A) 1
B) 1+sin21
C) 1+cos21
D) ½
Correct Answer: B
Solution :
\[f(\theta )={{\cos }^{2}}(\cos \theta )+{{\sin }^{2}}(\sin \theta )\] Now, put\[q=p/2\And 0\] \[f(0)={{\cos }^{2}}1+0={{\cos }^{2}}1\] \[f(\pi /2)=1+{{\sin }^{2}}1\]Since, \[{{\sin }^{2}}1>{{\cos }^{2}}1\] \[[\because \sin \theta >\cos \theta \forall \frac{\pi }{4}<\theta <\frac{\pi }{2}\And 1>\frac{\pi }{4}]\] Now, seeing the options 1 + sin21 is greater than all other options.
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