Assertion (A) The function $f(x)=x-\log \left(\frac{1+x}{x}\right), x>0$ has no maximum.

Reason (R) If a function $f(x)$ is strictly increasing in an interval $(a, b)$, then at any point in $(a, b) f^{\prime}(x) \neq 0$

The correct option among the following is

If the tangent and normal drawn to the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$ at $P\left(\theta=\frac{\pi}{2}\right)$ cuts the $X$-axis at $A$ and $B$ respectively, then the area (in sq. units) of $\triangle P A B$ is

$x_1, x_2 \in \mathbf{N}$. If a line having slope 2 is a tangent to the curve $y=x^4-6 x^3+13 x^2-10 x+5$ at points $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right)$, then $x_1 x_2+y_1 y_2=$

Consider the following statements

Statement I If $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\ldots .+\frac{a_n}{n+1}=0$, where $a_0, a_1, \ldots, a_n$ are real numbers, then the polynomial $a_0+a_1 x+a_2 x^2+\ldots .+a_n x^n$ has a zero in the interval $(0,1)$.

Statement II If $f:[a, b] \rightarrow \mathbf{R}$ is continuous on $[a, b]$ and $f$ is differentiable in $(a, b)$, where $a>0$ and if $\frac{f(a)}{a}=\frac{f(b)}{b}$, then there exists $c \in(a, b)$, such that $c f^{\prime}(c)=f(c)$.

Which one of the following options is true?