Question

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^2y}{dx^2} + y $ is equal to

• A. -1
• B. 28
• C. 27
• D. 1

Hint:

Use determinant properties to simplify the expression before differentiation.

Answer 1

The Correct Option is A

Given \( y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix} \) 

Applying \( C_3 \rightarrow C_3 - C_1 \), we get \( y(x) = \begin{vmatrix} \sin x & \cos x & \cos x + 1 27 & 28 & 0 \\ 1 & 1 & 0 \end{vmatrix} \)

Expanding the determinant, we have \( y(x) = -(\cos x + 1) \begin{vmatrix} 27 & 28 1 & 1 \end{vmatrix} \) \( y(x) = -(\cos x + 1) (27 - 28) \) \( y(x) = \cos x + 1 \) 

Differentiating with respect to x, we get \( \frac{dy}{dx} = -\sin x \) 

Differentiating again with respect to x, we get \( \frac{d^2y}{dx^2} = -\cos x \) 

Therefore, \( \frac{d^2y}{dx^2} + y = -\cos x + \cos x + 1 = 1 \)