If a, b, c in mathbb{R} are such that 4a + 2b | vedclass

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(A) Given that $ax^2 + bx + c = 0$ has no real roots,the discriminant $D = b^2 - 4ac < 0$. Since $ax^2 + bx + c$ has no real roots,the expression $f(x

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greater than $ab$

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(A) Given that $ax^2 + bx + c = 0$ has no real roots,the discriminant $D = b^2 - 4ac Since $ax^2 + bx + c$ has no real roots,the expression $f(x) = ax^2 + bx + c$ must have the same sign for all $x \in \mathbb{R}$. Given $4a + 2b + c > 0$,we have $f(2) > 0$,which implies $a > 0$ and $f(x) > 0$ for all $x \in \mathbb{R}$. Since $f(x) > 0$ for all $x$,we have $f(0) = c > 0$ and $f(1) = a + b + c > 0$. We are given $f(2) = 4a + 2b + c > 0$. Now,consider $f(x) = ax^2 + bx + c$. Since $a > 0$,$f(x) > 0$ for all $x$. Specifically,$f(1) = a + b + c > 0 \implies a + c > -b$. Also,$f(0) = c > 0$. Consider the expression $(c+a)(c+b)$. Since $a > 0$ and $f(x) > 0$,we have $f(x) = a(x - \alpha)(x - \bar{\alpha}) > 0$. Using the property $f(x) > 0$,we can evaluate $f(1) = a+b+c > 0$ and $f(0) = c > 0$. From $ax^2+bx+c > 0$,we know $b^2 $(c+a)(c+b) = c^2 + bc + ac + ab$. Since $b^2  ab$.

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

If $a, b, c \in \mathbb{R}$ are such that $4a + 2b + c > 0$ and $ax^2 + bx + c = 0$ has no real roots,then the value of $(c + a)(c + b)$ is

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