Given the equation 4x^2 + 4(a - 1)x + (1 - 2a | vedclass

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(B) For the quadratic equation $4x^2 + 4(a - 1)x + (1 - 2a) = 0$,the sum of roots is $\sin 	heta + \cos 	heta = -\frac{4(a - 1)}{4} = 1 - a$ .....$(

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$\frac{1}{2}$

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(B) For the quadratic equation $4x^2 + 4(a - 1)x + (1 - 2a) = 0$,the sum of roots is $\sin 	heta + \cos 	heta = -\frac{4(a - 1)}{4} = 1 - a$ .....$(1)$The product of roots is $\sin 	heta \cdot \cos 	heta = \frac{1 - 2a}{4}$ .....$(2)$Squaring equation $(1)$,we get $1 + 2 \sin 	heta \cos 	heta = (1 - a)^2$.Substituting $(2)$ into this,$1 + 2(\frac{1 - 2a}{4}) = (1 - a)^2 \Rightarrow 1 + \frac{1 - 2a}{2} = 1 - 2a + a^2$.Multiplying by $2$,$2 + 1 - 2a = 2 - 4a + 2a^2 \Rightarrow 2a^2 - 2a - 1 = 0$.Solving for $a$,$a = \frac{2 \pm \sqrt{4 - 4(2)(-1)}}{2(2)} = \frac{2 \pm \sqrt{12}}{4} = \frac{1 \pm \sqrt{3}}{2}$.Since $\sin 	heta + \cos 	heta = 1 - a$ and $\sin 	heta, \cos 	heta > 0$,we must have $1 - a > 0$,so $a Then $\sin 	heta \cos 	heta = \frac{1 - 2(\frac{1 - \sqrt{3}}{2})}{4} = \frac{1 - 1 + \sqrt{3}}{4} = \frac{\sqrt{3}}{4}$.So $\sin 2	heta = 2 \sin 	heta \cos 	heta = \frac{\sqrt{3}}{2}$. Thus $2	heta = 60^\circ$ or $120^\circ$,so $	heta = 30^\circ$ or $60^\circ$.If $	heta = 30^\circ$,$\sin 	heta = 0.5$. If $	heta = 60^\circ$,$\sin 	heta = \frac{\sqrt{3}}{2}$.Maximum value of $a + \sin 	heta = \frac{1 - \sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{1}{2}$.

Given the equation $4x^2 + 4(a - 1)x + (1 - 2a) = 0$ has roots $\sin \theta$ and $\cos \theta$ where $0 < \theta < \frac{\pi}{2}$,then the maximum value of $(a + \sin \theta)$ is:

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