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

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

Similar Questions

If $f: R \rightarrow R$ is defined by $f(x) = \frac{1}{2 - \cos 3x}$ for each $x \in R$, then the range of $f$ is

The domain of the function $f: R \rightarrow R$ defined by $f(x) = \sqrt{x^{2}-7x+12}$ is

If $f(x) = \frac{x^2 - 1}{x^2 + 1}$ for every real number $x$,then the minimum value of $f$ is:

Let $f:[2, \infty) \rightarrow R$ be the function defined by $f(x)=x^{2}-4x+5$. Then the range of $f$ is:

Let $f(x) = \sqrt{\frac{x+1}{x+3}}$ and $g(x) = \sqrt{\frac{2-x}{x+3}}$ be two real-valued functions. Then the domain of $f/g$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module