If f(x)=`|{:(x,x^(2),x^(3)),(1,2,3),(0,1,x):}|` `underset(x to1)limf(x)` is equal to

Options

A. -1
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2 & 3 \\ 0 & 1 & x \end{vmatrix} \]

and then find the limit as \( x \) approaches 1.

### Step 1: Calculate the Determinant

We will use the determinant formula for a 3x3 matrix:

\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

- \( a = x \), \( b = x^2 \), \( c = x^3 \)
- \( d = 1 \), \( e = 2 \), \( f = 3 \)
- \( g = 0 \), \( h = 1 \), \( i = x \)

Substituting these values into the determinant formula:

\[ f(x) = x \begin{vmatrix} 2 & 3 \\ 1 & x \end{vmatrix} - x^2 \begin{vmatrix} 1 & 3 \\ 0 & x \end{vmatrix} + x^3 \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants

1. For the first determinant:

\[ \begin{vmatrix} 2 & 3 \\ 1 & x \end{vmatrix} = (2 \cdot x) - (3 \cdot 1) = 2x - 3 \]

2. For the second determinant:

\[ \begin{vmatrix} 1 & 3 \\ 0 & x \end{vmatrix} = (1 \cdot x) - (3 \cdot 0) = x \]

3. For the third determinant:

\[ \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = (1 \cdot 1) - (2 \cdot 0) = 1 \]

### Step 3: Substitute Back into the Determinant Expression

Now substituting back into the expression for \( f(x) \):

\[ f(x) = x(2x - 3) - x^2(x) + x^3(1) \]

This simplifies to:

\[ f(x) = x(2x - 3) - x^3 + x^3 \]

The \( -x^3 \) and \( +x^3 \) cancel out:

\[ f(x) = 2x^2 - 3x \]

### Step 4: Find the Limit as \( x \to 1 \)

Now we need to evaluate:

\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} (2x^2 - 3x) \]

Substituting \( x = 1 \):

\[ f(1) = 2(1)^2 - 3(1) = 2 - 3 = -1 \]

### Final Answer

Thus, the limit is:

\[ \lim_{x \to 1} f(x) = -1 \]