Question

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

• A. $sin9x+cos9x$
• B. $9cos9x-10sin10x$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

We have, ${\pi }^2=9.86$ (nearly)
$\therefore \cos \left[-{\pi }^2\right]x=\cos [-9.86]x$
$=\cos (-10)x=\cos 10x$
$\sin \left[{\pi }^2\right]x=\sin [9.86]x=\sin 9x$
$\therefore f(x)=\sin \left[{\pi }^2\right]x+\cos \left[-{\pi }^2\right]x$
$f(x)=\sin 9x+\cos 10x$
On differentiating w.r.t. $x$, we get
$\Rightarrow f^{\prime }(x)=9\cos 9x-10\sin 10x$