If x is complex,the expression frac{x^2+34x-7 | vedclass

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

Question

(C) Let $y = \frac{x^2+34x-71}{x^2+2x-7}$.Then,$x^2+34x-71 = y(x^2+2x-7)$.Rearranging the terms,we get $x^2(y-1) + x(2y-34) + (71-7y) = 0$.Since $x$ i

Vedclass · Accepted Answer

$a=5, b=9$

Vedclass · Answer

(C) Let $y = \frac{x^2+34x-71}{x^2+2x-7}$.Then,$x^2+34x-71 = y(x^2+2x-7)$.Rearranging the terms,we get $x^2(y-1) + x(2y-34) + (71-7y) = 0$.Since $x$ is a complex number,the discriminant $D$ of this quadratic equation in $x$ must be less than or equal to $0$ for the expression to take all complex values,but for the range of the rational function,we analyze the condition $D \leq 0$:$D = (2y-34)^2 - 4(y-1)(71-7y) \leq 0$.Expanding this,we get $4(y-17)^2 - 4(-7y^2 + 78y - 71) \leq 0$.Dividing by $4$,we have $(y^2 - 34y + 289) + 7y^2 - 78y + 71 \leq 0$.$8y^2 - 112y + 360 \leq 0$.Dividing by $8$,we get $y^2 - 14y + 45 \leq 0$.Factoring the quadratic,$(y-5)(y-9) \leq 0$.Thus,$5 \leq y \leq 9$.However,for the expression to take all values in the interval $(a, b)$,we identify $a=5$ and $b=9$.

If $x$ is complex,the expression $\frac{x^2+34x-71}{x^2+2x-7}$ takes all values which lie in the interval $(a, b)$,find the value of $a$ and $b$.

Similar Questions

The set of all real values of the expression $\frac{x^2-x+2}{x^2+x-2}$ for all $x \in R-\{-2, 1\}$ is

The set of all real values $a$ for which $-1 < \frac{2 x^2+a x+2}{x^2+x+1} < 3$ holds for all real values of $x$ is

The sum of all the rational roots of the equation $6x^6-25x^5+31x^4-31x^2+25x-6=0$ is

$E_1: a+b+c=0$,if $1$ is a root of $ax^2+bx+c=0$. $E_2: b^2-a^2=2ac$,if $\sin \theta, \cos \theta$ are the roots of $ax^2+bx+c=0$. Which of the following is true?

Solve $|x - 2| + |x - 1| = x - 3$

Vedclass Test Series

Exam Paper Generator

Online Exam Module