⇒Derivative

Correct Option is (C)

$\begin{array}{rl}& \tan y=\frac{x\sin \alpha }{1-x\cos \alpha }\\ & \therefore y={\tan }^{-1}\left(\frac{x\sin \alpha }{1-x\cos \alpha }\right)\\ & \therefore \frac{dy}{dx}=\frac{1}{1+{\left(\frac{x\sin \alpha }{1-x\cos \alpha }\right)}^2}\frac{d}{dx}\left(\frac{x\sin \alpha }{1-x\cos \alpha }\right)\\ & =\frac{1}{\frac{1-2x\cos \alpha +x^2{\cos }^2\alpha +x^2{\sin }^2\alpha }{(1-x\cos \alpha {)}^2}}\\ & \times \frac{(1-x\cos \alpha )\sin \alpha +(x\sin \alpha )\cos \alpha }{(1-x\cos \alpha {)}^2}\\ & =\frac{\sin \alpha -x\sin \alpha \cos \alpha +x\sin \alpha \cos \alpha }{1+2(-\cos \alpha )x+x^2}\\ & =\frac{\sin \alpha }{x^2+2(-\cos \alpha )+1}\\ & =\frac{m}{x^2+2nx+1}\text{... [Given]}\\ & \Rightarrow \mathrm{n}=-\cos \alpha \text{ and }\mathrm{m}=\sin \alpha \\ & \Rightarrow m^2+n^2=1\end{array}$

Correct Option is (B)

According to the given condition,

$\begin{array}{ll}& {\mathrm{f}}^{\prime }(1)=0\text{ and }{\mathrm{f}}^{\prime }(2)=0\\ & \mathrm{f}(x)=\alpha \log x+\beta x^2+x\\ \therefore & {\mathrm{f}}^{\prime }(x)=\frac{\alpha }{x}+2\beta x+1\\ \therefore & {\mathrm{f}}^{\prime }(-1)=0\Rightarrow \alpha +2\beta =1\text{.... (i)}\\ & \text{ and }{\mathrm{f}}^{\prime }(2)=0\Rightarrow \alpha +8\beta =-2\text{.... (ii)}\end{array}$

$\therefore$ From (i) and (ii), we get

$\begin{array}{rl}& \beta =\frac{-1}{2}\text{ and }\alpha =2\\ \therefore & {\alpha }^2+2\beta =4-1=3\end{array}$

Correct Option is (A)

$\begin{array}{rl}{\mathrm{f}}^{\prime }(x)=3^x\log 3& \Rightarrow {\mathrm{f}}^{\prime }(0)=\log 3\\ {\mathrm{g}}^{\prime }(x)=4^x\log 4& \Rightarrow {\mathrm{g}}^{\prime }(0)=\log 4\\ \therefore \frac{{\mathrm{f}}^{\prime }(0)-{\mathrm{g}}^{\prime }(0)}{1+{\mathrm{f}}^{\prime }(0){\mathrm{g}}^{\prime }(0)}& =\frac{\log 3-\log 4}{1+(\log 3)(\log 4)}\\ & =\frac{\log \left(\frac{3}{4}\right)}{1+(\log 3)(\log 4)}\end{array}$

Correct Option is (C)

$\begin{array}{rl}& f(x)=\frac{1-x+x^2}{1+x+x^2}\\ \therefore & f^{\prime }(x)=\frac{(1+x+x^2)(-1+2x)-(1-x+x^2)(1+2x)}{{(1+x+x^2)}^2}\\ =& \frac{(-1+2x-x+2x^2-x^2+2x^3)-(-1+2x-x-2x^2+x^2+2x^3)}{{(1+x+x^2)}^2}\\ =& \frac{-2+2x^2}{{(1+x+x^2)}^2}\end{array}$

If ${\mathrm{f}}^{\prime }(x)=0$, then $\frac{-2+2x^2}{{(1+x+x^2)}^2}=0\Rightarrow x^2=1$ $\Rightarrow x=\pm 1$

$\therefore \mathrm{f}(x)$ at $x=1$ is $\frac{1}{3}$ and $\mathrm{f}(x)$ at $x=-1$ is 1.

$\therefore$ Minimum value of $\mathrm{f}(x)$ is $\frac{1}{3}$.

Correct Option is (C)

$\begin{array}{rl}& y=\frac{\log \tan x}{\log \sin x}\\ & \frac{dy}{dx}=\frac{(\log \sin x)\left(\frac{1}{\tan x}\right)\cdot {\sec }^{2}x-(\log \tan x)\left(\frac{1}{\sin x}\right)(\cos x)}{(\log \sin x{)}^2}\\ & \text{ At }x=\frac{\pi }{4}\\ & {\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{4}}=\frac{\log \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{1}\right)(\sqrt{2}{)}^2-(\log 1)\left(\frac{\sqrt{2}}{1}\right)\left(\frac{1}{\sqrt{2}}\right)}{{[\log \left(\frac{1}{\sqrt{2}}\right)]}^2}\\ & =\frac{-2\times \frac{1}{2}(\log 2)-0}{\frac{1}{4}(\log 2{)}^2}\dots [\because \log 1=0]\\ & =\frac{-4}{\log 2}\end{array}$