Question ID: #1118
If $y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}, x \in \mathbb{R}$, then $\frac{d^{2}y}{dx^{2}}+y$ is equal to
- (1) -1
- (2) 28
- (3) 27
- (4) 1
Solution:
$$y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$$
$$C_{3} \rightarrow C_{3} – C_{1}$$
$$y(x) = \begin{vmatrix} \sin x & \cos x & \cos x + 1 \\ 27 & 28 & 0 \\ 1 & 1 & 0 \end{vmatrix}$$
$$y(x) = (\cos x + 1) \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} – 0 + 0$$
$$y(x) = (\cos x + 1)(27 \cdot 1 – 28 \cdot 1)$$
$$y(x) = (\cos x + 1)(-1)$$
$$y(x) = -1 – \cos x$$
$$\frac{dy}{dx} = \sin x$$
$$\frac{d^{2}y}{dx^{2}} = \cos x$$
$$\frac{d^{2}y}{dx^{2}} + y = \cos x + (-1 – \cos x)$$
$$= -1$$
Ans. (1)
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