The real number $k$ for which the equation $2{ x }^{ { 2 } }+3{ x }+{ k }=0$ has two distinct real roots in $[ 0,1 ].$
- lies between $1$ and $2 .$
- lies between $2$ and $3 .$
- lies between $-1$ and $0 .$
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does not exist
Let f(x) = 2x^2 + 3x + k. For roots in [0, 1], the discriminant D = 9 - 8k >= 0, so k <= 9/8. Also, the vertex -b/2a = -3/4 is not in [0, 1]. Since the parabola opens upward and the vertex is to the left of the interval, the function is increasing on [0, 1]. For roots in [0, 1], we need f(0) >= 0 and f(1) <= 0. f(0) = k >= 0. f(1) = 2 + 3 + k = 5 + k <= 0, so k <= -5. These conditions (k >= 0 and k <= -5) are contradictory.
For the quadratic equation 2x^2 + 3x + k = 0 to have roots in the interval [0, 1], the product of the roots given by k/2 must be positive, requiring k > 0. The sum of the roots is -3/2, meaning their average is -3/4, which is negative and cannot fall within the strictly non-negative interval [0, 1]. Since the sum of the roots cannot lie in [0, 1], it is impossible for both distinct roots to exist in this interval for any real number k. Therefore, such a real number k does not exist.