Let S be the set of all real roots of the equ | vedclass

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(D) Let $3^{x} = t$,where $t > 0$.The equation becomes $t(t-1) + 2 = |t-1| + |t-2|$.$t^{2} - t + 2 = |t-1| + |t-2|$.Case-$I$: $0 < t < 1$.$t^{2} - t +

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is a singleton.

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(D) Let $3^{x} = t$,where $t > 0$.The equation becomes $t(t-1) + 2 = |t-1| + |t-2|$.$t^{2} - t + 2 = |t-1| + |t-2|$.Case-$I$: $0 $t^{2} - t + 2 = (1 - t) + (2 - t) = 3 - 2t$.$t^{2} + t - 1 = 0$.$t = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.Since $t > 0$,$t = \frac{\sqrt{5}-1}{2} \approx 0.618$,which satisfies $0 Case-$II$: $1 \leq t $t^{2} - t + 2 = (t - 1) + (2 - t) = 1$.$t^{2} - t + 1 = 0$.The discriminant $D = (-1)^{2} - 4(1)(1) = -3 Case-$III$: $t \geq 2$.$t^{2} - t + 2 = (t - 1) + (t - 2) = 2t - 3$.$t^{2} - 3t + 5 = 0$.The discriminant $D = (-3)^{2} - 4(1)(5) = 9 - 20 = -11 Thus,the only valid value is $t = \frac{\sqrt{5}-1}{2}$.Since $3^{x} = \frac{\sqrt{5}-1}{2}$,there is exactly one real value for $x$,which is $x = \log_{3}\left(\frac{\sqrt{5}-1}{2}\right)$.Therefore,$S$ is a singleton set.

Let $S$ be the set of all real roots of the equation $3^{x}(3^{x}-1)+2=|3^{x}-1|+|3^{x}-2|$. Then $S$

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