A) \[1\le A\le 2\]
B) \[\frac{3}{4}\le A\le 1\]
C) \[\frac{13}{16}\le A<1\]
D) \[\frac{3}{4}\le A\le \frac{13}{16}\]
Correct Answer: B
Solution :
We have\[A={{\sin }^{2}}\theta +{{\cos }^{4}}\theta \] ?(i) \[\Rightarrow \] \[A=1-{{\cos }^{2}}\theta +{{\cos }^{4}}\theta \] \[=\frac{3}{4}+{{\left( {{\cos }^{2}}\theta -\frac{1}{2} \right)}^{2}}\] Clearly,\[A\ge 3/4\] for all real values of \[\theta .\] Again from Eq. (i) \[A={{\sin }^{2}}\theta +(1-si{{n}^{2}}\theta )co{{s}^{2}}\theta \] \[={{\sin }^{2}}\theta +{{\cos }^{2}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta \] \[=1-{{(sin\theta cos\theta )}^{2}}\] Clearly,\[A\le 1\] for all real values of \[\theta .\] Hence, \[\frac{3}{4}\le A\le 1.\]
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