Multiple choice

The set of values of $a$ for which one root of the equation $x^{2}+(2a+1)x+(a-1)=0$ is greater than $1$ and the other root is less than zero is

  1. $a < -\displaystyle \frac{1}{3}$
  2. $0 < a <1$
  3. $a > 1$
  4. $a \geq 1$
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A Correct answer
Explanation

Let f(x) = x^2 + (2a+1)x + (a-1). For one root > 1 and one root < 0, the function must satisfy f(1) < 0 and f(0) < 0. f(1) = 1 + 2a + 1 + a - 1 = 3a + 1 < 0 => a < -1/3. f(0) = a - 1 < 0 => a < 1. Intersection is a < -1/3.

AI explanation

For the quadratic function f(x) = x^2 + (2a + 1)x + (a - 1) to have one root greater than 1 and the other less than 0, the graph must pass below the x-axis at these points. Therefore, f(1) and f(0) must both be negative. Evaluating f(1) gives 1 + 2a + 1 + a - 1 < 0, which simplifies to 3a + 1 < 0, meaning a < -1/3. Evaluating f(0) gives a - 1 < 0, or a < 1. The intersection of these two conditions is a < -1/3.

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