To find the domain of the function f(x)=cos−1(34x+2[x]), where [x] denotes the greatest integer function, we use the property that the inverse cosine function cos−1(y) is defined for y∈[−1,1].
Therefore, we set up the inequality:
−1≤34x+2[x]≤1
Multiplying the entire inequality by 3:
−3≤4x+2[x]≤3
Let x=[x]+{x}, where [x]∈Z is the integer part and {x}∈[0,1) is the fractional part of x. Substituting this into the inequality gives:
−3≤4([x]+{x})+2[x]≤3
−3≤6[x]+4{x}≤3
Now, we analyze the inequality for different integer values of [x]:
-
For [x]=0:
−3≤6(0)+4{x}≤3⟹−3≤4{x}≤3
Since 0≤{x}<1, this reduces to:
0≤4{x}≤3⟹0≤{x}≤43
Since x=0+{x}, we get:
x∈[0,43]
-
For [x]=−1:
−3≤6(−1)+4{x}≤3⟹−3≤−6+4{x}≤3
Adding 6 across the inequality:
3≤4{x}≤9⟹43≤{x}≤49
Combining this with the constraint 0≤{x}<1:
43≤{x}<1
Since x=−1+{x}, we get:
x∈[−1+43,−1+1)⟹x∈[−41,0)
-
For [x]≥1:
6[x]+4{x}≥6(1)+0=6>3
There are no valid solutions for x.
-
For [x]≤−2:
6[x]+4{x}<6(−2)+4(1)=−8<−3
There are no valid solutions for x.
Combining the valid intervals from cases 1 and 2:
x∈[−41,0)∪[0,43]=[−41,43]
Thus, the domain of f(x) is [α,β]=[−41,43], which gives:
α=−41,β=43
We are required to find the value of 12(α+β):
12(α+β)=12(−41+43)=12(42)=12×21=6
Correct Option: A