A) \[>2\]
B) 2
C) 1
D) 0
Correct Answer: C
Solution :
Clearly, \[x\ge 4\](Since \[\sqrt{x-4}\]is real) so that \[\sqrt{x}\] is also real. Again, if \[\cos \,(\pi \sqrt{x})<1\]then \[\cos \,(\pi \sqrt{x-4)}>1\] and if \[\cos \,(\pi \sqrt{x})>1,\]then \[\cos \,\,(\pi \sqrt{x-4})<1\] (since this product = 1). But both of these are not possible (since \[cos\,\,\theta \] cannot be greater than 1). \[\therefore \,\,\cos (\pi \sqrt{x-4})=1\] and \[\cos (\pi \sqrt{x})=1\] \[\therefore \,\,x-4=0\] or \[x=0\] But \[x=0\] is not possible, \[\therefore \,\,x=4\] is the only solution.
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