Q. 395.0( 1 Vote )
Which of the foll A. f (x) = x3
B. f (x) = x + 2
C. f (x) = 2x + 1
D. f (x) = x2 + 1
Answer :
A function is bijective iff it is one-one and onto.
Option A. f (x) = x3
Let f(x1) = f(x2)
⇒ x13= x23
⇒ x1= x2
⇒ f is one one
Let f(x) = y, y ∈ Z
⇒ y = x3
⇒ x = y1/3 but y1/3∉ Z
⇒ f is not onto
Thus, f is not bijective.
Option B. f (x) = x + 2
Let f(x1) = f(x2)
⇒ x1+2 = x2+2
⇒ x1= x2
⇒ f is one one
Let f(x) = y, y ∈ Z
⇒ y = x + 2
⇒ x = y – 2
⇒ for each y ∈ Z there exists x ∈ Z (domain) such that f(x) = y.
⇒ f is onto
Thus, f is bijective.
Option C. f (x) = 2x + 1
Let f(x1) = f(x2)
⇒ 2x1+1 = 2x2+1
⇒ x1= x2
⇒ f is one one
Let f(x) = y, y ∈ Z
⇒ y = 2x + 1
⇒ y - 1 = 2x
⇒
We observe that if we put y=0, then .
Thus, y = 0 ∈ Z does not have pre image in Z (domain)
⇒ f is not onto.
Thus, f is not bijective.
Option D. f (x) = x2 + 1
let f(x1) = f(x2)
⇒ x12 + 1 = x22 + 1
⇒ x12 = x22
⇒ x1 = ± x2
⇒ x1= x2 and x1= - x2
For e.g., f(-1) = |-1| = 1 and f(1) = |1| = 1
⇒ f is not one-one.
Since, f is not one one it cannot be bijective.
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