FunctionHard
Question
Given the function f(x) such that 2f(x) + xf
- 2f
= 4 cos2
+ x cos
, then which one of the following is correct ?
Options
A.f(2) + f(1/2) = 1
B.f(1) = -1 but the values of f(2), f(1/2) cannot be determined
C.f(2) + f(1) = f(1/2)
D.f(2) + f(1) = 0
Solution
put x = 1
2f(1) + 1 f(1) - 2f( | √2 sin
| ) = -1
⇒ 3 f (1) - 2 f (1) = -1 ⇒ f (1) = -1
Now put x = 2
2 f(2) + 2 f
- 2f (1) = 4 cos2 π + 2 cos 
⇒ 2 f(2) + 2f
- 2 f (1) = 4
⇒ f(2) + f
= 1 ......... (i)
Now put x = 1/2 we get
4f
+ f (2) = 1 .............. (ii)
from (i) and (ii)
f
= 0 & f (2) = 1
2f(1) + 1 f(1) - 2f( | √2 sin
⇒ 3 f (1) - 2 f (1) = -1 ⇒ f (1) = -1
Now put x = 2
2 f(2) + 2 f
⇒ 2 f(2) + 2f
⇒ f(2) + f
Now put x = 1/2 we get
4f
from (i) and (ii)
f
Create a free account to view solution
View Solution FreeMore Function Questions
If f : Q → Q, f(x)= 2x and g : Q → Q, g(x) = x + 2, then (fog)-1(20) equals-...The inverse of the function f(x) = + 2 is given by -...If f(x) = , then range of f(x) is...Let (x0, y0) be solution of the following equations (2x)ln2 = (3y)ln3 3lnx = 2lny Then x0 is...The function f(x) = log is...