Multiple choice

If one root of the quadratic equation $2x^{2}-2kx+k-4=0$ is smaller than $1$ & other is greater than $2$, then the complete set of values of $k$ is

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

Let f(x) = 2x^2 - 2kx + k - 4. For one root < 1 and other > 2, the conditions are f(1) < 0 and f(2) < 0. f(1) = 2 - 2k + k - 4 = -k - 2 < 0 => k > -2. f(2) = 8 - 4k + k - 4 = 4 - 3k < 0 => k > 4/3. Intersection is k > 4/3.

AI explanation

For the roots to fall in the specified intervals, the quadratic expression must be negative at x equals 1 and negative at x equals 2, since the coefficient of x squared is positive. Evaluating at x equals 1 gives 2 minus 2k plus k minus 4, which simplifies to negative k minus 2 less than 0, meaning k is greater than negative 2. Evaluating at x equals 2 gives 8 minus 4k plus k minus 4, which simplifies to 4 minus 3k less than 0, meaning k is greater than 4 divided by 3. The intersection of these conditions is k greater than 4 divided by 3. The result is the interval from 4/3 to infinity.

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