A) \[{{\pi }^{2}}\]
B) \[2{{\pi }^{2}}\]
C) \[3{{\pi }^{2}}\]
D) None of these
Correct Answer: B
Solution :
[b] \[f(x)={{({{\sin }^{-1}}x)}^{2}}+2\pi {{\cos }^{-1}}x+{{\pi }^{2}}\] \[={{({{\sin }^{-1}}x)}^{2}}+2\pi \left( \frac{\pi }{2}-{{\sin }^{-1}}x \right)+{{\pi }^{2}}\] \[={{({{\sin }^{-1}}x)}^{2}}+{{\pi }^{2}}-2\pi {{\sin }^{-1}}x+{{\pi }^{2}}\] \[{{({{\sin }^{-1}}x-\pi )}^{2}}+{{\pi }^{2}}\] \[{{f}_{\min }}={{\left( \frac{\pi }{2}-\pi \right)}^{2}}+{{\pi }^{2}}={{\pi }^{2}}+\frac{{{\pi }^{2}}}{4}=\frac{5{{\pi }^{2}}}{4}\] \[{{f}_{\max }}={{\left( -\frac{\pi }{2}-\pi \right)}^{2}}+{{\pi }^{2}}={{\pi }^{2}}+\frac{9{{\pi }^{2}}}{4}=\frac{13{{\pi }^{2}}}{4}\] \[\therefore \,\,\,\,\,{{f}_{\max }}-{{f}_{\min }}=\frac{8{{\pi }^{2}}}{4}=2{{\pi }^{2}}\]
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