Question

If y = sin x + cos x and y_n denotes the n^th order derivative of y with respect to x, then at x = $\frac{\pi}{2}$, the value of

$\left| \begin{matrix} y_{n} & y_{n + 1} & y_{n + 2} \\ y_{n + 3} & y_{n + 4} & y_{n + 5} \\ y_{n + 6} & y_{n + 7} & y_{n + 8} \end{matrix} \right|$ is equal to –

• A. 0
• B. –1
• C. 2
• D. 1

Answer 1

Answer: (A)

0

Answer 2

y_n = sin$\left( \frac{n\pi}{2} + x \right)$ + cos$\left( \frac{n\pi}{2} + x \right)$

y_n+2 = – y_n similarly y_n+5 + y_n+3 = 0

and y_n+8 + y_n+6 = 0