Match the statements/expressions in Column I with the open...

(a) Given, $(x-3)^2 \cdot y^{\prime}+y=0$ $$\begin{array}{rlrl} \Rightarrow& \frac{d y}{d x}& =-\frac{y}{(x-3)^2}\\ \Rightarrow & & \int \frac{d y}{y}& =-\int \frac{d x}{(x-3)^3}\\ \Rightarrow & & \ln y & =\frac{1}{(x-3)}+\lnC \\ \Rightarrow & & y & =C e^{\frac{1}{x-3}}, C ≠ 0 \end{array}$$\therefore$Domain of$y$is$x \in R-\{3\}$Aliter Given differential equation is homogeneous linear differential equation and has$x=3$as a singular point, hence$x=3$cannot be in domain of solution. (b) Let$I=\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5) d x$Let$x-3=t \Rightarrow d x=d t\therefore I=\int_{-2}^2(t+2)(t+1) t(t-1)(t-2) d t\because$Integrand is an odd function.$\therefore \quad I=0$Aliter Let$I=\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5) d x$Using,$\int_a^b f(x) d x=\int_a^b f(a+b-x) d xI=\int_1^5(5-x)(4-x)(3-x)(2-x)(1-x) d x$On adding Eqs. (i) and (ii) we get 2I$=0 \Rightarrow I=0$(c) Let$f(x)=\cos ^2 x+\sin x\Rightarrow f^{\prime}(x)=-2 \cos x \sin x+\cos x=\cos x(1-2 \sin x)=0$(say) Sign scheme for first derivative$$\\ Points of local maxima are$\frac{\pi}{6}, \frac{5 \pi}{6}$. Aliter$y=\cos ^2 x+\sin x$\Rightarrow \quad y=$\frac{5}{4}-\left(\sin$x-$\frac{1}{2}\right)^$2 For$y$to be maximum,$$\begin{aligned}& \left(\sinx-\frac{1}{2}\right)^2=0 \Rightarrow \sin x=\frac{1}{2}\\ \Rightarrow \quad & x=n \pi+(-1)^n \frac{\pi}{6},n \in I \end{aligned}$$(d) Let$y=\tan ^{-1}(\sin x+\cos x)$\Rightarrow \quad$\frac{d y}{d x}=\frac{\cos x-\sin x}{1+(\sin x+\cos x)^2}$ \\ Clearly, by graph, \cos x \gt \sin x is true for option(s).