Question

If $f(x) =\sin [{\pi}^2] x + \cos [-\pi^2] x$ then $f '(x)$ is, here $[\pi^2]$ and $[-\pi^2]$ greatest integer function not greater than its value

• A. $sin \,9x + cos\, 9x$
• B. $9\, cos\, 9x - 10\, sin\, 10x$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

We have, $\pi^{2}=9.86 $ (nearly)
$\therefore \cos \left[-\pi^{2}\right] x =\cos [-9.86] \,x $
$=\cos (-10) \,x=\cos 10 \,x $
$\sin \left[\pi^{2}\right] x =\sin [9.86] x=\sin \,9 \,x $
$\therefore f(x) =\sin \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x $
$f(x) =\sin 9 x+\cos \,10\, x$
On differentiating w.r.t. $x$, we get
$\Rightarrow f'(x)=9\, \cos \,9 x-10 \sin \,10 \,x$