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 $t(t-1) + 2 = |t-1| + |t-2|$,which simplifies to $t^{2} - t + 2 = |t-1| + |t-2|$.Case-$I$: $0 <

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is a singleton.

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(D) Let $3^{x} = t$,where $t > 0$.The equation becomes $t(t-1) + 2 = |t-1| + |t-2|$,which simplifies to $t^{2} - t + 2 = |t-1| + |t-2|$.Case-$I$: $0 $t^{2} - t + 2 = -(t-1) - (t-2) = -t + 1 - t + 2 = 3 - 2t$.$t^{2} + t - 1 = 0$.$t = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.Since $t > 0$,we have $t = \frac{\sqrt{5}-1}{2}$. This is in the range $(0, 1)$.$3^{x} = \frac{\sqrt{5}-1}{2} \Rightarrow x = \log_{3}\left(\frac{\sqrt{5}-1}{2}\right)$,which is one real solution.Case-$II$: $1 \leq t $t^{2} - t + 2 = (t-1) - (t-2) = t - 1 - t + 2 = 1$.$t^{2} - t + 1 = 0$.The discriminant $D = (-1)^{2} - 4(1)(1) = 1 - 4 = -3 Case-$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,there is only one real value for $t$,which gives exactly one real value for $x$. Therefore,$S$ is a singleton.

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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