Multiple choice

If $\alpha ,\beta $ are the roots of the equation ${ x }^{ 2 }-3x+a=0,a\in R$ and $\alpha <1<\beta $, then $a$ belong to

  1. $\displaystyle \left( -\infty ,\frac { 9 }{ 4 } \right) $
  2. $\left( -\infty ,2 \right) $
  3. $\left( 2,\infty \right) $
  4. $\displaystyle \left( \frac { 9 }{ 4 } ,\infty \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the quadratic f(x) = x^2 - 3x + a, if the roots alpha and beta satisfy alpha < 1 < beta, then f(1) must be less than 0. f(1) = 1 - 3 + a = a - 2. So, a - 2 < 0, which means a < 2.

AI explanation

For the quadratic equation x^2 - 3x + a = 0 with roots alpha and beta, the condition that 1 lies strictly between the roots is f(1) < 0. Evaluating the function at x = 1 gives 1^2 - 3(1) + a < 0, which simplifies to a - 2 < 0. Therefore, a must belong to the interval (negative infinity, 2).

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