Matching Statements in Column I with Values in Column II

To match the statements in Column I with the values in Column II, we evaluate each statement step-by-step:

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(A) Number of solutions of $x e^{\sin x} - \cos x = 0$ in $\left(0, \frac{\pi}{2}\right)$

Let the function be:
$f(x) = x e^{\sin x} - \cos x$

Evaluating the function at the endpoints:

• At $x = 0$:
  $f(0) = 0 \cdot e^0 - \cos 0 = -1 < 0$
• At $x = \frac{\pi}{2}$:
  $f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} e^{\sin(\pi/2)} - \cos\left(\frac{\pi}{2}\right) = \frac{\pi}{2}e > 0$

Now, taking the first derivative of $f(x)$ with respect to $x$:
$f'(x) = e^{\sin x} + x \cos x \, e^{\sin x} + \sin x$

For all $x \in \left(0, \frac{\pi}{2}\right)$, we have:
$x > 0, \quad \sin x > 0, \quad \cos x > 0, \quad \text{and } e^{\sin x} > 0$

Thus, $f'(x) > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$, which means that $f(x)$ is strictly increasing in this interval.

By the Intermediate Value Theorem and monotonicity, $f(x) = 0$ has exactly $1$ solution in $\left(0, \frac{\pi}{2}\right)$.

$\implies \mathbf{(A) \to (p)}$

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(B) Value(s) of $k$ for which the planes intersect in a straight line

The equations of the three planes passing through the origin are:

1. $kx + 4y + z = 0$
2. $4x + ky + 2z = 0$
3. $2x + 2y + z = 0$

For three planes passing through the origin to intersect in a straight line, the homogeneous system must have non-trivial solutions with rank $2$, meaning the determinant of the coefficient matrix must vanish:

$$\begin{vmatrix} k & 4 & 1 \\ 4 & k & 2 \\ 2 & 2 & 1 \end{vmatrix} = 0$$

Applying elementary row operations:

• $R_1 \to R_1 - R_3$: $[k - 2, \, 2, \, 0]$
• $R_2 \to R_2 - 2R_3$: $[0, \, k - 4, \, 0]$

The determinant simplifies to:

$$\begin{vmatrix} k - 2 & 2 & 0 \\ 0 & k - 4 & 0 \\ 2 & 2 & 1 \end{vmatrix} = 1 \cdot \left((k - 2)(k - 4) - 0\right) = (k - 2)(k - 4) = 0$$

Thus, $k = 2$ or $k = 4$.

• For $k = 2$, the normal vectors are $(2, 4, 1)$, $(4, 2, 2)$, and $(2, 2, 1)$, no two of which are parallel, so the three planes intersect along a unique line.
• For $k = 4$, the second plane becomes $4x + 4y + 2z = 0 \iff 2x + 2y + z = 0$, which is identical to the third plane, and intersects the first plane $4x + 4y + z = 0$ along a line.

Hence, the possible values of $k$ are $2$ and $4$.

$\implies \mathbf{(B) \to (q), (s)}$

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(C) Value(s) of $k$ for which $|x - 1| + |x - 2| + |x + 1| + |x + 2| = 4k$ has integer solution(s)

Let $g(x) = |x - 1| + |x - 2| + |x + 1| + |x + 2|$.

Notice that $g(x)$ is an even function:

• For $x \ge 2$, all terms inside the absolute values are non-negative:
  $g(x) = (x - 1) + (x - 2) + (x + 1) + (x + 2) = 4x$
• For $x \in [-1, 1]$:
  $g(x) = (1 - x) + (2 - x) + (x + 1) + (x + 2) = 6$

For $x$ to be an integer solution:

• If $x = 0$ or $x = \pm 1$: $g(x) = 6 \implies 4k = 6 \implies k = 1.5$ (not an option).
• If $x = \pm 2$: $g(2) = 4(2) = 8 \implies 4k = 8 \implies k = 2$.
• If $x = \pm 3$: $g(3) = 4(3) = 12 \implies 4k = 12 \implies k = 3$.
• If $x = \pm 4$: $g(4) = 4(4) = 16 \implies 4k = 16 \implies k = 4$.
• If $x = \pm 5$: $g(5) = 4(5) = 20 \implies 4k = 20 \implies k = 5$.

Therefore, any integer $k \ge 2$ gives an integer solution $x = \pm k$. From the given options in Column II, the possible values are $2, 3, 4, 5$.

$\implies \mathbf{(C) \to (q), (r), (s), (t)}$

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(D) Value of $y(\ln 2)$ given $y' = y + 1$ and $y(0) = 1$

Rewrite the differential equation:
$\frac{dy}{dx} - y = 1$

The integrating factor is:
$I.F. = e^{\int -1 \, dx} = e^{-x}$

Multiplying both sides by $e^{-x}$:
$\frac{d}{dx}\left(y e^{-x}\right) = e^{-x}$

Integrating both sides:
$y e^{-x} = -e^{-x} + C \implies y(x) = C e^x - 1$

Using the initial condition $y(0) = 1$:
$1 = C e^0 - 1 \implies C = 2$

Thus, the particular solution is:
$y(x) = 2e^x - 1$

Evaluating at $x = \ln 2$:
$y(\ln 2) = 2e^{\ln 2} - 1 = 2(2) - 1 = 3$

$\implies \mathbf{(D) \to (r)}$

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Final Matching:

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