Theory of Equations 5 Question 1
1. The number of real roots of the equation $5+\left|2^{x}-1\right|=2^{x}\left(2^{x}-2\right)$ is
(2019 Main, 10 April II)
(a) 1
(b) 3
(c) 4
(d) 2
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Solution:
- Given equation $5+\left|2^{x}-1\right|=2^{x}\left(2^{x}-2\right)$
Case I
If $2^{x}-1 \geq 0 \Rightarrow x \geq 0$,
then $5+2^{x}-1=2^{x}\left(2^{x}-2\right)$
Put $2^{x}=t$, then
$$ \begin{aligned} & \quad 5+t-1=t^{2}-2 t \Rightarrow t^{2}-3 t-4=0 \\ & \Rightarrow t^{2}-4 t+t-4=0 \Rightarrow t(t-4)+1(t-4)=0 \\ & \Rightarrow \quad t=4 \text { or }-1 \Rightarrow t=4 \quad\left(\because t=2^{x}>0\right) \\ & \Rightarrow \quad 2^{x}=4 \Rightarrow x=2>0 \\ & \Rightarrow x=2 \text { is the solution. } \end{aligned} $$
Case II
If $2^{x}-1<0 \Rightarrow x<0$,
then $5+1-2^{x}=2^{x}\left(2^{x}-2\right)$
Put $2^{x}=y$, then $6-y=y^{2}-2 y$
$\Rightarrow \quad y^{2}-y-6=0 \quad \Rightarrow \quad y^{2}-3 y+2 y-6=0$
$\Rightarrow(y+2)(y-3)=0 \Rightarrow y=3$ or -2
$\Rightarrow \quad y=3\left(\right.$ as $\left.y=2^{x}>0\right) \Rightarrow 2^{x}=3$
$\Rightarrow x=\log _2 3>0$
So, $x=\log _2 3$ is not a solution.
Therefore, number of real roots is one.
