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The number of real roots of the equation $e^{x^1}+\log x+x-2=0, x \neq 0$, is
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Verified Answer
The correct answer is:
1
We have,
$$
e^{x-1}+\log x+x-2=0
$$
Let $\quad f(x)=e^{x-1}+\log x+x-2$
$$
f^{\prime}(x)=e^{x-1}+\frac{1}{x}+1
$$
$$
f^{\prime}(x)>0, \forall x \in R
$$
$f(x)$ is monotonic increasing function.
Hence, $f(x)$ has only one real roots.
$$
e^{x-1}+\log x+x-2=0
$$
Let $\quad f(x)=e^{x-1}+\log x+x-2$
$$
f^{\prime}(x)=e^{x-1}+\frac{1}{x}+1
$$
$$
f^{\prime}(x)>0, \forall x \in R
$$
$f(x)$ is monotonic increasing function.
Hence, $f(x)$ has only one real roots.
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