Question

If $f\left(x\right)=\sin \left[{\pi }^2\right]x+\cos \left[{-\pi }^2\right]x$ then $f^{\prime }\left(x\right)$ is, here $\left[{\pi }^2\right]$ and $\left[{-\pi }^2\right]$ 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)

$9\cos 9x-10\sin 10x$

$f\left(x\right)=\sin \left[{\pi }^2\right]x+cos[-{\pi }^2]x$

$\pi$ has the value $3.14$ and so ${\pi }^2$ will be around $9.86$

$\Rightarrow \left[{\pi }^2\right]=9$ and $[-{\pi }^2]=-10$ since $\left[\right]$ denotes the greatest integer less than the value.

$\Rightarrow f\left(x\right)=\sin 9x+\cos (-10)x=\sin 9x+\cos 10x$

$\Rightarrow f^{\prime }\left(x\right)=9\cos 9x-10\sin 10x$