Matching Analysis of Differential Equation Definite Integral and Extreme Values

To solve each part of Column I and match it with the correct interval(s) in Column II:

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Part (A)

The given differential equation is:
$(x-3)^2 \frac{dy}{dx} + y = 0$

For non-zero solutions, separating the variables gives:
$\frac{dy}{y} = -\frac{dx}{(x-3)^2}$

Integrating both sides:
$\ln |y| = \frac{1}{x-3} + C \implies y(x) = A e^{\frac{1}{x-3}} \quad (A \neq 0)$

The function $y(x)$ is defined and differentiable on any interval that does not include the point of singularity $x = 3$. Hence, the domain of definition of a solution is either $(-\infty, 3)$ or $(3, \infty)$.

An interval in Column II is contained in the domain of definition if it does not contain $x = 3$:

• (p) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \approx (-1.57, 1.57) \subset (-\infty, 3)$ $\implies$ Matched
• (q) $\left(0, \frac{\pi}{2}\right) \approx (0, 1.57) \subset (-\infty, 3)$ $\implies$ Matched
• (r) $\left(\frac{\pi}{8}, \frac{5\pi}{4}\right) \approx (0.39, 3.93)$, which contains $x = 3$ $\implies$ Not matched
• (s) $\left(0, \frac{\pi}{8}\right) \approx (0, 0.39) \subset (-\infty, 3)$ $\implies$ Matched
• (t) $(-\pi, \pi) \approx (-3.14, 3.14)$, which contains $x = 3$ $\implies$ Not matched

Thus, (A) $\to$ (p), (q), (s).

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Part (B)

We evaluate the integral:
$I = \int_1^5 (x-1)(x-2)(x-3)(x-4)(x-5) \, dx$

Substitute $t = x - 3$, so $dx = dt$. The limits become $t = -2$ to $t = 2$:
$I = \int_{-2}^2 (t+2)(t+1)t(t-1)(t-2) \, dt = \int_{-2}^2 t(t^2-1)(t^2-4) \, dt = \int_{-2}^2 (t^5 - 5t^3 + 4t) \, dt$

Since the integrand is an odd function of $t$ and the interval $[-2, 2]$ is symmetric about $0$:
$I = 0$

An interval contains the value of the integral if it contains $0$:

• (p) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ contains $0$ $\implies$ Matched
• (q) $\left(0, \frac{\pi}{2}\right)$ does not contain $0$ $\implies$ Not matched
• (r) $\left(\frac{\pi}{8}, \frac{5\pi}{4}\right)$ does not contain $0$ $\implies$ Not matched
• (s) $\left(0, \frac{\pi}{8}\right)$ does not contain $0$ $\implies$ Not matched
• (t) $(-\pi, \pi)$ contains $0$ $\implies$ Matched

Thus, (B) $\to$ (p), (t).

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Part (C)

Let $f(x) = \cos^2 x + \sin x = 1 - \sin^2 x + \sin x$.
Differentiating with respect to $x$:
$f'(x) = -2\sin x \cos x + \cos x = \cos x (1 - 2\sin x)$

For critical points:
$f'(x) = 0 \implies \cos x = 0 \quad \text{or} \quad \sin x = \frac{1}{2}$

Evaluating the second derivative at $\sin x = \frac{1}{2}$:
$f''(x) = -\sin x (1 - 2\sin x) - 2\cos^2 x$
$f''(x)\Big|_{\sin x = \frac{1}{2}} = -2\cos^2 x = -2\left(1 - \frac{1}{4}\right) = -\frac{3}{2} < 0$

Hence, points of local maximum occur where $\sin x = \frac{1}{2}$, which are:
$x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$

In the interval $(-\pi, \pi)$, the local maximum points are $x = \frac{\pi}{6} \approx 0.5236$ and $x = \frac{5\pi}{6} \approx 2.6180$.

• (p) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ contains $\frac{\pi}{6}$ $\implies$ Matched
• (q) $\left(0, \frac{\pi}{2}\right)$ contains $\frac{\pi}{6}$ $\implies$ Matched
• (r) $\left(\frac{\pi}{8}, \frac{5\pi}{4}\right)$ contains both $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ $\implies$ Matched
• (s) $\left(0, \frac{\pi}{8}\right)$ does not contain $\frac{\pi}{6}$ ($\text{since } \frac{\pi}{8} < \frac{\pi}{6}$) $\implies$ Not matched
• (t) $(-\pi, \pi)$ contains both $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ $\implies$ Matched

Thus, (C) $\to$ (p), (q), (r), (t).

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Part (D)

Let $g(x) = \tan^{-1}(\sin x + \cos x)$. Since $\tan^{-1}(u)$ is a strictly increasing function for all $u \in \mathbb{R}$, $g(x)$ is increasing wherever $h(x) = \sin x + \cos x$ is increasing:
$h'(x) = \cos x - \sin x = \sqrt{2}\cos\left(x + \frac{\pi}{4}\right)$

$h'(x) > 0$ when:
$-\frac{\pi}{2} < x + \frac{\pi}{4} < \frac{\pi}{2} \implies -\frac{3\pi}{4} < x < \frac{\pi}{4} \quad (\text{for } k=0)$

Therefore, $g(x)$ is increasing on any interval contained within $\left(-\frac{3\pi}{4}, \frac{\pi}{4}\right)$:

• (s) $\left(0, \frac{\pi}{8}\right) \subset \left(-\frac{3\pi}{4}, \frac{\pi}{4}\right)$, where $h'(x) > 0$ for all $x \in \left(0, \frac{\pi}{8}\right)$ $\implies$ Matched
• The other intervals (p), (q), (r), and (t) all extend beyond $\frac{\pi}{4}$ where $h'(x) < 0$, so $g(x)$ is not strictly increasing on those entire intervals.

Thus, (D) $\to$ (s).

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Final Matching:

• (A) $\to$ (p), (q), (s)
• (B) $\to$ (p), (t)
• (C) $\to$ (p), (q), (r), (t)
• (D) $\to$ (s)