A) Both statements I and II are true.
B) Both statements I and II are false.
C) Statement I is true and statement II is false.
D) Statement I is false and statement II is true.
Correct Answer: A
Solution :
\[{{\sin }^{-1}}x\in \left[ -\frac{\pi }{2},\frac{\pi }{2} \right]\] \[\Rightarrow \]\[-\frac{3\pi }{4}\le \left( {{\sin }^{-1}}x-\frac{\pi }{4} \right)\le \frac{\pi }{4}\] \[0\le {{\left( {{\sin }^{-1}}x-\frac{\pi }{4} \right)}^{2}}\le \frac{9}{16}{{\pi }^{2}}\] ?(1) Statement II is true \[{{({{\sin }^{-1}}x)}^{3}}+{{({{\cos }^{-1}}x)}^{3}}=a{{\pi }^{3}}\] \[\Rightarrow \]\[({{\sin }^{-1}}x+{{\cos }^{-1}}x)\left[ {{({{\sin }^{-1}}x+{{\cos }^{-1}}x)}^{2}}-3{{\sin }^{-1}}x{{\cos }^{-1}}x \right]\]\[=a{{\pi }^{3}}\] \[\Rightarrow \]\[\frac{{{\pi }^{2}}}{4}-3{{\sin }^{-1}}x{{\cos }^{-1}}x=2a{{\pi }^{2}}\] \[\Rightarrow \]\[{{\sin }^{-1}}x\left( \frac{\pi }{2}-{{\sin }^{-1}}x \right)=\frac{{{\pi }^{2}}}{12}(1-8a)\] \[\Rightarrow \]\[{{\left( {{\sin }^{-1}}x-\frac{\pi }{4} \right)}^{2}}=\frac{{{\pi }^{2}}}{12}(8a-1)+\frac{{{\pi }^{2}}}{16}\] \[\Rightarrow \]\[{{\left( {{\sin }^{-1}}x-\frac{\pi }{4} \right)}^{2}}=\frac{{{\pi }^{2}}}{48}(32a-1)\] Putting this value in equation (1) \[0\le \frac{{{\pi }^{2}}}{48}(32a-1)\le \frac{9}{16}{{\pi }^{2}}\] \[\Rightarrow \]\[0\le 32a-1\le 27\] \[\frac{1}{32}\le a\le \frac{7}{8}\] Statement-I is also true.
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