If a continuous function $$f$$ defined on the real line $$R$$, assumes 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 value is negative then the equation $$f(x)=0$$ has a root in $$R$$.
Consider $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

The line $$y=x$$ meets $$y = k{e^x}$$ for $$k \le 0$$ at

If a continuous function $$f$$ defined on the real line $$R$$, assumes 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 value is negative then the equation $$f(x)=0$$ has a root in $$R$$.
Consider $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

For $$k>0$$, the set of all values of $$k$$ for which $$k{e^x} - x = 0$$ has two distinct roots is

If a continuous function $$f$$ defined on the real line $$R$$, assumes 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 value is negative then the equation $$f(x)=0$$ has a root in $$R$$.
Consider $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

The positive value of $$k$$ for which $$k{e^x} - x = 0$$ has only one root is

Let $$F(x)$$ be an indefinite integral of $$si{n^2}x.$$

STATEMENT-1: The function $$F(x)$$ satisfies $$F\left( {x + \pi } \right) = F\left( x \right)$$
for all real $$x$$. because

STATEMENT-2: $${\sin ^2}\left( {x + \pi } \right) = {\sin ^2}x$$ for all real $$x$$.