Let `f(x)=|{:(,1//x,logx,x^(n)),(,1,-1//n,(-1)^(n)),(,1,a,a^(2)):}|` where `("where"f^(n)(x) "donotes "n^(th)` derivative of f(x). (A) `f^(n)(1)"is indepent of a"` (B) `f^(n)(1)"is indepent of n"` (C) `f^(n)(1)"is indepent a or n"` (D) `y=a(x-f^(n)(1))` represent a straight line through the origin

Options

A. `f^(n)(1)"is indepent of a"`
B. `f^(n)(1)"is indepent of n"`
C. `f^(n)(1)"is indepent a of n"`
D. `y=a(x-f^(n)(1))` represent a straight line through the origin

Answer

To solve the given problem, we need to analyze the determinant \( f(x) \) and find the \( n^{th} \) derivative \( f^{(n)}(1) \). The determinant is given as:

\[ f(x) = \begin{vmatrix} \frac{1}{x} & 1 & \log x \\ \frac{1}{n} & (-1)^n & -1 \\ 1 & a & a^2 \end{vmatrix} \]

### Step 1: Calculate the Determinant

We will calculate the determinant using the formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our determinant:

\[ f(x) = \frac{1}{x} \begin{vmatrix} (-1)^n & -1 \\ a & a^2 \end{vmatrix} - 1 \begin{vmatrix} \frac{1}{n} & -1 \\ 1 & a^2 \end{vmatrix} + \log x \begin{vmatrix} \frac{1}{n} & (-1)^n \\ 1 & a \end{vmatrix} \]

Calculating each of the \( 2 \times 2 \) determinants:

1. \( \begin{vmatrix} (-1)^n & -1 \\ a & a^2 \end{vmatrix} = (-1)^n a^2 + a = a((-1)^n + 1) \)
2. \( \begin{vmatrix} \frac{1}{n} & -1 \\ 1 & a^2 \end{vmatrix} = \frac{1}{n} a^2 + 1 = \frac{a^2 + n}{n} \)
3. \( \begin{vmatrix} \frac{1}{n} & (-1)^n \\ 1 & a \end{vmatrix} = \frac{1}{n} a + (-1)^n = \frac{a + (-1)^{n+1} n}{n} \)

Substituting these back into the determinant:

\[ f(x) = \frac{1}{x} \cdot a((-1)^n + 1) - 1 \cdot \frac{a^2 + n}{n} + \log x \cdot \frac{a + (-1)^{n+1} n}{n} \]

### Step 2: Find the \( n^{th} \) Derivative

Now we need to compute the \( n^{th} \) derivative \( f^{(n)}(x) \) and evaluate it at \( x = 1 \).

1. **First Term**: The first term \( \frac{1}{x} \) has the \( n^{th} \) derivative:
\[ f_1^{(n)}(x) = \frac{(-1)^n n!}{x^{n+1}} \]

2. **Second Term**: The second term is a constant with respect to \( x \), so its \( n^{th} \) derivative is \( 0 \).

3. **Third Term**: The third term \( \log x \) has the \( n^{th} \) derivative:
\[ f_2^{(n)}(x) = \frac{(-1)^{n+1} (n-1)!}{x^n} \]

Combining these, we find:

\[ f^{(n)}(x) = \frac{(-1)^n n!}{x^{n+1}} + 0 + \frac{(-1)^{n+1} (n-1)!}{x^n} \]

### Step 3: Evaluate at \( x = 1 \)

Now we evaluate \( f^{(n)}(1) \):

\[ f^{(n)}(1) = (-1)^n n! + (-1)^{n+1} (n-1)! \]

### Step 4: Analyze Independence

1. **Independence of \( a \)**: Since \( f^{(n)}(1) \) does not contain \( a \), it is independent of \( a \).
2. **Independence of \( n \)**: The expression \( f^{(n)}(1) \) does depend on \( n \) since it contains \( n! \) and \( (n-1)! \).
3. **Independence of both \( a \) and \( n \)**: As established, it is independent of \( a \) but dependent on \( n \).
4. **Straight Line Representation**: The equation \( y = a(x - f^{(n)}(1)) \) represents a straight line through the origin when \( f^{(n)}(1) = 0 \).

### Conclusion

Thus, the correct options are:
- (A) \( f^{(n)}(1) \) is independent of \( a \)
- (B) \( f^{(n)}(1) \) is dependent on \( n \)
- (D) \( y = a(x - f^{(n)}(1)) \) represents a straight line through the origin.