FunctionHard
Question
If(x) = cos[π2]x + cos[-π2]x, where [x] stands for the greatrst integer funiton, then
Options
A.f 
B.f(π) = 1
C.f(-π) = 0
D.f 
Solution
Since, f(x) = cos[π2]x + cos[-π2]x
⇒ f(x) = cos(9)x + cos(-10)x (using [π2] = 9 and [-π2] = - 10)
∴
= cos 
+ cos 5π = - 1
f(π) = cos 9π + cos10π = - 1 + 1= 0
f(-π) = cos 9π + cos10π = - 1 + 1 = 0
= cos 
+ cos 
Hence, (a) and (c) are correct options.
⇒ f(x) = cos(9)x + cos(-10)x (using [π2] = 9 and [-π2] = - 10)
∴
= cos + cos 5π = - 1f(π) = cos 9π + cos10π = - 1 + 1= 0
f(-π) = cos 9π + cos10π = - 1 + 1 = 0
= cos + cos Hence, (a) and (c) are correct options.
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