The real number $k$ for which the equation $2x^2 + 3x + k = 0$ has two distinct real roots in the interval $[0, 1]$.

All the values of $k$ such that the quadratic expression $2kx^2 - (4k+1)x + 2$ is negative for exactly three integral values of $x$,lie in the interval

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+2ax+(3a+10) = 0$ such that $\alpha < 1 < \beta$. Then the set of all possible values of $a$ is:

If one root of $x^2+px-q^2=0$, where $p$ and $q$ are real, is less than $2$ and the other is greater than $2$, then:

If both the roots of the quadratic equation $x^2 - 2kx + k^2 + k - 5 = 0$ are less than $5$,then $k$ lies in the interval:

If non-zero real numbers $b$ and $c$ are such that $\min \,f(x) > \max \,g(x)$,where $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 - 2cx + b^2$ for $x \in R$; then $\left| \frac{c}{b} \right|$ lies in the interval