A) \[\frac{3}{4}\le A\le \frac{13}{16}\]
B) \[\frac{3}{4}\le A\le 1\]
C) \[\frac{13}{16}\le A\le 1\]
D) \[1\le A\le 2\]
Correct Answer: B
Solution :
We have, \[A={{\sin }^{2}}x+{{\cos }^{4}}x=1-{{\cos }^{2}}x+{{\cos }^{4}}x\] \[=1+{{\left( {{\cos }^{2}}x-\frac{1}{2} \right)}^{2}}-\frac{1}{4}\] \[=\frac{3}{4}+{{\left( {{\cos }^{2}}x-\frac{1}{2} \right)}^{2}}\ge \frac{3}{4}\] Clearly \[\frac{3}{4}\le A\le 1\]
You need to login to perform this action.
You will be redirected in
3 sec
