013) x^{2012}-(2012) x^{2011}-16 x+8$, 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)

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