Question

If \( x \) is real, then the value of \[ \frac{x^2 + 34x - 71}{x^2 + 2x - 7} \] does not lie between:

• A. 8 and -5
• B. -5 and 9
• C. 0 and 9
• D. 5 and 9

Hint:

For rational expressions, examine the limits as \( x \to \infty \) and the points where the denominator is zero to determine the range of the function.

Answer 1

The Correct Option is D

Step 1: Simplify the expression.
We are given the expression: \[ f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7} \] Let’s examine the behavior of this expression. We first look for the conditions under which this expression could take extreme values. Begin by analyzing the values of \( f(x) \) for large and small values of \( x \).

Step 2: Find the limits as \( x \to \infty \) and \( x \to -\infty \).
As \( x \) becomes very large (positive or negative), the higher powers of \( x \) dominate both the numerator and denominator, so the expression approaches: \[ f(x) \approx \frac{x^2}{x^2} = 1 \] Thus, for large values of \( x \), the function approaches 1.

Step 3: Analyze the behavior of the expression at the roots of the denominator.
Now let’s check the values at which the denominator \( x^2 + 2x - 7 = 0 \). Solve for \( x \): \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-7)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 28}}{2} = \frac{-2 \pm \sqrt{32}}{2} = \frac{-2 \pm 4\sqrt{2}}{2} = -1 \pm 2\sqrt{2} \] Thus, the denominator is zero at \( x = -1 + 2\sqrt{2} \) and \( x = -1 - 2\sqrt{2} \).

Step 4: Conclusion.
From the above analysis, we conclude that the value of the expression does not lie between 5 and 9, hence the correct answer is option (D).