If `Delta = |(f(x),f(1/x)+f(x)), (1,f(1/x))|=0` where it is given `f(x) = a + bx^n` and `f(2) = 17` and `f(5) = K` then `K-620=`

Options

A. 126
B. 326
C. 428
D. 626

Answer

To solve the problem, we start with the given determinant:

\[ \Delta = \begin{vmatrix} f(x) & f(1/x) + f(x) \\ 1 & f(1/x) \end{vmatrix} = 0 \]

### Step 1: Calculate the determinant

The determinant of a 2x2 matrix is calculated as follows:

\[ \Delta = f(x) \cdot f(1/x) - (1) \cdot (f(1/x) + f(x)) \]

This simplifies to:

\[ \Delta = f(x) \cdot f(1/x) - f(1/x) - f(x) \]

### Step 2: Set the determinant to zero

Since we know that \(\Delta = 0\), we can set the equation to zero:

\[ f(x) \cdot f(1/x) - f(1/x) - f(x) = 0 \]

Rearranging gives us:

\[ f(x) \cdot f(1/x) = f(1/x) + f(x) \]

### Step 3: Substitute the function \(f(x)\)

Given \(f(x) = a + bx^n\), we can find \(f(1/x)\):

\[ f(1/x) = a + b(1/x)^n = a + \frac{b}{x^n} \]

Now substituting \(f(x)\) and \(f(1/x)\) into the equation:

\[ (a + bx^n) \left(a + \frac{b}{x^n}\right) = \left(a + \frac{b}{x^n}\right) + (a + bx^n) \]

### Step 4: Expand both sides

Expanding the left side:

\[ (a + bx^n) \left(a + \frac{b}{x^n}\right) = a^2 + a\frac{b}{x^n} + abx^n + b^2 \]

And the right side:

\[ \left(a + \frac{b}{x^n}\right) + (a + bx^n) = 2a + \frac{b}{x^n} + bx^n \]

### Step 5: Equate and simplify

Setting both sides equal gives us:

\[ a^2 + a\frac{b}{x^n} + abx^n + b^2 = 2a + \frac{b}{x^n} + bx^n \]

### Step 6: Collect like terms

Rearranging the equation leads to:

\[ a^2 + b^2 + (ab - 2a) + \left(a - 1\right)\frac{b}{x^n} + \left(ab - b\right)x^n = 0 \]

This must hold for all \(x\), which means the coefficients of \(x^n\) and \(\frac{1}{x^n}\) must separately equal zero.

### Step 7: Solve for \(a\) and \(b\)

From the constant term, we have:

1. \(a^2 + b^2 - 2a = 0\)
2. \(ab - b = 0\)

From the second equation, we can factor out \(b\):

\[ b(a - 1) = 0 \]

So either \(b = 0\) or \(a = 1\).

If \(b = 0\), then \(f(x) = a\), which contradicts \(f(2) = 17\). Thus, \(a = 1\).

Substituting \(a = 1\) into the first equation gives:

\[ 1 + b^2 - 2 = 0 \implies b^2 = 1 \implies b = 1 \text{ or } b = -1 \]

### Step 8: Determine \(f(x)\)

If \(b = 1\), then \(f(x) = 1 + x^n\). If \(b = -1\), then \(f(x) = 1 - x^n\).

Using \(f(2) = 17\):

1. For \(f(x) = 1 + x^n\):
\[ 1 + 2^n = 17 \implies 2^n = 16 \implies n = 4 \]

2. For \(f(x) = 1 - x^n\):
\[ 1 - 2^n = 17 \implies -2^n = 16 \text{ (not possible)} \]

Thus, \(f(x) = 1 + x^4\).

### Step 9: Find \(f(5)\)

Now, we compute \(f(5)\):

\[ f(5) = 1 + 5^4 = 1 + 625 = 626 \]

### Step 10: Calculate \(K - 620\)

Finally, we find \(K - 620\):

\[ K - 620 = 626 - 620 = 6 \]

Thus, the final answer is:

\[ \boxed{6} \]