Find the point of intersection of two lines and the number of ways to arrange girls on a bench with a condition. The final answer is B.

The two lines are given by:
$\mathbf{r_1} = 3\mathbf{i} - \mathbf{j} + \mathbf{k} + \lambda(\mathbf{i} + 2\mathbf{j} - \mathbf{k})$
$\mathbf{r_2} = 4\mathbf{i} + 4\mathbf{j} + \mathbf{k} + \mu(-\mathbf{i} + \mathbf{j} + 2\mathbf{k})$

For the lines to intersect, $\mathbf{r_1} = \mathbf{r_2}$.
Equating the components:
For $\mathbf{i}$: $3 + \lambda = 4 - \mu \quad (1)$
For $\mathbf{j}$: $-1 + 2\lambda = 4 + \mu \quad (2)$
For $\mathbf{k}$: $1 - \lambda = 1 + 2\mu \quad (3)$

From equation (1), $\lambda + \mu = 1$.
From equation (2), $2\lambda - \mu = 5$.

Add these two equations:
$(\lambda + \mu) + (2\lambda - \mu) = 1 + 5$
$3\lambda = 6$
$\lambda = 2$

Substitute $\lambda = 2$ into $\lambda + \mu = 1$:
$2 + \mu = 1$
$\mu = 1 - 2$
$\mu = -1$

Now, check these values in equation (3):
$1 - \lambda = 1 + 2\mu$
$1 - 2 = 1 + 2(-1)$
$-1 = 1 - 2$
$-1 = -1$
The values $\lambda = 2$ and $\mu = -1$ satisfy all three equations, so the lines intersect.

The correct option is B.
The final answer is $\boxed{\text{B}}$.

We need to arrange 5 girls on a bench, with the condition that two specific left-handed girls must be together.
Treat the two left-handed girls as a single unit.
Now we have 4 "units" to arrange: (LH1 LH2), G3, G4, G5.
The number of ways to arrange these 4 units is $4!$.
$4! = 4 \times 3 \times 2 \times 1 = 24$.

Within the unit of the two left-handed girls, they can be arranged in $2!$ ways (LH1 LH2 or LH2 LH1).
$2! = 2 \times 1 = 2$.

The total number of ways to arrange the 5 girls such that the two left-handed girls are together is the product of these two numbers:
Total ways = $4! \times 2! = 24 \times 2 = 48$.

The correct option is B.
The final answer is $\boxed{\text{B}}$.

We need to simplify the expression $\sin\left(x + \frac{\pi}{6}\right) - \sin\left(x - \frac{\pi}{6}\right)$.
Use the sum-to-product trigonometric identity:
$\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$
Here, $A = x + \frac{\pi}{6}$ and $B = x - \frac{\pi}{6}$.

First, find $\frac{A+B}{2}$:
$\frac{A+B}{2} = \frac{\left(x + \frac{\pi}{6}\right) + \left(x - \frac{\pi}{6}\right)}{2} = \frac{2x}{2} = x$

Next, find $\frac{A-B}{2}$:
$\frac{A-B}{2} = \frac{\left(x + \frac{\pi}{6}\right) - \left(x - \frac{\pi}{6}\right)}{2} = \frac{x + \frac{\pi}{6} - x + \frac{\pi}{6}}{2} = \frac{\frac{2\pi}{6}}{2} = \frac{\frac{\pi}{3}}{2} = \frac{\pi}{6}$

Substitute these into the identity:
$\sin\left(x + \frac{\pi}{6}\right) - \sin\left(x - \frac{\pi}{6}\right) = 2\cos(x)\sin\left(\frac{\pi}{6}\right)$
We know that $\sin\left(\frac{\pi}{6}\right) = \sin(30^\circ) = \frac{1}{2}$.
So, the expression becomes:
$2\cos(x)\left(\frac{1}{2}\right) = \cos(x)$

The correct option is C.
The final answer is $\boxed{\text{C}}$.

The given function is $y = \frac{4}{x^2 - 1}$.

To find vertical asymptotes, set the denominator equal to zero:
$x^2 - 1 = 0$
$(x-1)(x+1) = 0$
$x = 1$ or $x = -1$
So, there are two vertical asymptotes at $x = \pm 1$.

To find horizontal asymptotes, consider the limit as $x \to \pm\infty$:

$$\lim_{x \to \pm\infty} \frac{4}{x^2 - 1}$$

As $x \to \pm\infty$, $x^2 - 1 \to \infty$.
So, $\lim_{x \to \pm\infty} \frac{4}{x^2 - 1} = 0$.
Thus, there is a horizontal asymptote at $y = 0$.

The graph has two vertical asymptotes at $x = \pm 1$ and a horizontal asymptote at $y = 0$.

The correct option is D.
The final answer is $\boxed{\text{D}}$.

Given the differential equation $\frac{dy}{dx} - 2xy = 0$.
We can rewrite this as $\frac{dy}{dx} = 2xy$.
This is a separable differential equation.
$\frac{dy}{y} = 2x\,dx$

Integrate both sides:

$$\int \frac{1}{y}\,dy = \int 2x\,dx$$ $$\ln|y| = x^2 + C$$

Exponentiate both sides:

$$|y| = e^{x^2 + C}$$ $$y = \pm e^C e^{x^2}$$

Let $A = \pm e^C$, where $A$ is an arbitrary non-zero constant.

$$y = Ae^{x^2}$$

We are given the initial condition $y=1$ when $x=1$.
Substitute these values into the general solution:
$1 = Ae^{(1)^2}$
$1 = Ae^1$
$A = \frac{1}{e} = e^{-1}$

Substitute the value of $A$ back into the general solution:

$$y = e^{-1}e^{x^2}$$ $$y = e^{x^2 - 1}$$

The correct option is A.
The final answer is $\boxed{\text{A}}$.

We need to find the coefficient of $x^2$ in the binomial expansion of $\left(1 + \frac{1}{2}x\right)^5$.
The binomial theorem states that the general term in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
In this case, $a=1$, $b=\frac{1}{2}x$, and $n=5$.
We want the term with $x^2$, so we set $r=2$.

$T_{2+1} = T_3 = \binom{5}{2} (1)^{5-2} \left(\frac{1}{2}x\right)^2$
$T_3 = \binom{5}{2} (1)^3 \left(\frac{1}{2}\right)^2 (x)^2$
Calculate $\binom{5}{2}$:
$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$

Substitute this value back:
$T_3 = 10 \times 1 \times \frac{1}{4} \times x^2$
$T_3 = \frac{10}{4} x^2$
$T_3 = \frac{5}{2} x^2$

The coefficient of $x^2$ is $\frac{5}{2}$.

Wait, let me recheck the options. The options are $\frac{1}{32}$, $-\frac{1}{16}$, $\frac{1}{16}$, $-\frac{1}{32}$. My calculated coefficient $\frac{5}{2}$ is not among the options.
Let me re-read the question carefully. It says $\left(1 + \frac{1}{2}x\right)^5$.
My calculation is correct for this expression.

Let me check if the image is clear. The exponent is definitely 5.
Perhaps the question intended a different exponent or a different term.
Let me assume there might be a typo in the question or options.
If the question was $\left(1 + \frac{1}{2}x\right)^{-5}$, then the general term for $(1+y)^n$ is $\sum_{r=0}^{\infty} \binom{n}{r} y^r$.
For $n=-5$ and $y=\frac{1}{2}x$, the coefficient of $x^2$ would be $\binom{-5}{2} \left(\frac{1}{2}\right)^2$.
$\binom{-5}{2} = \frac{(-5)(-5-1)}{2!} = \frac{(-5)(-6)}{2} = \frac{30}{2} = 15$.
So, $15 \times \frac{1}{4} = \frac{15}{4}$. Still not matching.

What if the expression was $\left(1 + x\right)^{\frac{1}{2}}$? No, it's $\left(1 + \frac{1}{2}x\right)^5$.
Let me re-evaluate the calculation for $\binom{5}{2}$.
$\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$. This is correct.
$\left(\frac{1}{2}x\right)^2 = \frac{1}{4}x^2$. This is correct.
Coefficient is $10 \times \frac{1}{4} = \frac{10}{4} = \frac{5}{2}$. This is correct.

Given the options, there might be a misunderstanding of the question or a typo in the question/options.
Let's consider if the question was $\left(1 + \frac{1}{2x}\right)^5$. No, it's clearly $\frac{1}{2}x$.
What if the exponent was different?
If the exponent was, for example, $-2$, then $\binom{-2}{2} (\frac{1}{2})^2 = \frac{(-2)(-3)}{2} \times \frac{1}{4} = 3 \times \frac{1}{4} = \frac{3}{4}$.
If the exponent was $-1$, then $\binom{-1}{2} (\frac{1}{2})^2 = \frac{(-1)(-2)}{2} \times \frac{1}{4} = 1 \times \frac{1}{4} = \frac{1}{4}$.

Let's re-examine the image for the exponent. It looks like a 5.
Let's assume there's a typo in the question and it was meant to be $\left(1 + \frac{1}{2}x\right)^{-2}$ or something similar that would lead to one of the options.
However, I must solve the question as written.
My calculated answer is $\frac{5}{2}$. Since this is not among the options, I will state my calculated answer and note the discrepancy.

Let me check if I misread the options.
A: $\frac{1}{32}$
B: $-\frac{1}{16}$
C: $\frac{1}{16}$
D: $-\frac{1}{32}$

None of these match $\frac{5}{2}$.
I will proceed with the calculation as it is, and state the result.

The coefficient of $x^2$ is $\frac{5}{2}$.
Since $\frac{5}{2}$ is not among the options, there might be an error in the question or the provided options.
I will select the closest option if forced, but that's not ideal.
Let me re-read the question one more time. "The coefficient of $x^2$ in the binomial expansion of $\left(1 + \frac{1}{2}x\right)^5$ is".
The calculation is straightforward and seems correct.

Let's consider if the question was $\left(1 + \frac{1}{2x}\right)^5$. No, it's clearly $\frac{1}{2}x$.
What if the question was $\left(1 + x\right)^{\frac{1}{2}}$?
Then the expansion is $1 + \frac{1}{2}x + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}x^2 + \dots = 1 + \frac{1}{2}x + \frac{\frac{1}{2}(-\frac{1}{2})}{2}x^2 + \dots = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \dots$.
The coefficient of $x^2$ would be $-\frac{1}{8}$. Still not matching.

What if the question was $\left(1 + x\right)^{-2}$?
Then the expansion is $1 + (-2)x + \frac{(-2)(-3)}{2!}x^2 + \dots = 1 - 2x + 3x^2 + \dots$.
The coefficient of $x^2$ would be 3. Still not matching.

What if the question was $\left(1 + \frac{1}{2}x\right)^{-1}$?
Then the expansion is $1 + (-1)(\frac{1}{2}x) + \frac{(-1)(-2)}{2!}(\frac{1}{2}x)^2 + \dots = 1 - \frac{1}{2}x + 1 \cdot \frac{1}{4}x^2 + \dots = 1 - \frac{1}{2}x + \frac{1}{4}x^2 + \dots$.
The coefficient of $x^2$ would be $\frac{1}{4}$. Still not matching.

What if the question was $\left(1 + \frac{1}{2}x\right)^{-4}$?
Then the coefficient of $x^2$ would be $\binom{-4}{2} (\frac{1}{2})^2 = \frac{(-4)(-5)}{2} \times \frac{1}{4} = 10 \times \frac{1}{4} = \frac{5}{2}$. This is the same as my original answer.

Let's consider the possibility that the question is $\left(1 + x\right)^n$ and the coefficient of $x^2$ is one of the options.
If the coefficient of $x^2$ is $\frac{1}{16}$, then $\binom{n}{2} (\frac{1}{2})^2 = \frac{1}{16}$.
$\frac{n(n-1)}{2} \times \frac{1}{4} = \frac{1}{16}$
$\frac{n(n-1)}{8} = \frac{1}{16}$
$n(n-1) = \frac{8}{16} = \frac{1}{2}$.
$2n^2 - 2n - 1 = 0$. This does not give an integer $n$.

Let's assume the question is correct as written and my calculation is correct.
The coefficient of $x^2$ in the binomial expansion of $\left(1 + \frac{1}{2}x\right)^5$ is $\frac{5}{2}$.
Since this is not an option, I will state the calculated value.

The final answer is $\boxed{\frac{5}{2}}$.
(Note: This answer is not among the provided options, which suggests a potential error in the question or options.)

We need to find the coordinates of the points where the curves $y = 2x - 3$ and $y = x^2 - 2x$ intersect.
Set the two expressions for $y$ equal to each other:
$2x - 3 = x^2 - 2x$
Rearrange the equation to form a quadratic equation:
$0 = x^2 - 2x - 2x + 3$
$0 = x^2 - 4x + 3$

Factor the quadratic equation:
We need two numbers that multiply to 3 and add to -4. These are -1 and -3.
$(x-1)(x-3) = 0$
So, $x = 1$ or $x = 3$.

Now, find the corresponding $y$ values using $y = 2x - 3$:
For $x=1