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Question

The number of real roots of the equation $$x^{x-1}+x-2=0$$ is 


A
1
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B
2
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C
3
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D
4
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Solution

The correct option is A $$1$$
Let $$f(x)=e^{x-1}+x-2\Rightarrow f'(x)=e^{x-1}+1 > 0\forall x\in R$$
Also when $$x\to $$ $$\infty, f(x)\to \infty$$ and when $$x\to -\infty$$
Further $$f(x)$$ is continuous, hence its graph cuts $$x$$- axis only at one point.
Hence, equation $$f(x)=0$$ has only one root.
Alternative method:
Also $$e^{x-1}=2-x$$
As shown in the figure, graphs of $$y=e^{x-1}$$ and $$y=2-x$$ cuts at only one point. Hence, there is only one root.
1976644_1757495_ans_3e68ae1d132e4c948377ac24b53f8d5d.png

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