Trigonometric Identities Question 275
Question: The number of solutions of the equation $ \cos (\pi \sqrt{x-4}),\cos (\pi \sqrt{x})=1 $ is
Options:
A) $ >2 $
B) 2
C) 1
D) 0
Show Answer
Answer:
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.
BETA
