The zeroes of $f(x) = x^4 - 4x^3 + 4x^2+ c$, for real $c$

Question

Let $f(x) = x^4 - 4x^3 + 4x^2+ c$, where $c$ belongs to real numbers. Then

(A) $f(x)$ has infinitely many zeros in $(1, 2)$ for all $c$

(B) $f(x)$ has exactly one zero in $(1, 2)$ if $-1 < c < 0$

(C) $f(x)$ has double zeros in $(1, 2)$ if $-1 < c < 0$

(D) whatever be the value of $c, f(x)$ has no zero in $(1, 2)$

My attempt: $f'(x)=0$ for terminal values of $x$. Therefore $x=0,1,2$

$f(0)=c,f(1)=1+c, f(2)=c$, therefore $f$ is maximum is at $x=1$. Not sure how to proceed further. Hints are appreciated.

Answer 1

just use $f'(x)=4x(x-1)(x-2)$, clearly $f$ is srtictly decrasing on $(1,2)$, now see option $(B)$ $f(1)=c+1>0, f(2)=c<0$ so by IVT it has at least one root in $(1,2)$, but since the funtion is strictly monotone, so this root is unique.

Answer 2

Hint

$$f(x)=(x^2-2x)^2+c$$

Now $x^2-2x=(x-1)^2-1\ge-1$

Answer 3

For any $1<x<2$ we have: $$f(x)>c.$$ Also, for any $1<x<2$ by AM-GM we obtain: $$f(x)=(x(2-x))^2+c\leq\left(\frac{x+2-x}{2}\right)^4+x=1+c.$$ The equality does not occur.

Now, proof that $f$ is monotonic on $(1,2).$