A) The maximum value of y is 2
B) The minimum value of y is \[\frac{3}{4}\]
C) \[y\le \frac{1}{2}\]
D) \[y\ge \frac{5}{4}\]
Correct Answer: B
Solution :
\[y={{\cos }^{4}}x-{{\cos }^{2}}x+1={{\left( {{\cos }^{2}}x-\frac{1}{2} \right)}^{2}}+\frac{3}{4}\] |
\[\therefore \,{{y}_{\min .}}=\frac{3}{4}\] and y is maximum when \[{{\left( {{\cos }^{2}}x-\frac{1}{2} \right)}^{2}}\] is the maximum |
\[\therefore \,\,{{y}_{\max .}}=\frac{1}{4}+\frac{3}{4}=1\] |
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