To make the expression \(16a^2 – 12a\) a perfect square, we can complete the square by adding a term to it.

First, let’s factor out the common factor of 4:

\[16a^2 – 12a = 4(4a^2 – 3a)\]

Now, we want to complete the square for the expression inside the parentheses. The general form for a perfect square is \((x – y)^2 = x^2 – 2xy + y^2\). In our case, we have \(4a^2 – 3a\), so we can compare it to \(x^2 – 2xy\) to find the missing \(y^2\) term:

Comparing \(4a^2 – 3a\) to \(x^2 – 2xy\), we get:

– \(x = 2a\) (since \(x^2 = 4a^2\))
– \(-2xy = -3a\), so \(y = \frac{3}{4}\) (since \(x = 2a\))

Therefore, the missing \(y^2\) term is \(\left(\frac{3}{4}\right)^2 = \frac{9}{16}\).

However, remember that we factored out a 4 earlier, so we need to multiply this term by 4 to add it to the original expression:

\[4 \times \frac{9}{16} = \frac{9}{4}\]

So, the number that must be added to the expression \(16a^2 – 12a\) to make it a perfect square is \(\frac{9}{4}\).

Let’s denote the number of buffaloes as \(b\) and the number of ducks as \(d\).

Since each buffalo has 4 legs and each duck has 2 legs, the total number of legs in the group is \(4b + 2d\). The total number of heads, which is also the total number of animals, is \(b + d\).

According to the problem, the number of legs is 24 more than twice the number of heads. Therefore, we can write the equation:

\[4b + 2d = 2(b + d) + 24\]

Simplifying this equation:

\[4b + 2d = 2b + 2d + 24\]
\[2b = 24\]
\[b = 12\]

So, there are 12 buffaloes in the group.

The equation \(|x – 2| = 5\) can be solved by considering the two possible cases for the absolute value:

1. When \(x – 2 \geq 0\):
\[x – 2 = 5\]
\[x = 7\]

2. When \(x – 2 < 0\):
\[-(x – 2) = 5\]
\[x – 2 = -5\]
\[x = -3\]

Therefore, the solutions of the equation \(|x – 2| = 5\) are \(x = 7\) and \(x = -3\), which corresponds to option (b) \(-3, 7\).

To find the value of the Roman numeral MCDLXIV, we can break it down into its components:

– M = 1000
– CD = 400 (500 – 100)
– LX = 60 (50 + 10)
– IV = 4 (5 – 1)

Adding these values together, we get:

1000 + 400 + 60 + 4 = 1464

Therefore, the value of the numeral MCDLXIV is 1464, which corresponds to option (c) 1464.

Evaluation of the Integral Using Beta and Gamma Functions

Introduction

We are given the integral

\(I=\int_0^4 \frac{d x}{\sqrt{4 x-x^2}}\),

and we aim to evaluate it using the beta and gamma functions.

Solution

First, we rewrite the integral in a more convenient form:

\[
I = \int_0^4 \frac{d x}{\sqrt{x(4-x)}} = \int_0^4 x^{-1/2}(4-x)^{-1/2} d x = \int_0^4 x^{\frac{1}{2}-1}(4-x)^{\frac{1}{2}-1} d x
\]

Next, we apply the substitution \(x=4t\), which implies \(dx = 4dt\). The limits of integration change accordingly: when \(x=0\), \(t=0\), and when \(x=4\), \(t=1\). Therefore, the integral becomes:
\[
\begin{aligned}
I &= \int_0^1 (4t)^{\frac{1}{2}-1}(4-4t)^{\frac{1}{2}-1} \cdot 4 dt \\
&= \int_0^1 4^{\frac{1}{2}-1} t^{\frac{1}{2}-1} \cdot 4^{\frac{1}{2}-1}(1-t)^{\frac{1}{2}-1} \cdot 4 dt \\
&= 4^{\frac{1}{2}+\frac{1}{2}-1} \int_0^1 t^{\frac{1}{2}-1}(1-t)^{\frac{1}{2}-1} dt \\
&= 4^0 B\left(\frac{1}{2}, \frac{1}{2}\right)
\end{aligned}
\]

Using the definition of the beta function \(B(m, n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx\), we have:
\[
I = B\left(\frac{1}{2}, \frac{1}{2}\right)
\]

Now, using the relationship between the beta and gamma functions, \(B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\), we can express the integral in terms of gamma functions:
\[
I = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2} + \frac{1}{2}\right)} = \frac{\sqrt{\pi} \cdot \sqrt{\pi}}{1} = \pi
\]

Conclusion

Therefore, the value of the integral \(I=\int_0^4 \frac{d x}{\sqrt{4 x-x^2}}\) is \(\pi\).