The number of real roots of the equation 5 + | vedclass

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(D) Let $2^x = t$. Since $2^x > 0$,we must have $t > 0$.The equation becomes $5 + |t - 1| = t(t - 2) = t^2 - 2t$.Rearranging,we get $|t - 1| = t^2 - 2

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$1$

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(D) Let $2^x = t$. Since $2^x > 0$,we must have $t > 0$.The equation becomes $5 + |t - 1| = t(t - 2) = t^2 - 2t$.Rearranging,we get $|t - 1| = t^2 - 2t - 5$.Let $g(t) = |t - 1|$ and $f(t) = t^2 - 2t - 5$.We look for solutions where $t > 0$.Case $1$: $t \ge 1$.$t - 1 = t^2 - 2t - 5$ $\Rightarrow t^2 - 3t - 4 = 0$ $\Rightarrow (t - 4)(t + 1) = 0$.Since $t \ge 1$,we have $t = 4$. Thus $2^x = 4 \Rightarrow x = 2$.Case $2$: $0 $-(t - 1) = t^2 - 2t - 5$ $\Rightarrow -t + 1 = t^2 - 2t - 5$ $\Rightarrow t^2 - t - 6 = 0$ $\Rightarrow (t - 3)(t + 2) = 0$.Neither $t = 3$ nor $t = -2$ satisfy $0 Thus,there is only $1$ real root.

The number of real roots of the equation $5 + |2^x - 1| = 2^x(2^x - 2)$ is

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