The least integral value of
`f(x)=((x-1)^(7)+3(x-1)^(6)+(x-1)^(5)+1)/(x-1)^(5)`, `AAx gt 1` is (a) 8 (b) 6 (c) 12 (d) 18

Options

A. `8`
B. `6`
C. `12`
D. `18`

Explanation

To find the least integral value of the function

\[ f(x) = \frac{(x-1)^{7} + 3(x-1)^{6} + (x-1)^{5} + 1}{(x-1)^{5}} \]

for \( x > 1 \), we can follow these steps:

### Step 1: Substitute \( a = x - 1 \)

Since \( x > 1 \), we have \( a > 0 \). We can rewrite the function in terms of \( a \):

\[ f(x) = \frac{a^{7} + 3a^{6} + a^{5} + 1}{a^{5}} \]

### Step 2: Simplify the function

Now, simplifying the expression:

\[ f(a) = \frac{a^{7}}{a^{5}} + \frac{3a^{6}}{a^{5}} + \frac{a^{5}}{a^{5}} + \frac{1}{a^{5}} = a^{2} + 3a + 1 + a^{-5} \]

### Step 3: Analyze the function

We need to find the minimum value of

\[ f(a) = a^{2} + 3a + 1 + a^{-5} \]

for \( a > 0 \).

### Step 4: Use the AM-GM Inequality

We can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to the terms \( a^{2}, 3a, 1, a^{-5} \).

The AM-GM inequality states that the arithmetic mean of non-negative numbers is greater than or equal to the geometric mean.

Here, we have 4 terms:

- \( a^{2} \)
- \( a \)
- \( a \)
- \( a \)
- \( 1 \)
- \( a^{-5} \)

Thus, we can write:

\[ \frac{a^{2} + 3a + 1 + a^{-5}}{6} \geq (a^{2} \cdot a \cdot a \cdot a \cdot 1 \cdot a^{-5})^{1/6} \]

### Step 5: Calculate the geometric mean

Calculating the geometric mean:

\[ a^{2} \cdot a \cdot a \cdot a \cdot 1 \cdot a^{-5} = a^{2 + 3 - 5} = a^{0} = 1 \]

Thus, we have:

\[ \frac{a^{2} + 3a + 1 + a^{-5}}{6} \geq 1 \implies a^{2} + 3a + 1 + a^{-5} \geq 6 \]

### Step 6: Conclusion

This shows that \( f(a) \geq 6 \) for all \( a > 0 \). Therefore, the least integral value of \( f(x) \) is:

\[ \boxed{6} \]