Let f(x) = x^{4} - 4x^{3} + 4x^{2} + c, where | vedclass

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(B) Given, $f(x) = x^{4} - 4x^{3} + 4x^{2} + c$.Evaluating at the boundaries of the interval $(1, 2)$:$f(1) = 1^{4} - 4(1)^{3} + 4(1)^{2} + c = 1 - 4

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

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(B) Given, $f(x) = x^{4} - 4x^{3} + 4x^{2} + c$.Evaluating at the boundaries of the interval $(1, 2)$:$f(1) = 1^{4} - 4(1)^{3} + 4(1)^{2} + c = 1 - 4 + 4 + c = 1 + c$.$f(2) = 2^{4} - 4(2)^{3} + 4(2)^{2} + c = 16 - 32 + 16 + c = c$.According to the Intermediate Value Theorem, if $f(1) \cdot f(2) $f(1) \cdot f(2) = (1 + c)c$.For $f(1) \cdot f(2) Since $f'(x) = 4x^{3} - 12x^{2} + 8x = 4x(x^{2} - 3x + 2) = 4x(x - 1)(x - 2)$, we observe that $f'(x) = 0$ at $x = 0, 1, 2$.In the interval $(1, 2)$, $f'(x) Since the function is strictly decreasing on $(1, 2)$ and changes sign, it must have exactly one zero in $(1, 2)$ for $c \in (-1, 0)$.

Let $f(x) = x^{4} - 4x^{3} + 4x^{2} + c$, where $c \in R$. Then,

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