Q. The number of integral values of $k$ for which the equation $e^x=\frac{k}{x-3}$ has exactly two solutions is
Application of Derivatives
Solution:
$ e^x(x-3)=k$
Let $f(x)=e^x(x-3)$
$f^{\prime}(x)=e^x-1+(x-3) e^x=e^x(x-2)$
From the graph it is clear that, for exactly two roots $k \in\left(-e^2, 0\right)$
$\therefore$ number of integral values of $k$ is, 7 .
