Let f(x) = x^2 + 4x + 1. Then | vedclass

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

Question

(C) Given $f(x) = x^2 + 4x + 1$.For option $A$: The discriminant $D = b^2 - 4ac = 4^2 - 4(1)(1) = 16 - 4 = 12 > 0$. Since $D > 0$,the quadratic has tw

Vedclass · Accepted Answer

$f(x) \ge 1$ when $x \le -4$

Vedclass · Answer

(C) Given $f(x) = x^2 + 4x + 1$.For option $A$: The discriminant $D = b^2 - 4ac = 4^2 - 4(1)(1) = 16 - 4 = 12 > 0$. Since $D > 0$,the quadratic has two distinct real roots,so $f(x)$ takes both positive and negative values. Thus,$f(x) > 0$ is not true for all $x$.For option $B$: If $x \ge 0$,then $x^2 \ge 0$ and $4x \ge 0$. Thus $f(x) = x^2 + 4x + 1 \ge 1$. Specifically,$f(x) > 1$ for $x > 0$. This is a correct statement.For option $C$: $f(x) \ge 1 \implies x^2 + 4x + 1 \ge 1 \implies x^2 + 4x \ge 0 \implies x(x + 4) \ge 0$. This inequality holds when $x \le -4$ or $x \ge 0$. Thus,$f(x) \ge 1$ when $x \le -4$ is a correct statement.Note: Both $B$ and $C$ are mathematically correct statements based on the inequality $x(x+4) \ge 0$.

Let $f(x) = x^2 + 4x + 1$. Then

Similar Questions

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

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

What is the solution set of the inequality $x^2 + 2|x| - 15 \geq 0$?

The integral value of $k$ for which $x^2 - 2(4k - 1)x + 15k^2 - 2k - 7 > 0$ for all $x \in R$ is

The sum of the fourth powers of the roots of the equation $x^3+x+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module