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 $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.Rearranging the terms,we get $y(x^2 + 2x - 7) = x^2 + 34x - 71$.$x^2(y - 1) + x(2y - 34) - 7y + 71 =

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.Rearranging the terms,we get $y(x^2 + 2x - 7) = x^2 + 34x - 71$.$x^2(y - 1) + x(2y - 34) - 7y + 71 = 0$.For $x$ to be a real number,the discriminant $D = B^2 - 4AC \ge 0$.$D = (2y - 34)^2 - 4(y - 1)(-7y + 71) \ge 0$.$4(y - 17)^2 - 4(-7y^2 + 71y + 7y - 71) \ge 0$.$(y^2 - 34y + 289) - (-7y^2 + 78y - 71) \ge 0$.$y^2 - 34y + 289 + 7y^2 - 78y + 71 \ge 0$.$8y^2 - 112y + 360 \ge 0$.Dividing by $8$,we get $y^2 - 14y + 45 \ge 0$.$(y - 5)(y - 9) \ge 0$.Thus,the range is $y \in ( - \infty, 5] \cup [9, \infty)$.

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

Similar Questions

Let $A = \{1, 3, 5\}$ and $B = \{x : x \text{ is an odd natural number } < 6\}$. Which of the following is false?

Range of the function $f(x) = 9 - 7\sin x$ is

Consider the following relations:
$(1) \, A - B = A - (A \cap B)$
$(2) \, A = (A \cap B) \cup (A - B)$
$(3) \, A - (B \cup C) = (A - B) \cup (A - C)$
Which of these is/are correct?

The domain of the function $f(x) = \frac{1}{\sqrt{|x| - x}}$ is

The largest possible set of real numbers which can be the domain of $f(x) = \sqrt {1 - \frac{1}{x}} $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module