Consider the family of lines x(a + b) + y = 1 | vedclass

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(D) Given the equation $x^3 - 3x^2 + x + \lambda = 0$ with roots $a, b, c$.By Vieta's formulas,the sum of the roots is $a + b + c = 3$,which implies $

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$1/2$

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(D) Given the equation $x^3 - 3x^2 + x + \lambda = 0$ with roots $a, b, c$.By Vieta's formulas,the sum of the roots is $a + b + c = 3$,which implies $a + b = 3 - c$.The line equation is $x(a + b) + y = 1$,which can be written in intercept form as $\frac{x}{1/(a+b)} + \frac{y}{1} = 1$.The area $A$ of the triangle formed with the coordinate axes is $A = \frac{1}{2} 	imes |	ext{base}| 	imes |	ext{height}| = \frac{1}{2} 	imes \frac{1}{|a+b|} 	imes 1 = \frac{1}{2|3-c|}$.Since $c \in [1, 2]$,we have $3 - c \in [1, 2]$,so $a + b > 0$.Thus,$A = \frac{1}{2(3-c)}$.To maximize $A$,we need to maximize the denominator $2(3-c)$,which occurs when $c$ is at its maximum value in the interval $[1, 2]$.For $c = 2$,$A = \frac{1}{2(3-2)} = \frac{1}{2}$.Therefore,the maximum value of $A$ is $1/2$ sq. units.

Consider the family of lines $x(a + b) + y = 1$,where $a, b$ and $c$ are the roots of the equation $x^3 - 3x^2 + x + \lambda = 0$ such that $c \in [1, 2]$. If the given family of lines makes a triangle of area $A$ with the coordinate axes,then the maximum value of $A$ (in sq. units) will be:

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