Match the following

	Column – I		Column – II
(A)	$2\underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}{\left(\sin x\right)}^{\cos x}\left(\cos xcotx-\log {\left(sinx\right)}^{\sin x}\right)dx$	(p)	2
(B)	Area bounded by $-4y^2=x$ and $x-1=-5y^2$	(q)	1
(C)	Cosine of the angle of intersection of curves$y=3^{x-1}\log x$ and $y=x^x-1$  is	(r)	$6\ln 3$
(D)	Let $\frac{dy}{dx}=\frac{6}{x+y}$ where  $y\left(0\right)=0$ then value of $y$  when $x+y=12$ is	(s)	$\frac{4}{3}$

(A)       ${\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left(\sin x\right)}^{\cos x}(\cos x\cot x-\log {\left(\sin x\right)}^{\sin x})dx$

${\int }_0^{1}du$ Where ${\left(\sin x\right)}^{\cos x}=u=1$

$\left(A\right)\to \left(p\right)$

(B)      

Solving $y^2=-\frac{1}{4}x$ and $y^2=-\frac{1}{5}(x-1),$ we get intersection points as $(-4,\pm 1)$

$\therefore$ Required area $={\int }_{-1}^1[(1-5y^2)+4y]dy=2{\int }_0^1(1-y^2)dy=\frac{4}{3},$ $\left(B\right)\to \left(s\right)$

(C) By inspection, the point of intersection of two curves $y=3^{x-1}\log x$ and $y=x^x-1$ is $(1,0)$

For first curve $\frac{dy}{dx}=\frac{3^{x-1}}{x}+3^{x-1}\log 3\log x$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(1,0)}=1=m_2$

$\because m_1=m_2\Rightarrow$ Two curves touch each other

$\Rightarrow$ angle between them is $0°$

$\therefore \cos \theta =1,$

$\left(C\right)\to \left(q\right)$

(D) $\frac{dy}{dx}=\frac{6}{x+6}\Rightarrow \frac{dx}{dy}-\frac{1}{6}=\frac{y}{6}$

$\Rightarrow$ Solution is $x(I.F)=\int \left(\frac{y}{6}(I.F)\right)dy+c$

$\Rightarrow x{e}^{\raisebox{1ex}{$-y$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}=-y{e}^{\raisebox{1ex}{$-y$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}-6e^{\raisebox{1ex}{$-y$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}+c$

$e^{\raisebox{1ex}{$-y$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}(x+y+6)=c$

$\Rightarrow x+y+6=c{e}^{\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}$

$\Rightarrow x+y+6=6e^{y/6}$                    $\therefore \left(y\left(0\right)=0\right)$

$\Rightarrow 18=6e^{y/6}$                                  (using $x+y=12$)

$\Rightarrow y=6\ln 3\left(D\right)\to \left(r\right)$