The expression `(1)/( sqrt( 3x +1)) [ ( ( 1+ sqrt( 3x + 1))/(2))^(7) - (( 1- sqrt ( 3x +1))/(2))^(7)]` is a polynomial in x of degree eual to

Options

A. 3
B. 4
C. 2
D. 5

Answer

To solve the problem, we need to analyze the given expression step by step. The expression is:

\[ \frac{1}{\sqrt{3x + 1}} \left[ \left( \frac{1 + \sqrt{3x + 1}}{2} \right)^{7} - \left( \frac{1 - \sqrt{3x + 1}}{2} \right)^{7} \right] \]

### Step 1: Simplify the expression using the Binomial Theorem

We can use the Binomial Theorem to expand both terms in the brackets. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

For the first term \(\left( \frac{1 + \sqrt{3x + 1}}{2} \right)^{7}\):

Let \(a = \frac{1}{2}\) and \(b = \frac{\sqrt{3x + 1}}{2}\), then:

\[ \left( \frac{1 + \sqrt{3x + 1}}{2} \right)^{7} = \sum_{k=0}^{7} \binom{7}{k} \left(\frac{1}{2}\right)^{7-k} \left(\frac{\sqrt{3x + 1}}{2}\right)^{k} \]

For the second term \(\left( \frac{1 - \sqrt{3x + 1}}{2} \right)^{7}\):

Using the same \(a\) and \(b\):

\[ \left( \frac{1 - \sqrt{3x + 1}}{2} \right)^{7} = \sum_{k=0}^{7} \binom{7}{k} \left(\frac{1}{2}\right)^{7-k} \left(-\frac{\sqrt{3x + 1}}{2}\right)^{k} \]

### Step 2: Subtract the two expansions

When we subtract these two expansions, all even powers of \(\sqrt{3x + 1}\) will cancel out, and we will be left with only the odd powers:

\[ \left( \frac{1 + \sqrt{3x + 1}}{2} \right)^{7} - \left( \frac{1 - \sqrt{3x + 1}}{2} \right)^{7} = 2 \sum_{k \text{ odd}} \binom{7}{k} \left(\frac{1}{2}\right)^{7-k} \left(\frac{\sqrt{3x + 1}}{2}\right)^{k} \]

### Step 3: Identify the highest power of \(x\)

The highest power of \(\sqrt{3x + 1}\) will be when \(k = 7\) (the highest odd number less than or equal to 7). Thus, we have:

\[ \sqrt{3x + 1}^7 = (3x + 1)^{7/2} \]

### Step 4: Simplify the expression further

Now, we can write the expression as:

\[ \frac{1}{\sqrt{3x + 1}} \cdot \frac{2}{2^7} \cdot \text{(sum of odd powers)} \]

This simplifies to:

\[ \frac{1}{\sqrt{3x + 1}} \cdot \frac{1}{64} \cdot \text{(sum of odd powers)} \]

### Step 5: Determine the degree of the polynomial

The highest power of \(x\) in the sum of odd powers will be \(3x\) raised to the power of \(7/2\). Therefore, the degree of \(3x + 1\) is \(3\), and since we have \(\sqrt{3x + 1}\) in the denominator, the overall degree of the polynomial in \(x\) will be:

\[ \text{Degree} = \frac{7}{2} - \frac{1}{2} = 3 \]

### Final Answer

Thus, the expression is a polynomial in \(x\) of degree equal to **3**.

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