FunctionHard
Question
If f (x) = 2 [x] + cos x, then f: R → R is: (where [. ] denotes greatest integer function)
Options
A.one-one and onto
B.one-one and into
C.many-one and into
D.many-one and onto
Solution
f(x) = 2[x] + cos x
f(x) = cos x x ∈ [0, 1)
= 2 + cos x x ∈ [1, 2)
= 4 + cos x x ∈ [2, 3)
= 6 + cos x x ∈ [3, 4)
for x ∈ [0, 1) f′(x) = - ve
x ∈ [1, 2) f′(x) = - ve
x ∈ [2, 3) f′(x) = - ve
x ∈ [3, 4) f′(x) = + ve
⇒ function is not one-one
if x ∈ [0, 1) range : [1, cos 1)
x ∈ [1, 2) range : [2 + cos 1, 2 + cos 2)
not onto function
f(x) = cos x x ∈ [0, 1)
= 2 + cos x x ∈ [1, 2)
= 4 + cos x x ∈ [2, 3)
= 6 + cos x x ∈ [3, 4)
for x ∈ [0, 1) f′(x) = - ve
x ∈ [1, 2) f′(x) = - ve
x ∈ [2, 3) f′(x) = - ve
x ∈ [3, 4) f′(x) = + ve
⇒ function is not one-one
if x ∈ [0, 1) range : [1, cos 1)
x ∈ [1, 2) range : [2 + cos 1, 2 + cos 2)
not onto function
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