[ If y(x)=|[arrayccc x & x & x+ x+1 27 & 28 & 27 1 & 1 & 1 array]|, x R , then (d^2 y)/(d …

Question

$$
\text { If } y(x)=\left|\[\begin{array}{ccc} \sin x & \cos x & \sin x+\cos x+1 \ 27 & 28 & 27 \ 1 & 1 & 1 \end{array}$$\right|, x \in \mathbb{R} \text {, then } \frac{d\^2 y}{d x\^2}+y \text { is equal to }
\]

Accepted Answer

To solve for $ y(x) $, we start with the determinant of the given matrix:

$ \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \ 27 & 28 & 27 \ 1 & 1 & 1  \end{vmatrix} $

Using the formula for a $3 	imes 3$ determinant, we expand along the first row:

$ f(x) = \sin x \cdot \begin{vmatrix} 28 & 27 \ 1 & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} 27 & 27 \ 1 & 1 \end{vmatrix} + (\sin x + \cos x + 1) \cdot \begin{vmatrix} 27 & 28 \ 1 & 1 \end{vmatrix} $

Calculating the smaller $2 	imes 2$ determinants:

$ \begin{vmatrix} 28 & 27 \ 1 & 1 \end{vmatrix} = 28 \cdot 1 - 27 \cdot 1 = 1 $

$ \begin{vmatrix} 27 & 27 \ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 27 \cdot 1 = 0 $

$ \begin{vmatrix} 27 & 28 \ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 28 \cdot 1 = -1 $

Substituting these into the determinant calculation gives:

$ f(x) = \sin x \cdot 1 - \cos x \cdot 0 + (\sin x + \cos x + 1) \cdot (-1) $

$ f(x) = \sin x - (\sin x + \cos x + 1) $

$ f(x) = \sin x - \sin x - \cos x - 1 $

$ f(x) = -\cos x - 1 $

Now, calculate the derivatives:

First derivative $ f'(x) $:

$   f'(x) = \frac{d}{dx}(-\cos x - 1) = \sin x   $

Second derivative $ \frac{d^2 f}{dx^2} $:

$   \frac{d^2 f}{dx^2} = \frac{d}{dx}(\sin x) = \cos x   $

Finally, compute $ \frac{d^2 y}{dx^2} + y $:

$ \frac{d^2 f}{dx^2} + f(x) = \cos x - \cos x - 1 = -1 $

Thus, $ \frac{d^2 y}{dx^2} + y $ is equal to $-1$.

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