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 solve the problem, we start with the given determinant:

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

We need to first evaluate this determinant. The determinant of a \(3 \times 3\) matrix is calculated using:

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

For our matrix:

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

Substituting these into the formula, we calculate the determinant:

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

which simplifies to:

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

This simplifies to:

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

Further simplifying:

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

Finally:

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

So, we have:

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

Next, we need to calculate \(\frac{d^2y}{dx^2}\):

First derivative:

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

Second derivative:

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

Now, we calculate:

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

Which simplifies to:

\(= -1\)

This confirms that the answer is:

-1

Answer 2

Given the determinant function:

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

We need to find the expression \(\frac{d^2y}{dx^2} + y\).

1. First, calculate the determinant \( y(x) \) as a function of \( x \):

Using the determinant definition, we compute:

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

Expanding along the first row:

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

1. Calculate each of these determinants:

\(\begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} = (28)(1) - (27)(1) = 28 - 27 = 1\)

\(\begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} = (27)(1) - (27)(1) = 0\)

\(\begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} = (27)(1) - (28)(1) = 27 - 28 = -1\)

1. Substitute these results back in to get:

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

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

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

\(= - \cos x - 1\)

1. Next, compute the second derivative.
2. First derivative of \( y(x) \):

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

1. Second derivative of \( y(x) \):

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

1. Add \( \frac{d^2y}{dx^2} \) and \( y(x) \):

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

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

\(= -1\)

The value of \(\frac{d^2y}{dx^2} + y\) is therefore:

• -1