The real number $k$ for which the equation $2x^{2}+2x+k=0$ has two distinct real roots in $\left[ 0,1 \right]$
- Lies between $1$ and $2$
- Lies between $2$ and $3$
- Lies between $-1$ and $0$
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Does not exist
For 2x^2 + 2x + k = 0 to have distinct real roots, discriminant D = 4 - 8k > 0, so k < 0.5. For roots to be in [0, 1], let f(x) = 2x^2 + 2x + k. f(0) = k >= 0 and f(1) = 4 + k >= 0. Since k must be < 0.5 and k >= 0, roots would be negative. Thus, no such real k exists.
For the quadratic function f(x) = 2x^2 + 2x + k to have two distinct real roots in the interval [0, 1], its derivative f'(x) = 4x + 2 must be zero at the vertex, which is at x = -1/2. Since the vertex x-coordinate of -1/2 lies outside the interval [0, 1], the function is strictly increasing across this interval. A strictly increasing function can cross the x-axis at most once in the interval, making it impossible for the equation to have two distinct real roots in [0, 1] for any value of k. Therefore, the real number k does not exist.