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If a continuous $f$ defined on the real line $R$, assume positive and negative values in $R$, then the equation $f(x)=0$ has a root in $R$. For example, if it is known that a continuous function $f$ on $R$ is positive at some point and its minimum values is negative, then the equation $f(x)=0$ has a root in $R$.
Consider $f(x)=k e^x-x$ for all real $x$, where $k$ is real constant.Question:
For $k>0$, the set of all values of $k$ for which $k e^x-x=0$ has two distinct roots, is
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If a continuous $f$ defined on the real line $R$, assume positive and negative values in $R$, then the equation $f(x)=0$ has a root in $R$. For example, if it is known that a continuous function $f$ on $R$ is positive at some point and its minimum values is negative, then the equation $f(x)=0$ has a root in $R$.
Consider $f(x)=k e^x-x$ for all real $x$, where $k$ is real constant.Question:
For $k>0$, the set of all values of $k$ for which $k e^x-x=0$ has two distinct roots, is
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Verified Answer
The correct answer is:
$\left(0, \frac{1}{e}\right)$
$\left(0, \frac{1}{e}\right)$
For two distinct roots $1+\ln k < 0(k>0)$
$\ln k < -1 \quad k < \frac{1}{e}$
Hence, $k \in\left(0, \frac{1}{e}\right)$
$\ln k < -1 \quad k < \frac{1}{e}$
Hence, $k \in\left(0, \frac{1}{e}\right)$
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