Part (A): Domain of solution of a differential equation

1. Solve the differential equation: The given differential equation is $(x - 3)^2 y' + y = 0$. We can write it as $(x - 3)^2 \frac{dy}{dx} = -y$. This is a variable separable equation. $\frac{dy}{y} = -\frac{dx}{(x - 3)^2}$

2. Integrate both sides: $\int \frac{dy}{y} = -\int (x - 3)^{-2} dx$ $\ln|y| = -\frac{(x - 3)^{-1}}{-1} + C$ $\ln|y| = \frac{1}{x - 3} + C$

3. Find the explicit solution and its domain: The solution is $|y| = e^{\frac{1}{x-3} + C}$, which can be written as $y(x) = A e^{\frac{1}{x-3}}$ for a non-zero constant $A = \pm e^C$. The function $y(x)$ is defined whenever the exponent $\frac{1}{x-3}$ is defined. This requires the denominator $x-3 \neq 0$, so $x \neq 3$. The domain of definition for any specific non-zero solution must be a single connected interval. Therefore, the domain of a solution is either $(-\infty, 3)$ or $(3, \infty)$.

4. Check the intervals in Column II: We need to find intervals that are subsets of either $(-\infty, 3)$ or $(3, \infty)$.

   • (P) $( -\frac{\pi}{2}, \frac{\pi}{2} ) \approx (-1.57, 1.57)$. This is a subset of $(-\infty, 3)$. Match.
   • (Q) $( 0, \frac{\pi}{2} ) \approx (0, 1.57)$. This is a subset of $(-\infty, 3)$. Match.
   • (R) $( \frac{\pi}{8}, \frac{5\pi}{4} ) \approx (0.39, 3.92)$. This interval contains $x=3$. Not a subset. No match.
   • (S) $( 0, \frac{\pi}{8} ) \approx (0, 0.39)$. This is a subset of $(-\infty, 3)$. Match.
   • (T) $( -\pi, \pi ) \approx (-3.14, 3.14)$. This interval contains $x=3$. Not a subset. No match.

   Therefore, (A) matches (P), (Q), (S).

Part (B): Value of a definite integral

1. Analyze the integrand and interval: The integral is $I = \int_1^5 (x-1)(x-2)(x-3)(x-4)(x-5)dx$. Let $f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)$. The interval of integration is $[1, 5]$, which is symmetric about $x=3$.

2. Use symmetry property: Let's check the symmetry of $f(x)$ about the point $(3,0)$. Let $x = 3+u$. Then $dx=du$. The limits of integration become $u = 1-3 = -2$ and $u = 5-3 = 2$. The integrand becomes: $f(3+u) = ((3+u)-1)((3+u)-2)((3+u)-3)((3+u)-4)((3+u)-5)$ $f(3+u) = (u+2)(u+1)(u)(u-1)(u-2) = u(u^2-1)(u^2-4) = u^5 - 5u^3 + 4u$. This is an odd function of $u$, let's call it $g(u)$.

3. Evaluate the integral: The integral becomes $I = \int_{-2}^2 g(u) du = \int_{-2}^2 (u^5 - 5u^3 + 4u) du$. Since the integral of an odd function over a symmetric interval $[-a, a]$ is zero, we have $I=0$.

4. Check which intervals contain the value 0:

   • (P) $( -\frac{\pi}{2}, \frac{\pi}{2} )$. Contains 0. Match.
   • (Q) $( 0, \frac{\pi}{2} )$. Does not contain 0. No match.
   • (R) $( \frac{\pi}{8}, \frac{5\pi}{4} )$. Does not contain 0. No match.
   • (S) $( 0, \frac{\pi}{8} )$. Does not contain 0. No match.
   • (T) $( -\pi, \pi )$. Contains 0. Match.

   Therefore, (B) matches (P), (T). Note: The provided answer key suggests (S) is also a match, which is incorrect as the open interval $(0, \pi/8)$ does not contain 0. This appears to be an error in the question or options.

Part (C): Points of local maximum

1. Find the derivative: Let $f(x) = \cos^2x + \sin x$. Then $f'(x) = 2\cos x(-\sin x) + \cos x = \cos x (1 - 2\sin x)$.

2. Find critical points: Set $f'(x) = 0$. This gives $\cos x = 0$ or $1 - 2\sin x = 0 \implies \sin x = 1/2$.

3. Use the second derivative test: $f''(x) = -\sin x(1-2\sin x) + \cos x(-2\cos x) = -\sin x + 2\sin^2 x - 2\cos^2 x$. Using $\cos^2 x = 1-\sin^2 x$, $f''(x) = -\sin x + 4\sin^2 x - 2$.

   • If $\cos x = 0$, then $\sin x = \pm 1$. If $\sin x=1$, $f''(x) = -1+4-2=1>0$ (local minimum). If $\sin x=-1$, $f''(x) = -(-1)+4(-1)^2-2=1+4-2=3>0$ (local minimum).
   • If $\sin x = 1/2$, then $f''(x) = -1/2 + 4(1/2)^2 - 2 = -1/2 + 1 - 2 = -3/2 < 0$ (local maximum).

4. Identify points of local maximum and check intervals: Local maxima occur at $x$ where $\sin x = 1/2$. These points are $x = \frac{\pi}{6} + 2n\pi$ and $x = \frac{5\pi}{6} + 2n\pi$ for any integer $n$. We need to find intervals containing at least one of these points. Let's check for $x=\frac{\pi}{6} \approx 0.52$ and $x=\frac{5\pi}{6} \approx 2.62$.

   • (P) $( -\frac{\pi}{2}, \frac{\pi}{2} )$ contains $\frac{\pi}{6}$. Match.
   • (Q) $( 0, \frac{\pi}{2} )$ contains $\frac{\pi}{6}$. Match.
   • (R) $( \frac{\pi}{8}, \frac{5\pi}{4} ) \approx (0.39, 3.92)$ contains both $\frac{\pi}{6}$ and $\frac{5\pi}{6}$. Match.
   • (S) $( 0, \frac{\pi}{8} ) \approx (0, 0.39)$. $\frac{\pi}{6} > \frac{\pi}{8}$. No match.
   • (T) $( -\pi, \pi )$ contains both $\frac{\pi}{6}$ and $\frac{5\pi}{6}$. Match.

   Therefore, (C) matches (P), (Q), (R), (T).

Part (D): Interval of increasing function

1. Find the derivative: Let $f(x) = \tan^{-1}(\sin x + \cos x)$. The function is increasing when $f'(x) > 0$. $f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$.

2. Determine the condition for increasing function: The denominator is always positive. So, $f'(x) > 0$ when the numerator $\cos x - \sin x > 0$, which means $\cos x > \sin x$.

3. Solve the inequality: The inequality $\cos x > \sin x$ holds for $x$ in the intervals $(2n\pi - \frac{3\pi}{4}, 2n\pi + \frac{\pi}{4})$. For $n=0$, this is the interval $(-\frac{3\pi}{4}, \frac{\pi}{4}) \approx (-2.36, 0.785)$.

4. Check which interval is contained in an interval of increase:

   • (P) $( -\frac{\pi}{2}, \frac{\pi}{2} )$. Not a subset, as for $x \in [\pi/4, \pi/2)$, $\cos x \le \sin x$.
   • (Q) $( 0, \frac{\pi}{2} )$. Not a subset for the same reason.
   • (R) $( \frac{\pi}{8}, \frac{5\pi}{4} )$. Not a subset.
   • (S) $( 0, \frac{\pi}{8} )$. Since $0 < \frac{\pi}{8} < \frac{\pi}{4}$, this interval is a subset of $(-\frac{3\pi}{4}, \frac{\pi}{4})$. Match.
   • (T) $( -\pi, \pi )$. Not a subset.

   Therefore, (D) matches (S).

Conclusion

• (A) $\to$ (P), (Q), (S)
• (B) $\to$ (P), (T)
• (C) $\to$ (P), (Q), (R), (T)
• (D) $\to$ (S)

Comparing our findings with the given options: Option A: (A) $\to$(P), (Q), (S); (B) $\to$(P), (T), (S); (C) $\to$(P), (Q), (R), (T); (D) $\to$(S) This matches our results for (A), (C), and (D). For (B), it incorrectly includes (S). All other options have more significant mismatches.

• Option B is wrong on (D).
• Option C is wrong on (C) (it misses P).
• Option D is wrong on (A) (it misses Q and includes T).

Thus, Option A is the best fit, assuming an error in the question regarding the matching of (B) with (S).