FunctionHard
Question
Let f : R → R be a function defined by f(x) = x3 + x2 + 3x + sin x. Then f is:
Options
A.one - one and onto
B.one - one and into
C.many one and onto
D.many one and into
Solution
f(x) = x3 + x2 + 3x + sin x, x ∈ R
f′(x) = 3x2 + 2x + 3 + cos x
∵ 3x2 + 2x + 3 ≥
as a = 3 > 0 and D < 0
-1 ≤ cos x ≤ 1
so f′(x) > 0 ∀ x ∈ R
f(x) = + ∞
f(x) = - ∞
Hence f(x) is one-one and onto function (as f(x) is continuous function)
f′(x) = 3x2 + 2x + 3 + cos x
∵ 3x2 + 2x + 3 ≥
as a = 3 > 0 and D < 0-1 ≤ cos x ≤ 1
so f′(x) > 0 ∀ x ∈ R
f(x) = + ∞ f(x) = - ∞Hence f(x) is one-one and onto function (as f(x) is continuous function)
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