KCET · Maths · Differentiation
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
- A \(\sin 9 x+\cos 9 x\)
- B \(9 \cos 9 x-10 \sin 10 x\)
- C 0
- D \(-1\)
Answer & Solution
Correct Answer
(B) \(9 \cos 9 x-10 \sin 10 x\)
Step-by-step Solution
Detailed explanation
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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