A) 0
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
(c): \[\cos \theta +\frac{1}{\cos \theta }=2\] \[\Rightarrow \] \[{{\cos }^{2}}\theta +1=2\cos \theta \] \[\Rightarrow \] \[{{\cos }^{2}}\theta -2\cos \theta +1=0\] \[\Rightarrow \] \[{{\left( \cos \theta -1 \right)}^{2}}=0\] \[\Rightarrow \] \[\cos \theta =1\] \[\therefore \] \[\sec \theta =1\] \[\therefore \] \[{{\cos }^{3}}\theta +{{\sec }^{3}}\theta =1+1=2\] Mind of a mathematician A thoughtful student will straightaway see \[\left( \cos \theta +\frac{1}{\cos \theta } \right)\] as equivalent to \[x+\frac{1}{x}\] whose minimum value is attained when \[x=1\] and hence, \[x+\frac{1}{x}=2\] (min value)
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