If f(x) is a polynomial such that f(x) = [f'( | vedclass

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(C) Let $f(x)$ be a polynomial of degree $n$. Then the degree of $f'(x)$ is $n-1$. Comparing the degrees in $f(x) = [f'(x)]^2$, we get $n = 2(n-1)$, w

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$4$

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(C) Let $f(x)$ be a polynomial of degree $n$. Then the degree of $f'(x)$ is $n-1$. Comparing the degrees in $f(x) = [f'(x)]^2$, we get $n = 2(n-1)$, which implies $n = 2n - 2$, so $n = 2$. Let $f(x) = a(x-h)^2$. Then $f'(x) = 2a(x-h)$. Substituting into the given equation: $a(x-h)^2 = [2a(x-h)]^2 = 4a^2(x-h)^2$. For this to hold for all $x$, $a = 4a^2$, so $a = 1/4$ (since $a 
eq 0$). Given $f(2) = 0$, the vertex is at $h = 2$. Thus, $f(x) = \frac{1}{4}(x-2)^2$. Calculating $f(-2)$: $f(-2) = \frac{1}{4}(-2-2)^2 = \frac{1}{4}(-4)^2 = \frac{16}{4} = 4$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

If $f(x)$ is a polynomial such that $f(x) = [f'(x)]^2$ and $f(2) = 0$, then $f(-2) = \dots$

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