Question

If $x$ is complex, then the expression $\frac{x^2+34x-71}{x^2+2x-7}$ takes all values which lie in the interval $\left(a,b\right)$, where

• A. $a=-1,b=1$
• B. $a=1,b=-1$
• C. $a=5,b=9$
• D. $a=9,b=5$

Answer 1

The correct option is C

$a=5,b=9$

Explanation for the correct option.

Step 1. Form the inequality.

Let $y=\frac{x^2+34x-71}{x^2+2x-7}$.

Now cross multiply and form a quadratic equation in $x$.

$$\begin{gathered} y\left(x^2+2x-7\right)=x^2+34x-71 \\ \Rightarrow x^2\left(y-1\right)+x\left(2y-34\right)+71-7y=0 \end{gathered}$$

Now it is given that $x$ is complex, so the roots of the above quadratic equation are imaginary and so the value of the discriminant is less than $0$.

Thus using $b^2-4ac<0$ the inequality is:

${\left(2y-34\right)}^2-4\left(y-1\right)\left(71-7y\right)<0$

Step 2. Solve the inequality and find the range.

Expand the inequality ${\left(2y-34\right)}^2-4\left(y-1\right)\left(71-7y\right)<0$ and solve for $y$.

$$\begin{gathered} 4y^2-136y+1156-284y+28y^2+284-28y<0 \\ \Rightarrow 32y^2-448y^2+1440<0 \\ \Rightarrow y^2-14y+45<0 \\ \Rightarrow y^2-5y-9y+45<0 \\ \Rightarrow y(y-5)-9(y-5)<0 \\ \Rightarrow (y-5)(y-9)<0 \end{gathered}$$

So the solution of the inequality is $5<y<9$, but $y=\frac{x^2+34x-71}{x^2+2x-7}$.

So $5<\frac{x^2+34x-71}{x^2+2x-7}<9$.

Thus the value of $\frac{x^2+34x-71}{x^2+2x-7}$ lies in the interval $\left(5,9\right)$ and so $a=5,b=9$.

Hence, the correct option is C.