Q70 / 180 · KCET 2016 · Mathematics
If \(\mathrm{A}=\frac{1}{\pi}\left[\begin{array}{ll}\sin ^{-1}(\mathrm{x} \pi) & \tan ^{-1}\left(\frac{\mathrm{x}}{\pi}\right) \\ \sin ^{-1}\left(\frac{\mathrm{x}}{\pi}\right) & \cot ^{-1}(\pi \mathrm{x})\end{array}\right] \mathrm{B}=\left[\begin{array}{cc}-\cos ^{-1}(\mathrm{x} \pi) & \tan ^{-1}\left(\frac{\mathrm{x}}{\pi}\right) \\ \sin ^{-1}\left(\frac{\mathrm{x}}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]\) then \(\mathrm{A}-\mathrm{B}\) equal to
- AI
- B\( 0 \)
- C( \( 2 I \)
- Correct answerD\( \frac{1}{2} I \)
Solution
We have, \(A=\left[\begin{array}{ll}\frac{1}{\pi} \sin ^{-1}(x \pi) & \frac{1}{\pi} \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \frac{1}{\pi} \sin ^{-1}\left(\frac{x}{\pi}\right) & \frac{1}{\pi} \cot ^{-1}(\pi x)\end{array}\right] B=\left[\begin{array}{cc}-\frac{1}{\pi} \cos ^{-1}(x \pi) & \frac{1}{\pi} \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \frac{1}{\pi} \sin ^{-1}\left(\frac{x}{\pi}\right) & -\frac{1}{\pi} \tan ^{-1}(\pi x)\end{array}\right]\)
\(\therefore A-B=\left[\begin{array}{ll}
\frac{1}{\pi}\left(\sin ^{-1} x \pi+\cos ^{-1} x \pi\right) & \frac{1}{\pi}\left(\tan ^{-1} \frac{x}{\pi}-\tan ^{-1} \frac{x}{\pi}\right) \\
\frac{1}{\pi}\left(\sin ^{-1} \frac{x}{\pi}-\sin ^{-1} \frac{x}{\pi}\right) & \frac{1}{\pi}\left(\cot ^{-1} \pi x\right)+\tan ^{-1} \pi x
\end{array}\right]\)
\(\left[\begin{array}{cc}
\frac{1}{\pi} \cdot \frac{\pi}{2} & 0 \\
0 & \frac{1}{\pi} \cdot \frac{\pi}{2}
\end{array}\right]\left[\begin{array}{l}
\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\
\text { and } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}
\end{array}\right]\)
\(=\frac{1}{2}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\frac{1}{2} \mathrm{I}\)
Hence, the correct answer is option: 'D'.