Multiple choice

If $a,b$ are the roots of the equation $x^2-3x+k=0$, $k \ \epsilon \ R$ and $a < 1 < b$ then:

  1. $a\ \epsilon \ (-\infty,2)$
  2. $a\ \epsilon \ (-\infty,1)$
  3. $a\ \epsilon \left(-\infty, \dfrac{9}{4} \right)$
  4. $a\ \epsilon \left(2,\dfrac{9}{4} \right)$
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A Correct answer
AI explanation

Let the roots be alpha and beta. Since the coefficient of x is -3, their sum is alpha+beta=3. Substituting beta=3-alpha into the inequality alpha < 1 < beta gives alpha < 1 and 3-alpha > 1. Solving the second inequality 3-alpha > 1 yields -alpha > -2, which means alpha < 2. Combining this with the given condition alpha < 1, alpha is strictly less than 2, meaning alpha belongs to the interval from negative infinity to 2.

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