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If $f(x)=\sin \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x$, then $f^{\prime}(x)$ is, here $\left[\pi^{2}\right]$ and $\left[-\pi^{2}\right]$ greatest integer function not greater than its value
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Verified Answer
The correct answer is:
$9 \cos 9 x-10 \sin 10 x$
We have, $\pi^{2}=9.86 \quad$ (nearly)
$$
\begin{aligned}
\therefore \cos \left[-\pi^{2}\right] x &=\cos [-9.86] x \\
&=\cos (-10) x=\cos 10 x \\
\therefore \quad \sin \left[\pi^{2}\right] x &=\sin [9.86] x=\sin 9 x \\
\therefore \quad f(x) &=\sin \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x \\
f(x) &=\sin 9 x+\cos 10 x
\end{aligned}
$$
On differentiating w.r.t. $x$, we get
$$
\Rightarrow \quad f^{\prime}(x)=9 \cos 9 x-10 \sin 10 x
$$
$$
\begin{aligned}
\therefore \cos \left[-\pi^{2}\right] x &=\cos [-9.86] x \\
&=\cos (-10) x=\cos 10 x \\
\therefore \quad \sin \left[\pi^{2}\right] x &=\sin [9.86] x=\sin 9 x \\
\therefore \quad f(x) &=\sin \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x \\
f(x) &=\sin 9 x+\cos 10 x
\end{aligned}
$$
On differentiating w.r.t. $x$, we get
$$
\Rightarrow \quad f^{\prime}(x)=9 \cos 9 x-10 \sin 10 x
$$
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