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

To address the problem, we begin with the provided determinant:

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

The first step is to evaluate this determinant. The formula for a \(3 \times 3\) determinant is:

\(D = a(ei − fh) − b(di − fg) + c(dh − eg)\)

For the given matrix, the elements are:

• \(a = \sin x\), \(b = \cos x\), \(c = \sin x + \cos x + 1\)
• \(d = 27\), \(e = 28\), \(f = 27\)
• \(g = 1\), \(h = 1\), \(i = 1\)

Substituting these values into the determinant formula yields:

\(y(x) = \sin x (28 \times 1 - 27 \times 1) - \cos x (27 \times 1 - 1 \times 27) + (\sin x + \cos x + 1) (27 \times 1 - 28 \times 1)\)

Simplifying this expression:

\(= \sin x (28 - 27) - \cos x (27 - 27) + (\sin x + \cos x + 1) (27 - 28)\)

Further simplification leads to:

\(= \sin x (1) - 0 + (\sin x + \cos x + 1) (-1)\)

Continuing the simplification:

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

The final form of \(y(x)\) is:

\(y(x) = -\cos x - 1\)

Next, we need to compute the second derivative, \(\frac{d^2y}{dx^2}\).

The first derivative is:

\(\frac{dy}{dx} = \frac{d}{dx}(-\cos x - 1) = \sin x\)

The second derivative is:

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

Now, we compute \(\frac{d^2y}{dx^2} + y\):

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

This simplifies to:

\(= -1\)

Therefore, the result is:

-1