The range of f(x) = frac{x^2 + 34x - 71}{x^2 | vedclass

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

Question

(B) Let $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.$y(x^2 + 2x - 7) = x^2 + 34x - 71$$x^2(y - 1) + x(2y - 34) - 7y + 71 = 0$.For $x$ to be real,the dis

Vedclass · Accepted Answer

$( - \infty, 5] \cup [9, \infty)$

Vedclass · Answer

(B) Let $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.$y(x^2 + 2x - 7) = x^2 + 34x - 71$$x^2(y - 1) + x(2y - 34) - 7y + 71 = 0$.For $x$ to be real,the discriminant $D \ge 0$:$D = (2y - 34)^2 - 4(y - 1)(-7y + 71) \ge 0$$4(y - 17)^2 - 4(-7y^2 + 71y + 7y - 71) \ge 0$$(y^2 - 34y + 289) - (-7y^2 + 78y - 71) \ge 0$$8y^2 - 112y + 360 \ge 0$$y^2 - 14y + 45 \ge 0$$(y - 5)(y - 9) \ge 0$.Thus,$y \le 5$ or $y \ge 9$.The range is $( - \infty, 5] \cup [9, \infty)$.

$A$. $f(x)=\frac{\|x+2\|}{x+2}, x \neq-2$	$1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=\|[x]\|, x \in R$	$2$. $Z$
$C$. $h(x)=\|x-[x]\|, x \in R$	$3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$	$4$. $[0, 1)$
	$5$. $\{-1, 1\}$

The range of $f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$ is

Similar Questions

Consider the following lists.
$A$. $f(x)=\frac{|x+2|}{x+2}, x \neq-2$ $1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=|[x]|, x \in R$ $2$. $Z$
$C$. $h(x)=|x-[x]|, x \in R$ $3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$ $4$. $[0, 1)$
$5$. $\{-1, 1\}$

Let $[x]$ denote the greatest integer not more than $x$. If $A$ and $B$ are the domains of the functions $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ and $g(x)=\frac{x-[x]}{\sqrt{|x|+x}}$ respectively, then

The function $f: R \rightarrow R$ is defined by $f(x) = \cos^2 x + \sin^4 x$ for $x \in R$. Then $f(R)$ is equal to:

If $f:[2, \infty) \rightarrow R$ is defined by $f(x)=x^2-4x+5$, then the range of $f$ is

If $f: R \rightarrow A$ defined by $f(x) = \frac{1}{x^2+2x+2}$,$\forall x \in R$ is surjective,then $A =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module