number of real roots of $2012$ degree equation in $x\in \left[0,8^{\frac{1}{2011}}\right]$

Question

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

Options:

$(a)$ exactly one real root.

$(b)$ no real root.

$(c)$ atleast one and atmost two real roots.

$(d)$ atleast two real roots.

$\bf{Attempt:}$ Let $$f(x) = x^{2013}-x^{2012}-8x^2+8x+c$$

at $x=0, f(0) = 0$ and at $x=8^{\frac{1}{2011}},f\left(8^{\frac{1}{2011}}\right) = c$ Using Rolle,s Theorem $f(x)$ is Continuous in $x\in \left[0,8^{\frac{1}{2011}}\right]$ and Differentiable in $x\in \left(0,8^{\frac{1}{2011}}\right)$

and $f(0) = f\left(8^{\frac{1}{2011}}\right) = c$

so There exists at least one value of $x$ for which $f'(x)=0$ for $\left(0,8^{\frac{1}{2011}}\right)$

so equation $p(x) =0$ has at least one root in $\left(0,8^{\frac{1}{2011}}\right)$

But answer given as last option, could some help me, Thanks

Answer 1

You almost got the answer. Just note that $f(1)$ is also equal to $c$.

Namely: Since $f(0)=f(1)=f(8^\frac{1}{2011})$, apply Rolle's Theorem as you described to the intervals $[0,1]$ and $[1,8^\frac{1}{2011}]$.

Remark: Anyway, this is a repeated question, but the previous time this was asked, Rolle's Theorem was not used as a solution. I think this OP has given a better solution.