Let $$g(x) = {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}};0 < x < 2,m$$ and $$n$$ are integers, $$m \ne 0,n > 0$$, and let $$p$$ be the left hand derivative of $$|x - 1|$$ at $$x = 1$$. If $$\mathop {\lim }\limits_{x \to {1^ + }} g(x) = p$$, then

Let $$f(x)=2+\cos x$$ for all real $$x$$.

STATEMENT - 1 : For each real $$t$$, there exists a point $$c$$ in $$[t, t+\pi]$$ such that $$f^{\prime}(C)=0$$.

STATEMENT - 2 : $$f(t)=f(t+2 \pi)$$ for each real $$t$$.

If a continuous functions $$f$$ 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 $$f(x)=k e^{x}-x$$ for all real $$x$$ where $$k$$ is a real constant.

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

If a continuous functions $$f$$ 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 $$f(x)=k e^{x}-x$$ for all real $$x$$ where $$k$$ is a real constant.

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