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

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(B) From the given quadratic equation,the sum of roots is $\sin 	heta + \cos 	heta = -(a - 1) = 1 - a$ .....$(1)$The product of roots is $\sin 	het

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$\frac{1}{2}$

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(B) From the given quadratic equation,the sum of roots is $\sin 	heta + \cos 	heta = -(a - 1) = 1 - a$ .....$(1)$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,$1 + 2(\frac{1 - 2a}{4}) = (1 - a)^2 \Rightarrow 1 + \frac{1 - 2a}{2} = 1 - 2a + a^2$.Simplifying,$2 + 1 - 2a = 2 - 4a + 2a^2 \Rightarrow 2a^2 - 2a - 1 = 0$.Solving for $a$,$a = \frac{2 \pm \sqrt{4 - 4(2)(-1)}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{1 \pm \sqrt{3}}{2}$.Since $\sin 	heta + \cos 	heta = 1 - a > 0$ for $	heta \in (0, \frac{\pi}{2})$,we must have $1 - a > 0$,so $a Now,$\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}$.This gives $2	heta = 60^{\circ}$ or $120^{\circ}$,so $	heta = 30^{\circ}$ or $60^{\circ}$.If $	heta = 30^{\circ}$,$\sin 	heta = \frac{1}{2}$. 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 $\cos \theta$ $(0 < \theta < \frac{\pi}{2})$,then the maximum value of $(a + \sin \theta)$ is-

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