Question

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

A
1
B
2
C
3
D
4

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.Maths

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