\[
\text { Match the statements/expressions in Column I with the open intervals in Column II. }
\]

(a) Given, \((x-3)^2 \cdot y^{\prime}+y=0\)
\[
\begin{array}{rlrl}
\Rightarrow & \frac{d y}{d x} & =-\frac{y}{(x-3)^2} \\
\Rightarrow & & \int \frac{d y}{y} & =-\int \frac{d x}{(x-3)^3} \\
\Rightarrow & & \ln y & =\frac{1}{(x-3)}+\ln C \\
\Rightarrow & & y & =C e^{\frac{1}{x-3}}, C \neq 0
\end{array}
\]
\(\therefore\) 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, \(\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\) \(I=\int_1^5(5-x)(4-x)(3-x)(2-x)\) \((1-x) d x\)
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,
\[
\begin{aligned}
& \left(\sin x-\frac{1}{2}\right)^2=0 \Rightarrow \sin x=\frac{1}{2} \\
\Rightarrow \quad & x=n \pi+(-1)^n \frac{\pi}{6}, n \in I
\end{aligned}
\]
(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, \(\cos x>\sin x\) is true for option(s).