Assertion (A): 3x^2 - 16x + 4 -16 is satisf | vedclass

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(A) Given the inequality $3x^2 - 16x + 4 > -16$.This simplifies to $3x^2 - 16x + 20 > 0$.Let $f(x) = 3x^2 - 16x + 20$. Here $a = 3, b = -16, c = 20$.T

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$(A)$ is true,$(R)$ is true and $(R)$ is the correct explanation for $(A)$

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(A) Given the inequality $3x^2 - 16x + 4 > -16$.This simplifies to $3x^2 - 16x + 20 > 0$.Let $f(x) = 3x^2 - 16x + 20$. Here $a = 3, b = -16, c = 20$.The discriminant $D = b^2 - 4ac = (-16)^2 - 4(3)(20) = 256 - 240 = 16 > 0$.The roots of $3x^2 - 16x + 20 = 0$ are $x = \frac{16 \pm \sqrt{16}}{6} = \frac{16 \pm 4}{6}$,which gives $x = 2$ and $x = \frac{10}{3}$.Since $a > 0$,$f(x) > 0$ for $x \in (-\infty, 2) \cup (\frac{10}{3}, \infty)$.The interval $(0, \frac{10}{3})$ contains values like $x=1$ where $f(1) = 3 - 16 + 20 = 7 > 0$. Thus,$(A)$ is true.Reason $(R)$ states that $ax^2 + bx + c$ and $a$ have the same sign for some $x$ when $D > 0$. This is true because the parabola opens in the direction of $a$ outside the roots. Thus,$(R)$ is true and explains $(A)$.

Assertion $(A)$: $3x^2 - 16x + 4 > -16$ is satisfied for some values of real $x$ in $(0, \frac{10}{3})$.
Reason $(R)$: $ax^2 + bx + c$ and $a$ will have the same sign for some values of $x \in \mathbb{R}$ when $b^2 - 4ac > 0$.
The correct option among the following is

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Assertion $(A)$: $3x^2 - 16x + 4 > -16$ is satisfied for some values of real $x$ in $(0, \frac{10}{3})$.Reason $(R)$: $ax^2 + bx + c$ and $a$ will have the same sign for some values of $x \in \mathbb{R}$ when $b^2 - 4ac > 0$.The correct option among the following is