Match Column Statements Involving Functions Matrices Logarithms and Trigonometry

To solve this matching problem, we analyze each statement in Column I step-by-step:

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Statement (A)

We wish to find the minimum value of $f(x) = \frac{x^2+2x+4}{x+2}$.

Let $y = x+2$. Assuming $x > -2$ (so $y > 0$):
$\frac{x^2+2x+4}{x+2} = \frac{(y-2)^2 + 2(y-2) + 4}{y} = \frac{y^2 - 2y + 4}{y} = y + \frac{4}{y} - 2$

Applying the AM-GM inequality for $y > 0$:
$y + \frac{4}{y} \ge 2\sqrt{y \cdot \frac{4}{y}} = 4$

Thus,
$f(x) = y + \frac{4}{y} - 2 \ge 4 - 2 = 2$

The minimum value is $2$, which occurs when $y = 2 \implies x = 0$.

• Matching option for (A): (r)

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Statement (B)

We are given that $A$ is a symmetric matrix ($A^T = A$), $B$ is a skew-symmetric matrix ($B^T = -B$), and:
$(A+B)(A-B) = (A-B)(A+B)$

Expanding both sides:
$A^2 - AB + BA - B^2 = A^2 + AB - BA - B^2$
$-AB + BA = AB - BA \implies 2AB = 2BA \implies AB = BA$

Now, taking the transpose of the matrix product $AB$:
$(AB)^T = B^T A^T = (-B)(A) = -BA$

Since $AB = BA$, we have:
$(AB)^T = -AB$

We are given $(AB)^T = (-1)^k AB$, so:
$(-1)^k AB = -AB$

Assuming $AB \neq 0$:
$(-1)^k = -1$
This implies that $k$ must be an odd integer. From Column II, the odd integers listed are $1$ and $3$.

• Matching options for (B): (q), (s)

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Statement (C)

Given $a = \log_3(\log_3 2)$, we evaluate the term $3^{-a}$:
$3^{-a} = 3^{-\log_3(\log_3 2)} = 3^{\log_3\left(\frac{1}{\log_3 2}\right)} = \frac{1}{\log_3 2} = \log_2 3$

Substitute $3^{-a} = \log_2 3$ into the given inequality:
$1 < 2^{-k + \log_2 3} < 2$

Taking the logarithm to base $2$ across all parts of the inequality:
$\log_2 1 < -k + \log_2 3 < \log_2 2$
$0 < -k + \log_2 3 < 1$
$-\log_2 3 < -k < 1 - \log_2 3$

Multiplying by $-1$ reverses the inequality signs:
$\log_2 3 - 1 < k < \log_2 3$

Since $\log_2 3 \approx 1.585$:
$0.585 < k < 1.585$

Since $k$ is an integer, the unique solution is $k = 1$.
The problem states that integer $k$ "must be less than". Since $k = 1$, it is strictly less than $2$ and $3$.

• Matching options for (C): (r), (s)

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Statement (D)

Given the trigonometric equation:
$\sin\theta = \cos\phi$

Using the identity $\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)$:
$\cos\left(\frac{\pi}{2} - \theta\right) = \cos\phi$

The general solution to $\cos A = \cos B$ is $A = 2n\pi \pm B$, where $n \in \mathbb{Z}$:
$\frac{\pi}{2} - \theta = 2n\pi \pm \phi$

Rearranging the terms:
$\theta \pm \phi - \frac{\pi}{2} = -2n\pi$

Dividing by $\pi$:
$\frac{1}{\pi}\left(\theta \pm \phi - \frac{\pi}{2}\right) = -2n$

Since $n \in \mathbb{Z}$, the value $-2n$ must be an even integer. From Column II, the even integers are $0$ and $2$.

• Matching options for (D): (p), (r)

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Final Match Summary

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