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.
Create a free account to view solution
View Solution FreeMore Function Questions
The domain of the function f(x) = is-...Let $P = \left\{ \theta \in \lbrack 0,4\pi\rbrack:\tan^{2}\theta \neq 1 \right\}$ and $S = \left\{ a \in Z:2\left( \cos^...If f(x) = log x, then correct statement is-...If f : R → R, f(x) = x2 − 5x + 4 and g : R → R, g(x) = log x , then the value of (gof) (2) is -...If f : R → R, f(x) = x2 + 3, then pre- image of 2 under f is -...