`y(x)=|(sinx,cosx,sinx+cosx+1),(23,17,13),(1,1,1)|,x in RR,` then `(d^2y)/(dx^2)+y` is equal to :

Options

A. 6
B. 4
C. -10
D. 0

Explanation

To solve the problem, we need to find the expression for \( \frac{d^2y}{dx^2} + y \) where \( y(x) \) is given as the determinant:

\[ y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 23 & 17 & 13 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 1: Calculate the Determinant \( y(x) \)

Using the determinant formula for a 3x3 matrix, we can calculate \( y(x) \):

\[ y(x) = \sin x \begin{vmatrix} 17 & 13 \\ 1 & 1 \end{vmatrix} - \cos x \begin{vmatrix} 23 & 13 \\ 1 & 1 \end{vmatrix} + (\sin x + \cos x + 1) \begin{vmatrix} 23 & 17 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 17 & 13 \\ 1 & 1 \end{vmatrix} = 17 \cdot 1 - 13 \cdot 1 = 4 \)
2. \( \begin{vmatrix} 23 & 13 \\ 1 & 1 \end{vmatrix} = 23 \cdot 1 - 13 \cdot 1 = 10 \)
3. \( \begin{vmatrix} 23 & 17 \\ 1 & 1 \end{vmatrix} = 23 \cdot 1 - 17 \cdot 1 = 6 \)

Substituting these values back into the determinant:

\[ y(x) = \sin x \cdot 4 - \cos x \cdot 10 + (\sin x + \cos x + 1) \cdot 6 \]

Expanding this:

\[ y(x) = 4\sin x - 10\cos x + 6\sin x + 6\cos x + 6 \]

Combining like terms:

\[ y(x) = (4\sin x + 6\sin x) + (-10\cos x + 6\cos x) + 6 = 10\sin x - 4\cos x + 6 \]

### Step 2: Find the First Derivative \( \frac{dy}{dx} \)

Now we differentiate \( y(x) \):

\[ \frac{dy}{dx} = 10\cos x + 4\sin x \]

### Step 3: Find the Second Derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):

\[ \frac{d^2y}{dx^2} = -10\sin x + 4\cos x \]

### Step 4: Calculate \( \frac{d^2y}{dx^2} + y \)

Now we add \( y(x) \) and \( \frac{d^2y}{dx^2} \):

\[ \frac{d^2y}{dx^2} + y = (-10\sin x + 4\cos x) + (10\sin x - 4\cos x + 6) \]

Combining terms:

\[ \frac{d^2y}{dx^2} + y = (-10\sin x + 10\sin x) + (4\cos x - 4\cos x) + 6 = 6 \]

### Final Result

Thus, we find that:

\[ \frac{d^2y}{dx^2} + y = 6 \]