If $$\lambda $$ $$ \in $$ R is such that the sum of the cubes of the roots of the equation,
x² + (2 $$-$$ $$\lambda $$) x + (10 $$-$$ $$\lambda $$) = 0 is minimum, then the magnitude of the difference of the roots of this equation is :

The sum of all the real values of x satisfying the equation
2^{(x$$-$$1)(x2 + 5x $$-$$ 50)} = 1 is :

Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x$$-$$ 1 and it leaves remainder 6 when divided by x + 1; then :

If for a positive integer n, the quadratic equation

$$x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right)$$$$ + .... + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right)$$$$ = 10n$$

has two consecutive integral solutions, then n is equal to :