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

JEE · Math · previous-year question

  1. A.28
  2. B.27
  3. C.-1correct
  4. D.1

Answer

C. -1

Explanation

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 $3 \times 3$ 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 $2 \times 2$ 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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