⇒Application Of Derivative

Correct Option is (D)

$\begin{array}{rl}& \\ & \begin{array}{rl}& f(x)=\log (1+x)-\frac{2x}{2+x}\Rightarrow x\ne -2\\ & \therefore f^{\prime }(x)=\frac{1}{(1+x)}-\left[\frac{(2+x)(2)-(2x)(1)}{(2+x{)}^2}\right]\\ & =\frac{1}{1+x}-\left[\frac{4}{(2+x{)}^2}\right]=\frac{(x+2{)}^2-4(x+1)}{(x+2{)}^2(1+x)}\end{array}\end{array}$

When $f^{\prime }(x)>0$, we write

$\frac{x^2}{(x+2{)}^2(1+x)}>0$

Since $x^2>0$ and $(x+2{)}^2>0$, we write $(1+x)>0\Rightarrow x>-1$

Correct Option is (A)

$\begin{array}{rl}& f(x)={\cot }^{-1}x+x\\ & \therefore f^{\prime }(x)=\frac{-1}{1+x^2}+1=\frac{-1+1+x^2}{1+x^2}=\frac{x^2}{1+x^2}\\ & \text{ Here }{x}^2\ge 0\Rightarrow \frac{x^2}{1+x^2}\ge 0\end{array}$

Hence $\mathrm{f}(\mathrm{x})$ is always increasing.

Correct Option is (B)

$\begin{array}{rl}& f(x)=2x^3-15x^2-144x-7\\ & f^{\prime }(x)=6x^2-30x-144\\ & \text{ When }{f}^{\prime }(x)<0,\text{ we get }{x}^2-5x-24<0\\ & \therefore (x-8)(x+3)<0\Rightarrow -3<x<8\end{array}$