$$ \text { Match the statements/expressions in Column I with the open intervals in Column II. } $$
$$
\text { Match the statements/expressions in Column I with the open intervals in Column II. }
$$
(A) p,q,t, (B) q,t, (C) r,s, (D) p,s
(A) p,q,s, (B) p,t, (C) p,q,r,t, (D) s
(A) p,q,s, (B) q,t, (C) r,s,t, (D) p,s
(A) p,q,t, (B) p,t, (C) p,q,r,t, (D) s
Solution
(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)
