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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

If $x^2-4ax+5+a>0$ for all $x \in R$ whenever $a \in (\alpha, \beta)$,then $4\beta+\alpha=$

The set of all solutions of the inequation $x^2 - 2x + 5 \leq 0$ in $R$ is

Given that $x$ is a real number satisfying $\frac{5x^{2}-26x+5}{3x^{2}-10x+3} < 0$, then

If $(2k-1)x^2 - 2(3k-2)x + 4k > 0$ for every $x \in R$,then the sum of all possible integral values of $k$ is

The set of all real numbers satisfying the inequation $x^2-|x+2|+x>0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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