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(C) Let the quadratic equation be $ax^2 + bx + c = 0$.The sum of the roots is given by $-\frac{b}{a}$ and the product of the roots is given by $\frac{

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(C) Let the quadratic equation be $ax^2 + bx + c = 0$.The sum of the roots is given by $-\frac{b}{a}$ and the product of the roots is given by $\frac{c}{a}$.When the constant term is copied incorrectly,the sum of the roots remains unchanged because the sum depends only on the coefficients of $x^2$ and $x$.Given the roots are $5$ and $9$,the sum of the roots is $5 + 9 = 14$.Thus,$-\frac{b}{a} = 14$,which implies $b = -14a$.Another student copied the constant term $c = 12$ and the coefficient of $x^2$ as $a = 4$ correctly.Substituting $a = 4$ into $b = -14a$,we get $b = -14(4) = -56$.The correct equation is $4x^2 - 56x + 12 = 0$.The sum of the roots $s = -\frac{b}{a} = -\frac{-56}{4} = 14$.The product of the roots $p = \frac{c}{a} = \frac{12}{4} = 3$.The discriminant $\Delta = b^2 - 4ac = (-56)^2 - 4(4)(12) = 3136 - 192 = 2944$.Finally,calculating the required value: $\frac{\Delta}{3p + s} = \frac{2944}{3(3) + 14} = \frac{2944}{9 + 14} = \frac{2944}{23} = 128$.

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 $a =$	$(IV)$ $3$
	$(V)$ $4$

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