Multiple choice

The value of $a$, for which one root of the equation $(a-5)x^{2}-2ax+(a-4)=0$ is smaller then $1$ an the other is greater then $2$ is

  1. $a \in\ (5,24)$
  2. $a \in\ \left(\dfrac {20}{3},\infty\right)$
  3. $a \in\ (5,\infty)$
  4. $a \in\ (-\infty,\infty)$
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A Correct answer
Explanation

For a quadratic f(x) = (a-5)x^2 - 2ax + (a-4), if one root is < 1 and the other > 2, then for a-5 > 0 (a > 5), f(1) < 0 and f(2) < 0. f(1) = a-5-2a+a-4 = -9 < 0 (always true). f(2) = (a-5)(4) - 4a + a - 4 = 4a - 20 - 4a + a - 4 = a - 24 < 0, so a < 24. Thus 5 < a < 24.

AI explanation

For the smaller root to be less than 1, substitute x = 1 into the function to get f(1) less than 0, yielding (a - 5) - 2a + (a - 4) less than 0, which simplifies to a greater than 5. For the larger root to be greater than 2, substitute x = 2 to get f(2) less than 0, yielding 4(a - 5) - 4a + (a - 4) less than 0, which simplifies to a less than 24. Therefore, the value of a must be in the interval from 5 to 24.

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